zbirka zadataka iz termodinamike
strana 1
KVAZISTATI^KE (RAVNOTE@NE)
PROMENE STAWA IDEALNIH GASOVA
2/2/ Vazduh (idealan gas), 2)q2>3!cbs-!w2>1/5416!n40lh*!kvazistati~ki (ravnote`no) mewa stawe do
3)q3>7!cbs-!w3>w2!*/ Odrediti:
a) temperaturu vazduha u karakteristi~nim ta~kama procesa
b) razmewenu toplotu )r23* i zapreminski rad )x23*
c) promenu unutra{we energije )∆v*- entalpije )∆i* i entropije )∆t* vazduha
d) skicirati proces na qw!i!Ut dijagramu
a)
U2 =
q2 ⋅ w 2 3 ⋅ 21 6 ⋅ 1/5416
>
>411!LSh
398
U3 =
q3 ⋅ w 3
7 ⋅ 21 6 ⋅ 1/5416
>
>:11!L
Sh
398
b)
r23 = d w ⋅ (U3 − U2 ) > 1/83 ⋅ (:11 − 411) >!543!
x23!>!1!
lK
lh
lK
lh
c)
lK
lh
lK
= d q ⋅ (U3 − U2 ) > 2/11 ⋅ (:11 − 411) >!711!
lh
q3
w3
7
lK
= g (q- w ) = d w mo
− d q mo
> 1/83 ⋅ mo >1/8:2!
q2
w2
3
lhL
∆v23 = d w ⋅ (U3 − U2 ) > 1/83 ⋅ (:11 − 411) >!543!
∆i23
∆t23
!
d)
q
U
3
3
2
2
w
dipl.ing. @eqko Ciganovi}
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 2
2/3/ Dva kilograma kiseonika (idealan gas) po~etnog stawa 2)q>2!cbs-!U>484!L*- usled interakcije sa
toplotnim ponorom stalne temperature, mewa svoje toplotno stawe kvazistati~ki (ravnote`no)
politropski )o>1/9* do stawa 3)!w3> 1/6 ⋅ w 2 */ Skicirati proces u qw!i!Ut koordinatnom sistemu i
odrediti:
a) mehani~ke veli~ine stawa kiseonika )q-!w-!U* u karakteristi~nim ta~kama
b) koli~inu toplote )lK* koju radno telo preda toplotnom ponoru kao i zapreminski rad koji pri tom
izvr{i nad radni telom )lK*
c) promenu entropije izolovanog termodinami~kog sistema u najpovoqnijem slu~aju
q
U
3
2
o>1/9
o>1/9
2
3
w
UUQ
t
a)
w2 =
S h U2
q2
=
q2 w 3
=
q 3 w 2
U3 =
371 ⋅ 484
2 ⋅ 21 6
o
⇒
>1/:7:9!
n4
lh
w
q 3 = q2 ⋅ 2
w3
w 3 = 1/6 ⋅ 1/:7:9 >1/595:!
n4
lh
o
= 2 ⋅ 21 6 ⋅ 3 1/9 > 2/85 ⋅ 21 6 Qb
q 3 ⋅ w 3 2/85 ⋅ 21 6 ⋅ 1/595:
=
>435/62!L
Sh
371
b)
1/9 − 2/5
lK
o−κ
⋅ (435/62 − 484) >−:5/66!
⋅ (U3 − U2 ) > 1/76 ⋅
1/9 − 2
lh
o −2
= n ⋅ r23 = 3 ⋅ (− :5/66) >−29:/2!lK
r23 = d w ⋅
R 23
2
2
lK
⋅ (484 − 435/62) >!−74/15!
⋅ (U2 − U3 ) > 1/37 ⋅
1/9 − 2
o −2
lh
= n ⋅ x 23 = 3 ⋅ (− 74/15 ) >237/19!lK
x23!>! S h ⋅
X23
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 3
c)
∆Ttjtufn!>!∆Tsbeop!ufmp!,!∆Tupqmpuoj!qpops!>!///!>!−1/65!,!1/69!>!1/15!
U
q
∆Tsbeop!ufmp!>!∆T23!> n ⋅ d q mo 3 − S h mo 3 >!
U2
q2
435/62
2/85
lK
> 3 ⋅ 1/:2 ⋅ mo
− 1/37 ⋅ mo
>−1/65!
L
484
2
∆Tupqmpuoj!qpops!>!−!
lK
L
!
R 23
−29:/2
lK
>!−!
>1/69!
L
435/62
UUQ
2/4/ Kiseonik (idealan gas) n>21!lh, mewa stawe kvazistati~ki izobarski i pri tom se zagreva od
temperature U2>411!L!do!U3>:11!L. Kiseonik dobija toplotu od dva toplotna izvora stalnih
temperatura. Odrediti:
a) promenu entropije izolovanog termodinami~kog sistema ako su temperature toplotnih izvora
UUJ2>711!L!i UUJ3>:11
b) temperaturu toplotnog izvora 2!)UUJ2* tako da promena entropije sistema bude minimalna kao i
minimalnu promenu entropije sitema u tom slu~aju
U
3
B
UJ3
UJ2
2
t
b*
∆Ttjtufn!>!∆TSU!,!∆TUJ2!,!∆TUJ2!>!///!>!21!−!5/66!−!4/14!>!3/53!
U
q
∆TSU!>! n ⋅ d qmo 3 − S hmo 3
U2
q2
:11
lK
> 21 ⋅ 1/:2 ⋅ mo
>21
411
L
∆TUJ2!>!!−!
R 2B
3841
lK
>///>!−!
>−!5/66!
711
L
UUJ2
∆TUJ3!>!!−!
R B3
3841
lK
>///>!−!
>−!4/14!
L
:11
UUJ3
dipl.ing. @eqko Ciganovi}
lK
L
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
r2B = d q ⋅ (UB − U2 ) > 1/:2 ⋅ (711 − 411) >384!
strana 4
lK
lh
R 2B = n ⋅ r2B = 21 ⋅ 384 >3841!lK
r B3 = d q ⋅ (U3 − UB ) > 1/:2 ⋅ (:11 − 711) >384!
lK
lh
R B3 = n ⋅ r B3 = 21 ⋅ 384 >3841!lK
b)
dq (UB − U2 ) dq (U3 − UB )
U
q
−
∆Ttjtufn>!g!)!UB!*> n ⋅ dqmo 3 − Shmo 3 −
U
q
UB
U3
2
2
U
∂)∆T tjtufn *
2
= −n ⋅ d q 23 −
∂)UB *
UB U3
U2
∂)∆Ttjtufn *
2
−
=1
⇔
⇒
=1
3
∂)UB *
UB U3
UB = U2⋅U3 > :11 ⋅ 411 >62:/72!L
Pri temperaturi toplotnog izvora UB>!62:/72!L!promena entropije sistema ima minimalnu
vrednost i ona iznosi:
d q (62:/72 − U2 ) d q (U3 − 62:/72)
U
q
∆Tnjo> n ⋅ d qmo 3 − S hmo 3 +
+
U2
q2
62:/72
U3
:11 1/:2 ⋅ (62:/72 − 411) 1/:2 ⋅ (:11 − 62:/72)
lK
∆Tnjo> 21 ⋅ 1/:2 ⋅ mo
+
+
>28/7: L
411
62:/72
:11
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 5
2/5/!Tokom kvazistati~ke (ravnote`ne) politropske ekspanzije n>3!lh idealnog gasa, do tri puta ve}e
zapremine od po~etne, temperatura gasa opadne sa U2>!711!L!na!U3>444!L i izvr{i se zapreminski rad
211!lK. Da bi se proces obavio na opisani na~in, radnom telu se dovodi 31!lK toplote. Skicirati promene
stawa idealnog gasa na qw!i!Ut dijagramu i odredite specifi~ne toplotne kapacitete pri stalnom
pritisku (dq*! i pri stalnoj zapremini!)dw*!datog gasa.
prvi zakon termodinamike za proces od 1 do 2
⇒
R23!>!∆V23!,!X23
dw =
R 23 − X23
31 − 211
lK
!>
>!1/261!
lhL
n ⋅ (U3 − U2 )
3 ⋅ (444 − 711)
U2 w 3
=
U3 w2
U2
711
mo
U3
444
o=
+2=
+ 2 >2/646
w3
mo 4
mo
w2
mo
o −2
⇒
X23>!n!/!x23!>! n ⋅ S h ⋅
Sh!>!
!q
R23!>!n!/!dw!/!)!U3!−!U2!*!,!X23
2
⋅ (U3 − U2 ) ⇒
o −2
− 211 ⋅ (2/646 − 2)
lK
= 1/211!
lhL
3 ⋅ (444 − 711)
⇒
!U
!2
o>2/646
Sh =
X23 ⋅ (o − 2)
=
n ⋅ (U3 − U2 )
dq!>!dw!,!Sh!>!1/361!
lK
lhL
!2
o>2/646
!3
!3
!w
dipl.ing. @eqko Ciganovi}
!t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 6
2/6/!Dvoatomni idealan gas )o>3!lnpm* ekspandira kvazistati~ki adijabatski od U2>711!L!do!U3>411!L a
zatim se od wega izobarski odvodi toplota dok mu temperatura ne dostigne U4>361!L. Odrediti koliko se
zapreminskog rada dobije za vreme ekspazije )lK* i kolika se toplota odvede od gasa za vreme izobarskog
hla|ewa )lK*/
X23 = n ⋅ x 23 = n ⋅ d w (U2 − U3 ) = o ⋅ )Nd w * ⋅ (U3 − U2 )
X23 = 3 ⋅ 31/9 ⋅ (411 − 711) >−23591!lK
R 34 = n ⋅ r34 = n ⋅ dq (U4 − U3 ) = o ⋅ )Ndq * ⋅ (U4 − U3 )
R 34 = 3 ⋅ 3:/2 ⋅ (361 − 411) >!−3:21!lK
2/7/!Termodinami~ki sistem ~ine 21!lh kiseonika (idealan gas) kao radna materija i okolina stalne
temperature Up>1pD kao toplotni ponor. Kiseonik mewa svoje stawe od 2)q>2!NQb-!U>561pD* do 3)q>2
NQb-!U>38pD* na povratan na~in (povratnim promenama stawa).Skicirati promene stawa idealnog gasa u
Ut koordinatnom sistemu i odrediti razmewenu toplotu izvr{eni zapreminski rad.
!2
U
!3
Up
!C
!B
!t
drugi zakon termodinamike za proces 1−2:
∆Ttjtufn!>!∆Tsbeop!ufmp!,!∆Tplpmjob
⇒
∆Tsbeop!ufmp!>!−∆Tplpmjob!
U
q
n ⋅ d qmo 3 − S hmo 3
U2
q2
⇒
R23!>Up/! n ⋅ dqmo
R
= ! 23
UP
U3
q
− Shmo 3
U2
q2
411
R23!>!384! ⋅ 21 ⋅ 1/:2 ⋅ mo
>!−!3296/37!lK
834
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 7
prvi zakon termodinamike za proces 1−2
⇒
R23!>!∆V23!,!X23!
X23!>!R23!−!n!/!dw!/)U3!.U2*
X23!>! −3296/37 − 21 ⋅ 1/76 ⋅ (411 − 834 ) >676/35!lK
2/8/ Sedam kilograma azota (idealan gas) mewa svoje stawe, na povratan na~in, od stawa 2)q>6!cbs-!u>2pD*
do stawa 3, pri ~emu se dobija zapreminski rad X>2257!lK. Od okoline (toplotnog izvora) stalne
temperature Up>32pD, azotu se dovodi R>2511!lK toplote. Odrediti temperaturu i pritisak radne
materije (azot) na kraju procesa i skicirati promene stawa radnog tela na U−t dijagramu
prvi zakon termodinamike za proces od 1 do 2
U3 = 385 +
U3 = U2 +
⇒
R23!>!∆V23!,!X23
R 23 − X23
n ⋅ dw
2511 − 2257
>!434/14!L
8 ⋅ 1/85
drugi zakon termodinamike za proces od 1 do 2
∆Ttj!>!∆Tsu!,!∆Tp
2
q 3 = q2 ⋅ fyq
S h
⇒!
R 23
U3
−
n ⋅ U + d q mo U
P
2
U
q R
1!>! n ⋅ dqmo 3 − Shmo 3 − 23
U2
q2 UP
=
2 2511
434/19
q 3 = 6 ⋅ 21 6 ⋅ fyq
+ 2/15 ⋅ mo
−
= !1/:!cbs
385
1/3:8 8 ⋅ 3:5
U
!3
!B
!C
!Up
!2
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
zadaci za ve`bawe:
strana 8
)2/9/!−!2/21/*
2/9/ 3 mola troatomnog idealnog gasa stawa )q>:!cbs-!U>484!L*!kvazistati~ki (ravnote`no) politropski
ekspandira do stawa )w3>5/!w2-!q>2/:!cbs*/!Skicirati proces na qw!j!Ut dijagramu i odrediti:
a) eksponent politrope, o
b) promene unutra{nje energije )lK*- entalipje )lK* i entropije radnog tela )lK0L*
c) koli~inu toplote koja se preda radnom telu )lK*-!u ovom procesu
a) o>2/23
b) ∆V23>.6!lK-!∆I23>.7/5!lK-!∆T23>31/23!K0L
c) R23!>!7/75!LK
2/:/ Idealan gas (helijum) mase n>3/6!lh izobarski (ravnote`no) mewa svoje toplotno stawe pri ~emu
mu se entropija smawi za 7/6!lK0L. Po~etna temperatura gasa iznosi 311pD. Temperatura toplotnog
rezervoara koji u~estvuje u ovom procesu je konstantna i jednaka je ili po~etnoj ili krajwoj
temperaturi radnog tela. Odrediti promenu entropije toplotnog rezervoara.
∆TUS!>9/54!lK0L
2/21/ Termodinami~ki sistem ~ine 4!lh vazduha (idealan gas) kao radna materija i okolina stalne
temperature Up>36pD kao toplotni ponor. Radna materija mewa svoje toplotno stawe od stawa 2)q>1/2
NQb-!u>61pD* do stawa 3)u>6pD* na povratan na~in (povratnim promenama stawa). Pri tome se okolini
predaje 661!lK toplote. Odrediti:
a) pritisak radne materije na kraju procesa
b) utro{eni zapreminski rad )lK* u procesu 1−2
c) skicirati promene stawa radnog tela na Ut dijagramu
a) q3>6/16!cbs
b) X23>−!563/9!LK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 9
2/22/ Idealan gas )n>2!lh* mewa svoje toplotno stawe od 2)q>:!cbs-!w>1/2!n40lh*!do 3)q>2!cbs*/ Prvi put
promena se obavwa kvazistati~ki po liniji 2B3 (vidi sliku) pri ~emu je zavisnost pritiska od zapremine
linearna. Drugi put promena se obavqa kvazistati~ki linijom 2C3 po zakonu qw3>dpotu, pri ~emu se
radnom telu dovodi 31!lK!toplote. Odrediti:
a) dobijeni zapreminski rad )X23* du` promena 2B3!i!2C3
b) koli~inu toplote )R23* dovedenu gasu du` promena 2B3
q
2
B
C
3
w
a)
q
w 3 = w 2 ⋅ 2
q3
3
3
n4
:
= 1/2 ⋅ >!1/4!
lh
2
w3
)X23 * B = n ⋅
∫
q)w*ew = n ⋅
q2 + q 3
⋅ (w 3 − w 2 )
3
w2
)X23 * B = 2 ⋅
: ⋅ 21 6 + 2 ⋅ 21 6
⋅ )1/4 − 1/2* >211!/214!lK
3
w3
)X23 *C = n ⋅
∫
q)w*ew =n ⋅
w2
∫
L ⋅ w −3 ew = −n ⋅ L ⋅ w −2
w3
w2
2
2
= −n ⋅ L ⋅
−
w
w
2
3
2
2
)X23 *C = −2 ⋅ : ⋅ 21 4 ⋅
−
>71!/214!lK
1/4
1/2
napomena:!
L = q2 ⋅ w 23 = : ⋅ 21 6 ⋅ 1/23 >:!/214!!
dipl.ing. @eqko Ciganovi}
K ⋅ n4
lh3
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 10
b)
prvi zakon termodinamike za proces 2C3:
)R 23 *C = ∆V23 + )X23 *C !!!!!)2*
prvi zakon termodinamike za proces 2B3:!
)R 23 * B = ∆V23 + )X23 * B !!!!!)3*
oduzimawem prethodne dve jedna~ine )2* i )3*!!dobija se:
)R23 * B = )X23 * B − )X23 *C , )R 23 *C >211!−!71!,!31!>!71!lK
2/23/ Jedan kilogram vazduha (idealan gas) stawa 2)q>25!cbs-!U>434!L* kvazistati~ki ekspandira do
stawa 2. Tokom ekspanzije zavisnost pritiska od zapremine je linearna. U toku procesa vazduhu se dovede
R23>216!lK toplote i pri tom se dobije X23>211!lK zapreminskog rada. Odrediti temperaturu i pritisak
vazduha stawa 2.
prvi zakon termodinamike za proces 2−3:!
R 23 = n ⋅ d w )U3 − U2 * + X23
w2
X23 = n ⋅
∫
q)w*ew = n ⋅
U3 = U2 +
R 23 = ∆V23 + X23
R 23 − X23
216 − 211
>441!L
= 434 +
n ⋅ dw
2 ⋅ 1/83
q + q3
q2 + q 3
⋅ (w 3 − w 2 ) = n ⋅ 2
⋅ Sh
3
3
w3
U
U
⋅ 3 − 2
q 3 q2
X
n ⋅ S h ⋅ U2 ⋅ q 33 + )3 ⋅ 23 ⋅ q2 + S h ⋅ U2 ⋅ q2 − S h ⋅ U3 ⋅ q2 * ⋅ q 3 − S h ⋅ U3 ⋅ q2 ⋅ q2 = 1
n
b ⋅ q 33 + c ⋅ q 3 + d >1
b> 2 ⋅ 398 ⋅ 434 >:3812
211 ⋅ 21 4
c> 2 ⋅ 3 ⋅
⋅ 25 ⋅ 21 6 + 398 ⋅ 434 ⋅ 25 ⋅ 21 6 − 398 ⋅ 441 ⋅ 25 ⋅ 21 6 > 3/88 ⋅ 2122
2
d>− 2 ⋅ 398 ⋅ 441 ⋅ 25 ⋅ 21 6 ⋅ 25 ⋅ 21 6 >− 2/97 ⋅ 2128
:3812⋅ q33 + 3/88 ⋅ 2122 ⋅ q 3 − 2/97 ⋅ 2128 = 1
q3 =
− 3/88 ⋅ 2122 ±
(3/88 ⋅ 21 )
dipl.ing. @eqko Ciganovi}
22 3
⇒
+ 5 ⋅ :3812⋅ 2/97 ⋅ 2128
3 ⋅ :3812
>!6/76!cbs
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 11
2/24/ Dvoatomni idealan gas!u koli~ini o>31!npm!sabija se ravnote`no od po~etne zapremine W2>1/:
51821 4968
+
,
n4 do krajwe zapremine W3>1/3!n4. Promena stawa gasa odvija se po jedna~ini: q(W ) =
W
W3
pri ~emu je pritisak izra`en u Qb a zapremina u n4. Odrediti:
a) pritisak i temperaturu gasa na po~etku i kraju procesa
b) izvr{eni nad radnim telom kao i razmewenu toplotu tokom ovog procesa
a)
q2 =
51821 4968 51821 4968
+
>!5:46:!Qb
+
>
1/:
W2
1/: 3
(W2 )3
q3 =
51821 4968 51821 4968
+
>!3:8386!Qb
+
>
1/3
W3
1/3 3
(W3 )3
(
)
⇒
U2 =
q2 ⋅ W2
5:46: ⋅ 1/:
>378/24!L
=
o ⋅ NS h
1/13 ⋅ 9426
(
)
⇒
U3 =
q 3 ⋅ W3
3:8386 ⋅ 1/3
>468/63!L
=
o ⋅ NS h
1/13 ⋅ 9426
q2 ⋅ W2 = o ⋅ NS h ⋅ U2
q 3 ⋅ W3 = o ⋅ NS h ⋅ U3
(
(
)
)
b)
W3
X23 =
∫
W2
W3
q)W*eW >
∫
W3
4968
51821 4968
+
eW > 51281 ⋅ mo W −
3
W
W
W
W2
>
W2
W
4968 4968
1/3 4968 4968
51281 ⋅ mo 3 −
> 51281 ⋅ mo
+
−
+
>!−86529/3!K
W
W
W
1
/:
1/3
1/:
2
3
2
prvi zakon termodinamike za proces 2−3:
R 23 = ∆V23 + X23
R 23 = o ⋅ (N ⋅ d w ) ⋅ (U3 − U2 ) + X23
R23>! 1/13 ⋅ 31/9 ⋅ (468/63 − 378/24 ) − 86/53 >−49/36!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 12
2/25/ Dvoatomnan idealna gas )o>3!lnpm* kvazistati~ki mewa stawe od 2)U>411!L-!q>2!cbs*!do
3)U>:11!L* po zakonu prave linije u Ut koordinatnom sistemu. Pri tome se radnom telu saop{tava 711
lK!rada. Odrediti pritisak radne materije stawa 2 i skicirati proces na Ut djagramu.
R 23 = ∆V23 + X23
prvi zakon termodinamike za proces 2−3:
R 23 = o ⋅ (N ⋅ d w ) ⋅ (U3 − U2 ) + X23
R23>! 3 ⋅ 31/9 ⋅ (:11 − 411) − 711 >!35471!lK
R 23 =
U2 + U3
⋅ ∆T23
3
⇒
∆T23 =
3 ⋅ R 23
3 ⋅ 35471
lK
>!51/7!
=
U2 + U3
411 + :11
L
U
q
⇒
∆T23 = o ⋅ N ⋅ d q ⋅ mo 3 − NS h ⋅ mo 3
U2
q2
U3 ∆T23
:11 51/7
3:/2 ⋅ mo
−
N ⋅ d q ⋅ mo U − o
411
3
2
> 2 ⋅ 21 6 ⋅ fyq
q 3 = q2 ⋅ fyq
NS h
9/426
q3!>5/18!/216!Qb!>!5/18!!cbs
(
(
)
(
)
)
(
)
U
3
2
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 13
2/26/ Idealan gas sabija se kvazistati~ki od temperature U2>384!L do temperature U3>984!L po zakonu:
U
+ D (!L>−215!lK0L>dpotu!!i!D>dpotu). Odrediti nepovratnost ove promene stawa ,!)∆TtjT = L ⋅ mo
U2
lK0L*-!ako se toplota predaje izotermnom
toplotnom ponoru temperature UUQ>U2 i grafi~ki je predstaviti na Ut dijagramu
∆Ttjtufn!>!∆TSU!,!∆TUQ!>!///!>!−231/:!,!339/7!>218/8!
∆TSU!>T3!−!T2>///>!−231/:!
T3 = L ⋅ mo
U3
+D
U2
lK
L
kJ
K
T2 = L ⋅ mo
)3*
U2
+D
U2
)2*
Oduzimawem pretnodne dve jedna~ine dobija se:
U
984
lK
T 3 − T2 = L ⋅ mo 3 = −215 ⋅ mo
>−231/:!
L
U2
384
T3
∫
∆TUQ> −
U)T*eT
T
R 23
=−
Uuq
Uuq
= /// = L ⋅
T3
napomena:
∫
U3 − U2
984 − 384
lK
= −215 ⋅
>339/7
Uuq
384
L
T3
U)T*eT = /// =
T2
U2 ⋅ L
∫
U2
T −D
⋅ f L eT
= U2 ⋅ L
T −D
⋅f L
T2
T3 − D
⋅ )f L
−
T2 − D
f L *
T3
=
T2
= U2 ⋅ L ⋅ )f
U
mo 3
U2
− 2* = L ⋅ )U3 − U2*
Postupak grafi~kiog predstavqawa promene entropije sistema zasnovan je na jednakosti
povr{ina ispod:
1.
linije kojom predstavqamo promenu stawa radnog tela
2.
linije kojom predstavqamo promene stawa toplotnog ponora
Obe ove povr{ine predstavqaju razmewenu toplotu izme|u
radno tela i toplotnog ponora.
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 14
U
3
UUQ
2
∆TSU
t
∆TUQ
∆TTJ
zadaci za ve`bawe:
)2/27/!−!2/28/*
2/27/ Vazduh (idealan gas) kvazistati~ki mewa toplotno stawe od stawa 2)U>411!L*!do stawa!3)U>711!L*!i
pri tome je w2>3/!w3. Prvi put se promena vr{i po zakonu prave linije u Ut!kordinatnom sistemu, a drugi
put se od stawa 1 do stawa 2 dolazi kvazisati~kom politropskom promenom stawa. Odrediti koliko se
lK
lK
toplote )lK0lh*!dovede vazduhu u oba slu~aja .
)r23 *qsbwb = 246/2
- )r23 *qpmjuspqb = 23:/7
lh
lh
Sh ⋅ U
b
,
w3
)b>82/87!On50lh3-!c> 9/138 ⋅ 21 −5 n40lh!j!Sh>79/9!K0)lhL**-!lwb{jtubuj•lj!izotermski ekspandira pri
temperaturi od 1pD od w2>1/16!n40lh do w3>1/3!n40lh. Odrediti:
a) po~etni i krajwi pritisak gasa kao i dobijeni zapreminski rad tokom ekspanzije
b) po~etni i krajwi pritisak gasa kao i dobijeni zapreminski rad tokom ekspanzije kada bi navedeni
gas posmatrali kao idealan gas iste gasne konstante )Sh*
2/28/ Neki gas koji se pona{a saglasno jedna~ini stawa:
q)w* =
w −c
−
a) q2>4/64!cbs-!q3>1/:3!cbs-!x23>36/2:!lK0lh
b) q2>4/87!cbs-!q3>1/:5!cbs-!x23>37/15!lK0lh
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 15
NEKVAZISTATI^KE (NERAVNOTE@NE)
PROMENE STAWA IDEALNIH GASOVA
2/29/ Vazduh (idealan gas) stawa 2)q2>23!cbs-!U2>366pD* ekspandira nekvazistati~ki adijabatski sa
stepenom dobrote η fy
e >1/9 do stawa 3)q3>2!cbs*/!Odrediti:
a) temperaturu vazduha nakon ekspanzije
b) prira{taj entropije radnog tela usled mehani~ke neravnote`e
c) zakon nekvazistati~ke promene stawa u obliku qwn>jefn
U
q2
2
q3
B
3
3l
t
a)
U3L
q
= U2 ⋅ 3L
q2
η fy
e =
κ .2
κ
2
= 639 ⋅
23
1.4.2
1.4
>36:/7!L
U2 − U3
-!!!!! U3 = U2 + ηEfy ⋅ (U3L − U2 ) = 639 + 1/9 ⋅ (36:/7 − 639) >424/4!L
U2 − U3l
b)
qB
U
= q2 ⋅ B
U2
κ
κ .2
424/4
= 23 ⋅ 21 6 ⋅
639
2/5 .2
2/5
= 2/:4 ⋅ 21 6 Qb
U
q
2
K
∆t nfi = ∆t B3 = dqmo 3 − S hmo 3 = −398 ⋅ mo
>299/82!
qB
2/:4
lhL
UB
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 16
c)
!
S h ⋅ U2
w2 =
q2
w3 =
S h ⋅ U3
q3
=
398 ⋅ 639
=
23 ⋅ 21
6
>!1/2374!
398 ⋅ 424/4
2 ⋅ 21
6
n4
lh
>!1/9::3!
n4
lh
qw n = jefn
⇒
q2 ⋅ w 2n = q 3 ⋅ w n
3
q2
23
mo
q3
2
n=
>2/37
=
w3
1/9::3
mo
mo
1/2374
w2
⇒
qw 2/37 = jefn
mo
2/2:/ Kompresor proizvo|a~a B radi izme|u pritisaka qnjo!>2!cbs!i qnby>!:!cbs. Kompresor
proizvo|a~a C radi izme|u pritisaka qnjo!>2/6!cbs i!!qnby>!21!cbs. U oba slu~aja radni fluid je
vazduh (idealan gas) po~etne temperature!U2>41pD. Temperature vazduha na izlazu iz oba kompresora
su jednake. Odrediti koji je kompresor kvalitetniji sa termodinami~kog aspekta, predpostavqaju}i da
su kompresije adijabatske
Sa termodinami~kog aspekta kvalitetniji je ona kompresija kod koje je
1. na~in:
2. na~in:
ve}i stepen dobrote adijabatske kompresije
mawa promena entropije sistema
1. na~in:
q
U2 − U2 ⋅ 3L
q
2 B
B = U2 − U3LB =
ηe
U2 − U3
U2 − U3
q
U2 − U2 ⋅ 3L
q
U2 − U3LC
2
C
ηe =
=
U2 − U3
U2 − U3
dipl.ing. @eqko Ciganovi}
κ −2
l
κ −2
l
C
κ −2
q
l
U2 ⋅ 2 − 3L
q2
B
=
U2 − U3
q
U2 ⋅ 2 − 3L
q
2
=
U2 − U3
κ −2
l
C
)2*
)3*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 17
Deqewem prethodne jedna~ina (1) i (2) dobija se:
B
ηe
ηC
e
2/5 − 2
q κ −2
: 2/5
2 − 3L
κ
2−
q
2 B
2
= 2/32
=
=
2/5 − 2
q κ −2
3
L
2−
21 2/5
q κ
2−
2 C
2/6
Po{to je koli~nik stepena dobrote ve}i od 1 to zna~i da je stepen dobrote
kompresora proizvo|a~a B ve}i od stepena dobrote kompresora proizvo|a~a
C, pa je kompresor proizvo|a~a B kvalitetniji sa termodinami~kog aspekta.
Uo~iti da je zadatak mogao biti re{en i bez zadate temperature U2.
2. na~in:
(∆t tj )B
= d q ⋅ mo
U3 B
q
− S h ⋅ mo 3 B − ∆t p
U2B
q2B
)4*
(∆t tj )C
= d q ⋅ mo
U3C
q
− S h ⋅ mo 3C − ∆t p
U2C
q2C
)5*
Oduzimawem jedna~ina!)4*!j!)5*!epcjkb!tf;
(∆t tj )B − (∆t tj )C
q
q
= −S h ⋅ mo 3 B − mo 3C
q2C
q2B
(∆t tj )B < (∆t tj )C
⇒
K
21
:
> − 398 ⋅ mo − mo
>−97/24!
lhL
2/6
2
kompresor proizvo|a~a A je kvalitetniji
sa termodinami~kog aspekta.
U
q3
3
B
3l
q2
2
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 18
2/31/!Pet kilograma kiseonika (idealan gas) ekspandira nekvazistati~ki politropski po zakonu
qw2/2>jefn, od stawa 2)q2>8!cbs-!w2>1/23!n40lh* do stawa 3)U3>−2pD*/ Specifi~na toplota ove promene
stawa iznosi d23>−761!K0lhL. Skicirati proces na Ut dijagramu i odrediti:
a) prira{taj entropije radnog tela usled mehani~ke i usled toplotne neravnote`e
b) promenu entropije izolovanog termodinami~kog sistema ako je temperatura toplotnog izvora 484!L
U
q2
2
o
q3
n
B
3
3l
t
a)
q ⋅w
8 ⋅ 216 ⋅ 1/23
U2 = 2 2 =
= 434 L
Sh
371
[blpo!qspnfof!qwn!>jefn-!!usbotgpsnj|fnp!v!pcmjl;! U n ⋅ q2 − n = jefn
n
U 2− n
U2n ⋅ q22 − n = U3n ⋅ q32 − n
⇒
q 3 = q2 ⋅ 2
U
3
2/2
6 434 2 − 2/2
>2/17!cbs
q 3 = 8 ⋅ 21 ⋅
383
d − dw ⋅ κ
o−κ
⇒
o = 23
⇒
dolw!>!dlw
d23 = d w
o −2
d23 − d w
o=
−761 − 831 ⋅ 2/5
>2/32
− 761 − 831
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 19
o
2 / 32
o − 2
6 383 2 / 32 − 2
> 3/7 ⋅ 21 6 !Qb
= 8 ⋅ 21 ⋅
434
U
q
2/17
lK
∆Tnfi/ofs/>!∆TB3>! n ⋅ d qmo 3 − S hmo 3 > 6 ⋅ − 1/398 ⋅ mo
>2/3:!
UB
qB
3/7
L
U
q B = q2 ⋅ B
U
2
∆Tupq/ofs/>!∆T2B>! n ⋅ dqmo
UB
q
− Shmo B
U2
q2
lK
383
2/17
> 6 ⋅ 2 ⋅ mo
− 1/398 ⋅ mo
>1/54!
434
3
/
7
L
∆Tsbeop!ufmp!>!∆Tnfi/ofs/!,!∆Tupq/ofs/!>2/3:,1/54>2/83!
lK
L
b)
∆Ttjtufn!>!∆Tsbeop!ufmp!,!∆Tupqmpuoj!qpops!>!///!>!2/83!−!1/55!>2/39!
∆Tupqmpuoj!j{wps!>!−
lK
L
R 23
277/22
lK
=−
= −1/55
UUJ
484
L
R 23 = R 2B + R B3 >///>
R 2B = n ⋅ d w ⋅
o−κ
2/32 − 2/5
⋅ (UB − U2 ) > 6 ⋅ 1/83 ⋅
⋅ (383 − 434) >277/22!lK
2/32 − 2
o −2
2/32/!Termodinami~ki sistem sa~iwava n>6!lh azota (idealan gas) i okolina temperature Up>38pD.
Azot nekvazistati~ki politropski mewa toplotno stawe od stawa 2)q>21!cbs-!U>566pD* do stawa
3)U>98pD*/!Specifi~ni toplotni kapacitet promene stawa 1−2 iznosi d23>481!K0lhL a nepovratnost
procesa 1−2 iznosi ∆Ttj>2/56!lK0L. Skicirati proces na Ut dijagramu i odrediti stepen dobrote ove
promene stawa.
!U
!q2
!2
!o
q3
n
!B
!3
!3l
dipl.ing. @eqko Ciganovi}
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
⇒
dolw!>!dlw
o=
strana 20
d23 = d w
o−κ
o −2
⇒
o=
d23 − d w ⋅ κ
d23 − d w
481 − 851 ⋅ 2/5
>2/9
481 − 851
R23 = R2B + R B3 !>///>
R 2B = n ⋅ d w ⋅
2/9 − 2/5
o−κ
⋅ (471 − 839) >!−!791/9!lK
⋅ (UB − U2 ) > 6 ⋅ 1/85 ⋅
2/9 − 2
o −2
U
q R
∆TTJ = n ⋅ dq mo 3 − S h mo 3 − 23
U2
q2 U1
2 ∆T TJ
R 23
U
q 3 = q2 ⋅ fyq−
+
− d q mo 3 =
n ⋅ UP
U2
S h n
∆TTJ!>!∆TSU!,!∆Tp
⇒
2 2/56 791/9
471
q 3 = 21 ⋅ 21 6 ⋅ fyq−
−
− 2/15 ⋅ mo
= 2/59!!cbs
6 ⋅ 411
839
1/3:8 6
q
U2
= 2
U3l q 3l
q3l!>!q3
o−2
o
!!!!!!!!!⇒ !!!!!!!! U3l
⇒
dipl.ing. @eqko Ciganovi}
ηFY
E =
q
= U2 ⋅ 3l
q2
o−2
o
2/59
> 839 ⋅
21
2/9 −2
2/9
= 422/5!!L
U2 − U3
839 − 471
>1/99
=
U2 − U3l
839 − 422/5
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 21
2/33/! Kiseonik (idealan gas) sabija se nekvazistati~ki politropski od stawa
2)q>2!cbs-!U>384!L* do stawa 3)q>7!cbs-!U>554!L*/ U toku procesa sabijawa od kiseonika se odvodi
431!lK0lh toplote. Skicirati proces na Ut dijagramu i odrediti stepene dobrote ove promene stawa.
r23 = r2B + r B3 !>!−431!
r2B = d w ⋅
lK
lh
o−κ
⋅ (UB − U2 )
o −2
r2B
− dw ⋅ κ
UB − U2
o=
r2B
− dw
UB − U2
⇒
−431
− 1/76 ⋅ 2/5
554
− 384
>2/2
o=
− 431
− 1/76
554 − 384
q
U2
= 2
U3l q 3l
ηlq
E =
o−2
o
⇒
q
= U2 ⋅ 3l
q2
U3l
o−2
o
7
> 384 ⋅
2
2/2−2
2/2
>432/4!L
U2 − U3l
384 − 432/4
>1/39
=
U2 − U3
384 − 554
q3
!U
!3
!B
!3l
!o
!q2
n
!2
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 22
2/34/ Tokom nekvazistati~kog sabijawa n>4!lh butana (idealan gas) od stawa 2)q2>2!cbs-!U2>31pD* do
stawa 3)q3>41!cbs-!T3>T2*- spoqa{wa mehani~ka sila izvr{i rad od 961!lK. Tokom procesa radna
materija predaje toplotu toplotnom ponoru stalne temperature Uq>1pD. Skicirati proces u na Ut
dijagramu i odrediti:
a) promenu entropije termodinami~kog sistema tokom posmatrane promene stawa
b) stepen dobrote nekvazistati~ke kompresije
q3
!U
!B
!3
n>κ
!3l
!q2
!o
!2
t
zakon nkv. promene stawa 1−2:
U2 q2
=
U3 q 3
κ −2
κ
⇒
κ −2
q κ
U3 = U2 ⋅ 3
q
2
1.28 − 2
41 2/39
U3 = 3:4 ⋅
>!727/7!L
2
prvi zakon termodinamike za proces!2−3;
R23>∆V23,X23
R23 = n ⋅ dw ⋅ (U3 − U2) + X23 > 4 ⋅ 1/6 ⋅ (727/7 − 3:4) − 961 >−475/7!LK
drugi zakon termodinamike za proces!2−3;
∆TTJ!>!∆TSU!,!∆TUQ
lK
L
−475/7
R23
lK
>2/45!!
∆TUQ = −
=−
L
UUQ
384
lK
∆TTJ!>!2/45!,!1!>!2/45!
L
∆TSU!>!1!
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
b)
strana 23
R 23 = R 2B + R B3 !>!−475/7!lK
R 2B = n ⋅ d w ⋅
o−κ
⋅ (UB − U2 )
o −2
R 2B
− dw ⋅ κ
n ⋅ (UB − U2 )
o=
R 2B
− dw
n ⋅ (UB − U2 )
⇒
−475/7
− 1/6 ⋅ 2/39
4 ⋅ (727/7 − 3:4 )
>2/27
o=
− 475/7
− 1/6
4 ⋅ (727/7 − 3:4)
q
U2
= 2
U3l q 3l
ηlq
E =
o−2
o
⇒
U3l
q
= U2 ⋅ 3l
q2
o−2
o
41
> 3:4 ⋅
2
2/27 −2
2/27
>579/5!L
U2 − U3l
3:4 − 579/5
>1/65
=
U2 − U3
3:4 − 727/7
2/35/ [est kilograma troatomnog idealnog gasa mewa toplotno stawe nekvazistati~ki po zakonu
qwn>jefn!)n>κ* (tj. nekvazistati~ki izentropski) od stawa!2)q2>41!cbs-!U2>727/7!L* do stawa 3)q3>2
cbs*/ Tokom ove promene stawa specifi~na zapremina gasa se pove}a za 1/49!n40lh i pri tome se dobije
811!lK mehani~kog rada. Skicirati promenu stawa idealnog gasa na Ut dijagramu i odrediti:
a) koli~inu razmewene toplote tokom ove promene stawa
b) stepen dobrote ove nekvazistati~ke promene
c) porast entropije radnog tela usled mehani~ke neravnote`e )lK0L*
!q2
!U
!2
n>κ
!o
!B
q3
!3
!3l
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 24
b*
zakon nkv. promene stawa 1−2:
U2 q2
=
U3 q 3
κ −2
κ
⇒
κ −2
q κ
U3 = U2 ⋅ 3
q
2
1.28 − 2
2 2/39
>!3:4!L
U3 = 727/7 ⋅
41
jedna~ina stawa idealnog gasa za kraj procesa:
q2 ⋅ w2 = Sh ⋅ U2 !!!!!!)2*
q3 ⋅ w 3 = S h ⋅ U3 !!!!)3*
uslov zadatka:
w 3 − w 2 = 1/49
jedna~ina stawa idealnog gasa za po~etak procesa:!
n4
!)4*
lh
Re{avawem prethodnog sistema tri jedna~ine sa 3 nepoznate dobija se:
w 2 = 1/1396
N=
o=
n4
n4
K
- w 3 = 1/517
- S h = 249/69
lh
lh
lhL
SV
9426
lh
=
= 71
S h 249/69
lnpm
n
7
=
>1/2!lnpm
N 71
prvi zakon termodinamike za proces 1−2:
R 23 = ∆V23 + X23
R 23 = o ⋅ )Nd w * ⋅ (U3 − U2 ) + X23 = 1/2 ⋅ 3:/2 ⋅ (3:4 − 727/7 ) + 811
b)
R 23 = R 2B + R B3 !>!−352/79!lK
R 2B
o−κ
= n ⋅ dw ⋅
⋅ (UB − U2 )
o −2
⇒
>−352/79!lK
(Nd w )
R2B
−
⋅κ
n ⋅ (UB − U2 )
N
o=
R2B
(Nd w )
−
n ⋅ (UB − U2 )
N
−352/79
3:/2
−
⋅ 2/39
7 ⋅ (3:4 − 727/7 ) 71
o=
>2/49
3:/2
− 352/79
−
7 ⋅ (3:4 − 727/7 ) 71
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
q
U2
= 2
U3l q 3l
ηEfy =
o−2
o
⇒
strana 25
U3l
q
= U2 ⋅ 3l
q2
o−2
o
2
> 727/7 ⋅
41
2/49 −2
2/49
>352/8!L
U2 − U3
727/7 − 3:4
>1/97
=
U2 − U3l
727/7 − 352/8
c)
U
q
∆T nfi/ofs/ = n ⋅ ∆t B3 = n ⋅ d q mo 3 − S h mo 3
UB
qB
Zakon kvazistati~ke promene stawa!2−B;
qB
U
= q2 ⋅ B
U2
o
2
lK
> 7 ⋅ (− 249/69 ) ⋅ mo >1/69!
3
L
U2 q2
=
UB q B
o−2
o
2/49
o−2
3:4 2/49−2
= 41 ⋅ 21 6 ⋅
>!3!cbs
727/7
2/36/ Tri kilograma vazduha (idealan gas) stawa!2)q2>3!cbs-!U2>261pD*!mewa svoje toplotno stawe
nekvazistati~ki (neravnote`no) izotermski do stawa 3)q3>9!cbs*. Promena entropije radne materije
usled mehani~ke neravnote`e iznosi
∆tnfi>349!K0lhL. Odrediti:
a) dovedeni rad i odvedenu toplotu tokom ove promene stawa )X23-!R23*
b) stepen dobrote izotermske kompresije )ηelq*
qB
U
q3
q2
B
3>3l
2
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 26
a)
∆t nfi = ∆t B3 = d qmo
U3
q
− S hmo 3
UB
qB
⇒
∆t
q B = q 3 ⋅ fyq nfi
Sh
349
6
q B = 9 ⋅ 21 6 ⋅ fyq
> 29/4 ⋅ 21 !Qb
398
R 23 = R 2B + R B3 = /// = −917/4/4 lK
R 2B > n ⋅ U2 ⋅ S h ⋅ mo
q2
3
> 4 ⋅ 534 ⋅ 398 ⋅ mo
>−917/4!lK
qB
29/4
R 23 = ∆V23 + X23
prvi zakon termodinamike za proces 1−2:
X23>!R23!>−917/4!lK
b)
ηlq
E =
X23L
−615/:
= /// =
= 1/74
X23
− 917/4
X23L = n ⋅ S h ⋅ U ⋅ mo
q2
3
= 4 ⋅ 398 ⋅ 534 ⋅ mo >−615/:!lK
q 3L
9
2/37/ Vazduh (idealan gas) stawa!2)q2>:!cbs-!U2>261pD*!mewa svoje toplotno stawe nekvazistati~ki
(neravnote`no) izotermski do stawa 3)q3>2!cbs*. Promena entropije radne materije usled mehani~ke
neravnote`e )∆tnfi* i promena entropije radne materije usled toplotne neravnote`e )∆tupq* su jednake.
Odrediti stepen dobrote ove nekvazistati~ke izotermske ekspanzije )ηefy*/
q2
qB
U
q3
2
B
3>3l
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
∆tupq!>!∆tnfi
d q mo
⇒
strana 27
∆t2B!>!∆tB3
UB
q
U
q
− S h mo B > d q mo 3 − S h mo 3
U2
UB
q2
qB
⇒
q B = q2 ⋅ q 3
q B = : ⋅ 21 6 ⋅ 2 ⋅ 21 6 > 4 ⋅ 21 6 !Qb
X23L = n ⋅ S h ⋅ U ⋅ mo
q2
:
= 2 ⋅ 398 ⋅ 534 ⋅ mo >377/86!lK
q 3L
2
R 23 = R 2B + R B3 = /// = −917/4/4 lK
R 2B > n ⋅ U2 ⋅ S h ⋅ mo
q2
:
> 2 ⋅ 534 ⋅ 398 ⋅ mo >244/48!lK
qB
4
prvi zakon termodinamike za proces 1−2:
R 23 = ∆V23 + X23
X23>!R23!>244/48!lK
η fy
e =
X23
244/48
>
>1/6
377/86
X23l
zadaci za ve`bawe:
)2/38/−2/39/*
2/38/!Vazduh (idealan gas) po~etnog stawa 2)q>6!cbs-!w>1/337!n40lh* ekspandira nekvazistati~ki
politropski do stawa 3)q>2!cbs-!U>3:4!L*/!Tokom ove promene stawa od radne materije ka okolini se
odvede 27!lK0lh toplote. Odrediti stepen dobrote ove promene stawa kao i promenu entropije vazduha
lK
samo usled mehani~ke neravnote`e.
! η fy
e >1/73-!!∆tnfi>1/37!
lhL
2/39/!Vazduh (idealan gas) stawa 2)q2>1/3!NQb-!u2* mewa svoje toplotno stawe nekvazistati~ki
(neravnote`no) izotermski do stawa 3)q3?q2*/ Promena entropije radne materije usled mehani~ke
neravnote`e iznosi 349!K0lhL/ Stepen dobrote ove nekvazistati~ke promene stawa iznosi ηe>1/95.
Odrediti pritisak vazduha na kraju procesa )q3*/
q3 = 266/6 cbs
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 28
PRVI I DRUGI ZAKON TERMODINAMIKE
(ZATVOREN TERMODINAMI^KI SISTEM)
2/3:/ U vertikalno postavqenom cilindru, od okoline
adijabatski izolovanom, (slika), unutra{weg pre~nika e>711
nn, nalazi se vazduh (idealan gas) temperature 31pD. Sud je
zatvoren klipom zanemarqive mase, koji se mo`e kretati bez
trewa. Na klipu se nalazi teg mase nu>3111!lh. U polaznom
!∆{>411!nn
polo`aju ~elo klipa se nalazi na visini {>611!nn u odnosu na
dowu bazu cilindra. U cilindru se nalazi elektri~ni greja~
pomo}u kojeg se vazduhu dovodi toplota. Pritisak okoline
!{>611
iznosi qp>2!cbs. Odrediti:
a) koli~inu toplote koju greja~ treba da preda gasu tako da se
klip u procesu pomeri za ∆{>411!nn
b) vreme trajawa procesa ako snaga elektri~nog greja~a
,R23
iznosi 2/77!lX
c) rad koji bi izvr{io gas u cilndru ako bi se u trenutku dostizawa stawa 2 istovremeno iskqu~io
greja~ i skino teg sa klipa
d) skicirati sve procese sa radnim telom na qw i Ut dijagramu
a)
e3 ⋅ π
1/7 3 ⋅ π
⋅{ =
⋅ 1/6 = 1/2525 n 4
5
5
e3 ⋅ π
1/7 3 ⋅ π
⋅ ({ + ∆{ ) =
⋅ (1/6 + 1/4 ) = 1/3373 n 4
W3 =
5
5
W2 =
jedna~ina stati~ke ravnote`e za proizvoqan polo`aj klipa:
n ⋅h
3111 ⋅ :/92
q = q p + 3u
= 2 ⋅ 21 6 +
= 2/8 ⋅ 21 6 Qb
e ⋅π
1/7 3 ⋅ π
5
5
jedna~ina stawa idealnog gasa na po~etku procesa:!
n=
q ⋅ W2
2/8 ⋅ 21 ⋅ 1/2525
=
= 1/3: lh
S h ⋅ U2
398 ⋅ 3:4
jedna~ina stawa idealnog gasa na kraju procesa:!
U3 =
q ⋅ W2 = n ⋅ Sh ⋅ U2
6
q ⋅ W3 = n ⋅ Sh ⋅ U3
q ⋅ W3
2/8 ⋅ 21 6 ⋅ 1/3373
=
= 573 L
n ⋅ Sh
1/3: ⋅ 398
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 29
R 23 = ∆V23 + X23
prvi zakon termodinamike za proces 1−2:
R23> n ⋅ d w ⋅ (U3 − U2 ) + q ⋅ (W3 − W2 ) >
R23>! 1/3: ⋅ 1/83 ⋅ (573 − 3:4 ) + 2/8 ⋅ 21 6 ⋅ 21 −4 ⋅ (1/3373 − 1/2525 ) >5:/9!lK
b)
τ=
R 23
⋅
=
R 23
5:/9
= 41 t
2/77
c)
napomena:
U3 q3
=
U4 q4
κ −2
κ
proces 2−3 je kvazistati~ki adijabatski, q4>qp>!2!cbs
q
U4 = U3 ⋅ 4
q3
⇒
κ −2
κ
2⋅ 216
= 573 ⋅
2/8 ⋅ 216
2/5 −2
2/5
>4:8!L
R 34 = ∆V 34 + X34
prvi zakon termodinamike za proces 2−3:
X34> −n ⋅ d w ⋅ (U4 − U3 ) >− 1/3: ⋅ 1/83 ⋅ (4:8 − 573) >24/68!lK
q
U
2
3
3
4
2
w
dipl.ing. @eqko Ciganovi}
4
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 30
2/41/!Cilindar je napravqen prema navedenoj skici. Klip je optere}en tegom nepoznate mase i le`i na
osloncu A. U cilindru se nalazi azot stawa!2)q>3/6!cbs-!U>3:4!L*/!Dovo|ewem!23/6!lK!toplote
zapremina azota se udvostru~i. Pritisak okoline iznosi!!qp>!2!cbs-!masa klipa je zanemarqiva a klip
se kre}e bez trewa.!Odrediti:
a) masu tega
b) pri kojoj temperaturi azota u cilindru |e se pokrenuti klip
c) promenu potencijalne energije tega
E
e>291!nn
E>311!nn
n
{>461!nn
{
e
a)
2−3!proces u cilindru do pokretawa klipa!
3−4!proces u cilndru nakon pokretawa klipa!
W2 =
e3 ⋅ π
1/29 3 ⋅ π
⋅{ =
⋅ 1/46 = 1/119: n 4
5
5
jedna~ina stawa idealnog gasa na po~etku procesa:!
n=
)w>dpotu*
)q>dpotu*
q2 ⋅ W2 = n ⋅ S h ⋅ U2
q2 ⋅ W2 3/6 ⋅ 21 ⋅ 1/119:
=
= 1/1367 lh
S h ⋅ U2
3:8 ⋅ 3:4
6
W3!>!W2>1/119:!n4
W4 = 3 ⋅ W2 > 3 ⋅ 1/119: >1/1289!n4
jedna~ina stawa idealnog gasa na kraju procesa:
n ⋅ S h ⋅ U4
q 4 ⋅ W4 = n ⋅ S h ⋅ U4
⇒
! q4 =
W4
dipl.ing. @eqko Ciganovi}
)2*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 31
prvi zakon termodinamike za proces 1−3:
R 23 + R 34 = ∆V23 + ∆V 34 + X23 + X34 > n ⋅ d w ⋅ (U4 − U2 ) + q 4 ⋅ (W4 − W3 ) !!!!!!!)3*
kada jedna~inu (1) uvrstimo u jedna~inu (2) dobija se:
n ⋅ S h ⋅ U4
⋅ (W4 − W3 )
⇒
R 23 + R 34 > n ⋅ d w ⋅ (U4 − U2 ) +
W4
R 23 + R 34 + n ⋅ d w ⋅ U2
23/6 + 1/1367 ⋅ 1/85 ⋅ 3:4
=
1/1289 − 1/119:
W4 − W3
1/1367 ⋅ 1/85 + 1/1367 ⋅ 1/3:8 ⋅
n ⋅ d w + n ⋅ Sh ⋅
1/1289
W4
U4!>8:4/7!L
U4 =
q4 =
1/1367 ⋅ 3:8 ⋅ 8:4/7
> 4/5 ⋅ 21 6 Qb
1/1289
jedna~ina stati~ke ravnote`e za proizvoqan polo`aj za proces 2−3
n ⋅h
q − qp E3 ⋅ π
⇒
⋅
q 4 = q p + 3u
nu = 4
5
h
E ⋅π
5
nu =
6
4/5 ⋅ 21 − 2 ⋅ 21 6 1/3 3 ⋅ π
⋅
>879/7!lh
5
:/92
b)
jedna~ina stawa idealnog gasa za stawe 2:!
U3 =
q 3 ⋅ W3 = n ⋅ S h ⋅ U3
q 3 ⋅ W3
4/5 ⋅ 21 6 ⋅ 1/119:
=
= 4:9 L
n ⋅ Sh
1/1367 ⋅ 3:8
c)
∆Fq> n u ⋅ h ⋅ ∆{ > n u ⋅ h ⋅
dipl.ing. @eqko Ciganovi}
W4 − W3
3
E ⋅π
5
> 879/7 ⋅ :/92 ⋅
1/1289 − 1/119:
1/3 3 ⋅ π
5
>3247!K
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 32
2/42. Vertikalni cilindar zatvoren je klipom mase nl>:!lh, ~iji je hod ograni~en na kraju cilindra
(slika). U cilindru se nalazi dvoatoman idealan gas stawa!2)q>2/6!cbs-!U>561pD*/ Odrediti:
a) za koliko }e se spustiti klip (zanemariti trewe) dovo|ewem vazduha u mehani~ku i toplotnu
ravnote`u sa okolinom stawa P)q>2!cbs-!U>31pD*
b) koliko se toplote pri tome preda okolini do trenutka pokretawa klipa a koliko nakon
pokrtetawa klipa do trenutka dostizawa ravnote`e sa okolinom
Skicirati procese na qw i Ut dijagramu
nl
e>211!nn
{
{>911!nn
e
2−3!proces u cilindru do pokretawa klipa!
3−4!proces u cilndru nakon pokretawa klipa!
q
U
2
4
)w>dpotu*
)q>dpotu*
2
3
3
4
w
dipl.ing. @eqko Ciganovi}
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 33
a)
W2 =
e3 ⋅ π
1/23 ⋅ π
⋅{ =
⋅ 1/9 = 1/1174 n4
5
5
jedna~ina stawa idealnog gasa na po~etku procesa:!
o=
(
)
q2 ⋅ W2 = o ⋅ NS h ⋅ U2
6
q2 ⋅ W2
2/6 ⋅ 21 ⋅ 1/1174
=
= 2/68 ⋅ 21.5 lnpm
NSh ⋅ U2
9426 ⋅ 834
(
)
jedna~ina stati~ke ravnote`e za polo`aj klipa u stawu 2
n ⋅h
: ⋅ :/92
q3 = qp + l
= 2 ⋅ 21 6 +
> 2/2 ⋅ 21 6 !Qb
3
3
E ⋅π
1/2 ⋅ π
5
5
W3!>!W2>1/1174!n4-
q4>q3- !
U4>Up
(
W4 =
(
)
o ⋅ NS h ⋅ U4
q4
W3 − W4 =
=
2/68 ⋅ 21 −5 ⋅ 9426 ⋅ 3:4
e3 ⋅ π
⋅ ∆{
5
2 ⋅ 21 6
⇒
∆{ =
>1/1149!n4
W3 − W4
3
e ⋅π
5
b)
jedna~ina stawa idealnog gasa za stawe 2:!
U3 =
)
q 4 ⋅ W4 = o ⋅ NS h ⋅ U4
jedna~ina stawa idealnog gasa na kraju procesa:!
=
1/1174 − 1/1149
1/23 ⋅ π
5
=0.318 m
(
)
q 3 ⋅ W3 = o ⋅ NS h ⋅ U3
q 3 ⋅ W3
2/2 ⋅ 21 6 ⋅ 1/1174
>641/96!L
=
o ⋅ NS h
2 ⋅ 68 ⋅ 21 .5 ⋅ 9426
(
)
prvi zakon termodinamike za proces 1−2:
R23 > o ⋅ (Nd w ) ⋅ (U3 − U2 ) > 2/68 ⋅ 21
−5
R 23 = ∆V23 + X23
⋅ 31/9 ⋅ (641/96 − 834) >−1/74!lK
prvi zakon termodinamike za proces 1−3:
R 34 = ∆V34 + X34
R 34 > o ⋅ (Nd w ) ⋅ (U4 − U3 ) + q 4 ⋅ (W4 − W3 )
R 34 > 2/68 ⋅ 21 −5 ⋅ 31/9 ⋅ (3:4 − 641/96) + 2/2 ⋅ 21 6 ⋅ 21 −4 ⋅ (1/1149 − 1/1174)
R 34 >!−2/16!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 34
2/43/!Dvoatoman idealan gas stawa 2)q>2/3!NQb-!U>411!L-!W>1/2!n4*- nalazi se u vertikalno
postavqenom nepokretnom adijabatski izolovanom cilindru sa (bez trewa) pokretnim adijabatskim
klipom zanemarqive mase. Preostali prostor cilindra (iznad klipa) ispuwen je nekom te~nosti
(slika). Usled predaje toplote gasu (od greja~a), on se {iri do stawa 3)q>1/7!NQb-!W>1/33!n4*-!~ime
izaziva prelivawe odgovaraju}e koli~ine te~nosti preko ivica cilindra.
a) izvesti zakon promene stawa gasa u obliku q!>!g)W*
b) prikazati promenu stawa gasa u qW koordinatnom sistemu
c) odrediti zapreminski rad koji izvr{i gas pri ovoj promeni stawa kao i koli~inu toplote koja se u
ovom procesu preda gasu
a)
jedna~ina stati~ke ravnote`e za proizvoqan polo`aj klipa u cilindru:
q = qp + ρ ⋅ h ⋅ i
q = qp +
qp +
ρ ⋅ h ⋅ (Wdjmjoebs − W )
3
e π
5
ρ ⋅ h ⋅ Wdjmjoebs
3
e π
5
q = b−c⋅W -
= b >dpotu
⇒
q = qp +
⇒
q = qp +
ρ⋅h
e3 π
5
ρ ⋅ h ⋅ Wuf•optu
e3 π
5
ρ ⋅ h ⋅ Wdjmjoebs
3
e π
5
−
ρ⋅h
e3 π
5
⋅W
>c>dpotu
zavisnost pritiska od zapremine je linearna, a konstante b i
c odre|ujemo iz grani~nih uslova:
q2 = b − c ⋅ W2
q 3 = b − c ⋅ W3
dipl.ing. @eqko Ciganovi}
)2*
)3*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
c=
strana 35
q2 − q 3 2/3 ⋅ 21 7 − 1/7 ⋅ 21 7
Qb
> 6 ⋅ 21 7
=
W3 − W2
1/33 − 1/2
n4
b = q2 + c ⋅ W2 = 2/3 ⋅ 21 7 + 6 ⋅ 21 7 ⋅ 1/2 > 2/8 ⋅ 21 7 Qb
q = 2/8 ⋅ 21 7 − 6 ⋅ 21 7 ⋅ W !!analiti~ki oblik zavisnosti pritiska od zapremine
b)
q
2
3
w
c)
W3
X23 =
∫
q)W*eW =
w W2
q2 + q3
2/3 ⋅ 217 + 1/7 ⋅ 217
⋅ )W3 − W2* =
⋅ (1/33 . 1/2) >219!lK
3
3
jedna~ina stawa idealnog gasa na po~etku procesa:!
o=
(
)
(
)
q2 ⋅ W2 = o ⋅ NS h ⋅ U2
q2 ⋅ W2
2/3 ⋅ 21 7 ⋅ 1/33
=
= 5/9 ⋅ 21 .3 lnpm
NS h ⋅ U2
9426 ⋅ 411
(
)
jedna~ina stawa idealnog gasa za stawe 2:!
U3 =
q 3 ⋅ W3 = o ⋅ NS h ⋅ U3
q 3 ⋅ W3
1/7 ⋅ 21 7 ⋅ 1/33
=
>441/8!L
o ⋅ NS h
5/9 ⋅ 21 .3 ⋅ 9426
(
)
prvi zakon termodinamike za proces 1−2:
R23 > o ⋅ (Nd w ) ⋅ (U3 − U2 ) + X23 > 5/9 ⋅ 21
dipl.ing. @eqko Ciganovi}
−3
R 23 = ∆V23 + X23
⋅ 31/9 ⋅ (441/8 − 411) + 219 >249/7!lK
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 36
2/44/!Dvoatoman idealan gas, stawa 2)q2>1/7NQb-!U2>411!L-!W2>1/3!n4*- nalazi se u horizontalno
postavqenom nepokretnom cilindru sa (bez trewa) pokretnim klipom. Klip je preko opruge, linearne
karakteristike k, povezan sa nepokretnim zidom (slika). Predajom toplote gasu, on se dovodi do stawa
3)q3>2!NQb-!W3>1/5!n4*/ U po~etnom polo`aju opruga je rastere}ena.
b* izvesti zakon promene stawa gasa u obliku q>g)W*
b) prikazati promenu stawa idealnog gasa na qw dijagramu
c) odrediti zapreminski rad koji izvr{i gas pri ovoj promeni stawa kao i koli~inu toplote koja se u
ovom procesu preda gasu
∆y
3
2
b*
jedna~ina stati~ke ravnote`e za proizvoqan polo`aj klipa u cilindru:
q = qp +
q = qp −
k ⋅ ∆y
⇒
3
e π
5
k ⋅ W2
3
+
k
e3 π
e3 π
5
5
k ⋅ W2
qp −
= b >dpotu
3
e3 π
5
q = b+c⋅W-
3
q = qp +
k ⋅ (W − W2 )
e3 π
5
3
⇒
⋅W
k
e3 π
5
3
>c>dpotu
zavisnost pritiska od zapremine je linearna, a konstante b i
c odre|ujemo iz grani~nih uslova:
q2 = b + c ⋅ W2
q 3 = b + c ⋅ W3
dipl.ing. @eqko Ciganovi}
)2*
)3*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
c=
strana 37
q 3 − q2 2 ⋅ 21 7 − 1/7 ⋅ 21 7
Qb
> 3 ⋅ 21 7
=
W3 − W2
1/5 − 1/3
n4
b = q2 − c ⋅ W2 = 1/7 ⋅ 21 7 − 3 ⋅ 21 7 ⋅ 1/3 > 1/3 ⋅ 21 7 Qb
q = 1/3 ⋅ 21 7 + 3 ⋅ 21 7 ⋅ W !!analiti~ki oblik zavisnosti pritiska od zapremine
b)
q
3
2
w
c)
W3
X23 =
∫
q)W*eW =
w W2
q2 + q 3
1/7 ⋅ 21 7 + 2 ⋅ 21 7
⋅ )W3 − W2 * =
⋅ (1/5 . 1/3) >271!lK
3
3
jedna~ina stawa idealnog gasa na po~etku procesa:!
o=
(
)
(
)
q2 ⋅ W2 = o ⋅ NS h ⋅ U2
q2 ⋅ W2
1/7 ⋅ 21 7 ⋅ 1/3
=
= 5/9 ⋅ 21 .3 lnpm
NS h ⋅ U2
9426 ⋅ 411
(
)
jedna~ina stawa idealnog gasa za stawe 2:!
U3 =
q 3 ⋅ W3 = o ⋅ NS h ⋅ U3
q 3 ⋅ W3
2 ⋅ 21 7 ⋅ 1/5
=
>2113/3!L
o ⋅ NS h
5/9 ⋅ 21 .3 ⋅ 9426
(
)
prvi zakon termodinamike za proces 1−2:
R 23 = ∆V23 + X23
R23 > o ⋅ (Nd w ) ⋅ (U3 − U2 ) + X23 > 5/9 ⋅ 21 −3 ⋅ 31/9 ⋅ (2113/3 − 411) + 271 >972/2!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 38
2/45/!U vertikalnom cilindru (slika) unutra{weg pre~nika e>311!nnnalazi se o>!1/6!npm dvoatomnog idealanog gasa. Masa klipa je nl>51
lh. Klip je poduprt oprugom linearne karakteristike l. Po~etni
pritisak gasa je q2>2/16!cbs, a pritisak okoline iznosi qp>2!cbs. Plin
se hladi tako da u momentu rastere}ewa opruge postigne temperatura od
U3>364!L, pri ~emu se od gasa odvede 2/6!lK toplote. Zanemaruju}i trewe
klipa odrediti:
a) po~etnu temperaturu gasa
b) za koliko se podigao gas do momenta rastere}ewa opruge
R23
3
∆{
2
b*
jedna~ina stati~ke ravnote`e za klip u trenutku rastere}ewa
opruge:
q3 +
nl ⋅ h
3
e ⋅π
5
= qp
⇒
nl ⋅ h
q3 = qp −
3
e ⋅π
5
= 2 ⋅ 21 6 −
(
)
o ⋅ NS h ⋅ U3
q3
1/6 ⋅ 21 −4 ⋅ 9426 ⋅ 364
=
1/98 ⋅ 21 6
> 1/98 !cbs
(
)
>1/1232!n4
R 23 = ∆V23 + X23
prvi zakon termodinamike za proces 1−2:
R23 > o ⋅ (Nd w ) ⋅ (U3 − U2 ) +
1/3 3 ⋅ π
5
q 3 ⋅ W3 = o ⋅ NS h ⋅ U3
jedna~ina stawa idealnog gasa za stawe 2:!
W3 =
51 ⋅ :/92
q2 + q 3
⋅ (W3 − W2 )
3
)2*
(
)
jedna~ina stawa idealnog gasa za stawe 1:!
q2 ⋅ W2 = o ⋅ NS h ⋅ U2
kombinovawem jedna~ina (1) i (2) dobija se:
W2>1/1259!n4-!U2>484!L
W3
napomena:! X23 =
∫
q)W*eW =
)3*
q2 + q 3
⋅ )W3 − W2 * -!kao u prethodnom zadatku
3
w W2
b)
W2 − W3 =
e3 ⋅ π
⋅ ∆{
5
dipl.ing. @eqko Ciganovi}
⇒
∆{ = 5 ⋅
W2 − W3
3
e ⋅π
= 5⋅
1/1259 − 1/1232
1/3 3 ⋅ π
>97!nn
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 39
2/46/!U vertikalnom, toplotno izolovanom cilindru pre~nika e>311!nn sme{tena je opruga
zanemarqive zapremine (slika). Na oprugu je naslowen adijabatski klip mase nl>36!lh. U cilindru se
nalazi azot stawa 2)q2>2/16!cbs!U2>414!L*. U po~etnom trenutku udaqenost klipa od dna cilindra
iznosi {2>611!nn. Du`ina opruge (linearne karakteristike) u neoptere}enom stawu iznosi {p>711
nn. Dolivawem `ive )ρ>24711!lh0n4* iznad klipa, klip se spusti za ∆{>211!nn!(zanemariti trewe).
Pritisak okoline iznosi qp>2!cbs. Odrediti:
a) koliko je `ive doliveno )lh*
b) za koliko se pove}ala unutra{wa energija gasa
c) do koje bi temperature trebalo zagrejati azot tako da se klip vrati u po~etno stawe
(pretpostaviti da ne dolazi do isticawa `ive) i koliko bi toplote pri tom trebalo dovesti
e
e
∆z
∆{
{1
{2
{3
a)
jedna~ina stati~ke ravnote`e za polo`aj klipa u trenutku 1:
3
l ⋅ ({ p − {2 )
nl ⋅ h
q − q + nl ⋅ h ⋅ e ⋅ π
=
q2 +
q
l
=
+
⇒
p
2
p
e3 ⋅ π 5 ⋅ ({ p − {2 )
e3 ⋅ π
e3 ⋅ π
5
5
5
36 ⋅ :/92
1/3 3 ⋅ π
O
l = 2 ⋅ 21 6 − 2/16 ⋅ 21 6 +
⋅
>992/8!
3
(
)
5
1
/
7
1
/
6
⋅
−
n
1/3 ⋅ π
5
e3 ⋅ π
1/3 3 ⋅ π
⋅ {2 =
⋅ 1/6 = 1/1268 n 4
5
5
e3 ⋅ π
1/3 3 ⋅ π
⋅ ({ 2 − ∆{ ) =
⋅ (1/6 . 1/2) = 1/1237 n 4
W3 =
5
5
W2 =
q2 W3
=
q 3 W2
κ
⇒
dipl.ing. @eqko Ciganovi}
W
q 3 = q2 ⋅ 2
W3
κ
1/1268
= 2/16 ⋅ 21 6 ⋅
1/1237
2/5
= 2/54 ⋅ 21 6 Qb
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 40
jedna~ina stati~ke ravnote`e za polo`aj klipa u trenutku 2:
q3 +
l ⋅ ({ p − {2 + ∆{ )
3
e ⋅π
5
= qp +
nl ⋅ h
e3 ⋅ π
5
⇒
+ ρ Ih ⋅ h ⋅ z
l ⋅ ({ p − {2 + ∆{ ) nl ⋅ h
2
z=
q
q
− 3
+ 3 − p⋅
3
ρ
⋅h
e ⋅π
e ⋅π
Ih
5
5
2483/3 ⋅ 1/3 36 ⋅ :/92
2
6
6
z>
2
/
54
21
2
/
16
21
>3:3!nn
⋅
−
+
⋅
−
⋅
3
1/3 3 ⋅ π
24711
⋅ :/92
1/3 ⋅ π
5
5
nIh = ρ Ih ⋅
e3 ⋅ π
1/3 3 ⋅ π
⋅ z = 24711 ⋅
⋅ 1/3:3 >235/87!lh
5
5
b)
jedna~ina stawa idealnog gasa na po~etku procesa:!
n=
q2 ⋅ W2 2/16 ⋅ 21 ⋅ 1/1268
=
= 1/129 lh
S h ⋅ U2
3:8 ⋅ 414
jedna~ina stawa idealnog gasa na kraju procesa:!
U3 =
q2 ⋅ W2 = n ⋅ S h ⋅ U2
6
q 3 ⋅ W3 = n ⋅ S h ⋅ U3
q 3 ⋅ W3 2/54 ⋅ 21 ⋅ 1/1237
=
= 448/2 L
n ⋅ Sh
1/129 ⋅ 3:8
6
∆V23 = n ⋅ d w ⋅ (U3 − U2 ) > 1/129 ⋅ 1/85 ⋅ (448/2 − 414) >!1/56!lK
c)
uo~iti da je:
W4!>!W2-
q4>q3
jedna~ina stawa idealnog gasa za stawe 3:!
U4 =
q 4 ⋅ W4 = n ⋅ S h ⋅ U4
q 4 ⋅ W4 2/54 ⋅ 21 ⋅ 1/1268
=
= 531 L
n ⋅ Sh
1/129 ⋅ 3:8
6
prvi zakon termodinamike za proces 2−3:
R34> n ⋅ d w ⋅ (U4 − U3 ) + q 3 ⋅ (W4 − W3 ) >
R 34 = ∆V 34 + X34
R34>! 1/129 ⋅ 1/85 ⋅ (531 − 448/2) + 2/54 ⋅ 21 6 ⋅ 21 −4 ⋅ (1/1268 − 1/1237 ) >2/66!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 41
2/47/ Toplotno izolovan cilindar, sa pokretnim toplotno izolovanim klipom, podeqen je nepokretnom,
toplotno propustqivom (dijatermijskom) pregradom na dva dela (slika). U delu B nalazi se troatomni
idealan gas po~etnog staqa B)qB2>1/26!NQb-!WB2>1/6!n4-!UB2>911!L*- a u delu C dvoatomni idealan gas
po~etnog stawa C)qC2>1/6!NQb-!WC2>1/3!n4-!UC2>411!L*/ Odrediti zapreminu u delu B i pritisak u delu C
u trenutku uspostavqawa termodinami~ke ravnote`e.
B
uo~iti da je:
qB2>qB3
WC2>WC3
UB3>UC3
C
jedna~ina stawa idealnog gasa A na po~etku procesa:
q ⋅W
o B = B2 B2
q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB
⇒
NS h ⋅ UB
(
oB =
)
(
)
1/26 ⋅ 21 7 ⋅ 1/6
> 2/24 ⋅ 21 −3 lnpm
9426 ⋅ 911
jedna~ina stawa idealnog gasa!C!ob!po~etku procesa:
q ⋅W
oC = C2 C2
q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC
⇒
NS h ⋅ UC
(
oC =
)
(
)
1/6 ⋅ 21 7 ⋅ 1/3
> 5/12 ⋅ 21 −3 lnpm
9426 ⋅ 411
prvi zakon za promenu stawa radnih tela B i C u celom cilindru
R23!>!∆V23!,X23
1 = o B ⋅ (Nd w ) B ⋅ (UB3 − UB2 ) + oC ⋅ (Nd w )C ⋅ (UC3 − UC2 ) + q B ⋅ (WB3 − WB2 ) !!!!)2*
jedna~ina stawa ideal. gasa A u trenutku uspostavqawa toplotne ravnote`e:
q ⋅W
)3*
UB3 = B3 B3
q B3 ⋅ WB3 = o B ⋅ NS h ⋅ UB3
⇒
o ⋅ NS h
(
dipl.ing. @eqko Ciganovi}
)
(
)
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 42
kada se jedna~ina (2) stavi u jedna~inu (1) dobija se:
q ⋅W
q ⋅W
1 = oB ⋅ (Ndw )B ⋅ B3 B3 − UB2 + oC ⋅ (Ndw )C ⋅ B3 B3 − UC2 + q B ⋅ (WB3 − WB2)
oB ⋅ NSh
oB ⋅ NSh
(
WB 3 =
(
)
o B ⋅ (Nd w )B ⋅ UB2 + oC ⋅ (Nd w )C ⋅ UC2 + q B ⋅ WB2
(Nd w )C
(Nd w )B
⋅ q B3 +
⋅ q C3 + q B
NS h
NS h
(
WB 3 =
)
)
(
)
2/24 ⋅ 21 −3 ⋅ 3:/2 ⋅ 21 4 ⋅ 911 + 5/12 ⋅ 21 −3 ⋅ 31/9 ⋅ 21 4 ⋅ 411 + 1/26 ⋅ 21 7 ⋅ 1/6
3:/2 ⋅ 21 4
31/9 ⋅ 21 4
⋅ 1/26 ⋅ 21 7 +
⋅ 1/6 ⋅ 21 7 + 1/26 ⋅ 21 7 q B
9426
9426
WB3>1/416!n4
odavde se dobija:
vra}awem u jedna~inu!)3*;!
UB3 =
lK
lnpmL
(Nd w )C >31/9! lK
lnpmL
(Nd w )B >3:/2!
napomena:
1/26 ⋅ 21 7 ⋅ 1/416
2/24 ⋅ 21 .3 ⋅ 9426
>598!L!>!UC3
troatoman idealan gas
dvoatoman idealan gas
kedna~ina stawa ideal. gasa C u trenutku uspostavqawa toplotne ravnote`e;
oC ⋅ NS h ⋅ UC3
q C3 ⋅ WC3 = oC ⋅ NS h ⋅ UC3
⇒
q C3 =
WC3
(
q C3 =
)
(
)
5/12 ⋅ 21 .3 ⋅ 9426 ⋅ 598
> 9/2 ⋅ 21 6 !Qb
1/3
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 43
2/48/!Vertikalan, toplotno izolovan cilindar, zatvoren i sa gorwe i sa
dowe strane pokretnim klipovima (toplotno izolovanim, zanemarqivih
masa, koji se kre}u bez trewa), podeqen je nepropusnom, krutom i
nepokretnom pregradom na deo B i deo C (slika). Pregrada je zanemarqivog
toplotnog kapaciteta i pru`a zanemarqiv otpor kretawu toplote. U delu
B nalazi se dvoatoman idealan gas, a u delu C troatoman idealan gas. U
po~etnom polo`aju gas u delu B ima stawe B2)WB2>1/6!n4-!qB2>1/5!NQbUB2>411!L* gas u delu C u stawe C2)WC2>1/5!n4-!qC2>1/16!NQb-!UC2>411!L*/
Odrediti zapreminski rad koji treba obaviti pri sabijawu gasa u delu B,
da bi zapremina gasa u delu C bila dva puta ve}a.
)X23*B
B
C
jedna~ina stawa idealnog gasa A na po~etku procesa:
q ⋅W
o B = B2 B2
q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB
⇒
NS h ⋅ UB
(
oB =
)
(
)
1/5 ⋅ 21 7 ⋅ 1/6
> 9/13 ⋅ 21 −3 lnpm
9426 ⋅ 411
jedna~ina stawa idealnog gasa C na po~etku procesa:
q ⋅W
oC = C2 C2
q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC
⇒
NS h ⋅ UC
(
oC =
)
(
)
7
1/16 ⋅ 21 ⋅ 1/3
> 9/13 ⋅ 21−4 lnpm
9426 ⋅ 411
uslov zadatka:
dijatermijska pregrada:
WC3!>!3!/!WC2!>!1/9!n4UB3>UC3
prvi zakon termodinamike za proces u cilindru:
R23!>!∆V23!,)!X23!*B!,!)!X23!*C
1 = oB ⋅ (Ndw )B ⋅ (UB3 − UB2) + oC ⋅ (Ndw )C ⋅ (UC3 − UC2) + (X23 )B + qC ⋅ (WC3 − WC2)
)2*
jedna~ina stawa idealnog gasa C na kraju procesa:
(
)
qC3 ⋅ WC3 = oC ⋅ NSh ⋅ UC3
)3*
kombinovawem jedna~ina )2* i )3*-!sistem dve jedna~ine sa dve nepoznate,
dobija se:
UB3>712!L-!!)X23*B>−6:1/4!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 44
2/49/!U izolovanom i sa obe strane zatvorenom cilindru nalaze se dva idealna gasa me|usobno odeqena
bez trewa pomi~nim i toplotno propusnim klipom. Po~etni pritisak oba gasa iznosi!qB2>qB3>4!cbs/
U delu nalazi se kiseonik stawa B)UB2>3:4!L-!nB>1/2!lh*-!a u delu C nalazi se metan stawa!C)UC2>634
L-!nC>1/2!lh*/!Odrediti:
a) pritisak i temperaturu oba gasa trenutku uspostavqawa termodinami~ke ravnote`e
b) promenu entropije sitema koja nastaje u procesu koji po~iwe od zadatog po~etnog stawa i traje do
trenutka uspostavqawa termodinami~ke ravnote`e
B
C
b*
jedna~ina stawa idealnog gasa B (po~etak procesa) : q B2 ⋅ WB2 = n B ⋅ S hB ⋅ UB2
WB2 =
n B ⋅ S h ⋅ UB2
q B2
=
1/2 ⋅ 371 ⋅ 3:4
4 ⋅ 21 6
>1/1365!n4
jedna~ina stawa idealnog gasa B (kraj procesa): q C2 ⋅ WC2 = nC ⋅ S hC ⋅ UC2
WC2 =
nC ⋅ S hC ⋅ UC2
q C2
=
1/2 ⋅ 631 ⋅ 634
4 ⋅ 21 6
>1/1:17!n4
R 23 = ∆V23 + X23
prvi zakon termodinamike za proces u cilindru:
∆V23>1
⇒
V2 = V 3
V2 = n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC
V 3 = n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U +
U+ =
n B ⋅ d wB ⋅ UB2 + nC ⋅ d wC ⋅ UC2 1/2 ⋅ 1/76 ⋅ 3:4 + 1/2 ⋅ 2/93 ⋅ 634
=
>573/6!L
n B⋅ d wB + nC ⋅ d wC
1/2 ⋅ 1/76 + 1/2 ⋅ 2/93
jedna~ina stawa idealnog gasa B na kraju procesa:
q B3 ⋅ WB3 = n B ⋅ S hB ⋅ U +
⇒
WB3 =
n B ⋅ S hB ⋅ U +
q B3
)2*
jedna~ina stawa idealnog gasa C na kraju procesa:
q C3 ⋅ WC3 = nC ⋅ S hC ⋅ U +
dipl.ing. @eqko Ciganovi}
⇒
WC3 =
nC ⋅ S hC ⋅ U +
q C3
)3*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 45
deqewem jedna~ina!)2*!i!)3*!dobija se;! WB3 = WC3
n B ⋅ S hB
nC ⋅ S hC
)4*
jednake zapremine cilindra pre i posle procesa;!WB2!,WC2!>WB3!,WC3!!!!)5*
kada se jedna~ina!)4*!uvrsti u jedna~inu!)5*!dobija se i re{i po!WC3!dobija se;
WC3!>!
WB2 + WC2
1/1365 + 1/1:17
>1/1884!!n4
=
n B ⋅ S hB
1/2 ⋅ 371
+2
+2
1/2 ⋅ 631
nC ⋅ S hC
WB3 = 1/1884 ⋅
vra}awem u jedna~inu )4*!dobija se;
q B3 =
n B ⋅ S hB ⋅ U +
WB3
=
1/2 ⋅ 371
>1/1498!n4
1/2 ⋅ 631
1/2 ⋅ 371 ⋅ 573/6
> 4/2 ⋅ 21 6 !Qb!>!qC3
1/1498
b)
∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>!33/:!
∆Tplpmjob>!1!
K
L
K
L
(adijabatski procesi u oba cilindra)
∆TSU!>! ∆T B + ∆T C >!///>!51/8!−!41/8!>!33/:!
!∆TB!>g!)!q-!U*!> nB ⋅ dqmo
UB3
q
− Shmo B3 >
UB2
qB2
573/6
4/2 ⋅ 21 6
− 1/371 ⋅ mo
∆TB!> 1/2 ⋅ 1/:2 ⋅ mo
3:4
4 ⋅ 21 6
∆TC!>!g!)!q-!U*!> nC ⋅ dqmo
K
L
>51/8! K
L
UC3
q
− Shmo C3 >
UC2
qCB2
573/3
4/2⋅ 216
lK
>−41/89!
∆TC!>! 1/2⋅ 3/45 ⋅ mo
− 1/631 ⋅ mo
6
L
634
4
21
⋅
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 46
2/4:/!U zatvorenom, delimi~no adijabatski izolovanom (vidi sliku), horizontalnom cilindru nalazi se
o>3!lnpm dvoatomnog idealnog gasa. Pokretna adijabatska povr{ina (klip) deli cilindar na dva jednaka
dela )WB>WC!*/ Po~etno stawe idealnog gasa (u oba dela) odre|eno je istim veli~inama stawa q>2!cbsU>399!L. Dovo|ewem toplote kroz neizolovani deo cilindra (leva ~eona povr{ina) dolazi do kretawa
klipa (bez trewa) dok pritisak u delu C ne dostigne 5!cbs ( pri tome se usled kvazistati~nosti ne
naru{ava mehani~ka ravnote`a tj. i pritsak u delu B iznosi
5!cbs). Odrediti:
a) zapreminski rad koji izvr{i radno telo u delu B (levi deo cilindra)
b) koli~inu toplote koja se preda radnom telu u istom delu cilindra
R23
B
C
a)
jedna~ina stawa idealnog gasa u delu A na po~etku procesa:
q ⋅W
o B = B2 B2
q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB
⇒
NS h ⋅ UB
(
)
(
)
q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC
(
oC =
⇒
iz jedna~ina )2*!i!)3* se dobija:
uslov zadatka:
)
q C2 ⋅ WC2
NS h ⋅ UC
(
)
oB!>!oC
o>oB!,!oC
kombinovawem jedna~ina!)4*!i!)5*!dobija se:
)2*
)3*
)4*
)5*
oB>!2!lnpm-!oC>2!lnpm
promena stawa idealnog gasa u delu!C!je kvazistati~ka i adijabatska:
UC3
q
= UC2 ⋅ C3
q C2
κ −2
κ
5 ⋅ 21 6
= 399 ⋅
6
2 ⋅ 21
2/5 −2
2/5
= 539 L
prvi zakon termodinamike za proces u delu C; !!!!! (R 23 )C = (∆V23 )C + (X23 )C
(X23 )C
= −(∆V23 )C = −oC ⋅ (Nd w ) ⋅ (UC3 − UC2 ) > −2 ⋅ 31/9 ⋅ (539 − 399) >−3:23!lK
(X23 )B
= −(X23 )C >3:23!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 47
b)
jedna~ina stawa idealnog gasa u delu B na po~etku procesa:
WB2 =
(
)
o B ⋅ NS h ⋅ UB2
q B2
(
)
o B ⋅ NS h ⋅ UB3
q B3
(
)
q C2
(
)
o W ⋅ NS h ⋅ UC3
q C3
)
⇒
U
U
WB3 − WB2 = o ⋅ NS h ⋅ B3 − B2 !)8*
q B3 q B2
(
)
(
)
⇒
(
)
⇒
q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC2
)9*
q C3 ⋅ WC3 = oC ⋅ NS h ⋅ UC3
jedna~ina stawa idealnog gasa u delu C na kraju procesa:
WC3 =
(
q B3 ⋅ WB3 = o B ⋅ NS h ⋅ UB3
jedna~ina stawa idealnog gasa u delu C na po~etku procesa:
oC ⋅ NS h ⋅ UC2
⇒
)7*
oduzimawem (7*!−!)6*!dobija se;!
WC2 =
)
)6*
jedna~ina stawa idealnog gasa u delu B na kraju procesa:
WB3 =
(
q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB2
):*
U
U
WC2 − WC3 = o ⋅ NS h ⋅ C2 − C3
q C2 q C3
(
oduzimawem (9*!−!):*!dobija se;
)
!!)21*
iz ~iwenice da su leve strane jedna~ina!)8*!i!)21*!jednake dobija se:
UB3 UB2
U
U
> C2 − C3
−
q B3 q B2 q C2 q C3
⇒
U
U
U
UB3 = q B3 ⋅ B2 + C2 − C3
q B2 q c2 q C3
399
539
399
+
−
UB3 = 5 ⋅ 21 6 ⋅
>2987!L
6
6
2 ⋅ 21
5 ⋅ 21 6
2 ⋅ 21
prvi zakon termodinamike za proces u delu B; !!!!! (R 23 ) B = (∆V23 ) B + (X23 )B
(R23 )B
= o B ⋅ (Nd w ) ⋅ (UB3 − UB2 ) + X23 > 2 ⋅ 31/9 ⋅ (2987 − 399) + 3:23 >!46:53!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 48
2/51. U hermeti~ki zatvorenim i toplotno izolovanim cilindrima B i C , koji su razdvojeni
slavinom (vidi sliku) nalazi se po!n>5!lh vazduha (idealan gas) stawa 2B)q>21!cbs-!U>511!L*,
odnosno 2C)q>2!cbs-!U>511!L). U krajwem levom delu cilindra C nalazi se adijabatski klip koji mo`e
da se kre}e u cilindru, ali uz savladavawe sila trewa. Otvarawem slavine, klip se usled razlike
pritisaka kre}e i sa stepenom dobrote ηlq
e >1/9 sabija vazduh u cilindru C dok se ne uspostavi
mehani~ka ravnote`a. Skicirati procese sa radnim telom na zajedni~kom
Ut dijagramu i odrediti:
a) pritisak i temperaturu u cilindrima B i C u stawu mehani~ke ravnote`e
b) promenu entropije izolovanog termodinami~kog sistema od zadatog po~etnog stawa do stawa
mehani~ke ravnote`e izme|u vazduha u cilindrima B i C
B
C
a)
prvi zakon termodinamike za proces u delu A;! (R 23 ) B = (∆V23 ) B + (X23 )B !!)2*
prvi zakon termodinamike za proces u delu C;!! (R 23 )C = (∆V23 )C + (X23 )C !!!)3*
sabirawem jedna~ina (1) + (2) dobija se: (∆V23 )B = −(∆V23 )C ! ⇒
UB3 − UB2 = UC2 − UC3
ili
UB2 + UC2 = UB2 + UC3
)4*
napomena:
po{to su oba cilindra adijabatski izolovana od okoline
(R23 )B = (R23 )C =0.
zapreminski rad koji izvr{i radno telo B i zapreminski rad
koji se izvr{i nad radnim telom C su jednaki, ali suprotni
po znaku (X23 ) B = −(X23 )C
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 49
jedna~ina stawa idealnog gasa u delu B na po~etku procesa:
WB2 =
(
)
o B ⋅ NS h ⋅ UB2
q B2
(
)
o B ⋅ NS h ⋅ UB3
q B3
(
)
q C2
(
)
o W ⋅ NS h ⋅ UC3
q C3
)
⇒
U
U
WB3 − WB2 = o ⋅ NS h ⋅ B3 − B2 !)7*
q B3 q B2
(
)
(
)
⇒
(
)
⇒
q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC2
)8*
q C3 ⋅ WC3 = oC ⋅ NS h ⋅ UC3
jedna~ina stawa idealnog gasa u delu C na kraju procesa:
WC3 =
(
q B3 ⋅ WB3 = o B ⋅ NS h ⋅ UB3
jedna~ina stawa idealnog gasa u delu C na po~etku procesa:
oC ⋅ NS h ⋅ UC2
⇒
)6*
oduzimawem (5*!−!)5*!dobija se;!
WC2 =
)
)5*
jedna~ina stawa idealnog gasa u delu B na kraju procesa:
WB3 =
(
q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB2
)9*
U
U
WC2 − WC3 = o ⋅ NS h ⋅ C2 − C3
q C2 q C3
(
oduzimawem )8*!−!)9*!dobija se;
)
!!):*
iz ~iwenice da su leve strane jedna~ina!)7*!i!):*!jednake dobija se:
UB3 UB2
U
U
> C2 − C3
−
q B3 q B2 q C2 q C3
⇒
UB3 + UC3
U
U
> B2 + C2 !
qy
q B2 q C2
)21*
UB3 UC3
U
U
> B2 + C2
+
q B3 q C3
q B2 q C2
⇒
kada se u jedna~inu )21* uvrsti jedna~ina )4* dobija se:
UB2 + UC2
U
U
U + UC2
511 + 511
> B2 + C2
⇒
qy!>! B2
=
UB2 UC2
511
511
qy
q B2 q C2
+
+
6
q B2 q C2
21 ⋅ 21
2 ⋅ 21 6
qy!> 2/93 ⋅ 21 6 Qb
napomena:!
qB3!>!qC3>!qy!(uslov mehani~ke ravnote`e na kraju procesa)
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
q
UC3l = UC2 ⋅ C3l
q C2
UC3l − UC2
ηlq
e =
UC3 − UC2
UC3> 511 +
κ −2
κ
strana 50
2/93 ⋅ 21 6
= 511 ⋅
2 ⋅ 21 6
⇒
2/5 −2
2/5
UC3 = UC2 +
>585/7!L
UC3l − UC2
ηlq
e
585/7 − 511
>5:4/4!L
1/9
iz jedna~ine!)4*!!!!!⇒
UB3 = UB2 + UC2 − UC3 = 511 + 511 − 5:4/4 = 417/8 L
b)
∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>!2/15!
∆Tplpmjob>!1!
lK
L
lK
L
(adijabatski procesi u oba cilindra)
∆TSU!>! ∆T B + ∆T C >!///>!1/9:!,!1/26!>!2/15!
417/8
2/93 ⋅ 216
UB3
q
− 1/398 ⋅ mo
− Shmo B3 > 5 ⋅ 2⋅ mo
UB2
qB2
511
21 ⋅ 216
>1/9:! lK
L
5:4/4
2/93 ⋅ 21 6
UC3
q
− 1/398 ⋅ mo
− Shmo C3 > 5 ⋅ 2 ⋅ mo
UC2
qCB2
511
2 ⋅ 21 6
>1/26! lK
L
!∆TB!> nB ⋅ dqmo
∆TC!> nC ⋅ dqmo
lK
L
qB2
3C
U
3lC
qy
qC2
2B
2C
UB2>UC2
3B
3lB
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 51
2/52/!Geometrijski identi~ni, adijabatski i bez trewa poktretni klipovi hermeti~ki zaptivaju dva
horizontalno postavqena, toplotno izolovana, nepokretna cilindra. Klipovi su me|usobno spregnuti
preko sistema zup~astih letvi, odnosno preko fiksnog i bez trewa pokretnog zup~anika (slika). U
levom cilindru )B*- nalazi se 1/9!lh sumpor dioksida (idealan gas), a u desnom cilindru )C* 1/9!lh
kiseonika (idealan gas). U polaznom polo`aju, sumpor-dioksid se nalazi u stawu B2)qB2>1/23!NQbUB2>411!L*- a kiseonik u stawu C2!)qC2>1/19!NQb-!UC2>411!L*. Odrediti koli~inu elektri~ne
energije koju bi elektri~ni greja~ H trebao da preda sumpor-dioksidu, da bi se temperatura kiseonika
snizila do UC3>393!L/
H
qB
C
B
qbnc
QC
jedna~ina stawa idealnog gasa C na po~etku procesa:! q C2 ⋅ WC2 = nC ⋅ S hC ⋅ UC2
WC2 =
nC ⋅ S hC ⋅ UC2
q C2
=
1/9 ⋅ 371 ⋅ 411
1/19 ⋅ 21 7
>1/89!n4
U
q
zakon kvazistati~ke adijabatske promene stawa gasa!C;!!!! C2 = C2
q C3 UC3
q C3
U
= q C2 ⋅ C3
UC2
κ
κ −2
= 1/19 ⋅ 21 7
κ
κ −2
2/5
393 2/5 −2
⋅
>! 1/75 ⋅ 21 6 Qb
411
jedna~ina stawa idealnog gasa C!na kraju procesa:!!!! q C3 ⋅ WC3 = nC ⋅ S hC ⋅ UC3
WC3 =
nC ⋅ S hC ⋅ UC3
q C3
=
1/9 ⋅ 371 ⋅ 393
1/755 ⋅ 21 6
>1/:2!n4
prvi zakon termodinamike za proces u delu C;!!!!!! (R 23 )C = (∆V23 )C + (X23 )C
(X23 )C >! − nC ⋅ d wC ⋅ (UC3 − UC2 ) > −1/9 ⋅ 1/76 ⋅ (393 − 411) >!:/47!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 52
(X23 )B >! (X23 )C >!:/47!lX
jedna~ina stawa idealnog gasa A na po~etku procesa: q B2 ⋅ WB2 = n B ⋅ S hB ⋅ UB2
WB2 =
n B ⋅ S hB ⋅ UB2
q B2
=
1/9 ⋅ 241 ⋅ 411
1/23 ⋅ 21 7
>1/37!n4
WB3!−!WB2!>!WC3!−!WC2!
uslov jednakih promena zapremina:
WB3!>!WB2!,!WC3!−!WC2
WB3>1/37!,!1/:2!−!1/89!>!1/4:!n4
jedna~ina stati~ke ravnote`e za idealan gas!C!na kraju procesa:
qC3!>!qbnc!−!q{vq•bojl! ⇒
q{vq•bojl!>!qbnc!−!qC3!>!2!−!1/75!>!1/47!cbs
jedna~ina stati~ke ravnote`e za idealan gas!B!na kraju procesa:
qB3!>!qbnc!,!q{vq•bojl! ⇒
qB3!>!2!,!1/47!>!2/47!cbs
jedna~ina stawa idealnog gasa!B!na kraju procesa:!! q B3 ⋅ WB3 = n B ⋅ S hB ⋅ UB3
UB 3 =
q B3 ⋅ WB3 2/47 ⋅ 21 6 ⋅ 1/4:
>621!L
=
n B ⋅ S hB
1/9 ⋅ 241
prvi zakon termodinamike za proces u delu!B;!!!!! (R 23 ) B = (∆V23 ) B + (X23 )B
(R23 )B > n B ⋅ d wB (UB3 − UB2 ) + (X23 )B
dipl.ing. @eqko Ciganovi}
= 1/9 ⋅ 1/56 ⋅ (621 − 411) + :/47 >95/:7!lX
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 53
2/53/ U toplotno izolovanom spremniku zapremine W>1/4!n4- nalazi se idealan gas B)Sh>394!K0lhLdw>821!K0lhL-!q>2!cbs-!U>3:4!L*/ Gre{kom je u ovaj spremnik pu{tena izvesna koli~ina idealnog gasa
C tako da je nastala me{avina idealni gasova stawa 2)q>2/49!cbs-!U>431!L*/ Da bi se saznalo koji je
gas u{ao u spremnik izmerena je ukupna masa me{avine nB,nC>1/573!lh, a zatim je me{avina
zagrejana to temperature od U3>464!L. Za ovo zagrevawe je utro{eno R23>21/4!lK toplote. Odrediti
koli~inu )nC* i vrstu )Sh-!dw* dodatog gasa C.
q B ⋅ WB = n B ⋅ S hB ⋅ UB
jedna~ina stawa idealnog gasa A pre me{awa:
nB =
qB ⋅ W
2 ⋅ 21 6 ⋅ 1/4
=
= 1/473 lh
S hB ⋅ UB
394 ⋅ 3:4
nC!>)nB!!,!nC!*!−!nB!>!1/573!−!1/473!>1/2!lh
koli~ina dodatog gasa:
jedna~ina stawa me{avine idealnih gasova B,C pre zagrevawa, stawe (1):
2 q2 ⋅ W
⋅
− n B ⋅ S hB
S hC =
q2 ⋅ WB = n B ⋅ S hB + nC ⋅ S hC ⋅ U2
⇒
nC U2
(
S hC =
)
K
2 2/49 ⋅ 21 6 ⋅ 1/4
⋅
− 1/473 ⋅ 394 >37:/4!
1/2
431
lhL
prvi zakon termodinamike za proces zagrevawa me{avine: R 23 = ∆V23 + X23
R 23 = (n B ⋅ d wB + nC ⋅ d wC ) ⋅ (U3 − U2 )
d wC =
⇒
d wC =
2
nC
R23
⋅
− n B ⋅ d wB
U3 − U2
2 21/4 ⋅ 21 4
K
⋅
− 1/473 ⋅ 821 >662
1/2 464 − 431
lhL
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 54
2/54!U adijabatski izolovanom sudu sa nepropusnim i adijabatskim pregradnim zidom odvojeno je B)O3W>8!n4-!q>5!cbs-!U>394!L* od C)DP3-!W>!5!n4-!q>9!cbs-!U>684!L*/ Izvla~ewem pregradnog zida gasovi }e
se izme{ati. Odrediti:
a) temperaturu )U+* i pritisak )q+* dobijene me{avine
b) dokazati da je proces me{awa O3!i!DP3 nepovratan
!B
a)
!C
jedna~ina stawa idealnog gasa B pre me{awa: q B ⋅ WB = n B ⋅ S hB ⋅ UB
nB =
q B ⋅ WB
5 ⋅ 21 6 ⋅ 8
=
= 44/42 lh
S hB ⋅ UB
3:8 ⋅ 394
jedna~ina stawa idealnog gasa C pre me{awa: q C ⋅ WC = nC ⋅ S hC ⋅ UC
nC =
q C ⋅ WC
9 ⋅ 21 6 ⋅ 5
=
= 3:/66 lh
S hC ⋅ UC
29: ⋅ 684
prvi zakon termodinamike za proces me{awa:
∆V23>1
⇒
R 23 = ∆V23 + X23
V2 = V 3
V2 = n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC
V 3 = n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U +
U+ =
n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC
44/42 ⋅ 1/85 ⋅ 394 + 3:/66 ⋅ 1/77 ⋅ 684
=
>522!L
n B⋅ d wB + nC ⋅ d wC
44/42 ⋅ 1/85 + 3:/66 ⋅ 1/77
jedna~ina stawa me{avine idealnih gasova u trenutku uspostavqawa
toplotne ravnote`e: q + ⋅ (WB + WC ) = n B ⋅ S hB + nC ⋅ S hC ⋅ U +
q+ =
(n B ⋅ S hB + nC ⋅ S hC ) ⋅ U
WB + WC
dipl.ing. @eqko Ciganovi}
(
∗
=
)
(44/42 ⋅ 3:8 + 3:/66 ⋅ 29:) ⋅ 522
8+5
> 6/89 ⋅ 21 6 Qb
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 55
pritisak gasne me{avine q+ se mo`e odrediti i primenom
Daltonovog zakona q+!> q +B + q C+ pri ~emu q +B i q C+ imaju
slede}a zna~ewa:
napomena:
q +B − pritisak gasa B u gasnoj me{avini u trenutku
dostizawa toplotne ravnote`e
q C+ −
pritisak gasa C u gasnoj me{avini u trenutku
dostizawa toplotne ravnote`e
jedna~ina stawa idealnog gasa B u trenutku uspostavqawa toplotne
q +B ⋅ (WB + WC ) = n B ⋅ S hB ⋅ U +
ravnote`e:
q +B =
n B ⋅ S hB ⋅ U +
WB + WC
=
44/42 ⋅ 3:8 ⋅ 522
= 4/81 ⋅ 21 6 !Qb
8+5
jedna~ina stawa idealnog gasa C u trenutku uspostavqawa toplotne
q C+ ⋅ (WB + WC ) = nC ⋅ S hC ⋅ U +
ravnote`e:
q C+ =
nC ⋅ S hC ⋅ U +
WB + WC
=
3:/66 ⋅ 29: ⋅ 522
= 3/19 ⋅ 21 6 !Qb
8+5
b)
∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>23/95!
∆Tplpmjob>−!
R23
lK
= 1!
L
Up
lK
!?1
L
(sud izolovan od okoline)
∆TSU!>∆TB!,!∆TC>!///>!24/78!−!1/94!>23!/95!
lK
L
W + WC
U∗
+ S hB mo B
∆TB!> n B ⋅ g (U- w ) !> n B ⋅ d wB mo
U
WB
B
>
522
8+5
lK
∆TB!> 44/42 ⋅ 1/85 ⋅ mo
+ 1/3:8 ⋅ mo
>24/78!
394
8
L
W + WC
U∗
+ S hC mo B
∆TC!> nC ⋅ g (U- w ) !> nC ⋅ d wC mo
UC
WC
>
522
8+5
lK
∆TC!> 3:/66 ⋅ 1/77 ⋅ mo
+ 1/29: ⋅ mo
>!−1/94!
L
684
5
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 56
2/55. Toplotno izolovan sud podeqen je izolovanom pregradom na dva dela (slika). U delu B zapremine
WB>2/6!n4 nalazi se vodonik (idealan gas) stawa B)qB>1/3!NQb-!UB>3:4!L*/ U delu C zapremine
WC>1/5!n4, nalazi se azot stawa C)qC>1/4!NQb-!nC>2!lh*. U jednom trenutku sa pregrade se uklawa
izolacioni nepropusni sloj sa pregrade, ~ime ona postaje toplotno ne izolovana polupropustqiva
membrana, kroz koju mogu da prolaze samo molekuli vodonika. Odrediti
a) promenu entropije sistema tokom procesa koji po~iwe uklawawem sloja pregrade i traje do
uspostavqawa toplotne ravnote`e u sudu
b) pritisak u delu suda B i delu suda C na kraju ovog procesa
C
B
a)
jedna~ina stawa idealnog gasa B pre me{awa: q B ⋅ WB = n B ⋅ S hB ⋅ UB
nB =
q B ⋅ WB
1/3 ⋅ 21 7 ⋅ 2/6
=
= 1/36 lh
S hB ⋅ UB
5268 ⋅ 3:4
jedna~ina stawa idealnog gasa C pre me{awa: q C ⋅ WC = nC ⋅ S hC ⋅ UC
UC =
q C ⋅ WC
1/4 ⋅ 21 7 ⋅ 1/5
=
>515!L
S hC ⋅ nC
3:8 ⋅ 2
prvi zakon termodinamike za proces me{awa:
∆V23>1
⇒
R 23 = ∆V23 + X23
V2 = V 3
V2 = n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC
V 3 = n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U +
U+ =
n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC
1/36 ⋅ 21/5 ⋅ 3:4 + 2 ⋅ 1/85 ⋅ 515
>428/7!L
=
n B⋅ d wB + nC ⋅ d wC
1/36 ⋅ 21/5 + 2 ⋅ 1/85
∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>1/39!
∆Tplpmjob>−!
R23
lK
= 1!
Up
L
lK
!?1
L
(sud izolovan od okoline)
∆TSU!>∆TB!,!∆TC>!///>!1/57!−!1/29!>1/39!
dipl.ing. @eqko Ciganovi}
lK
L
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 57
W + WC
U∗
∆TB!> n B ⋅ g (U- w ) !> n B ⋅ d wB mo
+ S hB mo B
UB
WB
>
428/7
2/6 + 1/5
lK
+ 5/268 ⋅ mo
∆TB!> 1/36 ⋅ 21/5 ⋅ mo
>1/57!
L
3:4
2/6
W
U∗
+ S hC mo C
∆TC!> nC ⋅ g (U- w ) !> nC ⋅ d wC mo
U
WC
C
>
428/7
lK
∆TC!> 2 ⋅ 1/85 ⋅ mo
>!−1/29!
L
515
b)
jedna~ina stawa idealnog gasa B u trenutku uspostavqawa toplotne
ravnote`e:
q +B ⋅ (WB + WC ) = n B ⋅ S hB ⋅ U +
q +B =
n B ⋅ S hB ⋅ U +
WB + WC
q +B −
=
1/36 ⋅ 5268 ⋅ 428/7
= 2/85 ⋅ 21 6 !Qb
2/6 + 1/5
pritisak vodonika u sudu A i istovremeno parcijalni pritisak
vodonika gasa ugasnoj me{avini (vodonik +azot) u delu suda B u
trenutku dostizawa toplotne ravnote`e
jedna~ina stawa idealnog gasa C u trenutku uspostavqawa toplotne
q C+ ⋅ WC = nC ⋅ S hC ⋅ U +
ravnote`e:
q C+ =
nC ⋅ S hC ⋅ U +
q C+ −
WC
=
2 ⋅ 3:8 ⋅ 428/7
= 3/47 ⋅ 21 6 !Qb
1/5
parcijalni pritisak azota gasnoj me{avini (vodonik +azot) u delu
suda C u trenutku dostizawa toplotne ravnote`e
(qC )3 > q +B + q C+ = 2/85 ⋅ 21 6 , 3/47 ⋅ 21 6 > 5/2 ⋅ 21 6 Qb
(qC )3 −
pritisak u sudu C u trenutku dostizawa toplotne ravnote`e
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 58
2/56/ Adijabatski izolovan termodinami~ki sistem prikazan na slici ~ine:
− zatvoren rezervoar )B* stalne zapremine WB>1/4!n4, u kojem se nalazi
kiseonik (idealan gas) stawa B)qB>3/7!cbs-!UB>411!L*
− zatvoren vertikalni cilindar )C* sa bez trewa pokretnim klipom, u kojem se
nalazi nC>!2!lh metana (idealan gas) stawa C)qC>3!cbs-!UC>511!L*/
(pokretni klip svojom te`inom obezbe|uje stalan pritisak gasa)
Otvarawem ventila dolazi do me{awa gasova. Smatraju}i da pri me{awu gasova ne}e do}i do hemijske
reakcije (eksplozija) odrediti:
a) rad koji izvr{i klip za vreme procesa me{awa
b) promenu entropije ovog adijabatski izolovanog sistema za vreme procesa me{awa
b*
B
C
jedna~ina stawa idealnog gasa B pre me{awa: q B ⋅ WB = n B ⋅ S hB ⋅ UB
nB =
q B ⋅ WB
3/7 ⋅ 21 6 ⋅ 1/4
=
= 2 lh
S hB ⋅ UB
371 ⋅ 411
jedna~ina stawa idealnog gasa C pre me{awa: q C ⋅ WC = nC ⋅ S hC ⋅ UC
WC =
nC ⋅ S hC ⋅ UC
qC
=
2 ⋅ 631 ⋅ 511
3 ⋅ 21 6
>2/15!n4
prvi zakon termodinamike za proces me{awa: !!R23!>!∆V23!,!X23
[
]
1> n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U + − [n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC ] + X23
izra~unavawe zapreminskog rada:
)2*
X23!>! q C ⋅ W + − WB − WC !! )3*
jedna~ina stawa dobijene me{avine idealnih gasova:
q C ⋅ W + = n B ⋅ S hB + nC ⋅ S hC ⋅ U +
)4*
(
dipl.ing. @eqko Ciganovi}
)
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
)4*
W+ =
⇒
strana 59
(n B ⋅ S hB + nC ⋅ S hC ) ⋅ U +
qC
ovu jedna~inu uvrstimo u jedna~inu (2):
n B ⋅ S hB + nC ⋅ S hC ⋅ U +
X23 = q C ⋅
− WB − WC
qC
ovu jedna~inu uvrstimo u jedna~inu )2* odakle se nakon sre|ivawa dobija:
n B ⋅ d wB ⋅ UB + n C ⋅ d wC ⋅ UC + q C ⋅ (WB + WC )
U+ =
n B ⋅ d wB + n C ⋅ d wC + n B ⋅ S hB + n C ⋅ S hC
(
U+ =
)
2 ⋅ 1/76 ⋅ 411 + 2 ⋅ 2/93 ⋅ 511 + 3 ⋅ 21 6 ⋅ 21 −4 ⋅ (1/4 + 2/15 )
>477/6!L
2 ⋅ 1/76 + 2 ⋅ 2/93 + 2 ⋅ 1/37 + 2 ⋅ 1/63
(2 ⋅ 371 + 2 ⋅ 631) ⋅ 477/6 >2/54!n4
)4*
⇒
W+ =
)3*!
⇒
X23 = 3 ⋅ 21 6 ⋅ 21 −4 ⋅ [2/54 − 1/4 − 2/15 ] >29!lK
3 ⋅ 21 6
b)
∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>1/654!
∆Tplpmjob>−!
R23
lK
= 1!
Up
L
lK
L
(sud i cilidar izolovani od okoline)
∆TSU!>∆TB!,!∆TC>!///>!1/647!,!1/118!>1/654!
U+
W+
+ S hB mo
∆TB> n B ⋅ d wB mo
UB
WB
U∗
W+
∆TC> nC ⋅ d wC mo
+ S hC mo
UC
WC
dipl.ing. @eqko Ciganovi}
lK
L
> 2 ⋅ 1/76 ⋅ mo 477/6 + 1/37 ⋅ mo 2/54 >1/647 lK
L
411
1/4
= 2 ⋅ 2/93 ⋅ mo 477/6 + 1/63 ⋅ mo 2/54 >1/118! lK
L
511
2/15
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 60
2/57/ Termodinami~ki sistem prikazan na slici ~ine:
− zatvoren rezervoar )B* stalne zapremine WB>1/37!n4, u kojem se nalazi
kiseonik (idealan gas) stawa B)qB>5!cbs-!UB>511!L*
− zatvoren rezervoar )C* stalne zapremine WC>1/37!n4 u kojem vlada apsolutni
vakum
− okolina stalne temperature Up>399!L
Otvarawem ventila kiseonik se {iri i u toku procesa {irewa okolini preda
25/5!lK toplote.
a) odrediti pritisak kiseonika nakon {irewa
b) dokazati da je proces {irewa kiseonika nepovratan.
B
a)
C
jedna~ina stawa idealnog gasa B pre {irewa: q B ⋅ WB = n B ⋅ S h ⋅ UB2
nB =
q B ⋅ WB
5 ⋅ 21 6 ⋅ 1/37
=
= 2 lh
S hB ⋅ UB2
371 ⋅ 511
prvi zakon termodinamike za proces {irewa:
R 23 = n ⋅ d w (UB3 − UB2 ) ⇒
UB3 = UB2 +
R 23 = ∆V23 + X23
R 23
25/5
= 511 −
>491!L
n ⋅ dw
2 ⋅ 1/83
jedna~ina stawa idealnog gasa B nakon {irewa: q B3 ⋅ (WB + WC ) = n B ⋅ S h ⋅ UB3
q B3 =
n B ⋅ S h ⋅ UB 3
WB + WC
=
2 ⋅ 371 ⋅ 491
>! 2/: ⋅ 21 6 Qb
1/37 + 1/37
b)
∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>1/258,1/16!>!1/2:8!
∆Tplpmjob>−!
lK
!?!1
L
R 23
− 25/5
lK
=−
>1/161!
L
Up
399
U
q
∆TSU> n B ⋅ d q mo B3 − S hmo B3
UB2
QB2
dipl.ing. @eqko Ciganovi}
491
2/:
lK
> 2 ⋅ 1/:2 ⋅ mo
− 1/37 ⋅ mo
>1/258
L
511
5
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 61
2/58/!Adijabatski izolovan sud podeqen je nepropusnom i adijabatskom membranom na dva dela WB>1/4!n4
i WC>1/6!n4 (slika). U delu B nalazi se nB>1/6!lh kiseonika (idealan gas) temperature UB>411!L, a u
delu C!nC>2!lh sumpor-dioksida (idealan gas) temperature UC>641!L. U delu A kiseonik po~iwe da se
me{a ventilatorom snage
41!X/ Membrana je projektovana da pukne kada razlika pritisaka prema{i ∆q>73!lQb i u tom trenutku se
iskqu~uje ventilator. Odrediti:
a) vreme do pucawa membrane
c* temperaturu i pritisak nastale me{avine posle pucawa membrane, a po uspostavqawu
termodinami~ke ravnote`e
X23
!C
B
b*
q C ⋅ WC = nC ⋅ S hC ⋅ UC
jedna~ina stawa idealnog gasa C:
nC ⋅ S hC ⋅ UC
qC =
WC
=
2 ⋅ 241 ⋅ 641
> 2/49 ⋅ 21 6 Qb
1/6
∆q = q B3 − q C
uslov pucawa membrane:
6
4
6
q B3 = q C + ∆q = 2/49 ⋅ 21 + 1/73 ⋅ 21 > 3 ⋅ 21 !Qb
jedna~ina stawa idealnog gasa B neposredno pred pucawe membrane:
q ⋅W
3 ⋅ 21 6 ⋅ 1/4
UB 3 = B 3 B =
>572/6!L
q B3 ⋅ WB = n B ⋅ S hB ⋅ UB3
⇒
n B ⋅ S hB
1/6 ⋅ 371
prvi zakon termodinamike za proces u delu A (za vreme rada ventilatora)
R 23 = ∆V23 + XU23
⇒
XU23 = −∆V23
XU23 = −n B ⋅ d wB ⋅ (UB3 − UB2 ) = −1/6 ⋅ 1/76 ⋅ (572/6 − 411) >−63/6!lK
τ=
XU23
⋅
X U23
=
− 63/6
. 41 ⋅ 21 .4
dipl.ing. @eqko Ciganovi}
>!2861!t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 62
b)
prvi zakon termodinamike za proces me{awa:
∆V23>1
⇒
R 23 = ∆V23 + X23
V2 = V 3
V2 = n B ⋅ d wB ⋅ UB3 + nC ⋅ d wC ⋅ UC
V 3 = n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U +
U+ =
n B ⋅ d wB ⋅ UB3 + nC ⋅ d wC ⋅ UC
1/6 ⋅ 1/76 ⋅ 572/6 + 2 ⋅ 1/56 ⋅ 641
>612/4!L
=
n B⋅ d wB + nC ⋅ d wC
1/6 ⋅ 1/76 + 2 ⋅ 1/56
jedna~ina stawa me{avine idealnih gasova u trenutku uspostavqawa
toplotne ravnote`e: q + ⋅ (WB + WC ) = n B ⋅ S hB + nC ⋅ S hC ⋅ U +
q+ =
(n B ⋅ S hB + nC ⋅ S hC ) ⋅ U
(
∗
=
WB + WC
)
(1/6 ⋅ 371 + 2 ⋅ 241) ⋅ 612/4 > 2/74 ⋅ 21 6 Qb
1/4 + 1/6
2/59/!Adijabatski izolovan sud podeqen je nepropustqivom i adijabatskom membranom na dva dela WB>1/4
n4!i!WC>1/6!n4 (vidi sliku). U delu B nalazi se nB>1/6!lh kiseonika (idealan gas) temperature UB>411
L, a u delu C!nC>2!lh sumpor-dioksida (idealan gas) temperature UC>461!L/ Me{awe kiseonika se obavqa
ventilatorom pogonske snage 41!X, sumpor-dioksida ventilatorom pogonske snage 56!X. Membrana je
projektovana tako da pukne kada razlika pritisaka prema{i ∆q≥75/3!lQb i u tom trenutku se iskqu~uju
oba ventilatora. Odrediti:
a) vreme do pucawa membrane
b) temperaturu i pritisak nastale me{avine posle pucawa membrane, a po uspostavqawu
termodinami~ke ravnote`e
B
)XU23*B
dipl.ing. @eqko Ciganovi}
C
)XU23*B
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 63
b*
⋅
prvi zakon termodinamike za proces u delu!B;!!! (R 23 ) B = (∆V23 ) B + X U23 ⋅ τ
B
⋅
n B ⋅ d wB ⋅ (UB3 − UB2 ) = − X U23 ⋅ τ
B
)2*
⋅
prvi zakon termodinamike za proces u delu!C;! (R 23 )C = (∆V23 )C + X U23 ⋅ τ
C
⋅
nC ⋅ d wC ⋅ (UC3 − UC2 ) = − X U23 ⋅ τ
C
)3*
⋅
X U23
n ⋅ d ⋅ (UB3 − UB2 )
B
> B wB
!!!!!)4*
deqewem jedna~ina!)2*!i!)3*!dobija se;!
nC ⋅ d wC ⋅ (UC3 − UC2 )
⋅
X U23
C
jedna~ina stawa idealnog gasa!C!neposredno pred!pucawa membrane:
nC ⋅ S hC ⋅ UC3
q C3 ⋅ WC = nC ⋅ S hC ⋅ UC3
⇒
q C3 =
)5*
WC
jedna~ina stawa idealnog gasa!B!neposredno pred!pucawa membrane:
n B ⋅ S hB ⋅ UB3
q B3 =
)6*
q B3 ⋅ WB = n B ⋅ S hB ⋅ UB3
⇒
WB
oduzimawem jedna~ina!)6*!i )5*!dobija se:
n B ⋅ S hB ⋅ UB3 nC ⋅ S hC ⋅ UC3
q B 3 − q C3 =
−
WB
WC
uslov pucawa membrane:
∆q = q B3 − q C3
)7*
)8*
kombinovawem jedna~ina!)4*-!)7*!i!)8*!dobija se:
UB3!>!577/26!L
UC3!>!641!L
vra}awem UB3!u jedna~inu!)2*!ili UC3!u jedna~inu )3*!dobija se:
−n B ⋅ d wB ⋅ (UB3 − UB2 ) −1/6 ⋅ 1/76 ⋅ (577/26 − 411)
τ=
=
>2911!t
⋅
− 41
X U23
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 64
b)
prvi zakon termodinamike za proces me{awa:
∆V23>1
⇒
R 23 = ∆V23 + X23
V2 = V 3
V2 = n B ⋅ d wB ⋅ UB3 + nC ⋅ d wC ⋅ UC3
V 3 = n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U +
U+ =
n B ⋅ d wB ⋅ UB3 + nC ⋅ d wC ⋅ UC3
1/6 ⋅ 1/76 ⋅ 577/26 + 2 ⋅ 1/56 ⋅ 641
=
>614/3!L
n B⋅ d wB + nC ⋅ d wC
1/6 ⋅ 1/76 + 2 ⋅ 1/56
jedna~ina stawa me{avine idealnih gasova u trenutku uspostavqawa
toplotne ravnote`e: q + ⋅ (WB + WC ) = n B ⋅ S hB + nC ⋅ S hC ⋅ U +
(
q+ =
)
(n B ⋅ S hB + nC ⋅ S hC ) ⋅ U ∗ (1/6 ⋅ 371 + 2 ⋅ 241) ⋅ 614/3
> 2/75 ⋅ 21 6 Qb
=
WB + WC
dipl.ing. @eqko Ciganovi}
1/4 + 1/6
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
zadaci za ve`bawe:
strana 65
)2/5:/!−2/61/*
2/5:/ Verikalni cilindar unutra{weg pre~nika e>361!nn,
adijabatski izolovan od okoline, zatvoren je sa gorwe strane bez
trewa pokretnim adijabatskim klipom mase nl>61!lh. Klip na sebi
nosi oprugu zanemarqive te`ine, linearne karakteristike l>231
O0dn3 i u po~etnom polo`aju udaqen je od dna cilindra {>511!nn
(slika). Pritisak okoline iznosi qp>2!cbs/ U cilindru se nalazi
vazduh (idealan gas) temperature U2>3:4!L. Na oprugu se odozgo
po~iwe spu{tati teg nbtf!nU>411!lh/ Od trenutka kada teg dodirne
opruga , on po~inwe oprugu sa klipom potiskivati na dole,
istovremeno sabijaju}i oprugu i gas u cilindru. Odrediti
a) za koliko se spusti klip )∆{* a koliko sabije opruga )∆m* do
trenutka kada sila u u`etu postane jednaka nuli (stawe 2)
b) do koje temperature bi trebalo zagrejati vazduh stawa 2 da bi
klip vratili u prvobitni polo`aj i koliko toplote je za to
potrebno dovesti
nu
m
{
e
a) ∆{>21:!nnb) U>563/9!L-
∆m>356!nn
R>4/2!lK
2/61/ Cilindar je napravqen prema slici. Slobodno
pomi~ni klip zanemarqive mase, optere}en tegom mase
nU>311!lh, nalazi se u po~etnom polo`aju na kao na
slici. U cilndru se nalazi vazduh po~etne temperature
U2>634!L koji se hladi predaju}i kroz zidove cilindra
toplotu okolini stawa P)qp>2!cbs-!!Up>3:4!L* sve do
uspostavqawa toplotne ravnote`e sa okolinom.
Odrediti:
a) pri kojoj temperaturi vazduha u cilindru }e klip
dodirnuti oslonac (stawe 2)
b) pritisak gasa na po~etku i kraju procesa
c) koli~inu toplote koju vazduh preda okolini tokom
procesa 1−2−3
nu
∆{
E
e>291!nn
E>311!nn
{
e
{>461!nn
∆{>261!nn
a) U3!>456/7!L
b) q2!>2/73!cbs-!q4!>2/48!cbs
c) R24!>!−4/27!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 66
PRVI I DRUGI ZAKON TERMODINAMIKE
(OTVOREN TERMODINAMI^KI SISTEM)
2/62/ Vazduh (idealan gas) struji stacionarno kroz vertikalnu
cev visine 4/7!n, konstantnog popre~nog preseka, masenim
3
⋅
protokom od n >411!lh0i (slika). Cev je toplotno izolovana
od okoline, a u cevi je instaliran greja~ koji vazduhu predaje
toplotu. Stawe vazduha na ulazu u cev odre|eno je veli~inama
stawa 2)q2>2/3!cbs-!U2>3:4!L-!x>5/6!n0t*, a na izlazu 3)q3>2
cbs!U3>472!L*. Odrediti:
a) brzinu vazduha na izlazu iz cevi
c* toplotni protok koji greja~ saop{tava vazduhu
{3−{2
a)
2
jedna~ina stawa idealnog gasa na ulazu u cev:
q2 ⋅ w 2 = S h ⋅ U2
S h ⋅ U2
w2 =
q2
=
398 ⋅ 3:4
2/3 ⋅ 21 6
>1/8118!
n4
lh
q 3 ⋅ w 3 = S h ⋅ U3
jedna~ina stawa idealnog gasa na izazu iz cevi:
w3 =
S h ⋅ U3
q3
=
398 ⋅ 472
2 ⋅ 21 6
>2/1472!
n4
lh
x2 ⋅
jedna~ina kontinuiteta:
x 3 = x2 ⋅
e23 ⋅ π
e 3 ⋅π
x3 ⋅ 3
5
5
=
w2
w3
w3
n
2/1472
= 5/6 ⋅
>7/76!
w2
1/8118
t
b)
⋅
⋅
⋅
prvi zakon termodinamike za proces u cevi:! R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23
⋅
⋅
x 33 − x 23
+ n⋅ ({ 3 − {2 )
3
411
411 7/76 3 − 5/6 3
411
=
⋅ 2 ⋅ 21 4 ⋅ (472 − 3:4) +
⋅
+
⋅ 4/7 >6779!X
4711
4711
3
4711
⋅
⋅
R 23 = ∆ n⋅ d q ⋅ (U3 − U2 ) + n⋅
⋅
R 23
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 67
⋅
2/63/ U toplotno izolovanoj komori me{aju se tri struje idealnih gasova: kiseonik B) n >7!lh0t-!q>1/29
⋅
⋅
NQb-!u>361pD*- azot C) n >4!lh0t-!q>1/44!NQb-!u>6:1pD* i ugqen−monoksid D) n >3!lh0t-!q>1/49!NQbu>551pD*/ Pritisak dobijene sme{e na izlazu iz komore q+>1/2!NQb/!Zanemaruju}i promenu kineti~ke
energije kao i potencijalne energije, odrediti:
a) temperaturu )U+* i zapreminski protok ( W + ) dobijene sme{e
c* promenu entropije sistema za proces me{awa
B
C
me{avina
D
a)
prvi zakon termodinamike za proces u me{noj komori:
⋅
⋅
⋅
R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23
⇒
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
I2 = I3
I2 = n B ⋅ d qB ⋅ UB + nC ⋅ d qC ⋅ UC + nD ⋅ d qD ⋅ UD
I3 = n B ⋅ d qB ⋅ U + + nC ⋅ d qC ⋅ U + + nD ⋅ d qD ⋅ U +
⋅
+
U =
⋅
⋅
n B ⋅ d qB ⋅ UB + nC ⋅ d qC ⋅ UC + nD ⋅ d qD ⋅ UD
⋅
⋅
⋅
n B ⋅ d qB + nC ⋅ d qC + nD ⋅ d qD
U+ =
7 ⋅ 1/:2 ⋅ 634 + 4 ⋅ 2/15 ⋅ 974 + 3 ⋅ 2/15 ⋅ 824
= 76:/7!L
7 ⋅ 1/:2 + 4 ⋅ 2/15 + 3 ⋅ 2/15
jedna~ina stawa idealne gasne me{avine na izlazu iz komore za me{awe:
⋅
⋅
⋅
q + ⋅ W + = n B ⋅ S hB + nC ⋅ S hC + nD ⋅ S hD ⋅ U +
⇒
⋅
⋅
⋅
n B ⋅ S hB + nC ⋅ S hC + nD ⋅ S hD ⋅ U +
(7 ⋅ 371 + 4 ⋅ 3:8 + 3 ⋅ 3:8) ⋅ 76:/7
W+ =
=
+
q
2 ⋅ 21 6
W + = 31 !
n4
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 68
b)
jedna~ina stawa idealnog gasa B u nastaloj me{avini:
⋅
q +B
=
n B ⋅ S hB ⋅ U +
W
+
7 ⋅ 371 ⋅ 76:/7
>1/62 ⋅ 21 6 !Qb
31
=
jedna~ina stawa idealnog gasa C u nastaloj me{avini:
⋅
q C+ =
nC ⋅ S hC ⋅ U +
W
+
⋅
⋅
=
nD ⋅ S hD ⋅ U +
W
⋅
+
=
⋅
⋅
⋅
lX
L
⋅
∆ T SU > ∆ T B , ∆ TC , ∆ TD >!///>4/34!,!2/44!,!2/6:!>!7/26!
⋅
⋅
q +D ⋅ W + = nD ⋅ S hD ⋅ U +
3 ⋅ 3:8 ⋅ 76:/7
>1/31 ⋅ 21 6 !Qb
31
∆ Ttj > ∆ T SU , ∆ Tp >!///!>7/26!
⋅
⋅
q C+ ⋅ W + = nC ⋅ S hC ⋅ U +
4 ⋅ 3:8 ⋅ 76:/7
>1/3: ⋅ 21 6 !Qb
31
=
jedna~ina stawa idealnog gasa D u nastaloj me{avini:
q +D
⋅
q +B ⋅ W + = n B ⋅ S hB ⋅ U +
lX
L
⋅
R23
lX
∆ Tp = −
>1!
L
Up
(adijabatski izolovana komora za me{awe)
⋅
⋅
q+
U+
∆ T B = g (q- U ) = n B . d qB mo
− S hB mo B
UB
qB
=
⋅
76:/7
1/62 ⋅ 21 6
lX
∆ T B > 7 ⋅ 1/:2 ⋅ mo
− 1/37 ⋅ mo
= 4/34
7
L
634
1/29 ⋅ 21
⋅
⋅
q+
U+
∆ TC > g (q- U ) > nC /! d qC mo
− S hC mo C >
UC
q C
⋅
76:/7
1/3: ⋅ 21 6
lX
>2/44!
∆ TC > 4 ⋅ 2/15 ⋅ mo
− 1/3:8 ⋅ mo
7
L
974
1/44 ⋅ 21
⋅
⋅
q+
U+
∆ TD > g (q- U ) > nD /! d qD mo
− S hC mo D
UD
qD
>
⋅
76:/7
1/31 ⋅ 21 6
− 1/3:8 ⋅ mo
∆ TD > 3 ⋅ 2/15 ⋅ mo
824
1/49 ⋅ 21 7
dipl.ing. @eqko Ciganovi}
>2/6:! lX
L
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 69
2/64/ “Ludi nau~nik” tvrdi da je mogu}e, bez izmene toplote i/ili rada sa okolinom, struju vazduha stawa
⋅
2) n >2!lh0t-!q>4!cbs-!U>3:4!L* razdvojiti na dve struje. Struju 2 stawa 3)q>3/8!cbs-!U>444!L* i struju 3
stawa 4)q>3/8!cbs-!U>384!L*/ Dokazati da je “ludi nau~nik” u pravu. Zanemariti promene kineti~ke i
potencijalne energije vazduha.
prvi zakon termodinamike za proces u razdelnoj komori:
⋅
⋅
⋅
R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23
⋅
⋅
⋅
⇒
⋅
n2⋅ d q ⋅ U2 > n3 ⋅ d q ⋅ U3 + n4 ⋅ d q ⋅ U4
⋅
jedna~ina kontinuiteta:
⋅
I2 = I3
)2*
⋅
⋅
n2 = n3 + n4
)3*
⋅
kombinovawem jedna~ina!)2*!j!)3*!dobija se:
n3 >1/44!
lh ⋅
lh
-! n4 >1/78!
t
t
drugi zakon termodinamike za proces u razdelnoj komori:
⋅
⋅
⋅
∆ Ttj > ∆ T SU , ∆ Tp >!///!>36/7!
⋅
⋅
X
L
⋅
∆ T SU > ∆ T 3 , ∆ T 4 >!///>63/85!−38/25!>!36/7!
⋅
X
L
⋅
R23
X
∆ Tp = −
>1!
Up
L
(adijabatski izolovana komora za me{awe)
⋅
⋅
U
q
∆ T 3 = g (q- U ) = n3 . d q mo 3 − S h mo 3
U2
q2
⋅
444
3/8
X
∆ T 3 > 1/44 ⋅ 2 ⋅ mo
− 1/398 ⋅ mo
= 63/85
L
3:4
4
⋅
⋅
U
q
∆ T 4 = g (q- U ) = n4 . d q mo 4 − S h mo 4
U2
q2
=
⋅
384
3/8
X
∆ T 4 > 1/77 ⋅ 2 ⋅ mo
− 1/398 ⋅ mo
= −38/25
3:4
L
4
⋅
Kako je ∆ Ttj ?!1, ovaj proces je mogu} pa je “ludi nau~nik” u pravu.
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 70
!2
2/65/ [ire}i se u gasnoj turbini, tok vazduha (idealan gas), mewa
R23
⋅
svoje toplotno stawe od stawa 2) W 2>1/3!n40t- q>21!cbs-!U>691!L*!,
⋅
na ulazu u turbinu, do stawa 3) W 3>2/3!n40t-!q>2!cbs*- na izlazu iz
we. Tokom {irewa usled neidealnog toplotnog izolovawa turbine,
toplotni protok sa vazdu{nog toka na okolni vazduh iznosi 29!lX.
Zanemaruju}i promene kineti~ke i potencijalne energije vazdha ,
odrediti:
a) snagu turbine
b) dokazati da je proces u turbini nepovratan (temperatura okoline
iznosi Up>3:4!L)
XU23
!3
a)
⋅
jedna~ina stawa idealnog gasa na ulazu u turbinu:
⋅
q2 ⋅ W 2 = n⋅ S h ⋅ U2
⋅
⋅
q ⋅ W 2 21 ⋅ 21 6 ⋅ 1/3
lh
n= 2
=
>2/3!
S h ⋅ U2
398 ⋅ 691
t
⋅
jedna~ina stawa idealnog gasa na izlazu iz turbine:
⋅
U3 =
q3 ⋅ W 3
2 ⋅ 21 6 ⋅ 2/3
>459/5!L
398 ⋅ 2/3
=
⋅
⋅
q 3 ⋅ W 3 = n⋅ S h ⋅ U3
Sh ⋅ n
prvi zakon termodinamike za proces u turbini:
⋅
⋅
⋅
R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23
⋅
⋅
⋅
⋅
⇒
⋅
X U23 = R23 − ∆ I23 = R23 − n⋅ dq ⋅ (U3 − U2) > −29 − 2/3 ⋅ 2 ⋅ (459/5 − 691) >371!lX
b)
⋅
⋅
⋅
∆ Ttj > ∆ T SU , ∆ Tp >!///!>72/5!,!292/5!>353/9!
⋅
X
?!1
L
⋅
X
− 29
R23
∆ Tp = −
=−
>72/5!
Up
3:4
L
⋅
⋅
⋅
U
q
∆ T SU = ∆ T23 = g (q- U ) = n . d q mo 3 − S h mo 3
U2
q2
⋅
∆ T SU > 2/3 ⋅ 2 ⋅ mo
X
459/5
2
− 1/398 ⋅ mo = 292/5
691
21
L
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 71
⋅
2/66/!U dvostepenoj gasnoj turbini ekspandira n >3!lh0t vazduha (idealan gas) od po~etnog pritiska
q2>71!cbs!do krajweg pritiska q5>21!cbs. Nakon ekspanzije u prvom stepenu turbine vazduh se uvodi u
me|uzagreja~ u kome se izobarski zagreva do temperature U2>U4>911!L/ Ekspanzije u oba stepena su
adijabatske i kvazistati~ke. Zanemariti promene potencijalne i kineti~ke energije. Skicirati
proces u qw i Ut koordinatnim sistemima i odrediti:
a) pritisak u me|uzagreja~u tako da snaga dvostepene turbine bude maksimalna
b) snagu dvostepene turbine u tom slu~aju
2
XU23
3
4
R34
XU45
5
q
U
2
3
2
4
3
5
4
5
w
t
a)
⋅
⋅
⋅
Q = X U23 + X U 45 = n⋅ d q ⋅ (U2 − U3 + U4 − U5 ) !>!///
κ
q2 U2 κ −2
=
q3 U3
κ
⇒
U κ −2
q3 = q2 ⋅ 3 >qy
U2
⇒
U κ −2
q4 = q5 ⋅ 4 >qy
U5
κ
q4 U4 κ −2
=
q5 U5
dipl.ing. @eqko Ciganovi}
)2*
κ
)3*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 72
kombinovawem jedna~ina (1) i (2) dobija se:
κ
κ
U κ −2
U κ −2
q2 ⋅ 3
> q5 ⋅ 4
U2
U5
U ⋅U q
U3 = 4 2 ⋅ 2
U5 q5
⇒!
κ
⇒
q2 U4 ⋅ U2 κ −2
=
q5 U5 ⋅ U3
⇒
2− κ
U4 ⋅ U2 q2 κ
Q = n⋅ dq ⋅ U2 −
⋅
+ U4 − U5
U5 q5
q
U5 = U4 ⋅ U2 ⋅ 2
q5
2− κ
κ
71 ⋅ 216
> 911 ⋅ 911 ⋅
21 ⋅ 216
911 ⋅ 911 71 ⋅ 216
U3 =
⋅
72:/44 21 ⋅ 216
=
U4 ⋅ U2
U5 ⋅ U3
2− κ
⋅
U ⋅U q κ
∂Q
4
2 2
2
= n⋅ dq ⋅
⋅
−
3
q
∂U5
U5
5
2− κ
U ⋅U q κ
4
2 2
⋅
− 2 = 1
3
U5
q5
⇔
κ −2
κ
2− κ
κ
⋅
∂Q
=1
∂U5
q2
q
5
2−2/5
2/5
2−2/5
2/5
q
U5 = U4 ⋅ U2 ⋅ 2
q5
⇒
⇒
>72:/44!L
κ
>72:/44!L-
2− κ
κ
2/5
U κ −2
72:/44 2/5 −2
>35/6!cbs
q3 = q2 ⋅ 3 > 71 ⋅ 216 ⋅
911
U2
b)
Qnby = 3 ⋅ 2⋅ (911 − 72:/44 + 911 − 72:/44) >833/79!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 73
⋅
2/67/!Kopresor, snage!3!lX-!usisava okolni vazduh (idealan gas) stawa!2)q>2!cbs-!U>3:4!L-! n >1/16
lh0t*!i adijabatski ga sabija do stawa!3/!Nakon toga se vazduh adijabatski prigu{uje do po~etnog
pritiska )q4>q2*. Prira{taj entropije vazduha za vreme adijabatske kompresije i adijabatskog
prigu{ivawa je jednak. Odrediti stepen dobrote adijabatske kompresije. Zanemariti promene
kineti~ke i potencijalne energije vazduha.
2
3
4
XU23
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
ograni~enom isprekidanom linijom:
⋅
⋅
⋅
R 24 = ∆ I24 + X U24 + ∆Fl24 + ∆Fq24
⋅
⋅
1 = n⋅ d q ⋅ (U4 − U2 ) + X U23
⇒
U4 = U2 −
XU23
⋅
= 3:4 +
n⋅ d q
3
>444!L
1/16 ⋅ 2
prvi zakon termodinamike za proces u prigu{nom ventilu :
⋅
⋅
⋅
R 34 = ∆ I34 + X U34 + ∆Fl34 + ∆Fq34
∆t23 = ∆t 34
U
q 3 = q2 ⋅ 3
U2
q
U2
= 2
U3l q 3l
ηlq
e =
⇒
d q mo
⋅
⇒
⋅
I3 = I 4
U3>U4
U
q
U3
q
− S h mo 3 = d q mo 4 − S h mo 4
U2
q2
U3
q3
dq
⇒
2
3⋅Sh
444 3⋅1/398
> 2/36 ⋅ 21 6 !Qb
= 2 ⋅ 21 6 ⋅
3:4
κ −2
κ
⇒
⇒
U3l
q
= U2 ⋅ 3l
q2
κ −2
κ
q3l>q3
2/36
= 3:4 ⋅
2
2/5 −2
2/5
>423/4!L
U2 − U3l
3:4 − 423/4
>1/59
=
U2 − U3
3:4 − 444
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 74
2/68/!!U turbo kompresorskoj stanici se vr{i dvostepena kvazistati~ka adijabatska kompresija.
Kompresor usisava 6!lnpm0i okolnog vazduha (idelan gas) stawa!2)q>2!cbs-!U>3:4!L* i sabija ga na
neki me|u pritisak pri kojem se daqe hladi do temperature okoline. Nakon toga se vazduh sabija na
kona~ni pritisak 5)q>:!cbs*. Zanemaruju}i promene potencijalne i kineti~ke energije, odrediti:
a) vrednost me|u pritiska )q3>q4*!pri kojem su snage potrebne za oba stepena sabijawa jednake
b) u{tedu u snazi u ovom procesu u odnosu na kompresiju bez me|uhla|ewa, tj. kada bi se kvazistati~ka
adijabatska kompresija od stawa 2)q>2!cbs-!U>3:4!L* do pritiska od :!cbs!vr{ila u jednom stepenu
5
XU45
3
4
R34
XU23
2
a)
prvi zakon termodinamike za proces u kompresoru niskog pritiska:
⋅
⋅
⋅
R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23
⋅
⋅
X U23 = n⋅ d q ⋅ (U2 − U3 ) !!!!!!!)2*
⇒
prvi zakon termodinamike za proces u kompresoru visokog pritiska:
⋅
⋅
⋅
R 45 = ∆ I45 + X U 45 + ∆Fl 45 + ∆Fq45
⇒
⋅
⋅
⋅
⋅
X U 45 = n⋅ d q ⋅ (U4 − U5 ) !!!!!!!)3*
Kombinovawem uslova zadatka X U23 = X U 45 !i!U2>U4!sa jedna~inama!)2*!i!)3*
dobija se!U3>U5/
κ
q2 U2 κ −2
=
q3 U3
κ
⇒
U κ −2
q3 = q2 ⋅ 3
U2
⇒
U κ −2
q4 = q5 ⋅ 4
U5
κ
q4 U4 κ −2
=
q5 U5
dipl.ing. @eqko Ciganovi}
)4*
κ
)5*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 75
q
U5 = U2 ⋅ 5
q2
deqewem jedna~ina!)4* i!)5*!dobija se:
κ −2
3κ
2/5 −2
: 3⋅2/5
U5 = 3:4 ⋅
>512/15!L!>!U3
2
2/5
512/15 2/5 −2
>! 4 ⋅ 21 6 Qb!>!q4
q 3 = 2 ⋅ 21 ⋅
3:4
6
⇒
iz jedna~ine!)4*!
b)
q
U5( = U2 ⋅ 5
q2
κ −2
κ
:
> 3:4 ⋅
2
2/5 −2
2/5
>659/:3!L
⋅
⋅
⋅
(
)
X U > o⋅ Nd q ⋅ (U2 − U3 + U4 − U5 ) =
⋅ -
⋅ -
⋅
(
)
⋅
6
⋅ 3:/2 ⋅ (3:4 − 512/15 ) >!−5/48!LX
4711
⋅
X U = X U25 (
potrebna snaga u drugom slu~aju:
X U > o⋅ Nd q ⋅ (U2 − U5 ( ) =
⋅
X U = X U23 + X U 45
potrebna snaga u prvom slu~aju:
6
⋅ 3:/2 ⋅ (3:4 − 659/:3) >!−21/45!LX
4711
⋅
⋅ -
⋅
∆ X U = X U − X U >−5/48,21/45!>!6/:8!lX
u{teda u snazi:
5′
U
zatvorena povr{ina!!)3−4−5−5′−3*!predstavqa u{tedu u
⋅
5
snazi ( ∆ X U *!koja je ostvarena dvostepenom
kompresijom (u odnosu na jednostepenu kompresiju)
3
4
2
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 76
2/69/!U dvostepenom kompresoru sa me|uhla|ewem, pri ustaqenim uslovima, sabija se neravnote`no
(nekvazistati~ki) i adijabatski!1/4!lh0t!azota (idealan gas), od polaznog stawa!2)q2>1/2!NQb-!U2>3:4!L*
do stawa!5)q5>1/7!NQb-!U5>561!L*/!Stepeni dobrote prilikom sabijawa u oba stepena su jednaki i iznose
ηJe = ηJJe = 1/9 /!Odrediti ukupnu snagu za pogon kompresora i toplotnu snagu koja se odvodi pri hla|ewu
(izme|u dve kompresije), ako se proces me|uhla|ewa odvija pri stalnom pritisku!q3>q4!>!1/4!NQb/
Zanemariti promene kineti~ke i potencijalne energije vazduha i prikazati sve procese na!Ut!dijagramu.
5
XU45
4
3
XU23
R34
2
κ −2
κ
κ −2
κ
U2 q2
=
U3l q3l
⇒
U3l
U2 − U3l
U2 − U3
⇒
U3 = U2 −
ηJe =
q
U4
= 4
U5l q 5l
ηJJe =
κ −2
κ
U4 − U5l
U4 − U5
q
= U2 ⋅ 3l
q2
U2 − U3l
q
= U4 ⋅ 5l
q4
= 3:4 −
κ −2
κ
⇒
U5l
⇒
U5l = U4 − ηJJe ⋅ (U4 − U5 )
kombinovawem jedna~ina!)2*!i!)3*!dobija se:
dipl.ing. @eqko Ciganovi}
ηJe
4
= 3:4 ⋅
2
!
2/5 −2
2/5
>512!L
3:4 − 512
= 539!L
1/9
)2*
)3*
U4>465!L-!U5l>542!L
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 77
⋅
⋅
⋅
R 34 = n⋅ (r34 )q=dpotu
odvedena toplota u fazi me|uhla|ewa )3−4*;
⋅
R 34 = n⋅ d q ⋅ (U4 − U3 ) = 1/4 ⋅ 2/15 ⋅ (465 − 539) >!−34/2!lX
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
⋅
⋅
⋅
⋅
⋅
⋅
R 34 = ∆ I25 + X U25 + ∆Fl25 + ∆Fq25
ograni~enom isprekidanom linijom:
⋅
X U25 = R 34 − ∆ I25 = R 34 − n⋅ d q ⋅ (U5 − U2 ) > −34/2 − 1/4 ⋅ 2/15 ⋅ (561 − 3:4)
⋅
X U25 >−83/2!lX
napomena:
ukupna snaga za pogon oba kompresora jednaka je zbiru snaga
potrebnih za pogon kompresora niskog pritiska i kompresora
⋅
⋅
⋅
visokog pritiska tj:! X U25 = X U23 + X U 45
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 78
PRVI I DRUGI ZAKON TERMODINAMIKE
(PUWEWE I PRA@WEWE REZERVOARA)
2/6:/!Vazduh (idealan gas) stawa!)q>2!cbs-!U>3:1!L) nalazi se u toplotno izolovanom rezervoaru
zapremine!W>1/7!n4/!Toplotno izolovanim cevovodom u rezervoar se uvodi vazduh (idealan gas) stawa
)q>21!cbs-!U>511!L). Tokom procesa u rezervoaru je stalno ukqu~en greja~ snage!3/6!lX/!Kada vazduh u
rezervoaru dostigne stawe!)q>5!cbs-!U>611!L*!prekida se dotok vazduha i iskqu~uje greja~. Odrediti
vreme trajawa procesa puwewa rezervoara kao i maseni protok vazduha koji se uvodi u rezervoar.
q>2!cbs-!U>3:1!L
q>5!cbs-!U>611!L
q>21!cbs-!U>511!L
po~etak:
kraj:
ulaz:
jedna~ina stawa idealnog gasa za po~etak:
nqpd =
q qpd ⋅ W
S h ⋅ Uqpd
=
2 ⋅ 21 6 ⋅ 1/7
>1/83!lh
398 ⋅ 3:1
q ls ⋅ W = nls ⋅ S h ⋅ Uls
jedna~ina stawa idealnog gasa za kraj:
nls =
q qpd ⋅ W = nqpd ⋅ S h ⋅ Uqpd
q ls ⋅ W
5 ⋅ 21 6 ⋅ 1/7
=
>2/78!lh
S h ⋅ Uls
398 ⋅ 611
nqp + nvm = nls + nj{
materijalni bilans procesa puwewa:
nvm = nls − nqp >2/78!−!1/83!>!1/:6!lh
prvi zakon termodinamike za proces puwewa:
R 23 − X23 = Vls − Vqp + Ij{ − Ivm
R 23 = nls ⋅ d w ⋅ Uls − nqp ⋅ d w ⋅ Uqp − nvm ⋅ d q ⋅ Uvm
R 23 = 2/78 ⋅ 1/83 ⋅ 611 − 1/83 ⋅ 1/83 ⋅ 3:1 − 1/:6 ⋅ 2 ⋅ 511 >81/97!lK
τ=
R 23
⋅
R 23
⋅
nvm =
=
81/97
>39/4!t
3/6
nvm 1/:6
h
=
>44/7!
τ
39/4
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 79
2/71/!U verikalnom toplotno izolovanom cilindru-!povr{ine popre~nog preseka B>1/2!n3-! nalazi se
vazduh (idealan gas) stawa )U>291pD-!n>1/16!lh), ispod toplotno izolovanog klipa mase koja odgovara
te`ini od!31!lO-!a na koji spoqa deluje atmosferski pritisak od!1/2!NQb!(slika). U cilindar se, kroz
toplotno izolovan cevovod, naknadno uvede vazduh stawa!)q>1/5!NQb-!U>651pD-!n>1/2!lh*!{to dovede
do pomerawa klipa (bez trewa)/!Zanemaruju}i promene kineti~ke i potencijalne energije uvedenog
vazduha odrediti koliki rad izvr{i vazduh nad okolinom kao i temperaturu vazduha u cilindru na
kraju procesa.
kraj
∆z
po~etak
ulaz
Gufh
= 2 ⋅ 21 6 +
po~etak:
U>564!L-!q> q p +
kraj:
ulaz:
q>4!cbs
q>5!cbs-!U>924!L-!n>1/2!lh
B
jedna~ina stawa idealnog gasa za po~etak:
Wqpd =
nqpd ⋅ S h ⋅ Uqpd
=
1/16 ⋅ 398 ⋅ 564
4 ⋅ 21 6
materijalni bilans procesa puwewa:
q qpd
31 ⋅ 21 4
>4!cbs-!n>1/16!lh
1/2
q qpd ⋅ Wqpd = nqpd ⋅ S h ⋅ Uqpd
>1/1328!n4
nqp + nvm = nls + nj{
nls = nqp + nvm >1/16!,!1/2!>!1/26!lh
q ls ⋅ Wls = nls ⋅ S h ⋅ Uls !!!)2*
jedna~ina stawa idealnog gasa za kraj:
prvi zakon termodinamike za proces puwewa:
R 23 − X23 = Vls − Vqp + Ij{ − Ivm
−X23 = nls ⋅ d w ⋅ Uls − nqp ⋅ d w ⋅ Uqp − nvm ⋅ d q ⋅ Uvm
)3*
Wls
X23 =
∫()
(
q W ⋅ eW = q qpd ⋅ Wls − Wqpd
)
!!!)4*
Wqpd
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 80
kada se jedna~ine )2*!i )4* uvrste u jedna~inu )3* dobija se:
(
)
− q qpd ⋅ Wls − Wqpd = nls ⋅ d w ⋅
Wls =
Wls =
q ls ⋅ Wls
− nqp ⋅ d w ⋅ Uqp − nvm ⋅ d q ⋅ Uvm
nls ⋅ S h
⇒
nqp ⋅ d w ⋅ Uqp + nvm ⋅ d q ⋅ Uvm + q qpd ⋅ Wqpd
q qpd +
dw
⋅ q lsbk
Sh
1/16 ⋅ 1/83 ⋅ 564 + 1/2 ⋅ 2 ⋅ 924 + 4 ⋅ 21 6 ⋅ 21 −4 ⋅ 1/1328
>1/1:9:!n4
1/83
6
6
−4
⋅ 4 ⋅ 21
4 ⋅ 21 ⋅ 21 +
398
)2*!!!⇒ ! Uls =
q ls ⋅ Wls
4 ⋅ 21 6 ⋅ 1/1:9:
=
>79:/3!L
nls ⋅ S h
1/26 ⋅ 398
(
)
)4*!!!⇒! ! X23 = qqpd ⋅ Wls − Wqpd = 4 ⋅ 216 ⋅ 21 −4 ⋅ (1/1:9: − 1/1328) >34/27!lK
2/72/!Kroz toplotno izolovan cevovod, unutra{weg pre~nika!e>21!nn*-!biva uveden azot!)O3-!idealan
gas*!stawa!)U>416!L-!q>1/7!NQb-!x>21!n0t) u toplotno izolovan rezervoar zapremine!W>1/7!n4!!u
kojem se ve} nalazi ugqen−dioksid!)DP3-!idealan gas) stawa!)q>1/2!NQb-!U>3:4!L*/!Ako se proces
puwewa prekida kada pritisak sme{e u rezevoaru dostigne!1/6!NQb-!odrediti:
a) temperaturu me{avine idealnih gasova u rezervoaru na kraju procesa puwewa
b) promenu entropije sistema za vreme procesa puwewa
c) vreme trajawa procesa puwewa rezervoara
po~etak:
kraj:
ulaz:
q>2!cbs-!U>3:4!L-!DP3
q>6!cbs-!DP3!+!O3
q>7!cbs-!U>416!L-!x>21!n0t-!!O3
a)
prvi zakon termodinamike za proces puwewa:
R 23 − X23 = Vls − Vqp + Ij{ − Ivm
x3
!!)2*
1 = nDP3 ⋅ dwDP3 + nO3 ⋅ dwO3 ⋅ Uls − nDP3 ⋅ dwDP3 ⋅ Uqp − nO3 ⋅ dqO3 ⋅ Uvm +
3
(
)
jedna~ina stawa me{avine idealnih gasova za zavr{etak puwewa:
q ls ⋅ W = )nDP3 ⋅ S hDP3 + nO3 ⋅ S hO3 * ⋅ Uls
!!!!
jedna~ina stawa idealnog gasa za )DP3*!po~etak:
!!!!!!!)3*
q qpd ⋅ W = nDP3 ⋅ S hDP3 ⋅ Uqpd
!!!!!!!)4*
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
⇒
)4*
nDP3 =
q qpd ⋅ W
S hDP3 ⋅ Uqpd
strana 81
=
2 ⋅ 21 6 ⋅ 1/7
>2/19!lh
29: ⋅ 3:4
kombinovawem jedna~ina!)2*!i!)3*!dobija se:!
Uls>494/7!L-! nO3 >2/:5!lh
b)
∆Ttjtufn!>!∆Tubeop!ufmp!,!∆Tplpmjob!>!///!>:47/4!
∆Tplpmjob>1!
K
L
K
! (sud adijabatski izolovan od okoline)
L
∆Tsbeop!ufmp!> ∆TDP3 + ∆TO3 >!///>!2:3!,!855/4!>!:47/4!
K
L
U
W
494/7
K
∆T DP3 >g)U-!W*> nDP3 ⋅ dwDP3 mo ls + ShDP3 mo = 2/19 ⋅ 1/77 ⋅ mo
>2:3!
L
Uqp
W
3:4
O3
q lsbk
U
∆TO3 >g)U-!Q*> nO3 ⋅ d qO3 mo ls − S hO3 mo
Uvm
q vm
>
494/7
4/79
K
− 1/3:8 ⋅ mo
∆TO3 > 2/:5 ⋅ 2/15 ⋅ mo
>///>!855/4!
L
416
7
jedna~ina stawa idealnog gasa )O3* za kraj:
3
qO
ls =
nO3 ⋅ ShO3 ⋅ Uls
W
=
O
q ls3 ⋅ W = nO3 ⋅ S hO3 ⋅ Uls
2/:5 ⋅ 3:8 ⋅ 494/7
> 4/79 ⋅ 216 Qb
1/7
c)
τ=
nO3
⋅
nO3
⋅
>///>
2/:5
6/3 ⋅ 21−4
nO3 = ρvmb{ ⋅ x ⋅
ρ vmb{ =
>484/2!t
e3π
1/123 π
lh
>///> 7/73 ⋅ 21 ⋅
> 6/3 ⋅ 21−4
5
5
t
q vmb{
lh
7 ⋅ 21 6
=
>7/73! 4
S hO3 ⋅ Uvmb{ 3:8 ⋅ 416
n
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 82
2/73/ U rezervoaru zapremine!W>1/4!n4!nalazi se azot stawa!)q>231!cbs-!U>411!L*/!Ventil na rezervoaru
se brzo otvori, ispusti se izvesna koli~ina azota u atmosferu a zatim se ventil ponovo zatvori, tako da
se mo`e smatrati da pri takvim uslovima nema razmene toplote izme|u rezervoara i okoline. Ne
posredno po zatvarawu ventila pritisak azota u rezervoaru iznosi!q>71!cbs/!Odrediti:
a) koli~inu azota koja je istekla iz rezervoara )lh* kao i temperaturu azota u rezervoaru neposredno
posle zatvarawa ventila
c* koli~inu toplote koju bi trebalo dovesti preostalom vazduhu u sudu da bi dostigao pritisak koji je je
imao pre pra`wewa )NK*
b*
Promena stawa azota koji se za vreme procesa pra`wewa nalazi u sudu je
kvazistai~ka adijabatska promena, pa se temperatura azota u sudu na kraju
procesa pra`ewa mo`e odrediti iz zakona kvazistati~ke adijabatske promene:
Uls qls
=
Uqp qqp
κ −2
κ
q
Uls = Uqp ⋅ ls
qqp
⇒
κ −2
κ
71
= 411 ⋅
231
jedna~ina stawa idealnog gasa za po~etak:
nqpd =
qqpd ⋅ W
Sh ⋅ Uqpd
=
>357/2!L
q qpd ⋅ W = nqpd ⋅ S h ⋅ Uqpd
231 ⋅ 216 ⋅ 1/4
>51/5!lh
3:8 ⋅ 411
jedna~ina stawa idealnog gasa za kraj:
nls =
2/5 −2
2/5
q ls ⋅ W = nls ⋅ S h ⋅ Uls
q ls ⋅ W
71 ⋅ 21 6 ⋅ 1/4
=
>35/74!lh
S h ⋅ Uls
3:8 ⋅ 357/2
nqp + nvm = nls + nj{
materijalni bilans procesa pra`wewa:
nj{ = nqp − nls >51/5!−!35/74!>!26/88!lh
b)
2!>!kraj
3
!
)n>35/74!lh-!q>71!cbs-!U>357/2!L*
)n>35/74!lh-!q>231!cbs-!U>@*
jedna~ina stawa idealnog gasa za stawe 2:
U3 =
q3 ⋅ W = n ⋅ Sh ⋅ U3
⇒
6
q3 ⋅ W 231 ⋅ 21 ⋅ 1/4
=
>5:3/2!L
n ⋅ Sh
35/74 ⋅ 3:8
R23 = n ⋅ (r23 )w =dpotu = n ⋅ dw ⋅ (U3 − U2) > 35/74 ⋅ 1/85 ⋅ (5:3/2 − 357/2) >5/59!NK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 83
ME[AVINE IDEALNIH GASOVA
2/74. Za me{avinu idealnih gasova, kiseonika )B*!i azota!)C), odredititi molsku masu!)Nn*-!gasnu
konstantu!)Shn*-!specifi~ne toplotne kapacitete pri stalnom pritisku!)dqn) i pri stalnoj zapremini
)dwn!*-!eksponent izentropske promene stawa!)κn*!kao i parcijalne pritiske komponenata!B!i!C!ako se
me{avina nalazi na
qn>2!cbs!i ako je sastav me{avine zadat na slede}i na~in:
b* hB>1/7-!hC>1/5
c* sB>1/3-!sC>1/9
b*
N hN =
2
hB
+
NB
Shn
hC
NC
=
2
1/7 1/5
+
43 39
(NSh ) = 9426 >385/7:!
=
Nn
41/38
>41/38!
lh
lnpm
K
lhL
d Qn = h B ⋅ d QB + hC ⋅ d QC = 1/7 ⋅ 1/:2 ⋅ 21 4 + 1/5 ⋅ 2/15 ⋅ 21 4 >:73!
d wn = d qn − S hn = :73 − 385/7: >798/42
κn =
d qt
d wn
qB = hB ⋅
=
K
lhL
K
lhL
:73
>2/5
798/42
NN
41/38
⋅ q n = 1/7 ⋅
⋅ 2 >1/68!cbs
NB
43
q C = q n − q B = 2 − 1/68 >1/54!cbs
b)
Nn = sB ⋅ N B + sC ⋅ NC = 1/3 ⋅ 43 + 1/9 ⋅ 39 = 39/9!
lh
lnpm
S hn =
(NS h )
Nn
=
9426
>399/83
39/9
K
lhL
dqn = sB ⋅
K
NB
N
43
39
⋅ dqB + sC ⋅ C ⋅ dqC = 1/3 ⋅
⋅ 1/:2 + 1/9 ⋅
⋅ 2/15 >2122/22
NN
NN
39/9
39/9
lhL
d wn = d qn − S hn = 2122/22 − 399/83 >833/4:
κn =
d qt
d wn
=
K
lhL
2122/22
>2/5
833/4:
q B = sB ⋅ q n = 1/3 ⋅ 2 >1/3!cbs
q C = q n − q B = 2 − 1/3 >1/9!cbs
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 84
2/75/!Me{avina idealnih gasova-!n>2!lh-!sastoji se od azota!)B), zapreminskog udela!51& i metana!)C*zapreminskog udela!)71%). Me{avina se zagreva od temperature!U2>411!L!do temperature!U3>711!L!na
dva na~ina. Prvi put je promena stawa kvazistati~ki izohorska, a drugi put se odvija kvazistati~ki po
zakonu prave linije u!Ut!koordinatnom sistemu. U oba slu~aja po~etna i krajwa stawa radnog tela su
jednaka. Skicirati promene stawa na Ut dijagramu i odrediti:
a) zapreminski rad )lK* du` promene!2−2 koja se odvija po zakonu prave linije
b) promenu entropije izolovanog sistema!)lK0L*!koji ~ine radna materija i toplotni izvor stalne
temperature!UUJ>U3!za slu~aj izohorske promene stawa
U
3
2
t
Nn = sB ⋅ N B + sC ⋅ NC = 1/5 ⋅ 39 + 1/7 ⋅ 27 = 31/9!
dwn = sB ⋅
lh
lnpm
NB
N
39
27
lK
⋅ dwB + sC ⋅ C ⋅ dqC = 1/5 ⋅
⋅ 1/85 + 1/7 ⋅
⋅ 2/93 >2/35!
NN
NN
31/9
31/9
lhL
a)
t3
R 23 = n ⋅
∫
U (t ) ⋅ et = n ⋅
U2 + U3
U + U3
⋅ ∆t23 = n ⋅ 2
3
3
t2
U
w
⋅ d wn ⋅ mo 3 + S hn mo 3
U
w2
2
711 + 411
711
R 23 = 2 ⋅
⋅ 2/35 ⋅ mo
>497/89!lK
3
411
∆V23 = n ⋅ ∆v23 = n ⋅ d wtn ⋅ (U3 − U2 ) = 2 ⋅ 2/35 ⋅ (711 − 411) >483!lK
prvi zakon termodinamike za proces 2−3 koji se odvija po pravoj liniji:
R23!>!∆V23!,!X23
dipl.ing. @eqko Ciganovi}
⇒
X23!>!R23!−!∆V23!>!497/89−483>25/89!lK
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 85
b)
∆TTJ!>!∆TSU!,!∆TUJ!>!///>!96:/6!−!731!>34:/6!
∆TSU!>n!/!∆t23!>! n ⋅ d wtnmo
∆TUJ!>! −
K
L
U3
711
K
= 2 ⋅ 2/35 ⋅ mo
>96:/6!
U2
411
L
483
R23
K
>!!−731!
>///> −
UUJ
711
L
R 23 = n ⋅ (r23 )w =dpotu = n ⋅ d wn ⋅ (U3 − U2 ) = 2 ⋅ 2/35 ⋅ (711 − 411) >483!lK
2/76/![upqa kugla zanemarqive mase unutra{weg pre~nika!e>2!n!sastavqena je od dve polovine koje
su tesno priqubqene (slika). U kugli se nalazi me{avina idealnih gasova vodonika )B*ugqen−dioksida )C* i azota )D* sastava sB>1/46-!sC>1/5!j!sD>1/36!stawa 2)q>1/3!cbs-!U>3:4!L*/ Na dowoj
polovini kugle obe{en je teret mase nU>5111!lh. Pritisak okoline iznosi qp>2!cbs. Odrediti
koliko toplote treba dovesti me{avini idealnih gasova u kugli da bi se polovine mogle razdvojiti.
Nn = sB ⋅ N B + sC ⋅ NC + sD ⋅ ND = 1/46 ⋅ 3 + 1/5 ⋅ 55 + 1/36 ⋅ 39 = 36/4!
S hn =
(NS h )
Nn
d wn = sB ⋅
=
lh
lnpm
K
9426
>439/77!
lhL
36/4
N
NB
N
⋅ d wB + sC ⋅ C ⋅ d qC + sD ⋅ D ⋅ d qD
NN
NN
NN
d wn = 1/46 ⋅
3
55
39
lK
⋅ 21/5 + 1/5 ⋅
⋅ 1/77 + 1/36 ⋅
⋅ 1/85 >1/:6!
36/4
36/4
36/4
lhL
zapremina lopte:
dipl.ing. @eqko Ciganovi}
W>
e 4 π 24 ⋅ π
=
>1/6347!n4
7
7
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 86
jedna~ina stawa me{avine idealnih gasova stawa!)2*;! q2 ⋅ W = n ⋅ S hn ⋅ U2
n=
q2 ⋅ W
1/3 ⋅ 21 6 ⋅ 1/6347
=
>1/22!lh
S hn ⋅ U2
439/77 ⋅ 3:4
jedna~ina stati~ke ravnote`e neposredno pred odvajawe dowe polovine
n ⋅h
q 3 + 3u
> q p !!!
!!⇒
(stawe 2):
e ⋅π
5
n ⋅h
5 ⋅ 21 4 ⋅ :/92
= 2 ⋅ 21 6 −
q3!>! q p − 3u
> 1/6 ⋅ 21 6 Qb
e ⋅π
23 ⋅ π
5
5
jedna~ina stawa me{avine idealnih gasova stawa!)3*;!!! q 3 ⋅ W = n ⋅ S hn ⋅ U3
U3 =
q3 ⋅ W
1/6 ⋅ 21 6 ⋅ 1/6347
=
>835/26!L
n ⋅ S hn
1/22 ⋅ 439/77
R23 = n ⋅ (r23 )w =dpotu = 1/22⋅ 1/:6 ⋅ (U3 − U2 ) = 1/33 ⋅ 1/:6 ⋅ (835/26 − 3:4) >56/4!lK
2/77/!Za situaciju u proto~nom ure|aju za me{awe (prikazanu na slici) koji radi pri stacionarnim
uslovima, odrediti da li se u sistem dovodi mehani~ka snaga ili se iz sistema mehani~ka snaga odvodi
i izra~unati wenu vrednost!)lX*/
⋅
R 23 = −21 lX
⋅
struja 1:
x2>311!n0t
U2>611pD
n2>21!u0i
X U23 = @
struja 2:
x3>1!n0t
U3>531pD
n3>7!u0i
molski sastav:
B;
DP3>61&
C;
O3>61&
molski sastav:
D;
O3>71&
E;
P3>51&
me{avina:
x+>41!n0t
U+>611pD
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 87
struja 1:
Nn2 = sB ⋅ NB + sC ⋅ NC = 1/6 ⋅ 55 + 1/6 ⋅ 39 = 47!
lh
lnpm
⋅
⋅
o2 =
n2
21 ⋅ 21 4
lnpm(B + C)
=
>388/89!
i
Nn2
47
⋅
⋅
lnpmB
i
⋅
⋅
lnpmC
oC = sC ⋅ o2 = 1/4 ⋅ 382/85 >249/9:!
i
o B = sB ⋅ o2 = 1/6 ⋅ 388/89 >249/9:!
struja 2:
Nn3 = sD ⋅ ND + sE ⋅ NE = 1/7 ⋅ 39 + 1/5 ⋅ 43 >3:/7!
lh
lnpm
⋅
n3
7 ⋅ 21 4
lnpm(D + E )
o3 =
=
>313/81!
Nn3
3:/7
i
⋅
⋅
⋅
lnpmD
i
⋅
⋅
lnpmE
oE = sE ⋅ o3 = 1/5 ⋅ 313/81 >92/19!
i
o D = sD ⋅ o 3 = 1/7 ⋅ 313/81 >232/73!
prvi zakon termodinamike za proces me{awa fluidnih struja:
⋅
⋅
⋅
R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23
⋅
⋅
⋅
⋅
⋅
X U23 = R 23 − ∆ I23 − ∆Fl23
⇒
⋅
∆ I23 = I3 − I2 >///>435:/88!−!4229/7:!>!242/19!lX
⋅
⋅
⋅
⋅
⋅
I2 > o B ⋅ Nd q B + oC ⋅ Nd q C ⋅ U2 + oD ⋅ Nd q D + oE ⋅ Nd q E ⋅ U3
⋅
249/9:
92/19
249/9:
232/73
I2 >
⋅ 48/5 +
⋅ 3:/2 ⋅ 884 +
⋅ 3:/2 +
⋅ 3:/2 ⋅ 7:4
4711
4711
4711
4711
(
)
(
)
(
)
(
)
⋅
I2 >4229/7:!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 88
⋅
⋅
⋅
⋅
⋅
I3 > o B ⋅ Nd q B + oC ⋅ Nd q C ⋅ U + + oD ⋅ Nd q D + oE ⋅ Nd q E ⋅ U +
⋅
249/9:
92/19
249/9:
232/73
I3 >
⋅ 48/5 +
⋅ 3:/2 ⋅ 884 +
⋅ 3:/2 +
⋅ 3:/2 ⋅ 884
4711
4711
4711
4711
(
)
(
)
(
)
(
)
⋅
I3 >435:/88!lX
2
3
∆Fl23 = ∆Fltusvkb
+ ∆Fltusvkb
>///>−65/42!,!1/86!>−64/67!lX
23
23
( )
⋅
⋅
x+
∆F tusvkb2 = o B ⋅ N B + oC ⋅ NC ⋅
l23
3
− (x 2 )3
3
3
3
249/9:
249/9:
41 − 311
>−65/42!lX
∆F tusvkb2 =
⋅ 55 +
⋅ 39 ⋅
l23
4711
3
4711
∆F tusvkb3
l23
( )
⋅
⋅
x+
= o D ⋅ ND + oE ⋅ NE ⋅
3
− (x 3 )3
3
3
3
92/19
232/73
41 − 1
∆F tusvkb3 =
⋅ 39 +
⋅ 43 ⋅
>,1/86!lX
l23
4711
3
4711
⋅
X U23 >!−21!−!242/19!,!65/42!>!−97/88!lX
⋅
Po{to je vrednost za X U23 negativan broj to zna~i da se u sistem dovodi
mehani~ka snaga.
⋅
2/78/ Vazduh (idealna gas) po~etne temperature!Uw2>61pD-!masenog protoka! n w>3!lh0t!zagreva se u
rekuperativnom razmewiva~u toplote na ra~un hla|ewa me{avine idealnih gasova DP3!i!TP3!od
⋅
Un2>511pD!do!Un3>351pD. Maseni protok me{avine idealnih gasova je! n n>4!lh0t-!a maseni udeo!DP3!u
me{avini je!91&/!Razmewiva~ toplote je toplotno izolovan od okoline. Pokazati da je proces razmene
toplote u razmewiva~u toplote nepovratan. Zanemariti promene kineti~ke i potencijalne energije
gasnih struja kao i padove pritiska gasnih struja pri strujawu kroz razmewiva~ toplote.
DP3!,!TP3
vazduh
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
hDP3 = 1/9
⇒
strana 89
hTP3 = 2 − hDP3 = 1/3
d qn = hDP3 ⋅ d qDP3 + hTP3 ⋅ d qTP3 = 1/9 ⋅ 1/96 + 1/3 ⋅ 1/69 >1/9!
lK
lhL
prvi zakon termodinamike za proces u razmewiva~u toplote:
⋅
⋅
⋅
R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23
⋅
⋅
⋅
⇒
⋅
⋅
I2 > I3
⋅
n w ⋅ d qw ⋅ Uw2 + nn ⋅ d qn ⋅ Un2 = n w ⋅ d qw ⋅ Uw3 + nn ⋅ d qn ⋅ Un3
⋅
Uw 3 =
⋅
⋅
n w ⋅ d qw ⋅ Uw2 + nn ⋅ d qn ⋅ Un2 − nn ⋅ d qn ⋅ Un3
⋅
n w ⋅ d qw
Uw3 =
⋅
3 ⋅ 2 ⋅ 434 + 4 ⋅ 1/9 ⋅ 784 − 4 ⋅ 1/9 ⋅ 624
>626!L
3 ⋅2
⋅
⋅
∆ T TJ = ∆ TSU + ∆ T plpmjob >//!/>1/39!
⋅
∆ T plpmjob = −
⋅
⋅
⋅ R 23
lX
=1
Up
L
lX
L
(razmewiva~ toplote je izolovan od okoline)
⋅
lX
!
L
q
626
lX
− S hw mo w3 > 3 ⋅ 2 ⋅ mo
>1/:4!
q w2
434
L
∆ TSU = ∆ T w + ∆ Tn = /// >1/:4!−!1/76!>!1/39!
/
/
⋅
U
∆ T w = n w ⋅ g (q- U ) = n w ⋅ d qw mo w3
Uw2
/
/
⋅
U
q
∆ T n = nn ⋅ g (q- U ) = nn ⋅ d qn mo n3 − S hn mo n3
Un2
q n2
napomena:
624
lX
> 4 ⋅ 1/9 ⋅ mo
>−1/76!
784
L
po{to je promena entropije sistema pozitivan broj
⋅
( ∆ T TJ > 1 ) to zna~i da je proces razmene toplote u
razmewiva~u toplote nepovratan
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 90
2/79/!!Me{avina idealnih gasova (kiseonik i ugqen−dioksid), pogowena kompresorom snage!2:!lXstruji kroz kanal. Usled neidealnog izolovawa kanala i kompresora okolini se predaje!2/39!lX
⋅
toplote. Zapreminski protok i temperatura me{avine na ulazu u kanal iznose! W 2>1/26!n40t!i
U2>486!L. Na izlazu iz kanala, pri pritisku!q>3!cbs-!zapreminski protok i temperatura me{avine
⋅
iznose! W 3>1/22!n40t!i!U3>586!L/!Zanemaruju}i promene kineti~ke i potencijalne energije me{avine
idealnih gasova, odrediti:
a) masene udele komponenata u me{avini
b) promeu entropije sistema u navedenom procesu, ako temperatura okoline iznosi Up>3:4!L
a)
R23
XU23
2
3
jednan~ina stawa me{avine idealnih gasova na izlazu iz kanala:
⋅
(
⋅
)
⇒
q 3 ⋅ W 3 = o⋅ NS h ⋅ U3
⋅
q3 ⋅ W 3
3 ⋅ 21 6 ⋅ 1/22
lnpm
o=
>!6/68/21−4!
=
NS h ⋅ U3
9426 ⋅ 586
t
⋅
(
)
prvi zakon termodinamike za proces u kanalu:
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23
(Ndq )n =
⋅
⋅
R 23 − X U23
⋅
o⋅ (U3 − U2 )
=
− 2/39 + 2:
6/68 ⋅ 21
−4
⋅
(
⇒!!!!!! R 23 = o⋅ Nd q
⋅ (586 − 486)
>42/92!
lK
lnpmL
(Ndq )n = sP3 ⋅ (Ndq )P3 + sDP3 ⋅ (Ndq )DP3
)2*
sP3 + sDP3 = 2
)3*
Kombinovawem jedna~ina!)2* i!)3*!dobija se:!
sP3 >1/7:-!!! sDP3 >1/42
hP3 =
sP3
⋅
)n ⋅ (U3 − U2 ) + X U23
sP3 ⋅ NP3
1/7: ⋅ 43
>1/7
=
⋅ NP3 + sDP3 ⋅ NDP3 1/7: ⋅ 43 + 1/42 ⋅ 55
hDP3 = 2 − hP3 = 2 − 1/7 >!1/5
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 91
b)
jedna~ina stawa me{avine idealnih gasova na ulazu u kanal:
⋅
(
⋅
⋅
)
q2 ⋅ W 2 = o⋅ NS h ⋅ U2
q2 =
⇒
(
)
o⋅ NS h ⋅ U2
⋅
W2
q2 =
6/68 ⋅ 21
⋅
⋅
−4
⋅ 9426 ⋅ 486
> 2/27 ⋅ 21 6 !Qb
1/26
⋅
∆ T TJ = ∆ TSU + ∆ T plpmjob >//!/>5/48!,!27/76!>32/13!
X
L
⋅
R 23
− 2/39
X
=−
>5/48!
Up
3:4
L
/
/
⋅
⋅
U
q
∆ T SU = ∆ T n > o⋅ g (q- U ) = o⋅ Nd qn ⋅ mo 3 − NS hn ⋅ mo 3
U2
q2
⋅
586
3
− 9/426 ⋅ mo
∆ T n > 6/68 ⋅ 21 −4 ⋅ 42/92 ⋅ mo
> 27/76
486
2/27
∆ T plpmjob = −
(
zadatak za ve`bawe:
)
(
)
X
L
)2/7:*
2/7:/!U ~eli~noj boci zapremine W>1/2!n4!nalazi se vazduh ) sP3 >1/32-! sO3 >1/8:* okolnog stawa
P)qp>2!cbs-!Up>3:4!L*/ Boca se puni ugqen−dioksidom. Odrediti:
a) koliko se lh!DP3!treba ubaciti u bocu, da bi molski udeo kiseonika u novonastaloj me{avini bio
6& i koliki je tada pritisak me{avine u boci pri temperaturi od 3:4!L
b) koliko toplote treba dovesti da se me{avina u boci zagreje na 564!L
a)
nDP3 > :/6 ⋅ 21 −4 !lh-!!!q>2/16!cbs
b) R23>25/76!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 92
POLUIDEALNI GASOVI
2/81/!Tokom kvazistati~ke promene stawa 2−2 kiseoniku (poluidealan gas) mase!n>1/4!lh, po~etnog
stawa!2)U2>484!L-!q2) predaje se toplota pri ~emu se kisonik zagreje do!U3>784!L. Specifi~ni
toplotni kapacitet kiseonika tokom ove promene stawa mewa se po zakonu:
d23
U
= 1/: + 2 ⋅ 21 −4
[lK 0)lhL]
[L ]
Od stawa 2 kiseonik kvazistati~ki izotermski mewa stawe do stawa!4!)q4>q2*/!Odrediti koli~inu
toplote!)lK*!koja se kisoniku preda:
b* tokom procesa!2−3
c* tokom procesa!3−4
a)
U3
R 23 = n ⋅
U3
∫()
d U ⋅ eU = n ⋅
U2
(
)
∫
1/: + 2 ⋅ 21 − 4 U ⋅ eU
U2
(
)
R 23 = n ⋅ 1/: ⋅ U3 − U2 + 6 ⋅ 21 − 5 ⋅ U33 − U23
(
)
R 23 = 1/4 ⋅ 1/: ⋅ (784 − 484) + 6 ⋅ 21 − 5 ⋅ 784 3 − 484 3 >239/18!lK
b)
U3
∆t23 =
∫
d(U ) ⋅ eU
=
U
U3
U2
∆t23 = 1/: ⋅ mo
U3
dq
U2
dq
U3
U2
=
−4
3
U
U2
+ 2 ⋅ 21 −4 ⋅ (U3 − U2 )
U2
784
lK
+ 2 ⋅ 21 −4 ⋅ (784 − 484) >1/94!
484
lhL
2
⋅ dq
U3 − U2
>2/129!
∫
(1/: + 2 ⋅ 21 U ) ⋅ eU = 1/: ⋅ mo U
U2
2
⋅ (1/:757 ⋅ 784 − 1/:329 ⋅ 484) >
⋅ U3 − dq ⋅ U2 =
12
U2
784 − 484
U3
lK
lhL
U3 U
q
⇒
∆t23 = d q mo 3 − S h mo 3 !!
U
U
q
2
2
2
q3
lK
784
S h mo
= 2/129 ⋅ mo
− 1/94 >−1/34
lhL
q2
484
dipl.ing. @eqko Ciganovi}
! S h mo
q3
= dq
q2
U3
U2
mo
U3
− ∆t23
U2
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
t4
R 34 = n ⋅
∫
t3
strana 93
U (t ) ⋅ et = n ⋅ U3 ⋅ ∆t 34 = n ⋅ U3 ⋅ d q
U
q
mo 4 − S hmo 4
U
q3
U3
3
U4
q3
= 1/4 ⋅ 784 ⋅ (− 1/34) >−57/55!lK
q2
R 34 = n ⋅ U3 ⋅ S hmo
2/82/!Zavisnost molarnog toplotnog kapaciteta od temperature za neki poluidealan gas, pri stalnom
D qN
U
pritisku, data je izrazom:
= 3:/3 + 5/18 ⋅ 21 −4
− 384
[K 0 npmL]
[L ]
b* odrediti koli~inu toplote koju treba predati gasu da bi se on zagrejao od polazne temperature
U2>3:1!L!do temperature!U3>684!L-!ako se predaja toplote vr{i pri stalnom pritisku!q>2/6!NQb-!a
posle izobarskog {irewa gas zauzima zapreminu!W3>1/9!n4
b) odrediti koli~inu toplote koju je potrebno predati istoj koli~ini istog gasa, da bi se on zagrejao od
iste polazne temperature!U2!do iste temperature!U3-!ako gas biva zagrevan pri stalnoj zapremini
b*
(
o=
)
q 3 ⋅ W3 = o ⋅ NS h ⋅ U3
jedna~ina stawa idealnog gasa stawa!3;!
q 3 ⋅ W3
2/6 ⋅ 21 7 ⋅ 1/9
=
>1/36!lnpm
NS h ⋅ U3
9426 ⋅ 684
(
)
U3
(R23 )q=dpotu
= o⋅
∫
U3
D qN (U ) ⋅ eU = o ⋅
U2
∫[
]
3:/3 + 5/18 ⋅ 21 −4 ⋅ (U − 384) ⋅ eU
U2
3
(R23 )q=dpotu = o ⋅ 3:/3 ⋅ (U3 − U2) + 5/18 ⋅ 21−4 ⋅ U3
− U23
− 5/18 ⋅ 21−4 ⋅ 384 ⋅ (U3 − U2)
3
3
(R23 )q=dpotu = 1/36 ⋅ 3:/3 ⋅ (684 − 3:1) + 5/18 ⋅ 21−4 ⋅ 684
− 3:13
− 2/22 ⋅ (684 − 3:1)
3
(R23 )q=dpotu >3/22!NK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 94
b)
U3
(R23 )w =dpotu
= o⋅
∫
U3
D wN (U ) ⋅ eU = o ⋅
U2
= o⋅
(
)]
D qN (U ) − NS h ⋅ eU
U2
U3
(R 23 )w =dpotu
∫[
∫[
(
)]
(
)
D qN (U ) − NS h ⋅ eU = (R 23 )q=dpotu − o ⋅ NS h ⋅ (U3 − U2 )
U2
(R 23 )w =dpotu
= 3/22 ⋅ 21 7 − 1/36 ⋅ 9426 ⋅ (684 − 3:1) = 2/63 NK
⋅
2/83/!Vazduh (poluidealan gas), masenog protoka! n w>1/3!lh0t, po~etne temperature!Uw>711pD-!pri
konstantnom pritisku, struji kroz adijabatski izolovanu cev u kojoj se hladi kiseonikom (poluidealan
gas) koji struji kroz cevnu zmiju, a zatim se i me{a sa jednim delom ovog kiseonika (slika).
Temperatura tako nastale me{avine iznosi UN>411pD-!a maseni udeo kiseonika u toj sme{i je!hC>1/6.
Temperatura kiseonika na ulazu u cev je!UL2>31pD-!a na izlazu iz cevi!UL3>311pD/!Pritisak
kiseonika je stalan. Odrediti maseni protok kiseonika kojim se vr{i hla|ewe vazduha )nB,nC*.
Zanemariti promene potencijalne i kineti~ke energije poluidealnih gasova.
kiseonik,!nB,nC-!3:4!L
vazduh,!nw-!984!L
me{avina,!nC!,nw-!684!L
kiseonik,!nB-!584!L
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 95
⋅
nC
hC =
⋅
⇒
⋅
n w + nC
d qw
Uw2=711p D
>2/161!
1
Ul2=31p D
d ql
1
lK
lhL
d qw
d ql
>1/:6!
1
Un =411p D
>2/131!
1
Ul 3 =311p D
lK
>1/:222!
lhL
Un =411p D
⋅
h ⋅ nw
1/6 ⋅ 1/3
lh
=
>1/3!
nC = C
2 − hC
2 − 1/6
t
⋅
d ql
lK
lhL
>1/:466!
1
lK
lhL
lK
lhL
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
⋅
⋅
⋅
⋅
⋅
⋅
! R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23
ograni~enom isprekidanom linijom:
I2 = I3
I2 = n w ⋅ d qw
⋅
⋅
I3 = n w ⋅ d qw
Uw2
Ul2
1
Un
1
⋅
⋅
⋅ Uw2 + n B + nC d ql
Ul 3
⋅
⋅ Un + n B ⋅ d ql
1
⋅
nB =
⋅
n w ⋅ d qw
⋅
⋅
⋅
⋅ Ul3 + nC ⋅ d ql
Un
⋅ Un
1
Un
⋅ Un − d qw
1
⋅
⋅ Uw2 + nC ⋅ d ql
Uw2
1
1
d ql
nB =
⋅ Ul2
Ul2
Ul 3
1
1
⋅ Ul2 − d ql
⋅ Ul2
Un
Ul2
1
1
⋅ Un − d ql
⋅ Ul3
lh
1/3 ⋅ (2/13 ⋅ 684 − 2/16 ⋅ 984) + 1/3 ⋅ (1/:6 ⋅ 684 − 1/:222⋅ 3:4 )
>1/17!
1/:222⋅ 3:4 − 1/:466 ⋅ 584
t
⋅
n B + nC = 1/3 + 1/17 >1/37!
dipl.ing. @eqko Ciganovi}
lh
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 1
VELI^INE STAWA REALNIH FLUIDA
2.1. Odrediti specifi~nu entalpiju, specifi~nu entropiju, specifi~nu zapreminu kao
i specifi~nu unutra{wu energiju vode stawa (p=1 bar, t=20oC).
kJ
kJ
m3
, sw = 0.296
, vw = 0.001001
kg
kgK
kg
priru~nik za termodinamiku (tabela 4.2.6. ′′iznad crte′′) strana 41−55
kJ
uw = hw − p . vw = 83.9 −1.105 .10−2.0.001001 = 83.8
kg
hw = 83.9
2.2. Odrediti specifif~nu entalpiju, specifi~nu entropiju, specifi~nu zapreminu
i specifi~nu unutra{wu energiju pregrejane vodene pare stawa (p=25 bar, t=360oC).
kJ
kJ
m3
, spp = 6.870
, vpp = 0.1117
kg
kgK
kg
priru~nik za termodinamiku (tabela 4.2.6. ′′ispod crte′′) strana 41−55
kJ
upp = hpp − p . vpp = 3146– −25 .105 .10−2 .0.1117 = 2866.75
kg
hpp = 3146
2.3. Odrediti specifi~nu unutra{wu energiju, specifi~nu entropiju i temperaturu
a) kqu~ale vode pritiska p=10 bar
b) suvozasi}ene vodene pare pritiska p=10 bar
a)
kJ
kJ
, s’′ = 2.138
, tK= 179.88oC
kg
kgK
priru~nik za termodinamiku (tabela 4.2.4.) strana 36−38
u′’= 761.6
b)
kJ
kJ
, s′′”= 6.587
, tK= 179.88oC
kg
kgK
priru~nik za termodinamiku (tabela 4.2.4.) strana 36−38
u”′′= 2583
2.4. Odrediti specifi~nu entalpiju i temperaturu vla`ne vodene pare stawa (x=0.95,
p=15 bar).
kJ
hx = h' +x ⋅ (h"−h' ) = ... = 844 .6 + 0.95 ⋅ (2792 − 844 .6 ) =2694
kg
kJ
kJ
h′’ = 844.6
,
h′′”= 2792
kg
kg
o
tx = tK = 198.28 C
priru~nik za termodinamiku (tabela 4.2.4.) strana 36−38
dipl.ing. @eqko Ciganovi}
[email protected]
zbirka zadataka iz termodinamike
strana 2
2.5. Odrediti specifi~nu entropiju i pritisak vla`ne vodene pare stawa (x=0.6,
t=200oC).
kJ
s x = s ' + x ⋅ (s"−s ' ) = ... = 2.3308 + 0 .6 ⋅ (6 .4318 − 2 .3308 ) =4.7806
kgK
kJ
kJ
s’′ = 2.3308
s′′”= 6.4318
kgK
kgK
px = pK = 15.551 bar
priru~nik za termodinamiku (tabela 4.2.5.) strana 39−40
2.6. Primewuju}i postupak linearne interpolacijeodrediti:
a) specifi~nu entalpiju pregrejane vodene pare stawa (p=1 bar, t=250oC)
b) specifi~nu entropiju pregrejane vodene pare stawa (p=7 bar, t=300oC)
c) specifi~nu entalpiju pregrejane vodene pare stawa (p=5 bar, t=350oC)
a)
kJ
tabela 4.2.6. strana 41−55, za p=1 bar i t=240oC=x1
kg
kJ
y2 = hpp= 2993
tabela 4.2.6. strana 41−55, za p=1 bar i t=260oC=x2
kg
y − y1
2993 − 2954
kJ
hpp= 2
⋅ (x − x1) + y1 =
⋅ (250 − 240 ) + 2954 = 2973.5
x2 − x1
260 − 240
kg
y1 = hpp= 2954
b)
kJ
tabela 4.2.6. strana 41−55, za p=6 bar=x1 i t=300oC
kgK
kJ
y2 = spp= 7.226
tabela 4.2.6. strana 41−55, za p=8 bar=x2 i t=300oC
kgK
y − y1
7.226 − 7.366
kJ
spp= 2
⋅ (x − x1) + y1 =
⋅ (7 − 6) + 7.366 = 7.296
x2 − x1
8 −6
kgK
y1 = spp= 7.366
c)
1.korak
kJ
kg
kJ
y2 = hpp= 3190
kg
y1 = hpp= 3148
h1 = hpp=
tabela 4.2.6. strana 41−55, za p1 =4 bar i t=340oC=x1 .
tabela 4.2.6. strana 41−55, za p1 =4 bar i t=360oC=x2 .
kJ
y 2 − y1
3190 − 3148
⋅ (x − x 1 ) + y 1 =
⋅ (350 − 340) + 3148 = 3169
x2 − x2
360 − 340
kg
dipl.ing. @eqko Ciganovi}
[email protected]
zbirka zadataka iz termodinamike
strana 3
2.korak
kJ
tabela 4.2.6. strana 41−55, za p2 =6 bar i t=340oC=x1 .
kg
kJ
y2 = hpp= 3185
tabela 4.2.6. strana 41−55, za p2 =6 bar i t=360oC=x2 .
kg
y − y1
3185 − 3143
kJ
h2 = hpp= 2
⋅ (x − x 1 ) + y 1 =
⋅ (350 − 340 ) + 3143 = 3164
x2 − x2
360 − 340
kg
y1 = hpp= 3143
3.korak
h − h1
3164 − 3169
kJ
h= 2
⋅ (p − p1 ) + h1 =
⋅ (5 - 4 ) + 3169 =3166.5
p2 − p1
6 −4
kg
2.7. Odrediti specifi~nu entalpiju i specifi~nu entropiju:
a) leda temperature t=−5oC
b) me{avine leda i vode (mw=2 kg, ml=3 kg) u stawu toplotne ravnote`e (t=0oC)
a)
hl = cl ⋅ (Tl − 273 ) − rl = 2 ⋅ (− 5 ) − 332 .4 = −342.4
sl = cl ⋅ ln
kJ
kg
Tl − 273
r
− 5 332 .4
kJ
− l = 2 ⋅ ln
−
=−1.25
273
273
273
273
kgK
b)
y=
mw
2
=
=0.4
mw + ml 2 + 3
(maseni udeo vode u me{avini vode i leda)
hy = hl + y ⋅ (hw − hl ) = ... = 0 + 0.4 ⋅ (0 + 332 .4 ) =−199.4
hw= 0
kJ
kg
kJ
kg
hL = −332.4
kJ
kg
s y = s l + y ⋅ (s w − s l ) = ... = 0 + 0.4 ⋅ (0 + 1 .22 ) =−0.732
kJ
kgK
kJ
kgK
r
332 .4
kJ
sL = − l = −
−1.22
273
273
kgK
sw= 0
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 4
2.8. Odrediti specifi~nu entalpiju i specifi~nu entropiju:
a) pregrejane vodene pare stawa (t=600oC, v=1 m3 /kg)
b) vla`ne vodene pare stawa (x=0.9, v=20 m3/kg)
a)
v=1 m3 /kg
h
t=600oC
h=3705 kJ/kg
s=8.46 kJ/kgK
kJ
hpp = 3705
kg
kJ
spp = 8.46
kgK
s
(upotrebom hs dijagrama za vodenu paru)
(upotrebom hs dijagrama za vodenu paru)
b)
h
v=20 m3 /kg
h=2325 kJ/kg
x=0.9
s=7.54 kJ/kgK
kJ
kg
kJ
sx = 7.54
kgK
hx = 2325
dipl.ing. @eqko Ciganovi}
s
(upotrebom hs dijagrama za vodenu paru)
(upotrebom hs dijagrama za vodenu paru)
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zbirka zadataka iz termodinamike
strana 5
2.9. Odrediti specifi~nu entalpiju pregrejane vodene pare stawa (u=2654 kJ/kg,
v=1.08 m3 /kg).
pretostavimo p= 4 bar:
⇒
3
m
kJ
=3846.35
hpp = f p = 4 bar, v = 1.08
kg
kg
provera pretpostavke:
h − u 3846 .35 − 2654
p=
=
=11.04 bar
v
1.08
pretostavimo p= 3 bar:
⇒
3
m
kJ
=3340.03
hpp = f p = 3 bar, v = 1 .08
kg
kg
provera pretpostavke:
h − u 3340 .03 − 2654
p=
=
=6.35 bar
v
1.08
pretostavimo p= 2 bar:
⇒
3
m
kJ
=2870
hpp = f p = 2 bar , v = 1 .08
kg
kg
provera pretpostavke:
h − u 2870 − 2654
p=
=
=2 bar
v
1 .08
kJ
ta~na vrednost iznosi: hpp=2870
kg
tabela 4.2.6. strana 41−55
(pretpostavka nije ta~na)
tabela 4.2.6. strana 41−55
(pretpostavka nije ta~na)
tabela 4.2.6. strana 41−55
(pretpostavka je ta~na)
2.10. Odrediti specifi~nu entalpiju i specifi~nu entropiju suvozasi}ene pare
amonijaka na T=300 K.
kJ
kJ
h”′′= 2246
s′′”= 9.993
kg
kgK
tabela 4.4.1. strana 62, za T=300 K
2.11. Odrediti specifi~nu entalpiju i specifi~nu entropiju pregrejane pare freona
12 stawa (p=6 bar, t=200oC).
kJ
kJ
hpp =789
spp = 1.889
kg
kgK
tabela 4.6.2. strana 79−81, za p=6 bar, t=200oC
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 6
PROMENE STAWA REALNIH FLUIDA
2.12. Vla`na para stawa 1(x=0.3, p=0.2 bar) izohorski se {iri do stawa 2(p=1.5 bar),
a zatim ravnote`no izentropski ekspandira do stawa 3(p3=p1 ). Skicirati promene
stawa vodene pare na Ts i pv dijagramu i odrediti razmewenu toplotu (kJ/kg) za
promenu stawa 1−2 i zapreminski rad (kJ/kg) za promenu stawa 2−3.
T
2
2
p
3
1
1
3
s
v
ta~ka 1:
u1 = ux =u’′” + x1 . (u′′” - u′’) = ...= 251 .38 + 0.3 ⋅ (2456 − 251.38 ) =912.8
u’′ =251.38
kJ
kg
u′′” =2456
kJ
kg
kJ
kg
v1 = vx = v’′ + x1 . (v′′” - v’′ ) = 0.0010171 + 0 .3 ⋅ (7.647 − 0.0010171) =2.29
v′’=0.0010171
m3
kg
v′′”=7.647
m3
kg
p2 =1.5 bar
m3
kg
m3
kg
ta~ka 2:
v2 = v1 =2.29
m3
m3
v′′”=1.159
kg
kg
ta~ka 2 nalazi se u oblasti pregrejane pare
provera polo`aja ta~ke 2:
v2 > v”′′
⇒
h2 = hpp =3448.1
kJ
,
kg
v′’=0.0010527
s2 = spp=8.5
kJ
kgK
u2 = upp = hpp –− p . vpp = 3448 .1 − 1 .5 ⋅ 10 5 ⋅ 10 −3 ⋅ 2.29 =3104.6
dipl.ing. @eqko Ciganovi}
kJ
kg
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zbirka zadataka iz termodinamike
strana 7
ta~ka 3:
s3 = s2 =8.5
kJ
kgK
p3 = p1 = 0.2 bar
kJ
kJ
s′′” =6.822
kgK
kgK
ta~ka 3 nalazi se u oblasti u pregrejane pare
provera polo`aja ta~ke 3:
s3 > s′′”
⇒
h3 = hpp = 2797.6
kJ
,
kg
s′’ =08321
v3 = vpp =14.61
m3
kg
u3 = upp = hpp –− p . vpp = 2797 .6 − 0.2 ⋅ 10 5 ⋅ 10− 3 ⋅ 14.61 =2505.4
kJ
kg
(q12 )v= const = u2 − u1 = 3104 .6 − 912 .8 =2191.8 kJ
(w23 )s= const
kg
kJ
= u2 − u3 = 3104 .6 − 2505 .4 = 599.2
kg
2.13. Pregrejana vodena para stawa 1(m=1 kg, p=0.05 MPa, t=270oC) predaje toplotu
izotermnom toplotnom ponoru, usled ~ega ravnote`no mewa svoje toplotno stawe: prvo
izohorski (1−2) do temperature 60oC, potom izotermski (2−3) do pritiska 0.1 MPa i
kona~no izobarski (3−4) do temperature 20oC. Odrediti promenu entropije izolovanog
sistema za slu~aj termodinmi~ki najpovoqnijeg temperaturskog nivoa toplotnog
ponora. Skicirati proces u Ts i pv koordinatnom sistemu.
1
T
1
p
3
3
2
4
4
s
dipl.ing. @eqko Ciganovi}
2
v
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zbirka zadataka iz termodinamike
strana 8
∆S SI =∆S RT + ∆S TP = .… ..
∆S RT =∆S14 = m ⋅ (s 4 − s1 ) =...
∆ STP = −m ⋅
∆ STP = −m ⋅
(q12 )v =const + (q 23 ) T =const + (q34 )p=const
Ttp
u2 − u1 + T2 ⋅ (s 3 − s 2 ) + h4 − h3
= ...
TTP
ta~ka 1:
p1 =0.5 bar,
t1 =270oC
tk=81.35oC
t1 > tk
m3
,
kg
kJ
s1 =spp= 8.423
,
kgK
v1 = vpp= 5
ta~ka 1 nalazi se u oblasti pregrejane pare
h1 =hpp = 3015
kJ
kg
u1 = upp=h1 − p1 . v1 = 2765
kJ
kg
ta~ka 2:
t2 =60oC
v2 =v1 =5
v′’= 0.0010171
m3
,
kg
v′ < v2 < v”′′
x2 =
m3
kg
m3
kg
ta~ka 2 nalazi se u oblasti vla`ne pare
v′′”=7.678
v 2 − v'
= 0.6512
v" − v'
u2 = ux =u’′” + x2 . (u′′” − u′’) =…...= 251 .1 + 0.6512 ⋅ (2456 − 251 .1) = 1684.29
u’′ =251.1
kJ
kg
u′′” =2456
kJ
kg
s2 = sx =s’′” + x2 . (s′′” − s′’) =…...= 0.8311 + 0.6512 ⋅ (7.9084 − 0 .8311) =5.43
s’′ =0.8311
kJ
,
kgK
dipl.ing. @eqko Ciganovi}
s′′”=7.9084
kJ
kg
kJ
kgK
kJ
kgK
[email protected]
zbirka zadataka iz termodinamike
strana 9
ta~ka 3:
p3 =1 bar
t3 =60oC
tk=99.64oC
t3 < tk
h3 = hw=251.1
kJ
,
kg
s3 = sw= 0.83
ta~ka 3 nalazi se u oblasti te~nosti
kJ
kgK
ta~ka 4:
p4 =1 bar
t4 =60oC
tk=99.64oC
t4 < tk
h4 =hw= 83.9
kJ
,
kg
s4 = sw=0.296
ta~ka 4 nalazi se u oblasti te~nosti
kJ
kgK
toplotni ponor:
TTP=T4 =293 K
(najpovoqniji termodinami~ki slu~aj)
∆S RT =∆S14 =1 ⋅ (0 .296 − 8.423 ) =...=−8.127
∆ STP = −1 ⋅
kJ
K
kJ
1684 .29 − 2765 + 333 ⋅ (0.83 − 5.43 ) + 83 .9 − 251 .1
= 9.487
K
293
∆S SI =∆S RT + ∆S TP = … − 8.127 + 9.487 =1.36
kJ
K
2.14. Vodi (m=10 kg) stawa 1(p=0.1 MPa, t=20oC) dovodi se toplota od izotermnog
toplotnog izvora, usled ~ega voda mewa svoje toplotno stawe: prvo izobarski (1−2) do
temperature 60oC, potom izotermski (2−3) do specifi~ne zapremine 5 m3 /kg i na kraju
izohorski (3−4) do pritiska 0.05 MPa. Skicirati proces u Ts koordinatnom sistemu i
odrediti promenu entropije adijabatski izolovanog sistema za slu~aj termodinmi~ki
najpovoqnijeg temperaturskog nivoa toplotnog izvora.
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 10
∆S SI = ∆S RT + ∆S TI = …...
⋅
∆S RT = ∆S14 = m ⋅ (s 4 − s1 ) =...
∆ STI = −m ⋅
∆ STI
(q12 )p=const + (q 23 )T =const + (q 34 ) v= const
TTI
h − h1 + T2 ⋅ (s 3 − s 2 ) + u 4 − u 3
=m⋅ 2
=...
TTI
ta~ka 1:
t1 =20oC
p1 =1 bar,
tk=99.64oC
h1 =hw= 83.9
t1 < tk
kJ
,
kg
ta~ka 2:
s1 = sw=0.296
t2 < tklj
kJ
kg
kJ
kgK
t2 = 60oC
p2 =1 bar
tk=99.64oC
h2 =hw = 251.1
ta~ka 1 nalazi se u oblasti te~nosti
ta~ka 2 nalazi se u oblasti te~nosti
s2 =sw = 0.83
kJ
kgK
ta~ka 3:
t3 = 60oC
v′’= 0.0010171
v3 = 5
m3
,
kg
v′ < v3 < v”′′
x3 =
m3
kg
m3
kg
ta~ka 3 nalazi se u oblasti vla`ne pare
v′′”=7.678
v 3 − v'
5 − 0.0010171
=
= 0.6512
v" −v'
7.678 − 0.0010171
kJ
kg
kJ
s3 = sx = s '+ x 3 ⋅ (s"−s ' ) = 0.8311 + 0.6512 ⋅ (7 .9084 − 0 .8311) =5.43
kgK
u3 =ux = u' + x3 ⋅ (u"−u' ) = 251 .1 + 0.6512 ⋅ (2456 − 251 .1) =1684.29
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 11
ta~ka 4:
p4 = 0.5 bar,
v′’= 0.0010299
v4 = v3 = 5
m3
kg
m3
kg
m3
kg
ta~ka 4 nalazi se u oblasti pregrejane pare
v′′”=3.239
v4 > v”′ ′
m3
,
kg
kJ
s1 =spp= 8.42
,
kgK
v1 = vpp= 5
h1 =hpp = 3015
kJ
kg
u 1 = upp=h1 − p1 . v1 = 2765
kJ
kg
T4 =Tpp= 270oC = 543 K
toplotni izvor:
TTI =T4 = 543 K
(najpovoqniji termodinami~ki slu~aj)
kJ
K
kJ
251.1 − 83.9 + 333 ⋅ (5.45 − 0.83 ) + 2782 .85 − 1689 .36
= −10 ⋅
=−51.6
K
542.5
∆S RT = ∆S14 =10 ⋅ (8.42 − 0.296 ) =81.24
∆ STIi
∆S SI = 81.24 − 51.6 = 29.64
kJ
K
4
T
TI
2
3
1
s
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 12
2.15. Jednom kilogramu leda stawa 1(p=1 bar T=−5 oC) dovodi se toplota od
toplotnog izvora konstantne temperature TTI=300oC tako da se na kraju izobarske
promene stawa (1−2) dobije suvozasi}ena vodena para (stawe 2). Odrediti promenu
entropije izolovanog sistema pri ovoj promeni stawa i grafi~ki je predstaviti na
Ts dijagramu.
∆S SI =∆S RT + ∆S TI = .… ..= 8.61 − 5.27 = 3.34
∆S RT =m . ∆s12=m . ( s2 − s1 )=...= 8.61
(q12 )p=const
kJ
K
(h2 − h1 )
kJ
K
= ... = −5.27
kJ
K
h1 = hl = cl ⋅ (TL − 273 ) − rl = 2 ⋅ (− 5 ) − 332.4 = −342.4
kJ
kg
∆ STI = −m ⋅
= −m ⋅
Tti
Tti
ta~ka 1:
s1 = s l = c l ⋅ ln
TL
273
−
rl
273
= 2 ⋅ ln
-5 + 273 332.4
kJ
−
= − 1.25
kgK
273
273
ta~ka 2:
h2 = h′′ = 2675
kJ
,
kg
s2 = s′′”=7.36
kJ
kgK
T
TI
2
1
∆S RT
∆S TI
s
jednake povr{ine
∆S SI
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 13
2.16. Te~an CO2, stawa 1(p=5 MPa, t=0oC), adijabatski se prigu{uje (h=idem) do
stawa 2(p=0.6 MPa). Grafi~ki predstaviti po~etno i krajwe stawe CO2 u Ts i hs
kJ
koordinatnom sistemu i odrediti prira{taj entropije CO2 tokom procesa 1−2 (
).
kgK
T
p1
h
p1
p2
p2
1
2
1
2
s
s
ta~ka 1:
p1 =50 bar,
h1 =− 94
kJ
,
kg
t1 =0oC
ta~ka 1 nalazi se u oblasti te~nosti
s1 = 3.1133
kJ
kgK
tabela 4.8.2. strana 93−98
ta~ka 2:
h2 =h1 = −94
p2 =6 bar,
kJ
kJ
, h′′”=142.7
kg
kg
h′ <h2 < h”′′
kJ
kg
h′’= −200
x2 =
tabela 4.8.1. strana 92
ta~ka 2 nalazi se u oblasti vla`ne pare
h2 − h'
−94 + 200
=
=0.3093
h"−h'
142 .7 + 200
s2 = sx = s '+ x 2 ⋅ (s "−s ' ) = 2.702 + 0.3093 ⋅ (4.260 − 2 .702 ) =3.184
s’′ =2.702
kJ
,
kgK
s′′”=4.260
kJ
kgK
∆s12 = s2 − s1 =...=3.184 − 3.1133 = 0.0707
dipl.ing. @eqko Ciganovi}
kJ
kgK
tabela 4.8.1. strana 92
kJ
kgK
[email protected]
zbirka zadataka iz termodinamike
strana 14
2.17. Freon 22 stawa 1(T=−30oC, x=1) nekvazistati~ki (neravnote`no) adijabatski
se komprimuje do stawa 2(p=6 bar). Prira{taj entropije freona tokom procesa
J
iznosi ∆s12 =51
. Predstaviti proces sa freonom na Ts i hs dijagramu i
kgK
odrediti stepen dobrote ove nekvazistati~ke adijabatske kompresije.
ta~ka 1:
h1 = h′′ = 691.92
kJ
kJ
, s1 = s′′= 1.7985
kg
kgK
tabela 4.7.1. strana 83
ta~ka 2k:
p2k=6 bar,
s′ = 1.024
kJ
kgK
kJ
s′′ = 1.74
kgK
s2k = s1 =1.7985
kJ
,
kgK
s2k > s′′
tabela 4.7.1. strana 83
ta~ka 2k nalazi se u oblasti pregrejane pare
h2k = hpp = 720.78
kJ
kg
ta~ka 2:
p2 =6 bar,
s2 = s1 +∆s12 =1.7895+0.051=1.8495
h2 = hpp = 739.36
kp
ηd =
kJ
kgK
kJ
kg
h1 − h2k
691 .92 − 720 .78
=
=0.61
h1 − h 2
691 .92 − 739 .36
2
2
T
h
2k
2k
1
1
s
dipl.ing. @eqko Ciganovi}
s
[email protected]
zbirka zadataka iz termodinamike
strana 15
(2.18. –− 2.19.)
zadaci za ve`bawe:
2.18. Kqu~ala voda temperature T1= 250oC mewa stawe ravnote`no:
− izotermski (1−2) do p2 = 4.5 bar
− zatim izohorski(2−3) do p= 2.2 bara
− i na kraju izobarski (3−4) do stawa 4(s4 =s1 )
Skicirati promene stawa vodene pare na pv i Ts dijagramu i odrediti razmewene
kJ
toplote (
) tokom procesa 1−2, 2−3 i 3−4.
kg
kJ
kJ
kJ
re{ewe:
q12 =2368.5
,
q23 =−846.7
,
q34 =−991.2
kg
kg
kg
2.19. Vodenoj pari stawa 1(T2 =100oC, x=0) ravnote`no se dovodi se toplota pri
~emi vodenu paru prevodimo u stawe 2(T=120oC, x=1). U procesu (1−2) temperatura
pare raste linerano u Ts kordinatnom sistemu. Nakon toga se vr{i neravnote`na
adijabatska ekspanzija (2−3) vodene pare (stepen dobrote nekvazistati~ke
adijabatske ekspanzije: η eks
d =0.9) do stawa 3(p=0.1 bar). Skicirati procese sa
vodenom parom na Ts dijagramu i odrediti dovedenu toplotu za proces 1−2 i dobijeni
tehni~ki rad za proces 2−3.
re{ewe:
q12 =2316.2
kJ
,
kg
dipl.ing. @eqko Ciganovi}
w T23 =402.4
kJ
kg
[email protected]
zbirka zadataka iz termodinamike
strana 16
PRVI I DRUGI ZAKON TERMODINAMIKE
(ZATVOREN TERMODINAMI^KI SISTEM)
2.20. U zatvorenom, adijabatski izolovanom, sudu zapremine V=7.264 m3 , nalazi se
me{avina m′=311 kg kqu~ale vode i m’′’ ′ =? suvozasi}ene vodene pare u stawu
termodinami~ke ravnote`e na p1 =0.95 bar. Vodenoj pari u sudu se dovodi toplota, od
toplotnog izvora stalne temperature TTI=300oC, tako da joj pritisak poraste na
p2 =68 bar. Skicirati procese sa vodenom parom na Ts i pv dijagramu i odrediti:
a) koliko je toplote dovedeno u procesu (MJ)
b) promenu entropije izolovanog termodinami~kog sistema za proces 1−2 (kJ/K)
a)
prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu
⇒
Q12 = ∆U12 + W12
Q12 = (m′ + m′′) . (u2 –− u1 )=...
ta~ka 1:
p1 =0.95 bar,
x1 =?
m
m3
v′′”=1.7815
kg
kg
V' = m'⋅v' = 311 ⋅ 0.00104205 =0.3241 m3
v’′ = 0.00104205
3
V′′” = V − V’′ =7.264 − 0.3241 = 6.9399 m3
m" =
V" 6.9399
=
=3.9 kg
v" 1.7815
x1 =
m"
3 .9
=
=0.0124
m"+m' 3 .9 + 311
u1 = ux =u’′ + x1 .(u′′” - u′’) = 411 .23 + 0 .0124 ⋅ (2504 − 411 .23 ) =437.21
kJ
u’′ = 411.23
,
kg
kJ
kg
kJ
u”′′ = 2504
kg
v1 =vx=v’ + x1. (v”′′ − v’′ )= 0. 00104205 + 0. 0124 ⋅ (1.7815 − 0. 00104205) =0.0231
s1 = sx =s’′ + x1 .(s′′” − s′’) = 1.2861 + 0.0124 ⋅ (7 .377 − 1 .2861) =1.362
kJ
s’′ = 1.2861
,
kgK
dipl.ing. @eqko Ciganovi}
m3
kg
kJ
kgK
kJ
s”′ ′ = 7.377
kgK
[email protected]
zbirka zadataka iz termodinamike
strana 17
ta~ka 2
p2 =68 bar,
v2 = v1 = 0.0231
v′’= 0.0013445
m3
,
kg
m3
kg
ta~ka 2 se nalazi u oblasti vla`ne pare
v′′”=0.028382
v′’ < v2 < v”′′
x2 =
m3
kg
v 2 − v'
0 .0231 − 0.0013445
=
=0.8046
v"− v'
0.028382 − 0 .0013445
u2 = ux =u’′ + x2 .(u′′” − u′’)= ...=1247 .52 + 0.8046 ⋅ (2582 − 1247 .52 ) =2321.24
u’′ = 1247.52
kJ
kg
u′′” = 2582
kJ
kg
kJ
kg
s2 = sx =s’′ + x1 .(s′′” − s′’) = ...= 3.103 + 0.8046 ⋅ (5.829 − 3.103 ) =5.2963
kJ
kgK
kJ
kJ
,
s”′ ′ = 5.829
kgK
kgK
= (311 + 3.9 ) ⋅ (2321 .24 − 437.21) =593.3 MJ
s’′ = 3.103
Q 12
b)
∆S SI =∆S RT + ∆S TI = …...=216.2
kJ
K
∆S RT = (m' +m' ' ) ⋅ (s 2 − s 1 ) = (311 + 3.9 ) ⋅ (5.2963 − 1.362 ) =1239.1
∆ STI = −
kJ
K
Q12
593.3 ⋅ 10 3
kJ
=−
=1035.43
TTI
573
K
T
p
2
2
1
1
s
dipl.ing. @eqko Ciganovi}
v
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zbirka zadataka iz termodinamike
strana 18
2.21. Izolovan zatvoren sud zapremine V=120 litara ispuwen je kqu~alom vodom i
suvozasi}enom parom u stawu termodinami~ke ravnote`e na pritisku p1 =1 bar. U
posudi se nalazi greja~ snage 5 kW . Dovo|ewem toplote nivo vode u sudu raste i kada
pritisak dostigne 50 bara, posuda je u celosti ispuwena te~nom fazom. Skicirati
proces na pv dijagramu i odrediti koliko dugo je trajalo dovo|ewe toplote.
p
K
2
1
vk
v
ta~ka 2:
p2 = 50 bar,
x2 =0
v2 = v′= 0.0012857
m=
m3
,
kg
u2 =u′= 1148
kJ
kg
V
120 ⋅ 10 −3
=
=93.33 kg
v 2 0.0012857
ta~ka 1:
p1 =1 bar,
v1 =v2 =0.0012857
v′’= 0.0010432
m3
,
kg
v′’ < v1 < v”′′
x1 =
m3
kg
m3
kg
ta~ka 1 se nalazi u oblasti vla`ne pare
v′′”=1.694
v 1 − v' 0.0012857 − 0.0010432
=
=0.0001
v" −v'
1.694 − 0 .0010432
u1 = ux =u’′ + x1 .(u′′” − u′’) = 417 .3 + 0.0001 ⋅ (2506 − 417 .3 ) =417.51
u’′ = 417.3
kJ
,
kg
dipl.ing. @eqko Ciganovi}
u”′′ = 2506
kJ
kg
kJ
kg
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zbirka zadataka iz termodinamike
strana 19
prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu
Q12 = ∆U12 + W12
⇒
Q12 =m . (u2 –− u1 )
Q12 = 93 .33 ⋅ (1148 − 417.51) =68.18 MJ
τ=
Q 12
⋅
=
Q 12
68.18 ⋅ 10 3
=13636 s
5
Uo~iti da se u ovom zadatku pojavquje fenomen podkriti~nih zapremina , tj.
izohorskim dovo|ewem toplote vla`noj pari v<vk ne dolazi do stvarawa
suvozasi}ene pare ve} kqu~ale te~nosti. U takvom slu~aju prilikom
dovo|ewa toplote vla`noj pari stepen suvo}e vodene pare ne raste monotono
do x=1, ve} raste do neke vrednosti x>x1 pa zatim monotono opada do x=0.
2.22. U vertikalno postavqenom cilindru povr{ine
popre~nog preseka A=0.1 m2 koji je po omota~u
izolovan nalazi se m=0.92 kg vode na temperaturi
od 10oC. Iznad vode je klip zanemarqive mase koji
ostvaruje stalni pritisak. Pritisak okoline
iznosi po=1 bar. U cilindru se nalazi greja~
toplotne snage 0.5 kJ/s. Zanemaruju}i trewe klipa o
zidove cilindra odrediti vreme potrebno da se
klip podigne za ?∆z=1.3 m. Predstaviti proces
dovo|ewa toplote na Ts dijagramu.
τ=
Q 12
⋅
∆z
=...
Q 12
prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu
Q12 = ∆U12 + W12
⇒
Q12 =m . (u2 − u1 )+ p ⋅ (V2 − V1 ) = ...
ta~ka 1:
p1 =1 bar,
h1 = hw = 42
t1 =10oC
kJ
,
kg
v1 = vw = 0.0010005
m3
kg
u1 = uw = h1 − p 1 ⋅ v 1 = 42 − 1 ⋅ 10 5 ⋅ 10 −3 ⋅ 0 .0010005 =41.9
kJ
kg
V 1 = m . v1 = 0.92 ⋅ 0.0010005 =0.0009 m3
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 20
ta~ka 2:
p2 =1 bar
v2 =?
V 2 = V1 + A. ∆z = 0.0009 + 0.1.1.3 = 0.1309 m3
v2 =
V2
0.1309
m3
=
=0.1423
m
0 .92
kg
v′’= 0.0010432
m3
,
kg
m3
kg
ta~ka 2 se nalazi u oblasti vla`ne pare
v′′”=1.694
v′’ > v2 > v′′”
x2 =
v 2 − v'
v" − v'
=
0.1423 - 0.0010432
=0.0834
1.694 - 0.0010432
u2 = ux =u′’ + x2 .(u′′” − u′’) = = 417 .3 + 0.0834 ⋅ (2506 − 417 .3 ) =591.5
u’′ = 417.3
kJ
,
kg
u”′′ = 2506
kJ
kg
kJ
kg
Q12 = 0.92 ⋅ (591 .5 − 41.9 ) + 1 ⋅ 10 5 ⋅ 10 −3 ⋅ (0 .1423 − 0.0009 ) =519.77 kJ
τ=
519.77
=1039.5 s
0 .5
T
p=const
1
2
s
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zbirka zadataka iz termodinamike
strana 21
2.23. Vertikalan cilindar (od okoline toplotno izolovan) unutra{weg pre~nika
d=250 mm zatvoren je sa gorwe strane pomi~nim (bez trewa) i od okoline
izolovanim klipom (zanemarqive mase) optere}enim sa dva tega masa: mT1=210 kg i
mT2=1800 kg. Po~etna udaqenost klipa od dna cilindra je z1 =300 mm. U cilindru
se nalazi 5 litara kqu~ale vode, a ostatak zapremine zauzima suvozasi}ena vodena
para. Pritisak okoline iznosi po=1 bar. Dovo|ewem toplote klip se podigne za
∆z=200 mm. Zatim se istovremeno te`i teg podigne dizalicom i skine sa klipa
({to dovodi do daqeg podizawa klipa) i iskqu~i greja~. Odrediti:
a) koli~inu toplote dovedenu u prvom delu procesa
b) izvr{eni zapreminski rad u drugom delu procesa
T1
T2
T2
T1
∆z
T1
∆z
z1
ta~ka 1:
p1 = p o +
(m T1 + m T2 ) ⋅ g = 1 ⋅ 10 5 + (210 + 1800 ) ⋅ 9.81 =5 bar
d2 π
4
0.25 2 π
4
d2 π
0.25 2 π
⋅ z1 =
⋅ 0 .3 =0.0147 m3
4
4
V′′ = V − V′ =0.0147 –− 0.0050 =0.0097 m3
V1 =
v′ =0.0010927 m3 ,
v′′=0.3747 m3
V'
0.0050
V' ' 0 .0097
=
=4.576 kg,
m' ' =
=
=0.026 kg
v' 0.0010927
v ' ' 0 .3747
m' '
0.026
x1 =
=
=0.0056
m' +m' ' 4.576 + 0.026
m' =
u1 = ux =u’′ + x1 .(u′′” − u′’) = 639 .4 + 0.0056 ⋅ (2562 − 639 .4 ) =650.17
kJ
u’′ = 639.4
,
kg
dipl.ing. @eqko Ciganovi}
kJ
kg
kJ
u”′′ = 2562
kg
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zbirka zadataka iz termodinamike
strana 22
ta~ka 2:
V 2 = V1 +
p2 = p1 = 5 bar,
d2 π
0.25 2 π
⋅ ∆z = 0.0147 +
⋅ 0.2 =0.0245 m3
4
4
0 .0245
m3
=0.0053
m
4 .602
kg
v′’ > v2 > v′′”
v2 =
x2 =
V2
=
v 2 − v'
v"− v'
ta~ka 2 se nalazi u oblasti vla`ne pare
0.0053 - 0.0010927
=0.0113
0.3747 - 0.0010927
=
u2 = ux =u’′ + x2 .(u′′” − u′’) = 639 .4 + 0.0113 ⋅ (2562 − 639.4 ) =661.13
kJ
kg
kJ
s2 = sx =s’′ + x2 .(s′′” − s′’) = ...= 1.86 + 0 .0113 ⋅ (6.822 − 1.86 ) =1.916
kgK
kJ
kJ
s’′ = 1.86
,
s”′ ′ = 6.822
kgK
kgK
ta~ka 3:
p3 = p o +
m T1 ⋅ g
2
= 1 ⋅ 10 5 +
210 ⋅ 9.81
2
d π
0.25 π
4
4
kJ
kJ
s’′ = 1.4184
, s”′ ′ = 7.2387
kgK
kgK
s′’ > s3 > s′′”
x3 =
s3 − s'
s" −s'
=1.42 bar,
kJ
kgK
s3 = s2 = 1.916
ta~ka 3 se nalazi u oblasti vla`ne pare
=
1.916 - 1.4184
=0.0855
7.2387 - 1.4184
u3 = ux =u’′ + x3 .(u′′” − u′’) = 461 .1 + 0.0855 ⋅ (2518 − 461 .1) =636.96
kJ
u’′ = 461.1
,
kg
dipl.ing. @eqko Ciganovi}
kJ
kg
kJ
u”′′ = 2518
kg
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zbirka zadataka iz termodinamike
strana 23
a)
prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu
Q12 = ∆U12 + W12
Q12 =m . (u2 –− u1 )+ p ⋅ (V2 − V1 )
⇒
Q 12 = 4 .602 ⋅ (661 .13 − 650 .17 ) + 5 ⋅ 10 5 ⋅ 10 − 3 ⋅ (0 .0245 − 0.0147 ) =55.34 kJ
b)
prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu
Q23 = ∆U23 + W23
⇒
W23 =−m . (u3 –− u2 )
W23 = −4 .602 ⋅ (636 .96 − 661 .13 ) =111.23 kJ
T
p1 =p2
2
1
p3
3
s
2.24. Vla`na vodena para stawa A(pA=0.11 MPa, x=0.443), koja se nalazi u toplotno
izolovanom sudu A, zapremine V A=0.55 m3 , razdvojena je ventilom od suvozasi}ene
vodene pare koja se pri istom pritisku (pB=pA) nalazi u toplotno izolovanom
cilindru B, zapremine V B=0.31 m3 (slika). Pri zako~enom (nepokretnom) klipu K
otvara se ventil i uspostavqa stawe termodinami~ke ravnote`e pare u oba suda
(stawe C). Po dostizawu tog ravnote`nog stawa, pokre}e se klip K, koji pri i daqe
otvorenom ventilu, kvazistati~ki sabija paru na pritisak p=2.4 MPa (stawe D).
Odrediti izvr{eni zapreminski rad ( za proces C−D) i prikazati sve promene u Ts
koordinatnom sistemu.
K
A
B
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
pA=1.1 bar
ta~ka A:
strana 24
xA=0.443
kJ
kg
uA= ux =u’′ + xA.(u′′” - u′’) = ...= 428. 79 + 0. 443 ⋅ (2509 − 428.79 ) =1350.32
u’′ =428.79
kJ
,
kg
u”′′ = 2509
kJ
kg
vA= vx= v’ + xA. (v”′′ − v′’)= ...= 0 .0010452 + 0. 443 ⋅ (1. 555 − 0 .0010452) =0.6894
VA
vA
m3
kg
0 .55
=
= 0.8 kg
0.6894
ta~ka B:
pB=1.1 bar
v’′ = 0.0010452
mA =
v′′”=1.555
m3
kg
m3
kg
xB=1
kJ
m3
,
vB = v′′”=1.555
kg
kg
0.31
=
0.2 kg
1.555
uB=u”′ ′= 2509
mB =
VB
vB
ta~ka C:
pC=1.1 bar
uC=?
prvi zakon termodinamike za proces me{awa u zajedni~kom sudu:
Q12 = ∆U12 + W12
⇒
U1 = U2
U1 = mA. uA + mB. uB
U2 = mA. uC + mB. uC
mA ⋅ u A + mB ⋅ uB 0.8 ⋅ 1350 .32 + 0 .2 ⋅ 2509
kJ
=
=1582.06
m A + mB
0.8 + 0.2
kg
kJ
kJ
u’′ = 428.79
,
u”′′ = 2509
kg
kg
u′’ < uC < u”′ ′
ta~ka 2 se nalazi u oblasti vla`ne pare
uC =
xC =
u C − u'
u"−u'
=
1582 .06 − 428 .79
=0.5544
2509 − 428 .79
sC= sx =s’′ + xC.(s′′” - s′’) = ...= 1.3327 + 0.5544 ⋅ (7 .238 − 1.3327 ) =4.606
s’′ = 1.3327
kJ
kgK
dipl.ing. @eqko Ciganovi}
s”′ ′ = 7.328
kJ
kgK
kJ
kgK
[email protected]
zbirka zadataka iz termodinamike
ta~ka D:
s’′ = 2.534
pD=24 bar
kJ
kgK
kJ
kgK
sD=sC = 4.606
s”′ ′ = 6.272
s′’ < sD < s”′ ′
xD =
strana 25
kJ
kgK
ta~ka D se nalazi u oblasti vla`ne pare
s D − s' 4 .606 − 2.534
=
=0.55543
s"− s'
6 .272 − 2.534
uD= ux =u’′ + xD.(u′′” - u′’) = ...= 948 .9 + 0.5543 ⋅ (2602 − 948 .9) =1865.21
u’′ = 948.9
kJ
,
kg
u”′′ = 2602
kJ
kg
kJ
kg
prvi zakon termodinamike za proces adijabatske kompresije (C−D):
QCD = ∆UCD + WCD
⇒
WCD= UC − UD
WCD= (m A + m B ) ⋅ (u C − uD ) = (0.8 + 0.2 ) ⋅ (1582 .06 − 1865 .21) = −283.15 kJ
T
pD
D
pA=pB=pC
A
C
B
s
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 26
2.25. U zatvorenom sudu zapremine V=2 m3 , nalazi se suvozasi}ena vodena para
stawa 1(p=10 bar). Tokom hla|ewa do stawa 2 od vodene pare odvede se 14.3 MJ
toplote. Odrediti promenu entropije sistema u najpovoqnijem slu~aju tokom
procesa hla|ewa pare.
p= 10 bar,
kJ
u1 = u″ = 2583
,
kg
ta~ka 1:
v1 = v″=0.1946
m3
,
kg
x=1
s1 = s″= 6.587
m=
kJ
kgK
V
2
=
=10.28 kg
v 1 0 .1946
m3
, u2 =?
kg
prvi zakon termodinamike za proces hla|ewa pare:
ta~ka 2:
v2 = v1 =0.1946
Q12 = ∆U12 + W12
⇒
Q12 = m. ( u2 –− u1 )
3
Q
14 .3 ⋅ 10
kJ
u2 = u1 + 12 = 2583 −
=1191.95
kg
m
10.28
pretpostavimo p2 =2.6 bar:
m3
m3
v′′”=0.6925
kg
kg
v 2 − v ' 0.1946 − 0 .0010685
x2 =
=
=0.28
v"− v '
0.6925 − 0 .0010685
kJ
kJ
u’′ = 540.63
,
u”′′ = 2539
kg
kg
v’′ = 0.0010685
kJ
kg
pretpostavka nije ta~na
u2 =ux = u′ + x2 . ( u″ - u′ )= 540 .63 + 0.28 ⋅ (2539 − 540 .63 ) =1100.17
pretpostavimo p2 =2.8 bar:
m3
m3
v′′”=0.6461
kg
kg
v − v' 0.1946 − 0.0010709
x2 = 2
=
=0.3
v"− v '
0.6461 − 0.0010709
kJ
kJ
u’′ = 551.1
,
u”′′ = 2541
kg
kg
v’′ = 0.0010709
kJ
kg
pretpostavka nije ta~na
u2 =ux = u′ + x2 . ( u″ - u′ )= 551 .1 + 0.3 ⋅ (2541 − 551 .1) =1148.07
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 27
pretpostavimo p2 =3.0 bar:
m3
m3
v′′”=0.6057
kg
kg
v 2 − v ' 0.1946 − 0 .0010733
x2 =
=
=0.32
v"− v '
0.6057 − 0 .0010733
kJ
kJ
u’′ = 561.1
,
u”′′ = 2543
kg
kg
v’′ = 0.0010733
kJ
kg
pretpostavka ta~na
u2 =ux = u′ + x2 . ( u″ - u′ )= 561 .1 + 0.32 ⋅ (2543 − 561 .1) = 1195.3
Obzirom da je pretpostavka ta~na to zna~i da je p2 =3 bar. Na osnovu
vrednosti pritiska p2 odre|uje se temperatura T2 =133.54oC.
s’′ = 1.672
kJ
,
kgK
s”′ ′ = 6.992
kJ
kgK
s2 = sx =s’′ + x2 . (s′′” − s′’) = 1.672 + 0 .32 ⋅ (6.992 − 1.672 ) =3.374
kJ
kgK
drugi zakon termodinamike za proces hla|ewa pare:
∆S SI = ∆S RT + ∆S TP = ...= 3.97
kJ
K
∆S RT = m . ∆s12 = m .( s2 −– s1 ) = 10 .28 ⋅ (3.374 − 6.587 ) = − 33.03
∆S TP = −
kJ
K
Q 12
− 14.3 ⋅ 10 3
kJ
=−
=35.17
TTP
133 .54 + 273
K
zadatak za ve`bawe:
(2.26.)
2.26. U zatvorenom sudu nalazi se 5 kg pregrejane vodene pare stawa 1(p1 =0.1 MPa, t1 ).
a) koliko iznosi temperatura pregrejane pare (t1 ) ako od we hla|ewem nastaje suva
vodena para specifi~ne entalpije h=2653 kJ/kg (stawe 2)
b) koliki }e biti stepen suvo}e (x3 ) vla`ne pare kada usled daqeg odvo|ewa toplote
temperatura vodene pare dostigne 50oC (stawe 3)
c) odrediti masu (kg) kqu~ale te~nosti (m’′’ ) i suvozasi}ene pare (m’′’ ′) stawa 3
a) t1 = 320oC
b) x3 = 0.227
c) m′’=3.87 kg,
m”′ ′=1.13 kg
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 28
PRVI I DRUGI ZAKON TERMODINAMIKE
(OTVOREN TERMODINAMI^KI SISTEM)
2.27. U adijabatski izolovanom ure|aju me{aju se suvozasi}ena vodena para stawa
⋅
1(p=0.4 MPa) i voda stawa 2(p=0.4 MPa, t=20oC, m w=1 kg/s ). Iz ure|aja izlazi voda
stawa 3(p=0.4 MPa, t=80oC). Zanemaruju}i promene kineti~ke i potencijalne energije
vodene pare, odrediti:
a) potrebnu koli~inu pare (kg/s)
b) promenu entropije sistema za proces me{awa (kW/K)
1
para
3
2
voda
a)
p1 =4 bar
x=1
kJ
kJ
h1 =h′′ = 2738
,
s1 =s′′ = 6.897
kg
kgK
ta~ka 1:
p2 =4 bar
t2 =20oC
kJ
kJ
h2 =hw = 84.1
,
s2 =sw =0.296
kg
kgK
ta~ka 2:
p2 =4 bar
t2 =80oC
kJ
kJ
h3 =hw = 335.1
,
s3 =sw =1.074
kg
kgK
ta~ka 3:
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu:
⋅
⋅
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⇒
H1 = H 2
⋅
⋅
⋅
⋅
mp ⋅ h1 + m w ⋅ h 2 = mp + m w ⋅ h3
⋅
m w ⋅ (h3 − h2 ) 1 ⋅ (335 .1 − 84 .1) ⋅
kg
mp =
=
= mp = 0.1
h1 − h 3
335 .1 − 2738
s
⋅
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 29
b)
⋅
⋅
⋅
∆ S SI = ∆ S RT + ∆ S o = ... = 0.196 + 0 = 0.196
kW
K
⋅
⋅
Q12
kW
∆ S o= −
=0
TO
K
⋅
⋅
⋅
⋅
⋅
∆ S RT = S izlaz − S ulaz= mp + m w
⋅
⋅
⋅ s 3 − mp ⋅ s 1 − m w ⋅ s 2
⋅
kW
∆ S RT = (0.1 + 1) ⋅ 1.074 − 0.1 ⋅ 6.897 − 1 ⋅ 0.296 =0.196
K
⋅
t
suvozasi}ene pare stawa 1(p=13 bar). Deo te pare se
h
koristi za potrebe nekog tehnolo{kog procesa, dok se drugi deo pare, nakon
prigu{ivawa do p2 , me{a u napojnom rezervoaru sa vodom stawa 2(p=2 bar, t=20oC).
Voda se iz napojnog rezervoara uvodi u toplotno izolovanu pumpu gde joj se pritisak
kvazistati~i povisi do pritiska u kotlu. Ako je toplotna snaga kotla 4.56 MW,
2.28. Kotao proizvodi m =7
⋅
odrediti maseni protok pare koja se koristi u tehnolo{kom procesu ( m w) kao i
snagu pumpe.
Q12
4
1
ka tehnolo{kom procesu
WT34
3
2
napojna voda
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 30
ta~ka 1:
p=13 bar,
kJ
h1 = h”″ = 2787
kg
x=1
p=2 bar,
kJ
h2 = hw = 84.0
kg
t=20oC
ta~ka 2:
p=13 bar,
ta~ka 4:
h4 =?
⋅
⋅
⋅
prvi zakon termodinamike za proces u kotlu: Q 12 = ∆ H12 + W T12
⋅
⋅
⋅
Q12 = m⋅ (h1 − h 4 )
h4 = 2787 −
⇒
4.56 ⋅ 10 3
7⋅
3
10
3600
=441.86
p= 2 bar,
ta~ka 3:
h3 = hw = 440.95
Q 12
h4 = h1 −
mp
kJ
kg
s4 = sw = 1.363
s3 = s4 = 1.363
kJ
kgK
kJ
kgK
kJ
kg
prvi zakon termodinamike za proces me{awa:
⋅
⋅
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⇒
H1 = H 2
⋅
⋅
⋅
⋅
mw ⋅ h2 + mp − m w ⋅ h1 = mp ⋅ h 3
⇒
⋅
t
440.95 − 2787
mw = 7 ⋅
= 6.08
h
84 − 2787
⋅
⋅
m w = mp ⋅
h3 − h1
h2 − h1
prvi zakon termodinamike za proces u pumpi:
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⋅
⋅
⇒
W T12 = − m w ⋅ (h4 − h3 ) = 6 .08 ⋅
dipl.ing. @eqko Ciganovi}
⋅
⋅
W T12 = − ∆ H12
10 3
⋅ (441 .86 − 440 .95 ) =1.54 kW
3600
[email protected]
zbirka zadataka iz termodinamike
strana 31
2.29. Voda stawa 1(p=2 bar, t=80oC) dostrujava kroz cev unutra{weg pre~nika d=40
mm brzinom 0.5 m/s. Prolaskom kroz delimi~no otvoren ventil se prigu{uje na
p2 =0.4 bar i ulazi u odvaja~ te~nosti (od okoline toplotno izolovan). Odrediti:
a) promenu entropije sistema za proces prigu{ivawa
b) snagu kompresora koji izbacuje parnu fazu iz suda u okolinu pritiska p4 =1 bar
c) snagu pumpe koja te~nu fazu iz suda prebacuje u parni kotao koji radi na
pritisku p4 =4 bar
napomena:
kompresije u kompresoru i pumpi su kvazistati~ke i adijabatske
kompresor
1
2
4
3
pumpa
T
3
4
1
2′
2
2′′
s
ta~ka 1:
h1 = hw =334.9
⋅
m=
p1 =1 bar,
kJ
,
kg
t1 =80oC
s1 =sw=1.074
(
kJ
,
kgK
1
d2π
1
40 ⋅ 10 −3
⋅w⋅
=
⋅ 0.5 ⋅
v1
4
0 .001028
4
dipl.ing. @eqko Ciganovi}
v1 =vw = 0.001028
) π =0.61
2
m3
kg
kg
s
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zbirka zadataka iz termodinamike
p2 =0.4 bar,
ta~ka 2:
h’”′ =317.7
x2 =
strana 32
kJ
,
kg
h 2 − h'
h"−h'
=
h2 =h1 = 334.9
h′′ = 2636
kJ
kg
kJ
kg
334 .9 − 317 .7
=0.0074
2636 − 317 .7
s2 = sx = s '+ x 2 ⋅ (s ' ' −s ' ) = 1.0261 + 0.0074 ⋅ (7.67 − 1.0261) =1.075
kJ
,
kgK
s’′” =1.0261
s′′ = 7.67
p3 =4 bar,
ta~ka 3:
kJ
kgK
s3 =s′= 1.0261
kJ
kgK
kJ
kg
h3 = hw= 318.5
p4 =1 bar,
ta~ka 4:
kJ
kgK
h4 = hpp= 3129.7
s3 =s′= 7.67
kJ
kgK
kJ
kg
a)
⋅
⋅
⋅
∆ SSI = ∆ SRT + ∆ S O =0.61
⋅
⋅
⋅
W
K
⋅
∆ SRT = Sizlaz − Sulaz = m⋅ (s 2 − s 1 ) = 0 .61 ⋅ (1 .075 − 1.074 ) =0.61
W
K
b)
prvi zakon termodinamike za proces u kompresoru:
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⋅
⇒
⋅
⋅
W T12 = − ∆ H12
⋅
W T 12 = − m⋅ x 2 ⋅ (h 4 − h' ' ) = −0.61 ⋅ 0 .0074 ⋅ (3129 .7 − 2636 ) =−2.23 kW
c)
prvi zakon termodinamike za proces u pumpi:
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⋅
⇒
⋅
⋅
W T12 = − ∆ H12
⋅
W T 12 = − m⋅ (1 − x 2 ) ⋅ (h3 − h' ) = − 0.61 ⋅ (1 − 0.0074 ) ⋅ (318 .5 − 317 .7 ) =−0.48 kW
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 33
2.30. U toplotno izolovan kompresor ulazi freon 12 (R12) stawa 1(p=1 bar, t=−20oC,
⋅
m =60 kg/h). Stawe freona 12 na izlazu iz kompresora je 2(p=8 bar), a snaga
kompresora iznosi 1 kW. Nakon kompresije freon se hladi i potpuno kondenzuje u
razmewiva~u toplote. Kao rashladni fluid u razmewiva~u toplote koristi se voda
stawa 4(p=1bar, t=10oC) koja se prolaskom kroz razmewiva~ toplote zagreje do stawa
5(p=1 bar, t=30oC). Skicirati promene stawa freona 12 na hs dijagramu i odrediti:
a) stepen dobrote adijabatske kompresije u kompresoru
b) potro{wu vode u razmewiva~u toplote (kg/h)
voda
4
5
3
2
WT12
1
freon
2
h
2k
1
3
s
ta~ka 1:
h1 =hpp
p=1 bar,
kJ
=647.4
kg
ta~ka 2k:
p2K=8 bar,
h2K=hpp =686.1
t=−20oC
s1 =spp=1.612
kJ
kgK
s2K=s1 =1.612
kJ
kgK
kJ
kg
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 34
p2 =1 bar,
ta~ka 2:
h2 =?
prvi zakon termodinamike za proces u kompresoru:
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⋅
⋅
W T12 = − ∆ H12
⇒
W T12
h2 = h1 −
mf
⋅
⋅
W T12 = − m f ⋅ (h2 − h1 )
h2 = 647 .4 −
⋅
⇒
−1
kJ
=707.4
60
kg
3600
p=8 bar,
kJ
h1 =h′ =531.45
kg
x=0
ta~ka 3:
a)
kp
ηd =
b)
h1 − h2K
647.4 − 686 .1
=
=0.645
h1 − h 2
647.4 − 707 .4
t=10oC
p=1 bar,
ta~ka 4:
h1 =hw =42
kJ
kg
(tabele za vodu)
p=1 bar,
t=30oC
kJ
h1 =hw =125.7
(tabele za vodu)
kg
ta~ka 5:
prvi zakon termodinamike za proces u razmewiva~u toplote:
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⋅
⋅
⇒
⋅
⋅
m f ⋅ h 2 + m w ⋅ h4 = m f ⋅ h3 + m w ⋅ h5
⋅
m w = 60 ⋅
⋅
⋅
H1 = H 2
⇒
⋅
⋅
mw = m f ⋅
h2 − h3
h5 − h 4
kg
707 .4 − 531 .45
=12.61
h
125 .7 − 42
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 35
2.31. Pregrejana vodena para stawa 1(p=7 bar, t=450oC) ekspandira adijabatski u
parnoj turbini sa stepenom dobrote η eks
d =0.6 do stawa 2(p=1 bar). Po izlasku iz
turbine para se u toplotno izolovanoj me{noj komori me{a sa vodom, masenog protoka
⋅
mw =2.3 kg/s stawa 3(p=1bar, t=14oC). Stawe voda na izlazu iz komore za me{awe je
4(p=1 bar, t=47oC). Skicirati procese u turbini i me{noj komori na Ts dijagramu i:
a) odrediti snagu turbine (kW)
b) dokazati da je proces me{awa pare i vode nepovratan
para
1
WT12
3
voda
2
4
T
T
1
4
3
2
2k
s
1 –− 2
2 –− 4
3−4
2
s
promena stawa pare u turbini
promena stawa pare u me{noj komori
promena stawa vode u me{noj komori
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 36
a)
t=450oC
p=7 bar,
kJ
h1 = hpp = 3374.75
kg
ta~ka 1:
p=1 bar,
ta~ka 2k:
h′=417.4
kJ
,
kg
h′′= 2675
h2k>h′′
h2k
s1 =spp =7.789
kJ
kgK
s2k = s1 =7.789
kJ
kgK
kJ
kg
ta~ka 2k se nalazi u oblasti pregrejane pare
kJ
= hpp = 2854.3
kg
p=1 bar,
ta~ka 2:
η eks
=
d
h1 − h2
h1 − h2k
⇒
η eks
d 0.6
h2 = h1 − η eks
d ⋅ (h1 − h 2k )
kJ
kg
h2 = 3374 .75 − 0.6 ⋅ (3374 .75 − 2854 .3) = 3062.48
s2 = spp = 8.19
kJ
kgK
p=1 bar,
kJ
h3 = hw = 58.6
kg
t=14oC
p= 1 bar,
kJ
h4 = hw = 196.74
kg
t=47oC
ta~ka 3:
ta~ka 4:
s3 = sw = 0.21
kJ
kgK
s4 = sw = 0.66
kJ
kgK
prvi zakon termodinamike za proces u me{noj komori:
⋅
⋅
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⇒
H1 = H 2
⋅
⋅
⋅
⋅
mw ⋅ h3 + mp h2 = mw + m p ⋅ h4
⇒
⋅
196.74 − 58 .6
kg
mp = 2.3 ⋅
= 0.11
3062 .48 − 196.74
s
dipl.ing. @eqko Ciganovi}
⋅
⋅
mp = m w ⋅
h4 − h 3
h2 − h4
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zbirka zadataka iz termodinamike
strana 37
prvi zakon termodinamike za proces u turbini:
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⋅
⋅
⋅
W T 12 = − ∆ H 12 = − mp ⋅ (h 2 − h1 )
⋅
W T 12 = −0.11 ⋅ (3062 .48 − 3374 .75 ) =34.35 kW
b)
⋅
⋅
⋅
∆ S SI = ∆ S RT + ∆ S o = ... = 0.196 + 0 = 0.196
⋅
kW
K
⋅
Q12
kW
∆ S o= −
=0
TO
K
⋅
⋅
⋅
⋅
⋅
⋅
⋅
∆ S RT = S izlaz − S ulaz= mp + m w ⋅ s 4 − mp ⋅ s 2 − m w ⋅ s 3
⋅
∆ S RT = (0.11 + 2.3 ) ⋅ 0.66 − 0.11 ⋅ 8.19 − 2.3 ⋅ 0 .21 =0.207
kW
K
2.32. Pregrejana vodena para stawa 1(p1 =70 bar, t1 =450oC) adijabatski ekspandira u
parnoj turbini do stawa 2(p2 =1 bar). Snaga turbine je 200 kW. Nakon ekspanzije para
se uvodi u kondenzator u kome se izobarski potpuno kondenzuje (stawe 3=kqu~ala voda).
Protok vode za hla|ewe kondenzatora je mw=5 kg/s, stawe vode na ulazu u kondenzator
je (p=1 bar, tw1 =20oC), a na izlazu iz kondenzatora je (p=1 bar, tw2 =45oC). Skicirati
promene stawa pare (1−2−3) na hs dijagramu i odrediti:
a) maseni protok vodene pare (kg/s)
b) stepen suvo}e vodene pare na izlazu iz turbine
c) stepen dobrote ekspanzije pare u turbini
para
1
WT12
3
2
voda
4
dipl.ing. @eqko Ciganovi}
5
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zbirka zadataka iz termodinamike
a)
strana 38
ta~ka 4:
p4 =1 bar,
kJ
h4 = hw =83.9
kg
t4 =20oC
ta~ka 5:
p5 =1 bar,
kJ
h4 = hw =188.4
kg
t5 =45oC
p1 =70 bar,
kJ
h1 = hpp =3284.75
,
kg
t1 =450oC
p= 1 bar,
kJ
h1 =h′ =417.4
kg
x=0
ta~ka 1:
ta~ka 3:
s1 =spp = 6.634
kJ
kgK
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
ograni~enom isprekidanom linijom:
⋅
⋅
⋅
0 = H2 − H1 + W T12
⋅
⇒
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⋅
⋅
⋅
⋅
⋅
0 = mp ⋅ h3 + m w ⋅ h5 − mp ⋅ h1 + m w ⋅ h4 + W T 12
⋅
m w ⋅ (h 5 − h 4 ) + W T12
5 ⋅ (188 .4 − 83 .9 ) + 200
kg
mp =
=
=0.25
h1 − h3
3284 .75 − 417 .4
s
⋅
b)
p2K =1 bar,
ta~ka 2K:
kJ
,
kgK
s′’ > s2K > s′′”
s 2k − s'
s"− s'
=
kJ
kgK
kJ
kgK
ta~ka 2K se nalazi u oblasti vla`ne pare
s′=1.3026
x 2K =
s2K=s1 = 6.634
s′′=7.36
6 .634 − 1.3026
=0.88
7.36 - 1.3026
h2K = hx = h' + x 2k ⋅ (h' '−h') = 417 .4 + 0.88 ⋅ (2675 − 417 .4 ) = 2404.09
h′=417.4
kJ
,
kgK
dipl.ing. @eqko Ciganovi}
h′′=2675
kJ
kg
kJ
kgK
[email protected]
zbirka zadataka iz termodinamike
strana 39
c)
p2 =1 bar,
ta~ka 2:
h2 =?
prvi zakon termodinamike za proces u turbini:
⋅
⋅
⋅
⋅
⋅
h2 = h1 −
W T 12
⋅
= 3284 .75 −
mp
η ex
d =
⋅
⋅
W T 12 = − ∆ H 12 = − mp ⋅ (h 2 − h1 )
Q 12 = ∆ H12 + W T12
200
kJ
=2484.75
0.25
kg
h1 − h2
3284 .75 − 2484 .75
=
=0.91
h1 − h2k
3284 .75 − 2404 .1
h
1
2
2K
3
s
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 40
2.33. U sekundarnom krugu atomskog reaktora proizvodi se qm=4000 t/h
suvozasi}ene vodene pare stawa 1(p=70bar). Proizvedena para se, prema skici, deli
na dve struje qm1 i qm2. Para masenog protoka qm1 kvazistati~ki adijabatski
ekspandira u turbini do stawa 2(p=8 bar). Vla`na para stawa 2 se u odvaja~u
te~nosti (toplotno izolovan od okoline) deli na dve struje qm3 (suvozasi}ena para
stawe 3) i qm4 (kqu~ala voda, stawe 4). Para stawa 3 se daqe pregreva (p=const) u
razmewiva~u toplote do stawa 5(T=250 oC) na ra~un toplote koju oslobodii para
qm1 kondenzacijom (p=const) do stawa 6(x=0). Skicirati stawa pare na Ts
dijagramu i odrediti:
a) masene protoke fluidnih struja, qm1 , qm2 , qm3 i qm4
b) snagu turbine
qm
1
qm2
qm1
pregreja~
pare
WT12
qm3
2
qm2
6
3
qm4
odvaja~
te~nosti
4
p1 =70 bar,
kJ
h1 = h′′ =2772
,
kg
ta~ka 1:
ta~ka 2:
qm3
p1 =8 bar,
5
x=1
s1 =s′′ = 5.814
s2 =s1 =5.814
kJ
kgK
kJ
kgK
kJ
kJ
,
s′′=6.663
kgK
kgK
s − s' 5 .814 − 2.046
x2 = 2
=
=0.8161
s" −s'
6.663 - 2.046
s′=2.046
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 41
ta~ka 3:
p1 =8 bar,
kJ
h3 = h′′ =2769
kg
x=1
ta~ka 4:
p1 =8 bar,
kJ
h4 = h′ =720.9
kg
x=0
ta~ka 5:
p1 =8 bar,
kJ
h5 = hpp =2947.5
kg
T=250oC0
p1 =70 bar,
kJ
h6 = h′ =1267.4
kg
x=0
ta~ka 6:
materijalni bilans ra~ve:
para koja napu{ta odvaja~ te~nosti:
qm = qm1 + qm2
qm3 = qm1 . x2
(1)
(2)
prvi zakon termodinamike za proces u razmewiva~u toplote:
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⇒
⋅
⋅
H1 = H 2
qm 2 ⋅ h1 + qm 3 ⋅ h3 = q m2 ⋅ h6 + q m3 ⋅ h 5
(3)
Kombinovawem jedna~ina (1), (2) i (3) dobija se:
t
t
t
qm1 =3646.9
,
qm2 =353.1
,
qm3 = 2976.2
h
h
h
materijalni bilans odvaja~a te~nosti:
t
qm4 = qm1 − qm3 = 670.7
h
qm1 = qm3 + qm4
⇒
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
ograni~enom isprekidanom linijom:
⋅
⋅
⋅
0 = H2 − H1 + W T12
⋅
⇒
⋅
⋅
⋅
Q 12 = ∆ H12 + W T12
⋅
W T 12 = [qm ⋅ h1 ] − [qm2 ⋅ h 6 + qm 3 ⋅ h5 + q m4 ⋅ h4 ]
W T12 = [[4000 ⋅ 2772 ] − [353 .1 ⋅ 1267 .4 + 2976 .2 ⋅ 2947 .5 + 670 .7 ⋅ 720.9 ]] ⋅
10 3
3600
⋅
W T12 =384.6 MW
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 42
(2.34.)
zadatak za ve`bawe:
2.34. U parno-turbinskom postrojewu (slika) vodena para masenog protoka m=1.2 kg/s
ekspandira u turbini visokog pritiska (TVP) od stawa 1(p=1 MPa, t=440oC) do stawa
2(p=0.5 MPa). Po izlasku iz turbine deo pare masenog protoka mA =0.4 kg/s me{a se
adijabatski sa vodom stawa (p= 5 bar, tw=20oC). Stawe vode na izlasku iz komore za
me{awe je (p= 5 bar, tw=45oC). Preostali deo pare se po izlasku iz turbine visokog
pritiska izobarski zagreva do stawa 3(t=400oC), a zatim ekspandira u turbini niskog
pritiska (TNP) do stawa 4(p=5 kPa). Ekspanzije u turbinama su adijabatske sa istim
stepenom dobrote (ηdex=0.9). Odrediti:
a) snagu turbina visokog i niskog pritiska (kW)
b) maseni protok vode u komori za me{awe (kg/s)
para
TVP
1
WT12
2
voda
5
mA
3
TNP
WT34
6
4
a)
b)
⋅
WT
12 =230.5
kW,
⋅
kg
mw =11.35
s
⋅
WT
34 =642.4
dipl.ing. @eqko Ciganovi}
kW
[email protected]
zbirka zadataka iz termodinamike
strana 43
PRVI I DRUGI ZAKON TERMODINAMIKE
(PUWEWE I PRA@WEWE REZERVOARA)
2.35. U adijabatski izolovan rezervoar zapremine V=30 m3 , u kojem se nalazi vla`na
vodena para stawa (p=1.2 bar, x=0.95), uvodi se jednim izolovanim cevovodom voda
stawa (p=8 bar, t=15oC), a drugim izolovanim cevovodom suva vodena para stawa (p=30
bar). Stawe radne materije u rezervoaru na kraju procesa puwewa je (p=6 bar, x=0,1).
Odrediti masu vode i masu suve pare uvedene u rezervoar.
p=1.2 bar,
po~etak:
x=0.95
upo~etak= ux =u’′ + xp.(u′′” − u′’) = 439 .28 + 0 .95 ⋅ (2512 − 439 .28 ) =2408.4
kJ
u’′ = 439.28
,
kg
kJ
kg
kJ
u”′′ =2512
kg
vpo~etak= v x= v’′’ + x p. (v”′′ − v′’)= 0. 0010472 + 0. 95 ⋅ (1.429 − 0. 0010472) =1.3576
v’′ = 0.0010472
mpo~etak =
m3
,
kg
V
=
vpo~etak
ulaz:
kraj:
v′′”=1.429
m3
kg
30
=22.1 kg
1.3576
p=30 bar,
x=1
h = h″=2804
p=8 bar,
t=15oC
h = h w = 62.8
p = 6 bar,
x=0.1
ukraj= ux =u’′ + xk.(u′′” − u′’) = 669.8 + 0.1 ⋅ (2568 − 669.8 ) =859.6
u’′ = 669.8
kJ
,
kg
m3
kg
u”′′ =2568
kJ
kg
kJ
kg
kJ
kg
kJ
kg
vkraj= v x= v’′’ + x k. (v”′′ − v′’)= 0. 0011007 + 0. 1 ⋅ (0. 3156 − 0.0011007) =0.0326
m3
kg
m3
m3
,
v′′”=0.3156
kg
kg
V
30
mkraj =
=
=920.25 kg
vkraj 0.0326
v’′ = 0.0011007
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 44
prvi zakon termodinamike za proces puwewa suda:
Q12 − W12 = Ukraj − Upo~etak − Hizlaz − Hulaz
0 = mkraj . ukraj − mpo~etak . upo~etak –− mw . hw − m″ . h″
(1)
zakon odr`awa mase za proces puwewa suda:
mpo~etak + mw + m″ = m kraj + mizlaz
(2)
kombinovawem jedna~ina (1) i (2) dobija se:
mw =
(
)
(
mk ⋅ ukraj − h' ' − mp ⋅ up − h' '
)=
h w − h' '
920.25 ⋅ (859. 6 − 2804) − 22. 21⋅ (2408. 4 − 2804)
mw =
= 649.55 kg
62.8 − 2804
m″ = mkraj − mpo~etak + mw = 920.25 − 22.21 − 649.55 = 248.49 kg
2.36. U verikalnom toplotno izolovanom cilindru, povr{ine popre~nog preseka
A=0.1 m2 , nalazi se 0.05 kg vodene pare temperature 180oC, ispod toplotno
izolovanog klipa mase koja odgovara te`ini od 20 kN, a na koji spoqa deluje
atmosferski pritisak od 0.1 MPa. U cilindar se, kroz toplotno izolovan cevovod,
naknadno uvede 0.1 kg vodene pare pritiska 0.4 MPa i temperature 540oC.
Zanemariti trewe klipa i odrediti:
a) specifi~nu entalpiju i temperaturu vodene pare u cilindru na kraju procesa
b) za koliko se podigao klip tokom ekspanzije
kraj
∆y
po~etak
ulaz
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 45
a)
po~etak:
pp = patm +
mT ⋅ g
20 ⋅ 10 3
= 1 ⋅ 105 +
= 3 ⋅ 10 5 Pa,
A
0 .1
t=180oC
up = hp − pp . vp =…...= 2824 − 3 ⋅ 105 ⋅ 10− 3 ⋅ 0.6838 =2618.86
hp = hpp = 2824
kJ
,
kg
mp=0.05 kg
p=4 bar,
kJ
hul = hpp =3572
,
kg
ulaz:
m3
kg
V p = mp ⋅ vp = 0.05 ⋅ 0.6838 =0.03419 m3
t=540oC
(pregrejana para)
mul=0.1 kg
mk = mp + mul =0.15 kg
prvi zakon termodinamike za proces puwewa:
(
)
p ⋅ Vp − Vk = mk ⋅ uk − mp ⋅ up − mul ⋅ hul
hk =
p ⋅ Vp + mp ⋅ up + mul ⋅ hul
mk
hk =3322.67
kJ
kg
vp=vpp = 0.6838
pk = pp = 3 bar,
kraj:
(pregrejana para)
=
Q12 − W12 = Uk − Up + Hiz − Hul
⇒ p ⋅ Vp = mk ⋅ hk − mp ⋅ up − mul ⋅ hul
3 ⋅ 10 5 ⋅ 10 − 3 ⋅ 0.03419 + 0 .05 ⋅ 2618 .86 + 0.1 ⋅ 3572
=
0.15
kJ
kg
kJ
,
kg
h′ < h k < h′′
h′=561.4
h′′ = 2725
kJ
kg
(stawe kraj je u pregrejanoj pari)
tk = tpp = 422.7oC,
vk = vpp =1.067
m3
kg
V k = mk ⋅ v k = 0.15 ⋅ 1.067 =0.16005 m3
b)
∆y =
Vk − Vp
A
=
0 .16005 − 0.03419
=1.26 m
0 .1
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 46
2.37. Pomi~nim klipom sa tegom koji se kre}e bez trewa odr`ava se konstantan
pritisak p=4 bar u vertikalnom cilindru u kojem se nalazi V=500 dm3 vode po~etne
temperature t=20oC (slika kao u prethodnom zadatku). Parovodom se u cilindar
postepeno uvodi mp=53 kg suvozasi}ene vodene pare pritiska p=6 bar, koja se pre
me{awa prigu{uje do pritiska od p=4 bara. Temperatura me{avine (voda) na kraju
procesa me{awa iznosi t=80oC. U toku me{awa usled neidealnog toplotnog izolovawa
okolini se predaje 1.5 kW toplote. Odrediti vreme trajawa procesa me{awa.
p= 4 bar,
po~etak:
vp = vw = 0.001001
3
m
,
kg
t=20oC
hp = hw = 84.1
(voda)
kJ
kg
up = uw = hp − pp ⋅ vp = 84. 1 − 4 ⋅ 105 ⋅ 10−3 ⋅ 0. 001001=83.7
mp =
Vp
vp
=
0 .5
=499.5 kg
0.001001
p= 6 bar,
kJ
hu = h′′ =2757
kg
ulaz:
p= 4 bar,
kraj:
kJ
kg
x=1
mu= 53 kg
t=80oC
uk = hk − pk . vk =…...= 335.1 − 4 ⋅ 105 ⋅ 10− 3 ⋅ 0.001028 =334.7
kJ
kg
kJ
m3
,
vkraj = vw = 0.001028
kg
kg
mk = mp + mu = 499.5 + 53 = 552.5 kg,
hk= hw = 335.1
V k = mk . vk = 552.5 ⋅ 0.001028 =0.568 m3
prvi zakon termodinamike za proces puwewa rezervoara:
Q12 − W12 = Uk − Up + Hiz − Hul
(
Q 12 = mk ⋅ uk − mp ⋅ up − mul ⋅ hul + p ⋅ Vk − Vp
)
Q12 = 552. 5 ⋅ 334. 7 − 499.5 ⋅ 83. 7 − 53 ⋅ 2757+ 4 ⋅105 ⋅10−3 ⋅ (0.568 − 0.5) =−2980.2 kJ
τ=
Q12
⋅
Q12
=
−2980 .2
=1987 s
− 1.5
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 47
2.38. U otvoren sud (slika) koji sadr`i sme{u ml=15 kg leda i mw=20 kg vode u stawu
termodinami~ke ravnte`e, uvedeno je mp=0.8 kg pregrejane vodene pare stawa (p=3
bar, t=340oC). Okolni vazduh stawa O(p=1bar, to=7oC), tokom ovog procesa sme{i u
sudu preda Q12 =320 kJ toplote. Zanemaruju}i promenu zapremine (tj. rad koji radno
telo vr{i nad okolinom), odrediti promenu entropije sistema tokom ovog procesa.
po~etak:
t=0oC,
y=
mw
20
=
=0.5714
m w + ml
20 + 15
up = uy ≅ hy= hl + y ⋅ (hw − hl ) = − 332 .4 + 0.5714 ⋅ 332 .4 = −142.47
sp = sl = s l + y ⋅ (s w − s l ) =
kJ
kg
−332 .4
332 .4
kJ
+ 0.5714 ⋅
=− 0.522
kgK
273
273
mp = ml + mw =20 + 15 =35 kg
p=3 bar,
t=340oC
(pregrejana para)
kJ
kJ
hu = hpp = 3150
,
su = spp = 7.835
,
mu = 0.8 kg
kgK
kg
ulaz:
p=1 bar,
kraj:
mk= mp + mu= 35 + 0.8 =35.8 kg
prvi zakon termodinamike za proces puwewa rezervoara:
Q12 − W12 = Uk − Up + Hiz − Hul
uk =
Q12 + mp ⋅ up + mul ⋅ h ul
mk
ul < u k < uw
yk =
Q 12 = mk ⋅ uk − mp ⋅ up − mul ⋅ hul
=
320 + 35 ⋅ (− 142.57 ) + 0 .8 ⋅ 3150
kJ
=−60.05
35 .8
kg
(stawe ″kraj″ je me{avina vode i leda)
u k − ul
− 60 .05 + 332 .4
=
=0.82
uw − ul
0 + 332 .4
sk = sy = s l + y ⋅ (s w − s l ) =
dipl.ing. @eqko Ciganovi}
−332 .4
332.4
kJ
+ 0 .82 ⋅
= − 0.219
kgK
273
273
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zbirka zadataka iz termodinamike
∆S SI = ∆S RT + ∆S O = ...=4.162 − 1.143 =3.019
strana 48
kJ
K
Q 12
320
kJ
=−
= −1.143
To
280
K
∆S RT = mp ⋅ s k − s p + mu ⋅ (s k − s u ) =
∆S O= −
(
)
∆S RT = 35 ⋅ (− 0.219 + 0 .522 ) + 0.8 ⋅ (−0 .219 − 7.835 ) =4.162
dipl.ing. @eqko Ciganovi}
kJ
K
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zbirka zadataka iz termodinamike
strana 49
2.39. Zatvoreni rezervoar zapremine V=10 m3 sadr`i kqu~alu vodu i suvu vodenu paru
u stawu termodinami~ke ravnote`e na p=20 bar. Te~nost zauzima polovinu zapremine
rezervoara. Iz rezervoara fluid mo`e isticati kroz ventil na vrhu i kroz ventil na
dnu rezervoara. Dovo|ewem toplote za vreme isticawa temperatura vla`ne pare u
rezervoaru se odr`ava stalnom. Odrediti koli~inu dovedene toplote ako je iz
rezervoara isteklo 300 kg fluida kroz:
a) dowi ventil
b) gorwi ventil
a)
b)
a)
p=20 bar,
po~etak:
vp =
up=?
V
10
m3
=...=
=0.0023
mp
4299 .74
kg
mp = m′ + m′′ = ... = 4249.53 + 50.21=4299.74 kg
V'
5
=…...=
= 4249.53 kg
v'
0 .0011766
V' '
5
m' ' =
=...=
= 50.21 kg
v' '
0 .09958
m' =
v′ = 0.0011766
m3
,
kg
v′′=0.09958
m3
kg
up= ux =u’′ + xp.(u′′” − u′’)=...= 906 .1 + 0.0117 ⋅ (2600 − 906 .1) =925.92
u’′ = 906.1
xp =
kJ
,
kg
u”′′ =2600
kJ
kg
kJ
kg
m' '
50 .21
=
= 0.0117
m'+ m' ' 4249 .53 + 50 .21
Tp = 212.37oC
dipl.ing. @eqko Ciganovi}
(temperatura kqu~awa za pritisak od p=20 bar)
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zbirka zadataka iz termodinamike
strana 50
Tk=Tp=212.37oC,
kraj:
uk=?
mk =mp − miz = 4299.74 − 300 = 3999.74 kg
V
10
m3
vk =
=
= 0.0025
mk
3999 .74
kg
v′’ < vk < v”′ ′
xk =
v kj − v'
v' ' − v'
(stawe kraj se nalazi u oblasti vla`ne pare)
=
0.0025 − 0.0011766
=0.0134
0.09958 − 0.0011766
uk = ux = =u’′ + xp.(u′′” − u′’)= 906 .1 + 0.0134 ⋅ (2600 − 906 .1) = 928.8
hiz= h′’ = 908.5
kJ
kg
kJ
kg
izlaz:
mizlaz = 300 kg,
napomena:
zbog polo`aja ventila iz suda isti~e kqu~ala voda
prvi zakon termodinamike za slu~aj pra`wewa suda:
Q 12 − W12 = U k − U p + Hiz − H ul
⇒
Q 12 = mk ⋅ u k − mp ⋅ up + miz ⋅ hiz
Q12 = 3999 .74 ⋅ 928 .8 − 4299 .74 ⋅ 925 .92 + 300 ⋅ 908 .5 =6293.25 kJ
b)
po~etak:
nema promena u odnosu na pod a)
kraj:
nema promena u odnosu na pod a)
izlaz:
miz = 300 kg,
napomena:
kJ
kg
zbog polo`aja ventila iz suda isti~e suva para
hiz= h′′ =2799
prvi zakon termodinamike za slu~aj pra`wewa suda:
Q 12 − W12 = U k − U p + Hiz − H ul
⇒
Q 12 = mk ⋅ u k − mp ⋅ up + miz ⋅ hiz
Q12 = 3999 .74 ⋅ 928 .8 − 4299 .74 ⋅ 925 .92 + 300 ⋅ 2799 =573443.25 kJ
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 51
2.40. U zatvorenom, toplotno izolovanom rezervoaru, zapremine
V=0.5 m3 nalazi se 30 kg vla`ne vodene pare. Kada, pri zagrevawu,
pritisak pare u rezervoaru dostigne vrednost p=5 MPa, biva
iskqu~en elektri~ni greja~ stalne snage i istovremeno otvoren
sigurnosni ventil na rezervoaru tako da jedan deo vodene pare
naglo istekne u okolinu. Po zatvarawu ventila pritisak vodene
pare u rezervoaru iznosi 3 MPa. Preostala vla`na para biva potom
dogrevana istim elektri~nim greja~em, stalne snage od 800 W.
Skicirati promene stawa vodene pare na Ts dijagramu i odrediti:
a) masu vla`ne pare u rezervoaru nakon zatvarawa sigurnosnog ventila
b) vreme nakon kojeg }e se sigurnosni ventil ponovo otvoriti
a)
p= 50 bar,
po~etak:
xp =
v p − v'
v' ' − v'
=
v′ = 0.0012857
vp =
V
0.5
m3
= 0.0167
=
mp
30
kg
0 .0167 − 0.0012857
=0.404
0 .03944 − 0.0012857
m3
,
kg
v′′=0.03944
m3
kg
sp= sx =s’′ + xp.(s′′” − s′’) = ...= 2.921 + 0.404 ⋅ (5.973 − 2 .921) =4.154
s’′ = 2.921
kraj:
xk =
kJ
kgK
s”′ ′ = 5.973
p=30 bar,
s k − s'
s' ' −s'
s’′ = 2.646
=...=
kJ
kgK
kJ
kgK
kJ
kgK
sk =sp =4.154
kJ
kgK
4.154 − 2 .646
=0.426
6.186 − 2.646
kJ
s”′ ′ = 6.186
kgK
vk= vx =v’′’ + xk. (v”′′ −” v′’)= 0. 0012163 + 0. 426 ⋅ (0. 06665 − 0.0012163) =0.0291
v’′ = 0.0012163
mk =
m3
,
kg
v′′”=0.06665
m3
kg
m3
kg
V
0. 5
=
=17.18 kg
vk 0.0291
uk= ux =u’′ + xk.(u′′” − u′’)=...= 1004.7 + 0.426 ⋅ (2604 − 1004. 7) =1686
u’′ = 1004.7
kJ
,
kg
dipl.ing. @eqko Ciganovi}
u”′′ =2604
kJ
kg
kJ
kg
[email protected]
zbirka zadataka iz termodinamike
strana 52
b)
m3
,
kg
ta~ka 1=kraj
p1 =30 bar,
v1 =0.0291
ta~ka 2:
p2 =50 bar,
v2 = v1 =0.0291
v′ = 0.0012857
m3
,
kg
v′ < v2 < v′′
x2 =
u2 = 1686
kJ
kg
m3
kg
m3
kg
(ta~ka 2 je vla`na para)
v′′=0.03944
v2 − v' 0.0291 − 0. 0012857
=0.729
v"−v ' 0. 03944 − 0.0012857
u2 = ux =u’′ + x2 .(u′′” − u′’)=...= 1148 + 0. 729 ⋅ (2597 − 1148 ) =2204.32
kJ
u’′ = 1148
,
kg
kJ
kg
kJ
u”′′ =2597
kg
prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu
Q12 = ∆U12 + W12
⇒
Q12 = mk . (u2 –- u1 )
Q12 = 17. 18 ⋅ (2204. 32 − 1686) =8904.74 kW
τ=
Q12
⋅
=
Q12
8904.74
=11131 s
0.8
≅ 3h
T
0 = po~etak
1 = kraj
0
2
1
s
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 53
2.41. Toplotno izolovan rezervoar zapremine V=20 m3 , sadr`i vodenu paru po~etnog
stawa P(p=2 MPa, T=553 K). Rezervoar je povezan sa toplotno izolovanom parnom
turbinom, u kojoj se odvija ravnote`no (kvazistati~ko) {irewe pare (slika). Pritisak
pare na izlazu iz turbine je stalan i iznosi piz=0.15 MPa, a proces se odvija dok
pritisak pare u rezervoaru ne opadne na pk=0.3 MPa. Zanemaruju}i prigu{ewe paare u
ventilu, odrediti koji izvr{i para tokom ovog procesa.
WT
piz
p=20 bar,
po~etak:
t=280oC
5
−3
up=hp –− pp. vp = ...= 2972 − 20 ⋅ 10 ⋅ 10
kJ
,
kg
hp= hpp = 2972
mp =
p=3 bar,
kJ
kgK
s′ < sk < s′′
sp= spp= 6.674
kJ
kgK
s k − s'
s' ' −s'
kJ
kgK
s”′ ′ = 6.992
=...=
vk = vx = 0.5694
sk= sp= 6.674
kJ
kgK
(vla`na para)
s’′ = 1.672
mk =
m3
,
kg
V
20
=
= 166.67 kg
vp 0. 12
kraj:
xk =
vp= vpp = 0.12
(pregrejana para)
kJ
⋅ 0.12 =2732
kg
6. 674 − 1. 672
=0.9402
6.992 − 1. 672
m3
kJ
, uk = ux = 2424.09
kg
kg
V
20
=
= 35.12 kg
vk 0.5694
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
strana 54
p=1.5 bar,
izlaz:
s’′ = 1.4336
kJ
,
kgK
siz = sp= 6.674
kJ
kgK
(vla`na para)
s”′ ′ = 7.223
s′ < siz < s′′
xiz =
kJ
kgK
siz − s' 6. 674 − 1.4336
=
=0.905
s' ' −s'
7. 223 − 1.4336
hiz= h′ + xiz . (h′′ − h′) =...= 467. 2 + 0.905 ⋅ (2693 − 467.2 ) =2481.55
h’′ = 467.2
kJ
,
kgK
h”′′ = 2693
kJ
kg
kJ
kgK
miz= mp −– mk = 166.67 − 35.12 = 131.55 kg
prvi zakon termodinamike za slu~aj pra`wewa suda:
Q 12 − W12 = U k − U p + Hiz − H ul
⇒
W12 = −mk ⋅ uk + mp ⋅ up − miz ⋅ hiz
W12 = −35. 12 ⋅ 2424. 09 + 166. 67 ⋅ 2732 − 131. 55 ⋅ 2481. 55 =43.76 MJ
2.42. U ispariva~u zapremine V=2 m3 , u kome se odvija proces isparavawa vode na
⋅
pritisku p=1 MPa, kontinualno se uvodi mul =10 kg/s kqu~ale vode pritiska p=1
MPa, a iz wega izvodi nastala suva para istog pritiska. Greja~ima, urowenim u
⋅
kqu~alu vodu u ispariva~u, vodi se predaje Q12 = 19.26 MW toplote. Ako se u po~etnom
trenutku u ispariva~u na pritisku p=1 MPa nalazila me{avina kqu~ale vode i suve
pare u stawu termodinami~ke ravnote`e, a kqu~ala voda pri tom zauzimala 1/10
zapremine ispariva~a, izra~unati vreme potrebno da kqu~ala voda ispuni ceo
ispariva~. Zanemariti razmenu toplote sa okolinom.
suva para
vla`na para
kqu~ala voda
+Q12
dipl.ing. @eqko Ciganovi}
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zbirka zadataka iz termodinamike
p=10 bar,
po~etak:
strana 55
x=?
m3
m3
, v′′=0.1946
kg
kg
V'
V' '
0.2
1.8
m' =
=…
=177.42 kg,
m' ' =
=
= 9.25 kg
v' 0 .0011273
v' ' 0 .1946
v′ = 0.0011273
mp= m′ + m′′ =177.42 + 9.25 =186.67 kg
xp =
m' '
9 .25
=
=0.0496
m'+m' ' 177.42 + 9.25
up= ux =u’′ + xp.(u′′” − u′’)=...= 761.6 + 0.0496 ⋅ (2583 − 761.6 ) =851.94
u’′ = 761.6
kJ
,
kg
u”′′ =2583
p = 10 bar,
kraj:
vk = v’′ = 0.0011273
mk =
3
m
,
kg
kJ
kg
kJ
kg
x=0
uk = u′’ = 761.6
kJ
kg
V
2
=
= 1774.15 kg
vk 0.0011273
p = 10 bar,
x=0
⋅
kJ
hul = h′’ =762.7
,
mul = mul ⋅ τ
kg
ulaz:
p = 10 bar,
kJ
hiz = h′′” = 2778
kg
izlaz:
x=1
prvi zakon termodinamike za slu~aj istovremenog puwewa i pra`wewa suda:
Q 12 − W12 = U k − U p + H iz − H ul
⋅
Q12 = mk ⋅ uk − mp ⋅ up + miz ⋅ hiz − mul ⋅ hul
⋅
Q 12 ⋅ τ = mk ⋅ uk − m p ⋅ up + m iz ⋅ hiz − mul ⋅ t ⋅ hul
zakon o odr`awu mase:
⋅
mp + mul ⋅ τ = mk + miz
dipl.ing. @eqko Ciganovi}
(1)
mp + mul= mk + miz
(2)
[email protected]
zbirka zadataka iz termodinamike
strana 56
kombinovawem jedna~ina (1) i (2) dobija se
⋅
⋅
⋅
Q 12 ⋅ τ = mk ⋅ uk − mp ⋅ up + mp + m ul ⋅ τ − mk ⋅ h iz − m ul ⋅ t ⋅ hul
mk ⋅ uk − mp ⋅ up + m p − m k ⋅ hiz
τ=
⋅
⋅
⋅
Q 12 − m ul ⋅ h iz + m ul ⋅ hul
(
τ=
)
1774.15 ⋅ 761 .6 − 186 .67 ⋅ 851 .94 + (186 .67 − 1774 .15 ) ⋅ 2778
19 .26 ⋅ 10 3 − 10 ⋅ 2778 + 10 ⋅ 762 .7
zadatak za ve`bawe:
=3603 s
(2.43.)
2.43. Kondenzacija pare vr{i se u prostoru zapremine V=2 m3 pri pritisku od 0.1
MPa. U posudu se kontinualno uvodi 100 kg/h suvozasi}ene vodene pare, a iz we
izvodi nastala kqu~ala voda istog pritiska piz=0.1 MPa. Ako se u po~etnom
trenutku u posudi na pritisku pp=0.1 MPa nalazila kqu~ala voda i suvozasi}ena
para u stawu termodinami~ke ravnote`e, pri ~emu je te~nost zauzimala 1/8
zapremine suda, odrediti vreme potrebno da te~nost ispuni 1/2 zapremine posude.
Toplotna snaga koja se razmewuje sa hladwakom iznosi 250 kW. Zanemariti predaju
toplote okolini.
− Q12
suva para
vla`na para
kqu~ala voda
re{ewe:
τ=5.43 s
dipl.ing. @eqko Ciganovi}
[email protected]
zbirka zadataka iz termodinamike
strana 1
3. MAKSIMALAN RAD, EKSERGIJA
4/2/!U zatvorenom rezervoaru nalazi se n>21!lh vazduha (idealan gas) stawa 2)q>2/7!cbs-!U>634!L*.
Stawe okoline odre|eno je sa P)q>2!cbs-!U>3:9!L*/ Odrediti koliko se najvi{e zapreminskog rada mo`e
dobiti dovo|ewem vazduha stawa 1 u ravnote`u sa okolinom stawa O (maksimalan rad, eksergija
zatvorenog termodinami~kog sistema). Dobijeni rad predstaviti na qw dijagramu.
w2 =
S h ⋅ U2
q2
=
398 ⋅ 634
2/7 ⋅ 216
=1/:492!
n4
lh
wp =
S h ⋅ Up
qp
=
398 ⋅ 3:9
2⋅ 216
=1/9664!
n4
lh
Xnby!>! n ⋅ [− ∆v2p + Up ⋅ ∆t2p − q p ⋅ ∆w 2p ]
Xnby!>! n ⋅ − d w ⋅ (Up − U2 ) + Up
U
q
⋅ d q mo p − S h mo p
U
q2
2
− q p ⋅ (w p − w 2 )
3:9
2
3
− 1/398 ⋅ mo
Xnby!>! 21 ⋅ − 1/83 ⋅ (3:9 − 634) + 3:4 ⋅ 2⋅ mo
− 2⋅ 21 ⋅ (1/9664 − 1/:492)
634
2/7
Xnby!>!2731!−!2364!,!94!>!561!lK
q
q
2
P
2
P
B
⊕
−
,
w
q
B
⊕
w
q
2
=
P
B
2
P
B
,
w
dipl.ing. @eqko Ciganovi}
w
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 2
4/3/ Termodinami~ki sistem se sastoji od zatvorenog suda u kojem se nalazi kiseonik (idealan gas) stawa
2)q>2!cbs-!u>511pD-!n>2!lh* i okoline stawa P)q>2!cbs-!u>31pD*/ Zapreminski udeo kiseonika u
okolnom vazduhu (idealan gas) iznosi sP3 >1/32. Odrediti:
a) da li se navedeni termodinami~ki sistem mo`e upotrebiti za dobijawe X>261!lK!sbeb
b) koliko bi trebalo da iznosi pritisak u sudu (q2*- uz ostale nepromewene uslove, da bi od sistema
mogli dobiti X>261!lK rada povratnim promenama stawa
a)
w2 =
S h ⋅ U2
(q p )P3
wp =
=
371 ⋅ 784
6
=2/86!
n4
lh
2 ⋅ 21
= sP3 ⋅ q p > 1/32 ⋅ 2 !>!1/32!cbs
q2
S h ⋅ Up
(q p )P3
=
371 ⋅ 3:4
1/32 ⋅ 21 6
=4/74!
n4
lh
[
Xnby!>! n ⋅ − ∆v2p + Up ⋅ ∆t2p − (q p )P ⋅ ∆w 2p
3
Xnby!>! n ⋅ − d w ⋅ (Up − U2 ) + Up
]
(q p )P3
U
⋅ d q mo p − S h mo
U2
q2
− (q p ) ⋅ (w p − w 2 )
P
3
3:4
1/32
3
Xnby!>! 2⋅ − 1/76 ⋅ (3:4 − 784) + 3:4 ⋅ 1/:2⋅ mo
− 1/37 ⋅ mo
− 1/32⋅ 21 ⋅ (2/86 − 4/74 )
784
2
Xnby!>!358!−!214/3!−!4:/6!>!215/4!lK
X!?!Xnby
⇒!
sistem se ne mo`e upotrebiti za dobijawe 261!lK rada,
jer najve}i mogu}i rad koji mo`emo dobiti (Eksergija
zatvorenog termodinami~kog sistema) iznosi 215/4!lK
b)
za povratne promene stawa va`i: X>Xnby!>261!lK
2
U
Xnby
2
⋅ d mo p −
+ d w ⋅ (Up − U2 ) + qp ⋅ (w p − w2 ) ⋅
q2 = (qp )P ⋅ fyq −
3
Sh q U2 n
Up
2
3:4 261
2
q2 = 1/32⋅ fyq −
⋅ 1/:2⋅ mo
−
+ 1/76 ⋅ (3:4 − 784) + 1/32⋅ 213 ⋅ (4/74 − 2/86) ⋅
784 2
3:4
1/37
q2!>2/92!cbs
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 3
3.3. U toplotno izolovanom rezervoaru zapremine W>31!n4 nalazi se vazduh (idealan gas) po~etnog
stawa 2)q>31!cbs-!u>381pD*. Rezervoar je povezan sa gasnom turbinom (slika) u kojoj se vazduh {iri
kvazistati~ki adijabatski. Pritisak na izlazu iz turbine je stalan i iznosi 4!cbs. Proces traje sve
dok pritisak u rezervoaru ne opadne na 9!cbs.
a) odrediti radnu sposobnost vazduha u rezervoaru (maksimalan rad) pre otvarawa ventila i
predstaviti je grafi~ki u qw i Ut koordinatnim sistemima ako je stawe okoline P)q>2!cbsu>31pD*
b) odrediti mehani~ki rad izvr{en u toku procesa (pri tome zanemariti proces prig{ivawa u
ventilu)
X23
!qj{mb{
a)
n2 =
w2 =
q2 ⋅ W
31 ⋅ 21 6 ⋅ 31
>
>367/78!lh
398 ⋅ 654
S h ⋅ U2
S h U2
q2
=
398 ⋅ 654
31 ⋅ 21 6
>!1/189!
n4
-!
lh
[
wP =
Xnby!>! n ⋅ − ∆v2p + Up ⋅ ∆t2p − (q p )P ⋅ ∆w 2p
3
S h UP
qP
=
398 ⋅ 3:4
2 ⋅ 21 6
>!1/952!
n4
lh
]
Up
q
− Sh mo p * − qp )w2 − w p *
Xnby = n2 ⋅ dw )U2 − Up * + Up )dq mo
U
q
2
2
Xnby!>! 367/78− 1/83 ⋅ (3:9 − 654) + 3:4 ⋅ 2 ⋅ mo
3:9
2
− 1/398 ⋅ mo − 2 ⋅ 213 ⋅ (1/952 − 1/189)
654
31
Xnby!>!56/38!!,!2:/64!!−2:/69!>!56/33!NK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 4
q
U
2
2
B
B
C
P
C
P
w
t
b)
Uj{mb{ = Uqpdfubl
q
⋅ j{mb{
q qpdfubl
κ −2
κ
4
= 654 ⋅
31
κ −2
2/5 −2
2/5
= 427 L
2/5 −2
q lsbk κ
9 2/5
Ulsbk = Uqpdfubl ⋅
= 654 ⋅
= 529 L
q qpdfubl
31
q lsbk ⋅ W 9 ⋅ 21 6 ⋅ 31
>244/48!lh,
nj{mb{>nqp•fubl!−!nlsbk>234/4!lh
nlsbk =
=
S h ⋅ Ulsbk
398 ⋅ 529
prvi zakon termodinamike za proces pra`wewa:
R 23 − X23 = Vlsbk − Vqp•fubl + Ij{mb{ − Ivmb{
X23 = −nlsbk ⋅ d w ⋅ Ulsbk + nqp•fubl ⋅ d w ⋅ Uqp•fubl − nj{mb{ ⋅ d q ⋅ Uj{mb{
X23 = −244/48 ⋅ 1/83 ⋅ 529 + 367/78 ⋅ 1/83 ⋅ 654 − 234/4 ⋅ 2 ⋅ 427 >32/36!NK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 5
4/5/!Klipni kompresor ravnote`no (kvazistati~ki) i politropski, sa eksponentom politrope o>2/4,
sabija okolni vazduh (idealan gas) stawa 1)q>2!cbs-!U>3:2!L*-!na pritisak q3>5!cbs, i puni toplotno
izolovan rezervoar zapremine W>21!n4. Toplotno stawe vazduha u rezervoaru na po~etku procesa puwewa
isto je kao i stawe okolnog vazduha 1/ Odrediti:
a) masu vazduha koju je potrebno ubaciti u rezervoar da bi pritisak vazduha u rezervoaru dostigao
vrednost od 4!cbsb
b) eksergiju vazduha (maksimalan rad) u rezervoaru u tom trenutku
2!⇒!4
2
3
X23
a)
U2 q2
=
U3 q 3
nqp•fubl =
o−2
o
q
U3 = U2 ⋅ 3
q2
⇒
q qp•fubl ⋅ W
S h ⋅ Uqp•fubl
>
o−2
o
5
> 3:2 ⋅
2
2/4 −2
2/4
>511/82!L
2 ⋅ 21 6 ⋅ 21
>22/:8!lh
398 ⋅ 3:2
prvi zakon termodiamike za proces puwewa rezervoara:
1= po~etak,
2 = ulaz,
3=kraj
R 23 − X23 = Vlsbk − Vqp•fubl + Ij{mb{ − Ivmb{
1 = nlsbk ⋅ d w ⋅ Ulsbk − nqp•fubl ⋅ d w ⋅ Uqp•fubl − nvmb{ ⋅ d q ⋅ Uvmb{
)2*
zakon odr`awa mase za proces puwewa rezervoara:
nqp•fubl + nvmb{ = nlsbk + nj{mb{
dipl.ing. @eqko Ciganovi}
)3*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 6
jedna~ina stawa idealnog gasa za zavr{etak puwewa:
q lsbk ⋅ W = nlsbk ⋅ S h ⋅ Ulsbk
(3)
kombinovawem jedna~ina (1) i (2) dobija se:
(
)
1 = nlsbk ⋅ d w ⋅ Ulsbk − nqp•fubl ⋅ d w ⋅ Uqp•fubl − nlsbk − nqp•fubl ⋅ d q ⋅ Uvmb{
)5*
kombinovawem jedna~ina )4*!i!)5* dobija se:
q lsbk ⋅ W
1 = nlsbk ⋅ d w ⋅
− nqp•fubl ⋅ d w ⋅ Uqp•fubl − nlsbk − nqp•fubl ⋅ d q ⋅ Uvmb{
S h ⋅ nlsbk
(
dw ⋅
q lsbk ⋅ W
nlsbk = nqp•fubl +
Sh
)
− nqp•fubl ⋅ d w ⋅ Uqp•fubl
d q ⋅ Uvmb{
1/83 ⋅
nlsbk = 22/:8 +
4 ⋅ 21 ⋅ 21
− 22/:8 ⋅ 1/83 ⋅ 3:2
398
>35/5:!lh
2 ⋅ 511/82
6
nvmb{>nlsbk!−!nqp•fubl!>!35/5:!−!22/:8!>23/63!lh
napomena:
Ulsbk>
q lsbk ⋅ W
S h ⋅ nlsbk
>
4 ⋅ 21 6 ⋅ 21
>537/94!L
398 ⋅ 35/5:
b)
okolina (ta~ka O) = ta~ka 1
w4 =
S h ⋅ U4
q2
=
398 ⋅ 537/94
4 ⋅ 21 6
kraj (ta~ka kraj) = ta~ka 3
=1/519!
S h ⋅ Up 398 ⋅ 3:2
n4
n4
-!!!!!!! w p =
=
=1/946!
lh
lh
qp
2 ⋅ 21 6
Xnby!>! nlsbk ⋅ [− ∆v 4p + Up ⋅ ∆t 4p − q p ⋅ ∆w 4p ]
Xnby!>! n ⋅ − d w ⋅ (Up − U4 ) + Up
U
q
⋅ d q mo p − S h mo p
U4
q4
Xnby!> 35/5: ⋅ − 1/83 ⋅ (3:2 − 537/94) + 3:2⋅ 2 ⋅ mo
− q p ⋅ (w p − w 4 )
3:2
2
− 1/398 ⋅ mo − 2 ⋅ 213 ⋅ (1/946 − 1/519)
537/94
4
Xnby!>!34:6!−!597!−53/8!>!2977/4!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 7
4/6/!Odrediti eksergiju struje vazduha (idealan gas) stawa 2)q2>2/7!cbs-!u2>361pD⋅
n >2!lh0t* i predstaviti je grafi~ki na qw dijagramu. Pod okolinom smatrati vazduh (idealan gas) stawa
P)qp>2!cbs-!up>36pD*/
⋅
⋅
Fy 2 = n⋅ (− ∆i2p + Up ⋅ ∆t2p ) >!///
⋅
⋅
q
U
Fy2 = n⋅ − dq )Up − U2* + Up )dq mo p − S h mo p *
q2
U2
⋅
⋅
3:9
2
Fy2 = 2⋅ − 2 ⋅ )3:9 − 634* + 3:9 ⋅ )2 ⋅ mo
− 1/398 ⋅ mo
*
634
2/7
⋅
Fy2 >!336!!−!238/5!>!:8/7!lX
q
q
2
P
2
,
=
P
⊕
−
w
w
q
2
,
P
w
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 8
⋅
4/7/ Eksergija toka vazduha (idealan gas), masenog protoka n >1/6!lh0t, koji struji sredwom brzinom
⋅
x>39!n0t, pri stawu vazduha u okolini P)qp>2!cbs-!Up>3:4!L), iznosi Fy 2 >94!lX. Promena
entropije okoline, koja bi nastala povratnim (reverzibilnim) promenama stawa vazduha (bez promene
⋅
brzine) na pritisak i temperaturu okoline iznosila bi ∆ T plpmjof>−!1/2!lX0L.
a) odredti pritisak i temperaturu vazduha stawa 1
b) grafi~ki prikazati u qw, koordinatnom sistemu eksergiju vazdu{nog toka, ne uzimaju}i u obzir
deo koji se odnosi na brzinu
a)
drugi zakon termodinamike za proces od stawa 1 do stawa O
⋅
⋅
⋅
∆ T tjtufn = ∆ T sbeop ` ufmp + ∆ T plpmjob
⋅
∆ T2p
⋅
⋅
lX
= − ∆ T plpmjob >1/2!
L
⋅
⇒
∆ t2p =
⋅
1 = ∆ T2p + ∆ T plpmjob
⋅
∆ T2p
⋅
>1/3!
n
lK
lhL
⋅
⋅
⋅
x3
Fy 2 = n⋅ (− ∆i2p + Up ⋅ ∆t2p + f l2 ) = n⋅ − dq )Up − U2* + Up ⋅ ∆t2p +
3
⋅
Fy 2
⋅
U2 = UP + n
− UP ∆t2P −
dq
⋅
⋅
U
∆ T 21!>! n⋅ )dq mo p
U2
x3
3
94
39 3
− 3:4 ⋅ 1/3 −
⋅ 21 −4
1
/
6
3
>511!L
> 3:4 +
2
⋅
Up ∆ T2p
dq mo U − ⋅
qp
2
n >6/:7!cbs
* !!⇒!! q2 = q p ⋅ fyq
− S h mo
Sh
q2
q
2
Fy2
!P
w
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 9
4/8/ U horizontalnoj cevi pre~nika e>311!nn ugra|en je greja~ stalne temperature UUJ>711!L. Stawe
vazduha u preseku 1 odre|eno je sa 2)q>3!cbs-!U>411!L-!x>31!n0t* a u preseku 2 sa 3)q>3!cbs-!U>511!L*.
Stawe okoline odre|eno je veli~iama stawa P)qp>2!cbs-!Up>3:1!L*. Cev je toplotno izolovana od
okoline. Odrediti:
a) snagu ugra|enog greja~a
b) eksergiju vazduha u preseku 1 i preseku 2
c) eksergijski stepen korisnosti procesa u cevi
⋅
+ R23
2
3
a)
w2 =
w3 =
S h ⋅ U2
q2
S h ⋅ U3
q3
>
398 ⋅ 411
3 ⋅ 21 6
>1/5416!
398 ⋅ 511
>
n4
lh
>1/685!
3 ⋅ 21 6
n4
lh
⋅
⋅
jedna~ina kontinuiteta:
n2 = n3
x2 ⋅ B2 x 3 ⋅ B 3
=
w2
w3
x 3 = x2 ⋅
⋅
n=
⇒
w3
1/685
n
> 31 ⋅
>!37/78!
1/5416
t
w2
e3 π
1/3 3 ⋅ π
31 ⋅
lh
5 =
5
>2/57!
t
w2
1/5416
x2 ⋅
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu:
⋅
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23 > n⋅ d q ⋅ (U3 − U2 ) + n⋅
⋅
R 23 = 2/57 ⋅ 2 ⋅ (511 − 411) + 2/57 ⋅
dipl.ing. @eqko Ciganovi}
31 3 − 37/78 3
3
x 23 − x 33
>
3
⋅ 21 −4 >256/88!lX
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 10
b)
⋅
⋅
Fy 2 = n⋅ (− ∆i2p + Up ⋅ ∆t2p + f l2 ) =
⋅
⋅
Fy2 = n⋅ − dq )Up − U2* + Up
⋅
U
q
⋅ d q mo p − S h mo p
U
q2
2
Fy2 > 2/57 ⋅ − 2⋅ )3:1 − 411* + 3:1 ⋅ 2⋅ mo
⋅
x 23
+
3
3:1
2 313
− 1/398 ⋅ mo +
⋅ 21−4 >95/88!lX
411
3
3
⋅
Fy 3 = n⋅ (− ∆i 3p + Up ⋅ ∆t 3p + f l3 ) =
⋅
⋅
Fy 3 = n⋅ − dq )Up − U3 * + Up
⋅
U
q
⋅ d q mo p − S h mo p
U3
q3
Fy3 > 2/57 ⋅ − 2⋅ )3:1 − 511* + 3:1 ⋅ 2⋅ mo
c)
⋅
⋅
Fy R = R 23 ⋅
x 33
+
3
3:1
2 313
− 1/398 ⋅ mo +
⋅ 21−4 >219/:8!lX
511
3
3
UUJ − Up
711 − 3:1
> 256/88 ⋅
>86/42!lX
711
UUJ
drugi zakon termodinamike za proces u otvorenom termodinami~kom sistemu:
⋅
⋅
⋅
∆ T tjtufn = ∆ Tsbeop ` ufmp + ∆ T UJ =.. .
⋅
⋅
⋅
q
U
∆ T sbeop ` ufmp = ∆ T23 > n⋅ d q mo 3 − S h mo 3
U2
q2
⋅
∆ T UJ
> 2/57 ⋅ 2 ⋅ mo 511 >1/53! lX
411
L
⋅
256/88
R 23
lX
>−1/35!
=−
>−
711
L
UUJ
⋅
∆ T tjtufn = 1/53 − 1/35 >!1/29!
⋅
lX
L
⋅
Fy h = Up ⋅ ∆ T tjtufn > 3:1 ⋅ 1/29 >63/3!lX
⋅
ηFy =
⋅
⋅
Fy2 + Fy R − Fy h
⋅
⋅
Fy 2 + Fy R
dipl.ing. @eqko Ciganovi}
>
95/88 + 86/42 − 63/3
>1/78
95/88 + 86/42
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 11
4/9. U neizolovanoj komori me{aju se, pri stacionarnim uslovima, dve struje idealnih gasova: kiseonika
⋅
⋅
B) n B>7!lh0t-!qB>1/29!NQb-!UB>634!L*!i azota!C) n C>4!lh0t-!qC>1/44!NQb-!UC>974!L*. U toku procesa
me{awa toplotni protok u okolni vazduh stawa )qp>1/2!NQb-!Up>3:4!L-!sP3>1/32-!sO3>1/8:* iznosi 511
lX. Pritisak me{avine na izlazu iz komore je q>!1/26!NQb. Odrediti brzinu gubitka eksergije u toku
procesa me{awa kao i eksergijski stepen korisnosti procesa u me{noj komori. Zanemariti promene
makroskopske potencijalne i kineti~ke energije.
prvi zadatak termodinamike za proces u otvorenom termodinami~kom sistemu:
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X U23
⋅
⋅
R 23 = I3 − I2
⋅
⋅
⋅
⋅
⋅
R23 = nB ⋅ dqB + nC ⋅ dqC ⋅ U+ − nB ⋅ dqB ⋅ UB − nC ⋅ dqC ⋅ UC
⋅
+
U =
⋅
⋅
R23 + n B ⋅ d qB ⋅ UB + nC ⋅ d qC ⋅ UC
⋅
⋅
−511 + 7 ⋅ 1/:2 ⋅ 634 + 4 ⋅ 2/15 ⋅ 974
>711!L
7 ⋅ 1/:2 + 4 ⋅ 2/15
>
n B ⋅ d qB + nC ⋅ d qC
jedna~ina stawa me{avine idealnih gasova na izlazu iz me{ne komore:
⋅
⋅
nB ⋅ ShB + nC ⋅ ShC ⋅ U +
⋅
⋅
W+ =
q+ ⋅ W+ = nB ⋅ ShB + nC ⋅ ShC ⋅ U+
⇒
+
q
(
)
7
⋅
371
+
4
⋅
3:8
⋅
711
n4
W+ =
>!:/915!
t
2/6 ⋅ 216
jedna~ina stawa idealnog gasa za kiseonik )B* na izlazu iz me{ne komore:
⋅
⋅
q+B ⋅ W+ = nB ⋅ ShB ⋅ U +
⇒
q+B =
nB ⋅ ShB ⋅ U +
W
+
>
7 ⋅ 371 ⋅ 711
> 1/:6 ⋅ 216 Qb
:/915
jedna~ina stawa idealnog gasa za azot )C* na izlazu iz me{ne komore:
qC+
+
⋅
⋅ W = nC ⋅ ShC ⋅ U
⋅
+
⇒
qC+
=
nC ⋅ ShC ⋅ U +
W
+
>
4 ⋅ 398 ⋅ 711
:/915 ⋅ 21−3
> 1/66 ⋅ 21 6 Qb
pritisak kiseonika )B* u okolnom vazduhu:
6
6
qP
B = sP3 ⋅ qp > 1/32 ⋅ 2 ⋅ 21 > 1/32 ⋅ 21 Qb
pritisak azota )C* u okolnom vazduhu:
6
6
qP
B = sP3 ⋅ qp > 1/8: ⋅ 2 ⋅ 21 > 1/8: ⋅ 21 Qb
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 12
drugi zakon termodinamike za proces me{awa gasova B!i!C :
⋅
⋅
⋅
∆ T tjtufn = ∆ T sbeop ` ufmp + ∆ T plpmjob =.. .
⋅
⋅
⋅
∆ Tsbeop ` ufmp = ∆ T B + ∆ TC >!///!>!2/86!,!1/57!>!3/32!
lX
L
⋅
⋅
q+
711
1/:6
lX
U+
− 1/37 ⋅ mo
∆ T B = n B ⋅ d qB mo
− S hB mo B > 7 ⋅ 1/:2 ⋅ mo
>2/86
U
q
L
634
2
/
9
B
B
⋅
⋅
q+
711
1/66
U+
lX
∆ TC = nC ⋅ d qC mo
− S hC mo C > 4 ⋅ 2/15 ⋅ mo
− 1/3:8 ⋅ mo
>1/57
L
UC
974
4/4
qC
⋅
⋅
∆ T plpmjob
−511
R 23
lX
=−
>−
>2/48!
3:4
L
Up
⋅
∆ T tjtufn = 3/32 + 2/48 >!4/69!
⋅
lX
L
⋅
Fy h = Up ⋅ ∆ T tjtufn > 3:4 ⋅ 4/69 >215:!lX
⋅
⋅
⋅
Fy 2 = Fy B + Fy C = ...
⋅
⋅
Fy B = n B ⋅ − d qB ⋅ (Up − UB ) + Up
U
qp
⋅ d qB ⋅ mo p − S hB ⋅ mo B
UB
qB
⋅
3:4
1/32
Fy B = 7 ⋅ − 1/:2 ⋅ (3:4 − 634) + 3:4 ⋅ 1/:2 ⋅ mo
− 1/37 ⋅ mo
>2422!lX
634
2/9
⋅
⋅
Fy C = nC ⋅ − d qC ⋅ (Up − UC ) + Up
U
qp
⋅ d qC ⋅ mo p − S hC ⋅ mo C
UC
qC
⋅
3:4
1/8:
Fy C = 4 ⋅ − 2/15 ⋅ (3:4 − 974) + 3:4 ⋅ 2/15 ⋅ mo
− 1/3:8 ⋅ mo
>2275!lX
974
4/4
⋅
Fy2 = 2422 + 2275 >!3586!lX
⋅
ηFy =
⋅
Fy2 − Fy h
⋅
Fy2
>
3586 − 215:
>1/68
3586
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 13
4/:. U suprotnosmernom razmewiva~u toplote pri qw>3!cbs, biva izobarski zagrevan tok vazduha (idealan
gas), od temperature Uw2>524!L do temperature Uw3>654!L, a tok vrelih gasova (sme{a idealnih gasova)
biva hla|ena od polaznog stawa H2)qh2>2/6!cbs-!Uh2>724!L* do stawa H3)qh3>2/4!cbs-!Uh3>@*/ Ako su
maseni protoci vazduha i vrelih gasova isti, a vreli gasovi imaju iste termofizi~ke osobine kao i vazduh
odrediti eksergijski stepen korisnosti procesa u ovom razmewiva~u toplote pri uslovima okoline
P)qp>2!cbs-!Up>3:4!L*/ Zanemariti promene makroskopske potencijalne i kineti~ke energije kao i
prisustvo hemijske neravnote`e.
Uh2
Uh3
Uw2
Uw3
prvi zadatak termodinamike za proces u otvorenom termodinami~kom sistemu:
⋅
⋅
⋅
⋅
⋅
⋅
⇒
R 23 = ∆ I23 + X U23
⋅
⋅
I2 = I3
⋅
n⋅ d q ⋅ Uw2 + n⋅ d q ⋅ Uh2 = n⋅ d q ⋅ Uw3 + n⋅ d q ⋅ Uh3
⇒
Uh3 = Uw2 + Uh2 − Uw3 > 524 + 724 − 654 >594!L
⋅
⋅
⋅
Fy vmb{ = Fy w2 + Fy h2 = ...
⋅
⋅
U
q
Fy w2 = n⋅ − d q (Up − Uw2 ) + Up d q mo p − S h mo p
U
q
w
w2
⋅
⋅
⋅
3:4
2
Fy w2 = n⋅ − 2 ⋅ (3:4 − 524) + 3:42 ⋅ mo
− 1/398 ⋅ mo = n⋅ 88/82 !lX
524
3
⋅
⋅
U
q
Fy h2 = n⋅ − d q Up − Uh2 + Up d q mo p − S h mo p
Uh2
q h2
(
)
⋅
⋅
⋅
3:4
2
Fy h2 = n⋅ − 2 ⋅ (3:4 − 724) + 3:4 ⋅ 2 ⋅ mo
− 1/398 ⋅ mo
> n⋅ 248/92!lX
724
2/6
⋅
⋅
⋅
Fy vmb{ = n⋅ (88/82 + 248/92) > n⋅ !326/63!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
⋅
strana 14
⋅
Fy hvcjubl = Up ⋅ ∆ T tj = ...
⋅
⋅
⋅
⋅
⋅
⋅
∆ Ttj = ∆ Tsu + ∆ Tp =…
∆ Tsu = ∆ T w + ∆ Th = …
⋅
⋅
⋅
⋅
Uw 3
q
654
lX
− S h mo w * = n⋅ 2 ⋅ mo
= n⋅ 1/385
L
524
Uw2
qw
⋅
⋅
Uh3
∆ T w > n /! )d q mo
∆ Th > n /! )d q mo
⋅
Uh2
− S h mo
⋅
⋅
⋅
⋅
594
2/4
lX
− 1/398 ⋅ mo
* = n⋅ 2 ⋅ mo
>− n⋅ 1/2:8
L
724
q h2
2/6
q h3
⋅
∆ Tsu > n⋅ 1/385!− n⋅ 1/2:8> n⋅ 1/188!
⋅
⋅
⋅
⋅
∆ Ttj > ∆ Tsu , ∆ Tp > n⋅ 1/188!
⋅
lX
L
lX
L
⋅
Fy hvcjubl = 3:4 ⋅ n⋅ 1/188 >33/67!lX
⋅
ηFy!>
⋅
Fy vmb{ − Fy h
⋅
Fy vmb{
=
326/63 − 33/67
>1/9:
326/63
⋅
3.10. Klipni kompresor kvazistati~ki politropski sabija n >1/6!lh0t vazduha (idealan gas) od stawa
2)q>211!lQb-!U>399!L*!do stawa 3)q>611!lQb-!U>513!L*/ Stawe okoline zadato je sa P)qp>211!lQbUp>399!lQb*. Odrediti:
a) snagu kompresora
b) eksergijski stepen korisnosti procesa
c) ako bi se kompresor hladio vodom koja bi pri tom mewala stawe pri stalnom pritisku q>211!lQb
od!Ux2>399!L!do!Ux3>414!L
a)
o−2
o
U2 q2
=
U3 q 3
⋅
⋅
X U23 = n⋅
mo
⇒
o>
mo
q2
q3
q2
U
− mo 2
q3
U3
mo
>
mo
211
611
211
399
− mo
611
513
>2/37
2/37
o
⋅ S h ⋅ (U2 − U3 ) > 1/6 ⋅
⋅ 1/398 ⋅ (399 − 513) >−8:/4!lX
o −2
2/37 − 2
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 15
b)
U
q
Fy2 = n⋅ (− ∆i2p + Up ⋅ ∆t2p ) > n⋅ − dq)Up − U2* + Up)dq mo p − Sh mo p * >!1!lX
U
q
⋅
⋅
⋅
2
2
napomena:!q2>qp-!U2>Uq
⋅
⋅
Fy hvcjubl = Up ⋅ ∆ T tj = ...
⋅
⋅
⋅
∆ Ttj = ∆ Tsu + ∆ Tp =…
⋅
⋅
U
q
∆ Tsu = n⋅ d q ⋅ mo 3 − S h ⋅ mo 3
U2
q2
513
611
X
> 1/6 ⋅ 2 ⋅ mo
− 1/398 ⋅ mo
>−75/32!
399
211
L
⋅
−33/2
X
R 23
>87/85!
∆ Tp = −
>///> −
399
L
Up
⋅
⋅
⋅
o−κ
2/37 − 2/5
⋅ (U3 − U2 ) > 1/6 ⋅ 1/83 ⋅
⋅ (513 − 399) >−33/2!lX
o −2
2/37 − 2
⋅
X
∆ Ttj >−75/32!,!87/85!>23/64!
L
! R 23 = n⋅ d w ⋅
⋅
Fyhvcjubl = 399 ⋅ 23/64 >4/7!lX
⋅
⋅
⋅
Fy vmb{ = Fy 2 + X U23 >8:/4!lX
⋅
ηFy!>
⋅
Fyvmb{ − Fyh
⋅
Fyvmb{
=
8:/4 − 4/7
>1/:6
8:/4
c)
⋅
⋅
R 23 = n x ⋅ (i x2 − i x3 ) !!!!⇒
lK
lh
lK
ix3!>!236/8!
lh
ix2!>!73/:6!
dipl.ing. @eqko Ciganovi}
⋅
nx
lh
−33/2
R 23
=
>
>!1/46!
73/:6 − 236/8
t
i x2 − i x3
)voda!q>211!lQb-!U>399!L*
)voda!q>211!lQb-!U>399!L*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 16
4/22/!U vertikalnom cilindru sa grani~nikom (slika) mo`e se
trewa) klip sa tegom. U po~etnom trenutku zapremina
n4 (ograni~ena klipom sa tegom), ispuwena je kqu~alom vodom
makroskopski razvijenom parom u stawu termodinami~ke
pritisku q>4!cbs!(vla`na para). Kqu~ala voda zauzima 1% od
zapremine cilindra. Maksimalna zapremina cilindra ispod
Wnby>3!n4. Odrediti termodinami~ki gubitak rada (gubitak
predaji toplote, od izotermnog toplotnog izvora, temperature
radnoj materiji u cilindru, ako je wena temperatura na kraju
zagrevawa 684!L. Temperatura okoline iznosi Up>411!L.
proces zagrevawa na Ut dijagramu. [rafirati na Ut dijagramu
predstavqa gubitak eksergije.
kretati (bez
cilindra W>1/7
i wenom
ravnote`e na
po~etne
klipa iznosi
eksergije) pri
UUJ!>734!L,
procesa
Predstaviti
povr{inu koja
ta~ka 1:
q2>4!cbs!
y2>@
n#
y2 =
= ///
n#+n(
1/6:5
W#
>///>
>1/:9!lh
n# =
1/7168
w#
W(
1/117
>6/6:!lh
n( =
= /// >
1/1121844
w(
n4
lh
W′!>!1/12/W!>!1/117!n4
W′!>!1/::/W!>!1/6:5!n4
w′>!1/1121844!
w′′>1/7168!
n4
lh
n#
1/:9
=
= 1/26
n#+n( 1/:9 + 6/6:
lK
lK
t2>!3/58!!
i2>!996/:5!!
lh
lhL
napomena:
n!>!n′!,!n′′!>!7/68!lh
y2 =
ta~ka 2:
q3>q2>4!cbs- w3>wnby!>
w′!?!w3!?!w′′
Wnby
n4
= 1/414
n
lh
(ta~ka 2 se nalazi u oblasti vla`ne pare)
1/414 − 1/1121844
w3 − w (
>
>1/6
1/7168 − 1/1121844
w #− w (
lK
lK
- v3!>vy!>2663/3!!
-! !!!u3>!ulmk>244/65pD!
i3!>iy!>!2754/3!!
lh
lh
y3 =
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 17
u>411pD!
ta~ka 3:
w′!>!1/1125147!
w4!?!w′′
w4!>w3>1/414!!
n4
lh
n4
n4
- !!!w′′!>!1/13275!
lh
lh
(ta~ka 3 se nalazi u oblasti pregrejane pare)
i4!>!iqq!>!4161!!
lK
lh
t4!>!tqq!>!8/29!!
lK
lhL
q4!>!qqq!>!9/6!!cbs
vrednosti i4-!t4 i q4 se moraju pro~itati sa it dijagrama za
vodenu paru
lK
v4!>!vqq>!i4!−!q4/!w4!>!4161!−!9/6/216!/!21−3/!1/414!>!38:3/6!
lh
napomena:
Fyhvcjubl!>!Uplpmjob!/!∆Ttjtufn!>!///!>!411!/!:/97!>!3:69!!lK
lK
L
lK
∆ T sbeop!ufmp> n ⋅ ∆t24 !> n ⋅ (t 4 − t2 ) > 7/68 ⋅ (8/29 − 3/58) >!41/:6!
L
)r23 *q=dpotu + )r34 * w =dpotu
R 23 + R 34
i3 − i2 + v 4 − v 3
∆S UJ = −
−n⋅
= −n ⋅
UUJ
UUJ
UUJ
2754/3 − 996/:5 + 38:3/6 − 2663/3
lK
>−32/1:!
∆TUJ!>! − 7/68 ⋅
L
734
∆Ttjtufn>∆Tsbeop!ufmp!,!∆Tupqmpuoj!j{wps!>!///>41/:6!−!32/1:!>!:/97!
U
4
2
3
Up
∆tsu
Fyh
t
∆tUJ
∆tTJ
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 18
4/23/!U razmewiva~u toplote vr{i se atmosfersko )q>2!cbs*, potpuno isparavawe kqu~ale vode i
istovremena potpuna kondenzacija suvozasi}ene vodene pare pri
q>4!cbs. Ukoliko toplotna snaga razmewiva~a toplote (interno razmewena toplota izme|u pare i
vode) iznosi 3/6!lX, izra~unati eksergijski stepen korisnosti procesa u ovom razmewiva~u toplote
pri stawu okoline P)qp>!2!cbs-!up>31pD*/
L2
!2
!3
!4
!5
suva para, q>4!cbs
kqu~ala voda,!q>2!cbs
L3
!U
!3
!2
!5
!4
!t
q>!4!cbs
lK
i2!>!i″!>!3836!
lh
ta~ka 1:
q>!4!cbs
lK
i3!>!i′!>!672/5!
lh
ta~ka 2:
q>!2!cbs
lK
i4!>!i′!>!528/5!
lh
ta~ka 3:
q>!2cbs
lK
i5!>!i″!>!3786!
lh
y>2
t2!>!t″!>!7/::3!
lK
lhL
y>1
t3!>!t′!>!2/783!
lK
lhL
y>1
t4!>!t′!>!2/4137!
lK
lhL
ta~ka 4:
dipl.ing. @eqko Ciganovi}
t5!>!t″!>!8/471!
lK
lhL
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 19
up!>!31pD
qp!>!2!cbs
lK
ip!>!!ix!>!94/:!
lh
ta~ka O:
tp!>!!tx!>!1/3:7!
lK
lhL
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
⋅
⋅
R 23 = ∆ I23 + X U23
ograni~enom konturom 1 (K1):
⋅
⋅
R 23 = nq ⋅ (i3 − i2 )
⇒
⋅
⋅
nq =
−3/6
R 23
lh
>
>2/26!
t
i3 − i2 672/5 − 3836
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
⋅
⋅
R 45 = ∆ I45 + X U245
ograni~enom konturom 2 (K2):
⋅
⋅
R 45 = n x ⋅ (i 5 − i 4 )
⇒
⋅
⋅
nx =
R 45
3/6
lh
>
>2/22!
t
i 5 − i 4 3786 − 528/5
⋅
ηFy =
⋅
Fy vmb{ − Fy hvcjubl
⋅
Fy vmb{
⋅
>
892/16 − 286/9
>1/88
892/16
⋅
Fy hvcjubl = Up ⋅ ∆ T tj >///> 3:4 ⋅ 1/7 >286/9!lX
⋅
⋅
⋅
⋅
⋅
∆ T tjtufn!>!∆ T sbeop!ufmp!,!∆ T plpmjob!>!///>1/7!
⋅
lX
L
∆ T sbeop!ufmp> T j{mb{!−! T vmb{!>!///>!21/1:!−!:/5:>1/7!
⋅
⋅
lX
L
⋅
lX
L
⋅
⋅
⋅
lX
T vmb{!>! nq ⋅ t2 + n x ⋅ t 4 > 2/26 ⋅ 7/::3 + 2/22 ⋅ 2/4137 >:/5:!
L
T j{mb{!>! nq ⋅ t 3 + n x ⋅ t 5 > 2/26 ⋅ 2/783 + 2/22 ⋅ 8/47 >21/1:!
⋅
⋅
⋅
Fy vmb{!>! Fy 2!, Fy 4!>!///>!892/16!,!53/92!>!934/97!lX
⋅
⋅
⋅
Fy 2> nq ⋅ (− ∆i2p + Up ⋅ ∆t2p ) >! nq ⋅ [i2 − i p + Up ⋅ (t p − t2 )]
⋅
Fy 2> 2/26 ⋅ [3836 − 94/: + 3:4 ⋅ (1/3:7 − 7/::3)] >892/16!lX
⋅
⋅
⋅
Fy 4!> n x ⋅ (− ∆i 4p + Up ⋅ ∆t 4p ) >! n x ⋅ [i 4 − i p + Up ⋅ (t p − t 4 )]
⋅
Fy 4> 2/22 ⋅ [528/5 − 94/: + 3:4 ⋅ (1/3:7 − 2/4137)] >53/92!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 20
⋅
3.13.!U ure|aj za pripremu kqu~ale vode (slika) uti~e suva para 2)q>23!cbs*, masenog protoka n 2>1/2
⋅
lh0t i voda stawa 3)q>5!cbs-!u>71pD*!masenog protoka n 3>@/ Prolaskom kroz razmewiva~ toplote suva
para se potpuno kondenzuje )stawe 3*. Nastali kondenzat se prigu{uje na pritisak q3(stawe 4), a zatim
izobarski me{a sa vodom (stawe 2). Toplotni gubici razmewiva~a toplote iznose 223!lX. Prestaviti
prosese u pojedina~nim ure|ajima (razmewiva~ toplote, prigu{ni ventil, me{na komora) na zasebnim Ut
dijagramima i odrediti:
⋅
a) maseni protok vode ( n 2) da bi iz ure|aja isticala kqu~ala voda pritiska q3 (stawe 6)
b) temperaturu vode stawa 6!)!u6!*
c) eksergijski stepen korisnosti ure|aja ako se okolina defini{e kao voda stawa P!)qp>2!cbs-!u>31pD*
!U
⋅
2 n2
razmewiva~ toplote
⋅
7
n3
3
6
q>23!cbs
q>5!cbs
4
2
7
4
5
6
!t
!U
!U
prigu{ni ventil
me{na komora
q>23!cbs
q>23!cbs
q>5!cbs
q>5!cbs
4
5
6
5
3
!t
2!−!4
4!−!5
5!,!3!>6!
6!−!7
!t
: promena stawa pare pri proticawu kroz razmewiva~ toplote (RT)
: promena stawa pare pri proticawu kroz prigu{ni ventil
: proces me{awa pare i vode u me{noj komori
: promena stawa me{avine pare i vode pri proticawu kroz RT
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 21
a)
Prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
ograni~enom isprekidanom linijom:
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
R 23 = I3 − I2 = n2 + n3 ⋅ i 7 − n2 ⋅ i2 − n3 ⋅ i3
⋅
⋅
R 23 = ∆ I23 + X U23
⇒
⋅
R 23 + n2 ⋅ (i2 − i 7 )
− 223 + 1/2 ⋅ (3896 − 715/8)
lh
>///>
n3 =
>1/4!
t
715/8 − 362/4
i 7 − i3
⋅
q2>23!cbs
lK
i2!>i′′>3896!
lh
y>2
q3>5!cbs
lK
i3!>ix>362/4!
lh
u3>71pD
ta~ka 1:
t2>t′′!>!7/634!
ta~ka 2:
t3>tx!>!1/94!
lK
lhL
lK
lhL
y>1
q7>5!cbs
lK
lK
t2>t′!>!2/888!
i7!>i′>715/8!
lh
lhL
ta~ka 6:
b)
⋅
⋅
⋅
R 23 = ∆ I23 + X U23
Prvi zakon termodinamike za proces me{awa:
⋅
⋅
⋅
⋅
⋅
⋅
⇒
I2 = I3
n2 ⋅ i 5 + n3 ⋅ i 3 = n2 + n3 ⋅ i 6
⋅
i6 =
⋅
n2 ⋅ i 5 + n3 ⋅ i3
⋅
⋅
>///>
n2 + n3
1/2 ⋅ 8:9/4 + 1/4 ⋅ 362/4
lK
>499/16!
lh
1/2 + 1/4
q7>5!cbs
lK
i5!>i′>8:9/4!
lh
ta~ka 4:
ta~ka 3:
i5>i4!>8:9/4!
ta~ka 5:
q>5!cbs
y>1
lK
lh
i>499/16!
lK
lh
u6!>!ux!>:4pD
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 22
c)
up>31pD
qp>2!cbs
lK
ip!>ix>94/:!
lh
ta~ka O:
tp>tx!>!1/3:7!
⋅
⋅
lK
lhL
⋅
Fy 2> n2 ⋅ (− ∆i2p + Up ⋅ ∆t2p ) >! n2 ⋅ [i2 − i p + Up ⋅ (t p − t2 )]
⋅
Fy 2> 1/2 ⋅ [3896 − 94/: + 3:4 ⋅ (1/3:7 − 7/634)] >98/77!lX
⋅
⋅
⋅
Fy 3> n3 ⋅ (− ∆i3p + Up ⋅ ∆t 3p ) >! n3 ⋅ [i3 − i p + Up ⋅ (t p − t 3 )]
⋅
Fy 3> 1/4 ⋅ [362/4 − 94/: + 3:4 ⋅ (1/3:7 − 1/94 )] >4/39!lX
⋅
⋅
⋅
Fy vmb{!>! Fy 2,! Fy 3!>98/77!,!4/39!>:1/:5!lX
⋅
⋅
Fy hvcjubl = Up ⋅ ∆ T tj >///> 3:4 ⋅ 1/2: >66/78!lX
⋅
⋅
⋅
⋅
⋅
∆ T tjtufn!>!∆ T sbeop!ufmp!,!∆ T plpmjob!>!///>−1/2:!,!1/49!>!1/2:!
lX
L
⋅
lX
L
⋅
⋅
⋅
lX
T j{mb{!>! n2 + n3 ⋅ t 7 > (1/2 + 1/4 ) ⋅ 2/888 >1/82!
L
∆ T sbeop!ufmp> T j{mb{!−! T vmb{!>!///>!1/82!−!1/:>−!1/2:!
⋅
⋅
⋅
T vmb{!>! n2 ⋅ t2 + n3 ⋅ t 3 > 1/2 ⋅ 7/634 + 1/4 ⋅ 1/94 >1/:1!
lX
L
⋅
R
−223
lX
>1/49!
∆ T plpmjob!>!−! 23 >−
L
3:4
Up
⋅
⋅
ηFy =
⋅
Fy vmb{ − Fy hvcjubl
⋅
Fy vmb{
dipl.ing. @eqko Ciganovi}
>
:1/:5 − 66/78
>1/4:
:1/:5
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 23
⋅
4/25/!U parnu turbinu ulazi n >21!lh0t vodene pare stawa 2)q>3!NQb-!u>471pD). Iz turbine se na
⋅
pritisku q3>1/5!NQb izdvaja, za potrebe nekog tehnolo{kog qspdftb-! n 3>3!lh0t!pare a preostali deo
nastavqa ekspanziju do stawa 4)q>1/27!NQb-!y>2*. Stepen dobrote adijabatske ekspanzije do izdvajawa
p
pare iznosi η23
e >2. Pod okolinom smatrati vodu stawa P)q>1/2!NQb-!u>31 D*. Skicirati promene
stawa vodene pare na Ts dijagramu i odrediti:
a) snagu turbine
b) eksergiju parnih tokova na ulazu u turbinu i oba izlaza iz turbine
c) eksergijski stepen korisnosti procesa u turbini
2
!U
2
⋅
3
X uvscjob
3
4l
!t
4
ta~ka 1:
q2>31!cbs
lK
i2!>iqq4267!
lh
ta~ka 2:
4
(pregrejana para)
lK
t2>tqq!>!7/:96!
lhL
q3>5!cbs
i3!>iqq!>3889/27!
u>471pD
!t3!>!t2!>7/:96!
lK
lhL
(pregrejana para)
lK
lh
q4>2/7!cbs
lK
i4!>i′′!>37:7!
lh
!y>2
qp>2!cbs
lK
ip!>ix>94/:!
lh
up>31pD
ta~ka 3:
ta~ka O:
dipl.ing. @eqko Ciganovi}
(suvo−zasi}ena para)
lK
t4!>t′′!>!8/313!
lhL
(voda)
lK
tp>tx!>!1/3:7!
lhL
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 24
a)
⋅
⋅
⋅
R 23 = ∆ I23 + X U23
Prvi zakon termodinamike za proces u turbini:
⋅
⋅
⋅
⋅
⋅
⋅ ⋅
X uvscjob = I2 − I3 = n⋅ i2 − n3 ⋅ i 3 − n− n3 ⋅ i 4
⋅
X uvscjob > 21 ⋅ 4267 − 3 ⋅ 3889/27 − (21 − 3) ⋅ 37:7 >5/55!NX
b)
⋅
⋅
⋅
Fy 2> n⋅ (− ∆i2p + Up ⋅ ∆t2p ) >! n⋅ [i2 − i p + Up ⋅ (t p − t2 )]
⋅
Fy 2> 21 ⋅ [4267 − 94/: + 3:4 ⋅ (1/3:7 − 7/:96 )] >22/23!NX
⋅
⋅
⋅
Fy 3> n3 ⋅ (− ∆i 3p + Up ⋅ ∆t 3p ) >! n3 ⋅ [i3 − i p + Up ⋅ (t p − t 3 )]
⋅
Fy 3> 3 ⋅ [3889/27 − 94/: + 3:4 ⋅ (1/3:7 − 7/:96)] >2/58!NX
⋅ ⋅
⋅ ⋅
Fy 4> n− n3 ⋅ (− ∆i 4p + Up ⋅ ∆t 4p ) >! n− n3 ⋅ [i 4 − i p + Up ⋅ (t p − t 4 )]
⋅
⋅
Fy 4> (21 − 3) ⋅ [37:7 − 94/: + 3:4 ⋅ (1/3:7 − 8/313)] >5/82!NX
c)
⋅
Bilans eksergije za proces u turbini:
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
Fy 2 = Fy 3 + Fy 4 + X uvscjob + Fy hvcjubl
⋅
Fy hvcjubl = Fy 2 − Fy 3 − Fy 4 − X uvscjob !>!22/23!−!2/58!−!5/82!−!5/55!>!1/6!NX
⋅
ηFy =
⋅
Fy vmb{ − Fy hvcjubl
⋅
Fy vmb{
⋅
>
22/23 − 1/6
>1/:7
22/23
⋅
Fy vmb{!>! Fy 2!>22/23!NX
napomena:
Do gubitka eksergije se moglo do}i i na uobi~ajen na~in
⋅
⋅
primenom Hpvz!−!Tupepmjoph! zakona:! Fy h = Up ⋅ ∆ T tjtufn
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 25
⋅
4/26/!Pregrejana vodena para stawa 2)q>!7!NQb-!U>844!L-! n >2!lh0t ) {iri se adijabatski u
dvostepenoj turbini sa me|uprigu{ivawem (slika), do krajweg stawa 5)U>424!L- vla`na para).
⋅
Stepeni dobrote u turbinama su: ηEUWQ = 2 i ηEUOQ = 1/99 . Deo pare, masenog protoka n B>1/4!lh0t, po
izlasku iz turbine visokog pritiska, pri pritisku q3>1/9!NQb odvodi se iz turbine, a preostala para
prolaskom kroz prigu{ni ventil adijabatski prigu{uje na pritisak q4>1/4!NQb. Prikazati procese u
it koordinatnom sistemu i odrediti snagu dobijenu na zajedni~kom vratilu kao i eksergijski stepen
korisnosti procesa u ovoj dvostepenoj turbini. Pod okolinom smatrati vodu stawa P)q>1/2!NQbU>3:4!L*/
!n
UWQ
UOQ
2
X
3
4
!nB
5
2
i
3
4
5L
5
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 26
u2!>571pD!
q2>!71!cbs
lK
lK
t2>tqq!>!7/86!
i2!>!iqq!>4435!!
lh
lhL
ta~ka 1:
q3>!9!cbs
ta~ka 2:
i3!>!iqq!>3919/7!!
q4>!4!cbs
t4>tqq!>!8/289!
lK
lhL
ta~ka 4K:
U5L!>!51pD
i4!>!i3!>3919/7!!
lK
lh
lK
lhL
(ta~ka 4K se nalazi u oblasti vla`ne pare)
t′!?!t5L!?!t′′
t − t(
y 5L = 5L
= 1/99
t#−t( U =51p D
ηEUWQ =
lK
lhL
lK
lh
ta~ka 3:
U5>51pD
ta~ka 4:
t3>t2!>!7/86!
i4 − i5
i 4 − i 5L
⇒
t5L>t4>!8/289!
i5L!>!iy!>!3396!
ηEUOQ = 1/99
i 5 = i 4 − ηEUWQ ⋅ )i 4 − j 5L * =
i5 = 3919/7 − 1/99 ⋅ )3919/7 − 3396* = 3459
i − i(
y5 = 5
= 1/:17
i#−i( U = 51p D
lK
lh
lK
lh
⇒
t5!>!ty!>8/65!
lK
lhL
qp>2!cbs
up>31pD
lK
lK
tp>tx>1/3:7!
ip>ix>94/:!
lh
lhL
ta~ka O:
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 27
Prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
⋅
⋅
ograni~enom isprekidanom linijom: R 23 = ∆ I23 + X U23
⋅
⋅
⋅
⋅
⋅
⋅ ⋅
X = I2 − I3 = n⋅ i2 − n B ⋅ i3 − n− n B ⋅ i 5
⋅
X > 2⋅ 4435 − 1/4 ⋅ 3917/7 − (2 − 1/4 ) ⋅ 3489 >948/9!lX
⋅
⋅
⋅
Fy 2> n⋅ (− ∆i2p + Up ⋅ ∆t2p ) >! n⋅ [i2 − i p + Up ⋅ (t p − t2 )]
⋅
Fy 2> 2 ⋅ [4435 − 94/: + 3:4 ⋅ (1/3:7 − 7/86 )] >245:/2!LX
⋅
⋅
Fy hvcjubl = Up ⋅ ∆ T tj >///> 3:4 ⋅ 1/66 >272/26!lX
⋅
⋅
⋅
⋅
⋅
∆ T tjtufn!>!∆ T sbeop!ufmp!,!∆ T plpmjob!>!///>1/57!
lX
L
⋅
∆ T sbeop!ufmp> T j{mb{!−! T vmb{!>!///>!8/41!−!7/86>1/66!
⋅
⋅
T vmb{!>! n⋅ t2 > 2⋅ 7/86 >7/86!
lX
L
lX
L
⋅
⋅
⋅ ⋅
lX
T j{mb{!>! n B ⋅ t 3 + n− n B ⋅ t 5 > 1/4 ⋅ 7/86 + (2 − 1/4 ) ⋅ 8/65 >8/41!
L
⋅
ηFy =
⋅
Fy vmb{ − Fy hvcjubl
⋅
Fy vmb{
⋅
>
245:/2 − 272/26
>1/99
245:/2
⋅
Fy vmb{!>! Fy 2!>245:/2!NX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 28
4/27/ Pregrejana vodena para )1/2!lh0t* stawa 2)q>91!cbs-!u>571pD* ulazi u parno turbisnki blok gde se
najpre kvazistati~ki adijabatski {iri u turbini visokog pritiska do stawa 3)q>21!cbs*. Zatim se
vodenoj pari stawa 2 u dogreja~u izobarski dovodi toplota od toplotnog izvora stalne temperature
UUJ>571pD sve do uspostavqawa toplotne ravnote`e (stawe 3). Nakon toga se para kvazistati~ki
adijabatski {iri u turbini niskog pritiska do stawa 5)q>2!cbs*. Pod okolinom smatrati vodu stawa
P)q>1/2!NQb-!U>3:4!L*/!Skicirati procese sa vodenom parom na it dijagramu i odrediti:
a) mehani~ku snagu parno turbinskog bloka kao i toplotnu snagu dogreja~a pare
b) ireverzibilnost procesa (gubitak eksergije) u parno turbinskom bloku
c) eksergijski stepen korisnosti procesa u parnoturbinsom bloku
2
⋅
X U 23
3
4
⋅
R 34
⋅
X U 45
5
i
2
3
4
5
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 29
u2!>571pD!
q2>!91!cbs
lK
lK
t2>tqq!>!7/699!
i2!>!iqq!>43:7!!
lh
lhL
ta~ka 1:
ta~ka 2:
q3>!21!cbs
i3!>!i′′!>3889!!
lK
lh
t3>t2!>!7/699!
lK
lhL
u4!>!571pD
q4>!21!cbs
lK
lK
t4>tqq!>!8/756!
i4!>!iqq!>44:3!!
lh
lhL
ta~ka 3:
q5!>!2!cbs
ta~ka 4:
i5!>!iqq!>!38:3/3!
t5>t4>!8/756!
lK
lhL
lK
lh
up>31pD
qp>2!cbs
lK
lK
tp>tx>1/3:7!
ip>ix>94/:!
lh
lhL
ta~ka O:
a)
Prvi zakon termodinamike za proces u turbini visokog pritiska:
⋅
⋅
⋅
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X U23
⇒
⋅
X U23 = n⋅ (i2 − i3 )
⋅
X U23 = 1/2 ⋅ (43:7 − 3889) >62/9!lX
Prvi zakon termodinamike za proces u turbini niskog pritiska:
⋅
R 45 = ∆ I45 + X U 45
⇒
⋅
X U 45 = n⋅ (i 4 − i 5 )
⋅
X U 45 = 1/2 ⋅ (44:3 − 38:3/3) >71!lX
⋅
⋅
⋅
X = X U23 + X U 45 >222/9!lX
⋅
Prvi zakon termodinamike za proces u dogreja~u pare:
⋅
⋅
⋅
R 34 = ∆ I34 + X U34
⇒
⋅
R 34 = n⋅ (i 4 − i3 )
⋅
R 34 = 1/2 ⋅ (44:3 − 3889) >72/5!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 30
b)
⋅
⋅
⋅
Js = Fy hvcjubl = Up ⋅ ∆ T tj >///> 3:4 ⋅ 33 >7/56!lX
⋅
⋅
⋅
⋅
⋅
∆ T tjtufn!>!∆ T sbeop!ufmp!,!∆ T upqmpuoj!j{wps!>!///>!−94/8!,!216/8!>!33!
⋅
∆ T sbeop!ufmp> T j{mb{!−! T vmb{!>!///>!875/6!−769/9>216/8!
⋅
X
L
X
L
⋅
X
L
⋅
⋅
X
T vmb{!>! n⋅ t2 > 1/2 ⋅ 7/699 >769/9!
L
T j{mb{!>! n⋅ t 5 > 1/2 ⋅ 8/756 >875/6!
⋅
R
72/5 ⋅ 21 4
X
>!−94/8!
∆ T upqmpuoj!j{wps!>!−! 34 >−
844
L
UUj
⋅
c)
⋅
⋅
⋅
Fy 2> n⋅ (− ∆i2p + Up ⋅ ∆t2p ) >! n⋅ [i2 − i p + Up ⋅ (t p − t2 )]
⋅
Fy 2> 1/2 ⋅ [43:7 − 94/: + 3:4 ⋅ (1/3:7 − 7/699)] >247/96!LX
⋅
⋅
Fy R = R 34 ⋅
⋅
ηFy =
UUJ − Up
844 − 3:4
>47/97!lX
> 72/5 ⋅
844
UUJ
⋅
⋅
Fy2 + Fy R − Fy h
⋅
⋅
Fy 2 + Fy R
zadatak za ve`bawe:
>
247/96 + 47/97 − 7/56
>1/:7
247/96 + 47/97
(3.17.)
⋅
4/28/ Kompresor usisava n =83!lh0i pare amonijaka stawa 2)q>366/:!lQb-!y>2* i sabija je adijabatski do
stawa 3)q>21!cbs-!U>511!L*/ Ako za okolinu smatramo amonijak stawa P)U>331!L-!y>1* odrediti
eksergijski stepen korisnosti procesa u kompresoru.
re{ewe:
ηFy>1/99
dipl.ing. @eqko Ciganovi}
⋅
⋅
⋅
)! X >7/13!lX-! Fy 2 = 5/45 lX-! Fy h >2/38!lX!*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 1
PRVI I DRUGI ZAKON TERMODINAMIKE
(KOMBINOVANI PROBLEMI)
5/2/!U toplotno izolovanom rezervoaru nalazi se!211!lh!vode!)dx>5/29!lK0)lhL**!po~etne temperature
Ux>3:4!L/!Komad bakra!)dDv>1/49!lK0)lhL**!mase!51!lh, po~etne temperature!UDv>469!L!i komad
gvo`|a!)dGf>1/57!lK0)lhL**!mase!31!lh, po~etne temperature!UGf>454!L-!naglo se unesu u rezervoar sa
vodom. U momentu uno{ewa u rezervoaru se ukqu~uje me{alica vode snage!711!X, koja radi dok se ne
uspostavi stawe termi~ke ravnote`e!U+>3::!L/!Odrediti:
a) vreme rada me{alice
b) promenu entropije izolovanog sistema tokom navedenog procesa
Dv
Gf
I3P
a)
prvi zakon termodinamike za proces u zatvorenom termodinami~kom
⇒
XU23>V2!−!V3
sistemu:
R23!>!∆V23!,!XU23
V2 = n x ⋅ d x ⋅ Ux + nDv ⋅ d Dv ⋅ UDv + nGf ⋅ d Gf ⋅ UGf
V2 = 211 ⋅ 5/29 ⋅ 3:4 + 51 ⋅ 1/49 ⋅ 469 + 31 ⋅ 1/57 ⋅ 454 >242182/3!lK
V 3 = n x ⋅ d x ⋅ U + + nDv ⋅ d Dv ⋅ U + + nGf ⋅ d Gf ⋅ U +
V 3 = 211 ⋅ 5/29 ⋅ 3:: + 51 ⋅ 1/49 ⋅ 3:: + 31 ⋅ 1/57 ⋅ 3:: >243388/7!lK
XU23!>!242182/3!−!243388/7!>!−2317/5!lK
τ=
XU23
− 2317/5
=
>3121/78!t
⋅
− 711 ⋅ 21 −4
X U23
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 2
b)
∆TTJ!>!∆TSU!,!∆TP!>///!>5/58!
lK
L
∆TSU!>!∆Tx!,!∆TDv!,!∆TGf!>///>9/58!−!3/85!−!2/37!!>!5/58!
U+
∆T x = n x ⋅
∫
lK
L
d x ⋅ eU
U+
3::
lK
= n x ⋅ d x mo
= 211 ⋅ 5/29 ⋅ mo
>9/58!
L
U
Ux
3:4
Ux
U+
∆T Dv = nDv ⋅
∫
d Dv ⋅ eU
U+
3::
lK
>−3/85!
= nDv ⋅ d Dv mo
= 51 ⋅ 1/49 ⋅ mo
U
UDv
469
L
∫
d Gf ⋅ eU
U+
3::
lK
>−2/37!
= nGf ⋅ d Gf mo
= 31 ⋅ 1/57 ⋅ mo
L
U
UGf
454
UDv
U+
∆T Gf = nGf ⋅
UGf
5/3/!U kalorimetarskom sudu, zanemarqivog toplotnog kapaciteta, nalazi se te~nost polazne
temperature UU>3:1!L!)stalnog toplotnog kapaciteta D>2/36!lK0L*/!U sud je unet bakarni uzorak mase
nC>1/26!lh!i polazne temperature!UC>484!L/!Zavisnost specifi~nog toplotnog kapaciteta za bakar
dC
U
od temperature data je izrazom:
!
/Tokom uspostavqawa
= 1/498 + 1/76: ⋅ 21 −5 ⋅
[lK 0 (lhL*)]
[L ]
toplotne ravnote`e u kalorimetru, okolini stalne temperature!Up>394!L-!predato je!6% od koli~ine
toplote koju je predao bakarni uzorak. Odrediti:
a) temperaturu u kalorimetarskom sudu u trenutku uspostavqawa toplotne ravnote`e
b) promenu entropije izolovanog sistema od polaznog stawa do stawa uspostavqawa toplotne
ravnote`e u kalorimetru.
a)
oznake koje se koriste u daqem tekstu re{ewa:
(R23 )C
(R23 )p
(R23 )U
US
koli~ina toplote koju bakar predaje te~nosti u sudu
koli~ina toplote koju bakar predaje okolini
koli~ina toplote koju te~nost prima od bakra
temperatura u kalorimetarskom sudu u trenutku
uspostavqawa toplotne ravnote`e
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 3
prvi zakon termodinamike za proces sa bakrom:
US
(R23 )C , (R23 )p > nC ⋅
∫[
]
1/498 + 1/76: ⋅ 21 −5 U ⋅ eU
UC
(R23 )C , (R23 )p > nC ⋅ 1/498 ⋅ (US − UC ) + 1/76: ⋅ 21 −5
US3 − UC3
3
)2*
prvi zakon termodinamike za proces sa te~no{}u
US
(R23 )U >
∫
D U ⋅ eU > D U ⋅ (US − UU )
)3*
UU
interno razmewena toplota izme|u bakra i te~nosti:
(R23 )C >− (R23 )U
)4*
(R23 )p > 1/16 ⋅ [(R23 )C + (R23 )p ]
uslov zadatka:
)5*
kombinovawem jedna~ina!)2*-!)3*-!)4*!j!)5*!dobija se:
US!>3:4/8!L-
(R23 )U >5736!K-
(R23 )C >−5736!K-
(R23 )p >−354!K
b)
K
L
K
∆TSU!>!∆TU!,!∆TC!>!///>!26/96!−!25/77!>!2/2:!
L
∆TTJ!>!∆TSU!,!∆TP!>!///>!2/2:!,!1/97!>!3/16!
US
∆T U =
∫
U
D(U ) ⋅ eU
3:4/8
K
>26/96!
= D U ⋅ mo S = 2/36 ⋅ mo
L
U
UU
3:1
UU
US
∆TC = nC ⋅
∫
d(U ) ⋅ eU
= nC ⋅
U
UC
US
∫
(1/498 + 1/76: ⋅ 21 U ) ⋅ eU =
.5
U
UC
U
K
n ⋅ 1/498 ⋅ mo S + 1/76: ⋅ 21−5 ⋅ (US − UC ) = /// = −25/77
UC
L
(R23 )p
354
K
∆T P = −
=−
>1/97!
L
UP
394
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 4
5/4/!Meteor temperature!U>4111!L-!brzinom x>21!ln0t ule}e u ledeni breg temperature!U>384!L/
Masa meteora je!nn>21!lh-!a specifi~ni toplotni kapacitet!dn>1/9!lK0lhL/!Odrediti:
a) koli~inu toplote koju meteor preda ledenom bregu
c* promenu entropije izolovanog sistema koji ~ine meteor i ledeni breg
d* masu otopqenog leda (toplota topqewa leda iznosi s>443/5!lK0lh)
meteor x>21!ln0t
ledeni breg
a)
prvi zakon termodinamike za proces koji se de{ava sa meteorom:
R23!>!∆V23!,!X23!,!∆Fl23
R23 = nn ⋅ dn ⋅ (Un3 − Un2) + n ⋅
R 23 = 21 ⋅ 1/9 ⋅ (384 − 4111) − 21 ⋅
(21 ⋅ 21 )
4 3
3
x33 − x23
3
⋅ 21 −4 >− 6/329 ⋅ 219 lK
b)
∆Ttjtufn = ∆Tnfufps + ∆Tmfe >///>−2:/286!,2:22/466>29:9/29!
∆Tnfufps = nn ⋅ dn mo
∆Tmfe =
lK
L
384
Un3
lK
> 21 ⋅ 1/9 ⋅ mo
>!−2:/286!
4111
Un2
L
R23 6/329 ⋅ 216
lK
>2:22/46!
>
Um
384
L
c)
R23 = nm ⋅ sm
⇒
dipl.ing. @eqko Ciganovi}
nm =
R23 6/329 ⋅ 216
>2681!lh
>
sm
443/5
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 5
5/5/!U rezervoar sa nx>61!lh!vode!)dx>5/2:!lK0lhL*-!temperature!Ux>394!L- uroni se zatvorena boca,
na~iwena od ~elika)d•>1/59!lK0lhL*-!sa!Wl>21!litara kiseonika (idealan gas). Masa ~eli~ne boce
zajedno sa kiseonikom je!n•!,!nC>!31!lh/!Pre urawawa temperatura kiseonika i boce je!Ul>U•>474
L-!a pritisak kiseonika u boci je!ql>26!NQb/!Sistem koji se sastoji od vode i boce sa kiseonikom
mo`e se smatrati izolovanim. Temperatura okolnog vazduha je!Up>3:4!L-!pritisak!qp>2!cbs-!a
zapreminski (molarni) udeo kiseonika u okolnom vazduhu je!32&/!Zanemaruju}i razmenu toplote sa
okolinom odrediti:
a) temperaturu u sudu u trenutku uspostavqawa toplotne ravnote`e
b) radnu sposobnost kiseonika u boci u trenutku postizawa toplotne ravnote`e
a)
prvi zakon termodinamike za proces koji po~iwe urawawem boce a zavr{ava
se uspostavqawem toplotne ravnote`e;
R23!>!∆V23!,!X23
⇒
V2!>!V3
V2 = n x ⋅ d x ⋅ Ux + n • ⋅ d • ⋅ U• + nl ⋅ d wL ⋅ UL
V 3 = n x ⋅ d x ⋅ U + + n • ⋅ d • ⋅ U + + nl ⋅ d wL ⋅ U +
U+ =
n x ⋅ d x ⋅ Ux + n• ⋅ d • ⋅ U• + nl ⋅ d wL ⋅ UL
>///
n x ⋅ d x + n • ⋅ d • + nl ⋅ d wL
jedna~ina stawa idealnog gasa za kiseonik u boci na po~etku procesa:
q ⋅W
26 ⋅ 21 7 ⋅ 21 ⋅ 21 −4
=
nl = l
q l ⋅ Wl = nl ⋅ S hl ⋅ Ul ⇒
>2/6:!lh
S hl ⋅ Ul
371 ⋅ 474
n•!>! (n • + nl ) −!nl!>!31!−!2/6:!>29/52!lh
U+ =
61 ⋅ 5/2: ⋅ 394 + 29/52 ⋅ 1/59 ⋅ 474 + 2/6: ⋅ 1/76 ⋅ 474
>397/7!L
61 ⋅ 5/2: + 29/52 ⋅ 1/59 + 2/6: ⋅ 1/76
napomena:
!
U+!−
temperatura u boci u trenutku uspostavqawa toplotne ravnote`e
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 6
b)
jedna~ina stawa idealnog gasa za kiseonik u boci u trenutku uspostavqawa
toplotne ravnote`e: q2 ⋅ Wl = nl ⋅ S hl ⋅ U2
q2 =
nl ⋅ S hL ⋅ U2
W
=
2/6: ⋅ 371 ⋅ 397/7
21 ⋅ 21 −4
>22/96!NQb
odre|ivawe pritiska kiseonika u okolnom vazduhu:
q lp = sL ⋅ q p >1/132!NQb
odre|ivawe radne sposobnosti kiseonika u boci u trenutku uspostavqawa
Xnby = n ⋅ (−∆v21 + Up ⋅ ∆t2p − q p ⋅ ∆w 2p )
toplotne ravnote`e:
qp
U
Xnby = nL ⋅ dlw ⋅ (U2 − UP ) + UP ⋅ dq ⋅ mo P − Sh ⋅ mo L + qLp ⋅ (w2 − w p ) = 761 lK
q2
U2
w2 =
wp =
S hl ⋅ U2
q2
S hl ⋅ Up
q lp
=
371 ⋅ 397/7
22/96 ⋅ 21 7
=
371 ⋅ 3:4
1/132 ⋅ 21 7
>!1/1174!!
n4
lh
>!4/7387!!
n4
lh
napomena:
U delu zadatka pod b) radi lak{e preglednosti veli~ine stawa kiseonika u
boci u trenutku uspostavqawa toplotne ravnote`e obele`ene su indeksom!2
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 7
⋅
5/6/!Radna materija u zatvorenom sistemu vr{i neki proces pri ~emu joj se u svakoj sekundi dovodi! R >4
⋅
lK0t!toplote i odvodi zapreminski rad! X )lK0t*-!koji se u toku vremena mewa po zakonu:
⋅
X 12
τ
= 3/5 ⋅
[lX]
[i]
){b!!1! < τ ≤ 2 i *
⋅
X 23
>,3/5
[lX]
){b!!τ!?!2!i*
⋅
b* odrediti brzinu promene unutra{we energije sistema-! ∆ V23 )lX*-!u trenutku!vremena!τ>1/7!i
b) odrediti promene unutra{we energije sistema-!∆V23!)lK*-!u toku prva dva ~asa
⋅
! X- [lX ]
τ-! [i]
2
3
a)
⋅
⋅
⋅
∆ V23 = R 23 − X 23 = 4 − 3/5 ⋅ τ = 4 − 3/5 ⋅ 1/7 = 2/67 lX
b)
2
X12 =
2
∫
X (τ) ⋅ eτ =
∫
X (τ) ⋅ eτ = 3/5 τ
1
3
X23 =
∫
3/5 ⋅ τ ⋅ eτ = 3/5
τ3
3
2
= 2/3 lXi
1
1
3
= 3/5 lXi
2
2
Xp3!>!X12!,!X23!>!4/7!lXi!>!23:71!lK
3
R 13 =
∫
⋅
R(τ) ⋅ eτ = 4 ⋅ 3 = 7 lXi = 32711 lK
1
∆V13>!R13!−!X13!>!9751!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 8
5/7/!Toplotne karakteristike neke radne materije zadate su zavisnostima:
q!/!w!>!B!/!U
v!>!C!/!U!,!D!/!U3!,!E
gde je!B>!3:8!K0)lhL*-!C>7:8!K0)lhL*-!D>!1/196!K0)lhL3*-!E!>!dpotu-!a!q-!w-!U!j!u su veli~ine stawa u
osnovnim jedinicama!TJ. Radna materija mewa svoje toplotno stawe kvazistati~ki adijabatski od stawa
2)q2!>!1/2!NQb-!U2>511!L*!do stawa!3!)U3>2451!L*/
b* izvesti jedna~inu kvazistati~ke adijabatske promene stawa radne materije u obliku:!q>g)U*
c* odrediti pritisak radne materije u stawu!3
b*
prvi zakon termodinamike za proces sa radnim telom (diferencijalni oblik)
δr = ev + q ⋅ ew
)2*
e(q ⋅ w ) = q ⋅ ew + w ⋅ eq )3*
diferencijal proizvoda:
kombinovawem jedna~ina!)2*!i!)3) sa toplotnim karakteristikama radne
materije dobija se:
1!>!ev!, e(q ⋅ w ) − w ⋅ eq
(
⇒
ev!,! e(q ⋅ w ) >! w ⋅ eq
)
eq
q
eq
e B ⋅ U + C ⋅ U + D ⋅ U3 + E = B ⋅ U ⋅
q
e C ⋅ U + D ⋅ U 3 + E + e(B ⋅ U ) = B ⋅ U ⋅
(
)
B +C
U 3D
q
⋅ mo
+
⋅ (U − U2 ) = mo
B
U2
B
q2
B + C + 3D ⋅ U eU eq
⋅
=
q
B
U
q = q2 ⋅ f
B +C U 3D
⋅mo + ⋅(U − U2 )
B
U2 B
q = 1/2 ⋅ 21 7 ⋅ f
⇒
q = 1/2 ⋅ 21
7
3:8 + 7:8
U 3⋅1/196
⋅mo
+
⋅(U − 511 )
3:8
511
3:8
⋅f
3:8 + 7:8
U 3⋅1/196
⋅mo
+
⋅(U − 511 )
3:8
511
3:8
b)
ako u izvedenu jedna~inu stavimo!U>U3>2451!K, kao i vrednosti za navedene
konstante!)B-!C!j!D*!dobija se:
q 3 = 1/2 ⋅ 21 7 ⋅ f
3:8+ 7:8 2451 3⋅1/196
⋅mo
+
⋅(2451− 511 )
511
3:8
3:8
>:/9!NQb
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 1
5. DESNOKRETNI KRU@NI PROCESI
6/2/ Koliko se korisnog (neto) rada mo`e najvi{e dobiti ako toplotni izvor temperature (UUJ=450 K)
predaje toplotnom ponoru temperature (UUQ=300 K) R=800 kJ toplote, ako se izme|u toplotnog izvora
i toplotnog ponora ukqu~i desnokretna toplotna ma{ina.
Xepcjkfo
!Repw
!!!UJ
Radno
telo
!Rpew
!!UQ
Xqplsfubokb
Najvi{e korisnog rada se mo`e dobiti ako desnokretna tolotna ma{ina
radi po Karnoovom desnokretnom ciklusu.
U
UUJ
UUQ
!t
R EPW + R PEW
U − UUQ
= UJ
R EPW
UUJ
η =ηK
⇒
Repw = − Rpew ⋅
UUJ
561
= 911 ⋅
= 2311!lK
411
U UQ
⇒
Xofup!>!Repw!,!Rpew!>!2311!−!911!>!511!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 2
6/3/ Radna supstanca vr{i potpuno povratni proces izme|u toplotnog izvora stalne temperature
UUJ>2111!L i toplotnog ponora promenqive temperature. Toplotni kapacitet toplotnog ponora
iznosi DUQ>311!lK0L, a temperatura toplotnog ponora se mewa od UUQ2>411!L do UUQ3>@ U toku
obavqawa kru`nog proces toplotni izvor je radnoj supstanci predao 211!NK toplote. Odrediti:
a) koristan rad kru`nog procesa
b) termodinami~ki stepen korisnosti kru`nog procesa
a)
(∆T tjtufn )23
= ∆T23 −
R epw
UUJ
(1)
(∆T tjtufn )45
= ∆T 45 −
R pew
(UUQ )mo
(2)
sabirawem jedna~ina (1) i (2) dobija se:
1=−
R epw
R pew
−
(UUQ )mo
UUJ
⇒
2 R epw
⋅
UUQ3 = UUQ2 ⋅ fyq
D UQ UUJ
1=−
R epw D UQ ⋅ (UUQ2 − UUQ3 )
−
UUQ2 − UUQ3
UUJ
U
mo UQ2
UUQ3
2 211 ⋅ 21 4
> 411 ⋅ fyq
⋅
311
2111
>5:5/7!L
R pew = D UQ ⋅ (UUQ2 − UUQ3 ) > 311 ⋅ (411 − 5:5/7 ) >−49/:!NX
Xlps!>!Repw!,!Rpew!>!211!−!49/:!>!72/2!lX
c*
ηC =
Xlps
72/2
>
>1/72
R EPW
211
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 3
6/4/ Proveriti koji od dva prikazana kru`na procesa A i B ima ve}i stepen korisnog dejstva, ako je ∆t
(vidi sliku) jednako za oba procesa. Pri kojoj temperaturi toplotnog ponora UUQ, a pri nepromeqenoj
temperaturi toplotnog izvora UUJ>600 K, bi stepen korisnog dejstva kru`nog procesa B bio dva puta
ve}i od stepena korisnog dejstva kru`nog procesa A. Svi procesi sa radnom telom su ravnote`ni.
B
U
∆t
2
600
C
U
3
3
600
∆t03
∆t
∆t03
100
100
4
t
ciklus A:
UUj = 711 K,
2
4
t
UUQ = 211 K
R epw = R 23 = UUJ ⋅ ∆T
R pew = R 34 + R 45 =
ηB =
ciklus B:
R EPW + R PEW
R EPW
UUj = 711 K,
R epw = R 23 + R 34 =
ηC =
UUJ + UUQ − ∆T UUJ + UUQ − ∆T UUJ + UUQ
⋅
+
⋅
=
⋅ (− ∆T)
3
3
3
3
3
U + UUQ
711 + 211
UUJ ⋅ ∆T + UJ
⋅ (− ∆T) 711 −
3
3
>1/53
>
>
UUJ ⋅ ∆T
711
R EPW + R PEW
R EPW
UUQ = 211 K
UUJ + UUQ ∆T UUJ + UUQ ∆T UUJ + UUQ
R pew = R 42 = UUQ ⋅ (−∆T)
⋅
+
⋅
=
⋅ ∆T
3
3
3
3
3
UUJ + UUQ
711 + 211
− 211
⋅ ∆T − UUQ ⋅ ∆T +
3
3
>1/82
>
>
UUJ + UUQ
711 + 211
⋅ ∆T
3
3
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
ηB =
UUJ −
UUJ + UUQ
3
UUJ
3 ⋅ η B = ηC
⇒
strana 4
UUJ + UUQ
− UUQ
3
ηC =
UUJ + UUQ
3
UUJ + UUQ UUJ + UUQ
UUJ −
− UUQ
3
3
3⋅
=
UUJ + UUQ
UUJ
3
⇒
UUQ >1!L
zadatak za ve`bawe:
(1.4.)
6/5/ Proveriti koji od dva prikazana kru`na procesa A i B ima ve}i stepen korisnog dejstva, ako je ∆t
(vidi sliku) jednako za oba procesa. U oba slu~aja temperatura toplotnog izvora iznosi UUJ>800 K, a
temperatura toplotnog ponora UUQ=300 K. Svi procesi sa radnom materijom su ravnote`ni.
B
1000
900
800
C
T, K
1000
900
UUJ
800
4
700
700
600
600
3
500
400
300
5
300
2
200
∆t03
100
s, J/(kgK)
η B =1/444
dipl.ing. @eqko Ciganovi}
3
5
UUQ
2
200
∆t03
100
0
s, J/(kgK)
∆t
∆t
re{ewe:
4
UUJ
500
400
UUQ
0
T, K
ηC >1/397
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 5
6/6/ Dvoatomni idealan gas obavqa proces gasnoturbinskog postrojewa koji se sastoji od dve izobare i dve
adijabate (Xulov ciklus). Stawe radnog tela na ulazu u kompresor je 1(q>2!cbs-!u>26pD), na izlazu iz
kompresora 2(q>6!cbs) i na ulazu u turbinu 3(u4>891pD). Stepen dobrote adijabatske kompresije je
fy
ηlq
e =0.83, a stepen dobrote adijabatske ekspanzije η e =0.85. Odrediti:
a) termodinami~ki stepen korisnog dejstva ovog ciklusa (η)
b) termodinami~ki stepen korisnog dejstva ovog ciklusa za slu~aj maksimalne mogu}e rekuperacije
toplote (η′)
a)
κ −2
2/5 −2
κ
U3L!>!U2 ⋅ q 3L > 399 ⋅ 6 2/5 >!567/25!L
q
2
2
U −U
567/25 − 399
U3!>!U2!,! 3L lq 2 !>! 399 +
>5:1/69!L
1/94
ηe
κ −2
2/5 −2
κ
> 2164 ⋅ 2 2/5 >!775/95!L
U5L!>!U4 ⋅ q 5L
q
6
4
fy
U5!>!U4!, η e ⋅ )U5l!−U4*!>!2164!, 1/96 ⋅ )775/95!−!2164*!>834/18!L
η!>!
U − U3 + U2 − U5
R epw + R pew
2164 − 5:1/69 + 399 − 834/18
>1/34
>///> 4
!>!
R epw
U4 − U3
2164 − 5:1/69
Repw!>!n!/!)r34*q>dpotu!>!n!/!dq!/!)!U4!−U3!*
Rpew!>!n!/!)r52*q>dpotu!>!n!/!dq!/!)!U2!−U5!*
Repw
!3
!4
X23
X45
!2
!5
Rpew
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 6
b)
Repw′
!3
X23
!2
!B
!C
R
E
K
U
P
E
R
A
T
O
R
!4
X45
!5
Rpew′
op{ti uslov rekuperacije:
U5!?!U3
uslov maksimalne rekuperacije:
UB!>!U3!
η′!>!
)UC!>!U5*
R epw (+R pew (
U − UC + U2 − UB
2164 − 834/18 + 399 − 5:1/69
>!1/4:
>///> 4
!>
R epw (
U4 − UC
2164 − 834/18
Repw′!>!n!/!)rC4*q>dpotu!>!n!/!dq!/!)!U4!−UC!*
Rpew′!>!n!/!)rB2*q>dpotu!>!n!/!dq!/!)!U2!−UB!*
U
4
U
4
5
!C
3
3L
5L
t
dipl.ing. @eqko Ciganovi}
5
3
2
bez rekuperacije
Rsfl
2
B
t
sa rekuperacijom
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 7
6/7/ Sa vazduhom (idealan gas) kao radnim telom izvodi se idealan Xulov desnokretni ciklus (sve
promene stawa su kvazistati~ke). Ekspaznzija je dvostepenom sa me|uzagrevawem radnog tela, a
kompresija je dvostepena sa me|uhla|ewem (slika). Ako je qnby>27!cbs, qnjo>2!cbs!i ako je stepen
q
q
q
q
kompresije u oba kompresora i stepen ekspanzije u obe turbine isti ( 8 = 2 = 3 = 5 ) i ako se
q7 q 9 q4 q6
toplota dovodi od toplotnog izvora temperatura UUJ>U3>U5>711!L- a predaje toplotnom ponoru
temperature UUQ>U7>U9>361!L, skicirati ciklus na Ut dijagramu i odrediti stepen korisnog dejstva
ciklusa (η).
2
3
5
9
8
4
7
6
3
U
5
6
2
9
dipl.ing. @eqko Ciganovi}
4
8
7
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
qnjo!>!q7>!q6!>!2!cbs
strana 8
qnby!>!q2>!q3!>!27!cbs
q3 q 5
=
q4 q6
⇒
q4>q5!>!5!cbs
q 8 q2
=
q7 q9
⇒
q2>q9!>!5!cbs
⇒
q
U2 = U9 ⋅ 2
q9
⇒
q
U4 = U3 ⋅ 4
q3
⇒
q
U6 = U5 ⋅ 6
q5
⇒
q
U8 = U7 ⋅ 8
q7
U2 q2
=
U9 q 9
U3 q3
=
U4 q 4
U5 q 5
=
U6 q 6
U7 q 7
=
U8 q 8
η!>!
κ −2
κ
κ −2
κ
κ −2
κ
κ −2
κ
κ −2
κ
27
= 361 ⋅
5
κ −2
κ
κ −2
κ
κ −2
κ
2/5 −2
2/5
5
= 711 ⋅
27
2
= 711 ⋅
5
5
= 361 ⋅
2
= 482/6!L
2/5 −2
2/5
2/5 −2
2/5
2/5 −2
2/5
= 514/8!L
= 514/8!L
= 482/6!L
U − U2 + U5 − U4 + U7 − U6 + U9 − U8
R epw + R pew
>!///> 3
!>!1/46
R epw
U3 − U2 + U5 − U4
Repw!>!n![)r23*q>dpotu!,!)r45*q>dpotu!]>!n!/!dq!/!)!U3!−!U2!,!U5!−!U4!*
Rpew!>!n![)r67*q>dpotu!,!)r89*q>dpotu!]>!n!/!dq!/!)!U7!−!U6!,!U9!−!U8!*
zadatak za ve`bawe:
(1.7.)
6/8/ U energetskom postrojewu za proizvodwu elektri~ne energije primewen je rekuperativni
desnokretni gasnoturbinski ciklus (Xulov ciklus) sa vazduhom (idealan gas) kao radnim telom. U
kompresoru se 311!lh0i radnog tela temperature 91pD adijabatski komprimuje od 1/4!NQb do 1/9!NQb, sa
stepenom dobrote ηelq>1/97. Dovo|ewem toplote radno telo se zatim zagreva do 891pD i odvodi u turbinu,
gde adijabatski ekspandira sa stepenom dobrote ηefy>1/:. Za vreme odvo|ewa toplote rekuperi{e se 91&
od koli~ine toplote koja bi se u najpovoqnijem slu~aju mogla rekuperisati. Skicirati proces u Ut
kooedinatnom sistemu i odrediti
a) stepen korisnog dejstva ciklusa
b) teorijsku snagu koja stoji na raspolagawu za pogon generatora, ako se turbina i kompresor nalaze na
istom vratilu
a)
b)
η!>!1/44
Q!>!6/6!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 9
6/9/ Dvoatomni idealan gas obavqa teorijski (idealan) Otov kru`ni proces izme|u temperatura
Unby>U4>UUJ>2111!L i Unjo>U2>UUQ>3:1!L. Odrediti stepen kompresije )ε>w20w3* tako da korisna
⋅
snaga motora bude najve}a kao i snagu motora ako je molski protok gasa kroz motor o >:!npm0t.
⋅
Xofup = R epw + R pew = ...= o⋅ (Nd w ) ⋅ (U4 − U3 + U2 − U5 )
⋅
Repw!>!n!/!)r34*w>dpotu!>! o⋅ (Nd w ) ⋅ (U4 − U3 )
⋅
Rpew!>!n!/!)r52*w>dpotu!>! o⋅ (Nd w ) ⋅ (U2 − U5 )
U2 w 3
=
U3 w 2
κ −2
U4 w 5
=
U5 w 4
κ −2
w
U3 = U2 ⋅ 2
w3
⇒
w
U5 = U4 ⋅ 4
w5
(
⋅
κ −2
⇒
Xofup = o⋅ (Nd w ) ⋅ U4 − U2 ⋅ ε κ −2 + U2 − U4 ⋅ ε2− κ
= U2 ⋅ ε κ −2
κ −2
= U4 ⋅ ε2− κ
)
[
]
⋅
∂Xofup
= o⋅ (Nd w ) ⋅ − U2 ⋅ (κ − 2) ⋅ ε κ −3 − U4 ⋅ (2 − κ ) ⋅ ε − κ
∂ε
∂Xofup
⇔
− U2 ⋅ (κ − 2) ⋅ ε κ −3 − U4 ⋅ (2 − κ ) ⋅ ε − κ = 1
=1
∂ε
2
2
U 3 κ −3 2111 3⋅2/5 −3
ε = 4
>
U2 ⋅ (κ − 2) ⋅ ε
= U4 ⋅ (κ − 2) ⋅ ε ,
>!5/8
3:1
U2
Qsj!tufqfov!lpnqsftjkf!ε>5/8!npups!ptuwbsvkf!obkwf~v!tobhv
κ −3
−κ
X
Xnby
ε>5/8
dipl.ing. @eqko Ciganovi}
ε
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 10
(
)
Xnby = : ⋅ 21 −4 ⋅ 31/9 ⋅ 2111 − 3:1 ⋅ 5/82/5 −2 + 3:1 − 2111 ⋅ 5/82−2/5 >51!lX
Nbltjnbmob!tobhb!npupsb!j{optj! Xnby >51!lX
U
4
UJ
3−4;!w>dpotu
5−2;!w>dpotu
3
5
UQ
2
t
6/:/ Radna materija (idealan gas) obavqa idealan Xulov kru`ni ciklus izme|u temperatura U3>UUJ>2144
L i U5>UUQ>3:2!L. Odrediti temperaturu radne materije posle kvazistati~ke izentropskog sabijawa u
kompresoru (U2), odnosno posle kvazistati~ke izentropske ekspanzije u turbini (U4), tako da koristan rad
(rad na zajedni~kom vratilu) ima maksimalnu vrednost.
U
3
UJ
2−3;!q>dpotu
4−5;!q>dpotu
2
4
5
UQ
t
Xlpsjtubo!>!Repw!,!Rpew!>!///!>n!/!dq!/!)!U3!−!U2!,!U5!−!U4!*!>!///
Repw!>!n!/!)r23*q>dpotu!>!n!/!dq!/!)!U3!−U2!*
Rpew!>!n!/!)r45*q>dpotu!>!n!/!dq!/!)!U5!−U4!*
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
U2 q2
=
U5 q 5
U ⋅U
U2 = 3 5
U4
κ −2
κ
U3 q3
=
U4 q 4
⊕
strana 11
κ −2
κ
q2
q5
⊕
q3
=
q4
U2
U
= 3
U5
U4
⇒
⇒
⇒
Xlpsjtubo!>!n!/!dq!/!)!U3!−
∂Xlpsjtop
= n ⋅ dq
∂U4
U3 ⋅ U5
,!U5!−!U4!*
U4
U ⋅U
⋅ 3 3 5 − 2 -!
U
4
U4 = U3 ⋅ U5 > 2144 ⋅ 3:2 >659/4!L
∂Xlpsjtop
=1
∂U4
!!!!!!! U2 =
U3 ⋅ U5
⇔
U43
− 2 >1
U3 ⋅ U5 2144 ⋅ 3:2
>!659/4!L
>
U4
659/4
X
Xnby
U4
U4>659/4
6/21/ Dvoatomni idealan gas obavqa realni desnokretni kru`ni proces gasnoturbinskog bloka (Xulov)
izme|u temperatura Unby>6:6pD i Unjo>26pD. Molski protok gasa kroz postrojewe iznosi 31!npm0t/!).
fy
Stepen dobrote adijabatske kompresije je ηlq
e =0.88, a stepen dobrote adijabatske ekspanzije η e =0.88.
Odrediti maksimalnu snagu gasnoturbinskog bloka pri datim uslovima.
4
U
5
3
3L
5L
2
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 12
Korisno dobijen rad ima najve}u vrednost (vidi prethodni zadatak) kada je:
U3l =
U2 ⋅ U4 >!611!L
U5l =
U2 ⋅ U4
!>!611!L
U3l
Me|utim u ovom zadatku su adijabatska kompresija i adijabatska ekspanzija
neravnote`ni (nekvazistati~ki) procesi. Uzimawem te ~iwenice u obzir
dobija se:
U3L − U2
U3!>!U2!,!
ηlq
e
!>! 399 +
611 − 399
>639/:!L
1/99
U5!>!U4!, η fy
e ⋅ )U5l!−U4*!>!979!, 1/99 ⋅ )611!−!979*!>655/27!L
Xlpsjtubo!>!Repw!,!Rpew!>!///
⋅
(
)
⋅
(
)
Repw!>!n!/!)r34*q>dpotu!>!n!/!dq!/!)!U4!−U3!*!> o⋅ Nd q ⋅ (U4 − U3 )
Rpew!>!n!/!)r52*q>dpotu!>!n!/!dq!/!)!U2!−U5!*!> o⋅ Nd q ⋅ (U2 − U5 )
⋅
(
)
Xlpsjtubo> o⋅ Ndq ⋅ (U4 − U3 + U2 − U5 ) >
Xlpsjtubo!>Xnby!>! 31 ⋅ 21 −4 ⋅ (3:/2) ⋅ (979 − 639/: + 399 − 655/27 ) >59/38!lX
zadatak za ve`bawe:
(1.11.)
6/22/ Sa troatomnim idealnim gasom obavqa se Eriksonov desnokretni kru`ni proses sa izme|u
temperatura Unby>UUj>711!L!i Unjo>UUQ>511!L. Odrediti stepen korisnog dejstva ovog ciklusa za
slu~aj maksimalno mogu}e rekuperacije toplote i {rafirati na Ut dijagramu povr{nu koja odgovara
rekuperisanoj toploti.
re{ewe:
η=0.33
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 13
6/23/ Nekom idealnom gasu dovodi se pri kvazistati~koj izotermskoj ekspanziji (2.3) R23>411!lK toplote
od izotermnog toplotnog izvora temperature UUJ>2111!L, pri ~emu entropija idealnog gasa poraste za
∆T23>1/6!lK0L. Pri kvazistati~koj promeni stawa (3.4) entropija idealnog gasa opada linearno u Ut
koordinatnom sistemu i pri tom se toplota predaje izotermnom toplotnom ponoru temperature UUQ>3:4
L sve dok se ne uspostavi stawe termodinami~ke ravnote`e. Od stawa (4) do po~etnog stawa (2) dolazi se
kvazistati~kom izentropskom kompresijom. Skicirati promene stawa idealnog gasa u Ut koordinatama i
odrediti:
a) stepen korisnog dejstva ovog kru`nog ciklusa
b) odrediti promenu entrpopije sistema (lK0L)
c) {rafirati na Ut dijagramu povr{inu ekvivalentnu korisno dobijenom radu
U
UJ
2
3
UQ
4
t
b*
η=
R epw + R pew
= /// = 1/37
R epw
)37&*
R epw = R23 = 411 lK
T3
R epw = R 23 =
∫
U)t*eT = U3 ⋅ ∆T23
⇒ U3 =
R epw
411
=
= 711 L
1/6
∆T23
T2
T4
R pew = R 34 =
∫
U)t*eT =
U + U4
U3 + U4
⋅ ∆T 34 = 3
⋅ )−∆T23 *
3
3
T3
R pew
711 + 3:4
=
⋅ )−1/6* = −334/4 lK
3
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 14
c*
∆T tjtufn = ∆Tsbeopufmp + ∆T UJ + ∆T UQ = /// = −1/4 + 1/87 = 1/57
∆T UJ = −
lK
lhL
R epw
411
lK
=−
= −1/41
UUJ
2111
L
∆T UQ = −
R pew
lK
− 334/4
=−
= 1/87
UUQ
3:4
L
d*
Xlpsjtubo!>!Repw!,!Rpew!>!Repw!−! R pew
Princip {rafirawja korisnog rada na Ut dijagramu je princip oduzimawa
povr{ina koje predstavqaju dovedenu (Repw* i odvedenu )Rpew*!toplotu.
U
U
2
3
3
4
t
t
−
Repw
R pew
U
2
3
>
4
t
Xlpsjtubo
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 15
6/24/ Radna susptancija (idealan gas) obavqa desnokretni kru`ni proces koji se sastoji iz izobarskog
dovo|ewa toplote, kvazistati~ke (ravnote`ne) adijabatske ekspanzije, izotermskog odvo|ewa toplote i
kvazistati~ke (ravnote`ne) adijabatske kompresije. Toplota se radnoj supstanciji predaje od dimnih
gasova (idealan gas), koji se pri tome izobarski hlade (Cp=1.06 kJ/K), od po~etne temperature!uh2>:11pD.
Od radne susptancije, okolini stalne temperature up>29pD- predaje se 411!lK!toplote na povratan na~in.
Odrediti stepen korisnosti ovog kru`nog procesa i skicirati ga u
Ut i qw koordinatnom sistemu.
U
q
2
3
3
2
5
4
5
4
t
w
Rpew!>!R45!>!−!411!lK
T5
R pew = R 45 =
∫
U)t*eT = U4 ⋅ ∆T 45
⇒ ∆T 45 =
R epw
− 411
lK
=
= −2/14
U4
3:2
L
T4
∆T23!>!−!∆T45!>!2/14!
(
(∆T tjtufn )23
(∆T
ejn o hbt
lK
L
= ∆T23 + ∆T ejno hbt
)
23
= n eh ⋅ S h ⋅ mo
)
q eh3
q eh2
∆T ejnoj hbt
UEH3 = UEH2 ⋅ fyq
D qEH
23
(
!!!!!!!⇒!!!!!!! ∆T ejno hbt
− D qEH ⋅ mo
UEH3
UEH2
)
23
= −∆T23 >2/14!
lK
L
⇒
> 2284 ⋅ fyq − 2/14 >554/:!L
2/17
Repw!>!R23!>−!REH!>! − D qEH ⋅ (UEH3 − UEH2 ) > − 2/17 ⋅ (554/: − 2284 ) >883/96!lK
η!>!
R epw + R pew
883/96 . 411
>1/72
>
R epw
883/96
zadatak za ve`bawe (1.14.)
6/25/ Vazduh (idealan gas) vr{i slede}i kru`ni proces. Od po~etnog stawa (U2>411!L) vr{i se
kvazistati~ka promena stawa po zakonu prave linije u Ut koordinatnom sistemu pri ~emu se radnom
telu dovodi toplota od toplotnog izvora stalne temperature UUJ>U3?U2, pri ~emu je w2>w3. Nakon toga
vr{i se kvazistati~ka izentropska ekspanzija do po~etne temperature. Kru`ni proces se zatvara
kvazistati~kom izotermom. Stepen korisnog dejstva ovog ciklusa iznosi η>1/36. Skicirati ciklus na
Ut dijagramu i odrediti promenu entropije izolovanog sistema za najpovoqniji polo`aj temperatura
toplotnog izvora i toplotnog ponora.
re{ewe:
∆ttjtufn!>!85!
dipl.ing. @eqko Ciganovi}
K
lhL
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 16
6/26/ Vazduh (idealan gas) u prvom slu~aju obavqa desnokretni Xulov kru`ni proces.
U drugom slu~aju pri istom odnosu pritisaka qnby0qnjo>4, istoj dovedenoj koli~ini toplote i istoj
Unby>:84!L, izentropska kompresija zamewuje se izotermskom kompresijom, pri temperaturi Unjo>4:6!L.
Sve promene stawa vazduha su ravnote`ne.
a) odrediti termodinami~ke stepene korisnosti kru`nih procesa za oba slu~aja
b) odrediti termodinami~ke stepena korisnosti, ako se u oba prethodna slu~aja obavqaju kru`ni
procesi sa potpunim regenerativnim zagrevawem radne materije
b*
U
4
U
4
Unby
Unby
3
3
5
5
Unjo
2′
2
t
q3!>!q4!>!qnby!
!
κ −2
κ
q
U5!>!U4! ⋅ 5
q4
q
U2!>!U3! 2
q3
κ −2
κ
q2!>!q5!>!qnjo
2/5 −2
2 2/5
= ! :84 ⋅
4
4
= 4:6 ⋅
2
dipl.ing. @eqko Ciganovi}
t
2/5 −2
2/5
= 821/:!L
= !399/7!L
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 17
prvi slu~aj:
ηJ!>!
r epw + r pew
689 − 553/4
= !1/38
= ///!>
689
r epw
repw!>!r34!>!dq!/!)!U4!−!U3!*!>! 2 ⋅ (:84 − 4:6 ) >689!!
lK
lh
rpew!>!r52!>!dq!/!)!U2!−!U5!*!> 2 ⋅ (399/7 − 821/: ) >−!533/4!
lK
lh
drugi slu~aj:
ηJJ!>!
689 − 551/5
repw + rpew
= !1/35
= ///!>
689
repw
lK
lh
lK
rpew!>!r52′!,!r2′3!>!///!>!−426/:!−!235/6!>−!551/5!!
lh
lK
r52′!>!dq!/!)!U2′!−!U5!*!>! 2 ⋅ (4:6 − 821/: ) >−!426/:!
lh
q
2
lK
r2′3!>!U2′!/!Sh!/!mo! 2( = 4:6 ⋅ 1/398 ⋅ mo >−235/6!
q3
4
lh
repw!>!r34!>!dq!/!)!U4!−!U3!*!> 2 ⋅ (:84 − 4:6 ) >!689!!
c*
u oba slu~aja maksimalna rekuperisana toplota je jednaka i iznosi:
lK
rsfl!>!r52′!>!−!427!
lh
prvi slu~aj:
r epw (+r pew (
= ///!>!1/6:
r epw (
lK
repw′!>!repw!.! q rek !>!373!
lh
ηJ′!>!
lK
rpew′!>!rpew!,! q rek !>!−!217!
lh
drugi slu~aj:
r epw (+r pew (
= ///!>!1/63
r epw (
lK
repw′!>!repw!.! q rek !>!373!
lh
ηJJ′!>!
dipl.ing. @eqko Ciganovi}
rpew′!>!rpew!,! q rek !>!−!217!
lK
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 18
6/27/ Ciklus gasne turbine koji radi sa troatomnim idealnim gasom kao radnim telom sastoji se iz
izotermskog kvazistati~kog sabijawa (2.3), izohorskog dovo|ewa toplote (3.4), adijabatske ekspanzije (4.
5) i izobarskog odvo|ewa toplote (5.2). Ako je odnos pritisaka )q40q5*>9/5 i )q30q2*>4/6- odrediti
termodinami~ki stepen korisnosti kru`nog procesa (η) za slu~aj da je adijabatska ekspanzija (4.5):
a) kvazistati~ka
b) nekvazistati~ka sa stepenom dobrote ekspanzije ηefy>1/:6
a)
!4
!U
!5l
!2
!3
!t
q3!>!4/6/q2
U3!>!U2
U3 q3
=
U4 q 4
U4 = U3 ⋅
⇒
q
U5l!>!U4! 5l
q4
κ −2
κ
q4
9/5 ⋅ q2
= U2 ⋅
= 3/5 ⋅ U2
q3
4/6 ⋅ q2
q2
= 3/5 ⋅ U2 ⋅
9/5 ⋅ q2
κ −2
κ
2
= 3/5 ⋅ U2 ⋅
9/5
2/39 −2
2/39
= 2/6!/U2
R epw = n ⋅ (r34 )w =dpotu = o ⋅ [(Nd w ) ⋅ (U4 − U3 )] !>!)Ndw*!/!2/5!/!U2
[
]
q
R pew = n ⋅ (r 5l2 )q = dpotu + (r23 )U = dpotu > o ⋅ Nd q ⋅ (U2 − U5l ) + NS h ⋅ U2 ⋅ mo 2
q3
(
(
)
(
)
)
2
R pew = o ⋅ Nd q ⋅ (− 1/6 ⋅ U2 ) + NS h ⋅ U2 ⋅ mo
!
4/6
+ R pew
R
η!>! epw
=!
R epw
)
⇒
(Nd w ) ⋅ 2/5 ⋅ U2 − (Ndq ) ⋅ 1/6 ⋅ U2 + (NSh ) ⋅ U2 ⋅ mo 2
4/6
(Nd w ) ⋅ 2/5 ⋅ U2
3:/2 ⋅ 2/5 − 48/5 ⋅ 1/6 + 9/426 ⋅ mo
η>!
(
3:/2 ⋅ 2/5
dipl.ing. @eqko Ciganovi}
!>
2
4/6 >1/396
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 19
b)
!4
!U
!5
!5l
!3
!2
!t
ηfy
e =
U4 − U5
!⇒
U4 − U5l
U5 = U4 − ηfy
e ⋅ (U4 − U5l ) ⇒
U5 = 3/5 ⋅ U2 − 1/:6 ⋅ (3/5 U2 − 2/6 ⋅ U2 ) !>!2/66!/U2
R epw = n ⋅ (r34 )w =dpotu = o ⋅ [(Nd w ) ⋅ (U4 − U3 )] !>!)Ndw*!/!2/5!/!U2
[
] (
q
R pew = n ⋅ (r 52 )q = dpotu + (r23 )U = dpotu > Nd q ⋅ (U2 − U5 ) + NS h ⋅ U2 mo 2 !
q
3
(
(
)
)
(
)
)
2
R pew = o ⋅ Nd q ⋅ (− 1/66 ⋅ U2 ) + NS h ⋅ U2 mo
!
4
/6
+ R pew
R
η!>! epw
=!
R epw
(Nd w ) ⋅ 2/5 ⋅ U2 − (Ndq )⋅ 1/66 ⋅ U2 + (NSh )⋅ U2 ⋅ mo 2
4/6
(Nd w ) ⋅ 2/5 ⋅ U2
3:/2 ⋅ 2/5 − 48/5 ⋅ 1/66 + 9/426 ⋅ mo
η>!
3:/2 ⋅ 2/5
dipl.ing. @eqko Ciganovi}
!>
2
4/6 >1/35
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 20
6/28/ Vazduh (idealan gas) obavqa desnokretni kru`ni proces koji se sastoji od dve kvazistati~ke
(ravnote`ne) izentrope i dve nekvazistati~ke (neravnote`ne) izoterme, izme|u temperatura Unjo>411!L
i Unby>911!L. Pritisak vazduha na kraju izotermske kompresije iznosi q2>1/2!NQb, a na kraju izotermske
ekspanzije iznosi q4>1/9!NQb. Temperature toplotnog izvora i toplotnog ponora su stalne i iznose
UUJ>961!L i UUQ>391!L. Odrediti:
a) termodinami~ki stepen korisnosti kru`nog procesa, ako promena entropije izolovanog sistema za
proces dovo|ewa toplote iznosi 71!K0)lhL*- odnosno promena entropije izolovanog sistema za proces
odvo|ewa toplote iznosi 211!K0lhL
b) promenu entropije izolovanog sistema (koji sa~iwavaju toplotni izvor, toplotni ponor, i radno
telo) za slu~aj da se sve promene stawa odvijaju kvazistati~ki (ravnote`no)
a)
U
UJ
3 U
nby
4
Unjo
2
5
UQ
t
repw + rpew
= ///
repw
repw!>!r34!>!///
η!>!
rpew!>!r52!>!///
q3!>!q2 ⋅ U3
U
2
κ
2/5
κ −2
> 2 ⋅ 21 6 ⋅ 911 2/5 −2 >!42/216!Qb
411
proces 2-3:
∆t34!>! dq mo
U4
q
9
K
= 49:!
− S h mo 4 = − 398 mo
42
U3
q3
lhL
(∆T tjtufn )34 !>!∆t34!−!
r 34
UUJ
r34> 961 ⋅ (49: − 71) >38:/76!
dipl.ing. @eqko Ciganovi}
⇒
r34!>!UUJ ⋅ (∆t 34 − (∆t tjtufn ) )
34
lK
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
q5!>!q4 ⋅ U5
U
4
strana 21
κ
2/5
κ −2
> 9 ⋅ 21 6 ⋅ 411 2/5 −2 >!1/37!/216!Qb
911
proces 4-1:
∆t52!>! d q mo
U2
q
2
K
= −49:!
− S h mo 2 = − 398 mo
!)uo~iti da je!∆t52>−∆t23*
1/37
U5
q5
lhL
(∆T tjtufn )52 >!∆t52!−! r52
⇒
Uuq
r52>! 391 ⋅ (−49: − 211) >−!247/:3!
r52!>!UUQ ⋅ (∆t 52 − (∆t tjtufn )52 )
kJ
kg
lK
lh
lK
rpew!>!r52!>!−!247/:3!
lh
repw!>!r34!>!38:/76!
η!>!
repw + rpew
38:/76 − 247/:3
!>!1/62
=
repw
38:/76
b)
repw′!>! (r 34 )U =dpotu > U3 ⋅ S h mo
rpew′!>! (r 52 )U =dpotu > U5 ⋅ Sh mo
q3
42
lK
= 911 ⋅ 398 ⋅ mo
= 422!
9
lh
q4
2
lK
q5
= −227!
= 391 ⋅ 398 ⋅ mo
1/37
lh
q2
∆t tjtufn = ∆t sbeopufmp + ∆t UJ + ∆t UQ = /// = −476/: + 525/4 = 59/5
∆t UJ = −
repw
422
K
=−
= −476/:
UUJ
961
lhL
∆t UQ = −
−227
rpew
K
=−
= 525/4
UUQ
391
lhL
dipl.ing. @eqko Ciganovi}
K
lhL
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 22
6/29/ Parno turbinsko postrojewe radi po idealnom Rankin−Klauzijus–ovom kru`nom procesu izme|u
qnjo>1/2!cbs!i qnby>21!cbs. U kondenzatoru, se rashladnoj vodi predaje toplota i pri tom se rashladna
⋅
voda n x >4!lh0t, zagreje od stawa B)q>2!cbs-!u>21pD* do stawa B!)q>2!cbs-!u>31pD*/ Snaga napojne pumpe
iznosi 0.2!lX. Skicirati kru`ni proces na Ut dijagramu i odrediti:
a) termodinami~ki stepen korisnog dejstva ciklusa
b) snagu turbine
2
U
3
3
2
5
4
C
5
4
B
t
a)
q>1/17!cbs-
ta~ka 4:
lK
i5!>!2:2/:!
lh
i2!>!2:4/72!!!
(kqu~ala te~nost)
lK
t5!>!1/75:3!
lhL
q!>21!cbs-
ta~ka 1:
y>1
t>1/75:3!
lK
)te~nost*
lhL
lK
lh
q!>2!cbs
ta~ka 3:
⋅
⋅
⋅
⋅
⋅
X 23 = − ∆ I23 > − nq ⋅ (i2 − i 5 ) > X q
⋅
nq =
⋅
⋅
R 23 = ∆ I23 + X 23
Prvi zakon termodinamike za proces u pumpi:
⇒
⋅
lh
−1/2
XQ
>
>6/: ⋅ 21 −3
t
i 5 − i2 2:2/: − 2:4/72
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 23
⋅
Prvi zakon termodinamike za proces u kondenzatoru:
⋅
⋅
I2 = I3
⋅
⋅
⋅
nq ⋅ i 5 + n x ⋅ (iC − i B )
⋅
⋅
>///>!
6/: ⋅ 21 −3 ⋅ 2:2/: + 4 ⋅ (95 − 53)
>3422/4!
6/: ⋅ 21 −3
nq
lK
t4!>8/3:6!
lh
)q>1/2!cbs-!i>3422/4!
ta~ka A:
q!>21!cbs-
u>21pD
⇒
ta~ka C:
q!>21!cbs-
u>31pD
⇒
ta~ka 2:
q!>21!cbs-
t>8/3:6!
i3!>!4266/6!!!
⋅
nq ⋅ i 4 + n x ⋅ i B > n q ⋅ i 5 + n x ⋅ i C
⇒
⋅
i4!>!
⋅
⋅
R 23 = ∆ I23 + X 23
lK
lh
lK
*
lh
lK
lhL
lK
lh
lK
iC!>!95!
lh
iB!>!53!
)pregrejana para*
lK
lh
a)
⋅
η=
⋅
R epw + R pew
⋅
>!///>!
R epw
⋅
⋅
⋅
⋅
⋅
⋅
285/8 − 236
>1/39
285/8
R epw = R 23 = n q ⋅ (i 3 − i2 ) > 6/: ⋅ 21 −3 ⋅ (4266/6 − 2:4/72) >285/8!lX
R epw = R 45 = n q ⋅ (i 5 − i 4 ) > 6/: ⋅ 21 −3 ⋅ (2:2/: − 3422/4 ) >−236!lX
b)
⋅
Prvi zakon termodinamike za proces u turbini:
⋅
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⋅
X 23 = − ∆ I23 > − nq ⋅ (i 4 − i 3 ) > − 6/: ⋅ 21 −3 ⋅ (3422/4 − 4266/6 ) >61!lX> X uvs
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 24
6/2:/ U parnom kotlu uz konstantan pritisak od q>51!cbs od vode temperature 41pD proizvodi se vodena
para temperature u>611pD. Ta para izentropski (ravnote`no) ekspandira u turbini do pritiska od
q>1/17!cbs, a zatim se odvodi u kondenzator. Napojna pumpa vra}a u kotao pothla|en kondenzat. Toplota
potrebna za proizvodwu pare u parnom kotlu obezbe|uje se hla|ewem dimnih gasova (idealan gas) od
po~etne temperature 2711pD do temperature od 311pD. Koli~ina dimnih gasova je 6611!lnpm0i- a wihov
zapreminski sastav 29&!DP3-!:&P3-!84&O3 . Skicirati promene stawa vodene pare na Ut dijagramu i
odrediti:
a) termodinami~ki stepen korisnog dejstva kru`nog procesa
c* snagu turbine
3
U
2
4
5
t
ta~ka 1:
q!>51!cbsu>41pD
lK
lK
- !t2>1/544!
i2!>!23:/4!
lhL
lh
)te~nost*
q>51!cbsu>611pD
lK
lK
- t3!>!8/198!
i3!>!4556!
lhL
lh
(pregrejana para)
ta~ka 2:
ta~ka 3:
q!>1/17!cbs-
y4!>!1/95-
i4!>!3291/6!!!
ta~ka 4:
q!>1/17!cbs-
i5!>!236/:!!!
lK
lhL
(vla`na para)
lK
lhL
(te~nost)
t4>!t3!>!8/198!
lK
lh
t5>!t2>!1/544!
lK
lh
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
η!>!
strana 25
repw + rpew
4426/8 − 3165/7
= ///!>
= !1/49
4426/8
repw
lK
lh
lK
rpew!>!r45!>i5!−!i4!>!236/:!−!3291/6>−3165/7!
lh
repw!>!r23!>!i3!!−!i2!>!4556!−!23:/4!>!4426/8!
c*
analiza dimnih gasova:
N eh = sDP3 ⋅ NDP3 + sP3 ⋅ NP3 + sO3 ⋅ NO3 > 1/29 ⋅ 55 + 1/1: ⋅ 43 + 1/84 ⋅ 39 >
Neh!>42/35!
d qeh =
lh
lnpm
(
)
2
⋅ sDP3 ⋅ NDP3 ⋅ d qDP3 + sP3 ⋅ NP3 ⋅ d qP3 + sO3 ⋅ NO3 ⋅ d qO3 >
N eh
2
lK
⋅ (1/29 ⋅ 55 ⋅ 1/96 + 1/1: ⋅ 43 ⋅ 1/:2 + 1/84 ⋅ 39 ⋅ 2/15 ) >1/:9!
42/35
lhL
⋅
6611
lh
⋅ 42/35 >58/8!
= o eh ⋅ N eh >
4711
t
d qeh =
⋅
n eh
⋅
prvi zakon termodinamike za proces u parnom kotlu:
⋅
⋅
⋅
⋅
nq ⋅ i2 + n eh ⋅ d qeh ⋅ Uh2 = nq ⋅ i 3 + n eh ⋅ d qeh ⋅ Uh3
⋅
nq =
⋅
n eh ⋅ d qeh ⋅ )Uh2 − Uh3 *
i3 − i2
>
⋅
⋅
⋅
⇒
⋅
⋅
⋅
58/8 ⋅ 1/:9 ⋅ )2711 − 311*
lh
>2:/85!
4556 − 23:/4
t
Prvi zakon termodinamike za proces u turbini:
⋅
⋅
R 23 = ∆ I23 + X 23
R 23 = ∆ I23 + X 23
⋅
X 23 = − ∆ I23 > − nq ⋅ (i 4 − i 3 ) > −2:/85 ⋅ (3291/6 − 4556 ) >36!NX> X uvs
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 26
6/31/ Parnoturbinsko postrojewe radi po Rankin-ovom kru`nom procesu. Stepen dobrote adijabatske
ekspanzije u turbini iznosi ηefy>1/9, a stepen dobrote adijabatske kompresije u pumpi iznosi ηelq>1/:7.
Stawe vodene pare na ulazu u turbinu je q>51!cbs i u>431pD, a pritisak u kondenzatoru kf!q>1/13!cbs.
Skicirati promene stawa vodene pare na Ut i it dijagramu i odrediti:
a) termodinami~ki stepen korisnog dejstva ciklusa (η)
b) termodinami~ki stepen korisnog dejstva Karnoovog ciklusa koji radi izme|u istih temperatura
toplotnog izvora i toplotnog ponora )ηL*
i
U
3
3
2L
2
2
4L
2L
5
t
4L 4
4
t
5
b*
q>51!cbs-
ta~ka 2:
i3>!4121!
lK
lh
t3>!7/557!
ta~ka 3k:
q!>1/13!cbs-
y4l!>!1/84-
i4l!>!2979/:!!!
ta~ka 3:
q!>1/13!cbs-
ηfy
e =
i3 − i 4
i3 − i4l
u>431pD
⇒
(pregrejana para)
lK
lhL
t>7/557!
lK
lhL
lK
lh
ηefy>1/9
i4!>!i3!−! ηfy
e (i3 − i 4l ) =
i4!>!4121!−! 1/9 ⋅ (4121 3 − 2979/: ) = 31:8/2!
ta~ka 4:
q>1/13!cbs-
lK
i5!>!84/63!
lh
dipl.ing. @eqko Ciganovi}
(vla`na para)
y>1
lK
! (vla`na para)
lh
(kqu~ala te~nost)
lK
t5!>!1/3:1:!
lhL
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
q!>51!cbs-
ta~ka 1k:
i2l!>!88/:7!!!
ta~ka 1:
i 5 − i2l
ηlq
e =
i 5 − i2
strana 27
t>1/3:1:!
lK
)te~nost*
lhL
lK
lh
ηelq>1/:7
i −i
lK
i2!>!i5!.− 5 lq 2l = 89/25!
lh
ηe
q!>51!cbs-
⇒
84/63 − 89/25
lK
= 89/25!
(te~nost)
1/:7
lh
+ rpew
r
3:42/: − 3134/7
= ///!>
= !1/42
η!>! epw
3:42/:
repw
lK
lK
rpew!>!r45!>i5!−!i4!>!−!3134/7!
repw!>!r23!>!i3!!−!i2!>!3:42/:!
lh
lh
i2!>!84/63!.−
c*
UUJ − UUQ
6:4 − 3:1/6
= 1/62
= ///!>
6:4
UUJ
UUJ!>!U3!>!6:4!L
UUQ!>!U4!>!U5!>!)Ulmk*q>1/13!cbs!>!3:1/6!L
ηL!>!
6/32/ Idealni Rankin-Klauzijusov ciklus obavqa se sa vodenom parom izme|u pritisaka qnjo>1/15!cbs i
qnby>51!cbs, sa pregrejanom vodenom parom (u>571pD*!na ulazu u turbinu. Za rekuperativno zagrevawe
napojne vode (u zagreja~u me{nog tipa), do temperature od uC>215/9pD, iz turbine se pri pritisku pe
⋅
q4>2/3!cbs oduzima deo qbsf!) n 4>291!lh0i*!(slika). Zanemaruju}i radove napojnih pumpi, skicirati
proces na Ut dijagramu i odrediti:
a) termodinami~ki stepen korisnosti ovog kru`nog procesa
b) snagu parne turbine
Repw
!2
!3
!C
!4
Xuvs
!B
!6
!5
Rpew
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 28
U
3
2
B
4
C
6
5
t
b*
u>571pD
q>51!cbs-
ta~ka 2:
(pregrejana para)
lK
lK
- t3!>!7/:76!
i3!>!4464!
lhL
lh
ta~ka 3:
q!>2/3!cbs-
y4!>!1/:5-
i4!>!3659/5!!!
ta~ka 4:
q>1/15!cbs-
y5!>!1/92-
i5!>!31:2/9!!!
t>7/:76!
lK
lhL
(vla`na para)
lK
lhL
(vla`na para)
lK
lh
t>7/:76!
lK
lh
q>1/15!cbsy>1
lK
lK
-t6!>!1/5336!
i6!>!232/53!
lhL
lh
ta~ka 5:
ta~ka A = ta~ka 5
q>!2/3!cbs
ta~ka B:
iC!>!54:/5!
(kqu~ala te~nost)
(jer se zanemaruje rad pumpe)
u>215/9pD
)kqu~ala te~nost)
lK
lh
ta~ka 1 = ta~ka B!
dipl.ing. @eqko Ciganovi}
(jer se zanemaruje rad pumpe)
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
⋅
η=
⋅
R epw + R pew
⋅
>!///>!
R epw
⋅
⋅
strana 29
2218/28 − 761/34
>1/52
2218/28
⋅
R epw = R 23 = n⋅ (i 3 − i2 ) > 1/49 ⋅ (4464 − 54:/5 ) >2218/28!lX
⋅
⋅
⋅ ⋅
Rpew = R 56 = n− n4 ⋅ (i6 − i5 ) > (1/49 − 1/16 ) ⋅ (232/53 − 31:2/9 ) >−761/34lX
prvi zakon termodinamike za proces u me{nom zagreja~u vode:
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⇒
⋅
⋅
I2 = I3
⋅
⋅
⋅
⋅
i − iB
⋅ ⋅
n− n 4 ⋅ i B + n 4 ⋅ i 4 = n⋅ iC
⇒
n = n4 ⋅ 4
i
C − iB
⋅
⋅
i − iB
3659/5 − 232/53
lh
>1/49!
> 1/16 ⋅
n = n4 ⋅ 4
54:/5 − 232/53
t
iC − i B
c*
⋅
⋅
⋅
prvi zakon termodinamike za proces u parnoj turbini: R 23 = ∆ I23 + X 23
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅ ⋅
X 23 = − ∆ I23 > I2 − I3 > n⋅ i 3 − n 4 ⋅ i 4 + n− n 4 ⋅ i 5 >567/:5!lX> X uvs
6/33/ Vodena para obavqa Rankin-Klauzijusov ciklus (slika kao u prethodnom zadatku)
izme|u pritisaka qnjo>1/2!cbs i qnby>2!cbs. U kotlu se voda zagreva i isparava. Suvozasi}ena vodena
para ulazi u turbinu gde ekspandira kvazistati~ki adijabatski. Pri ekspanziji se iz turbine oduzima
jedan deo pare na pritisku od q>1/4!cbs i koristi za rekuperativno zagrevawe napojne vode u me{nom
zagreja~u od temperature koja vlada u kondenzatoru do temperature od 7:/23pD. Zanemaruju}i radove
napojnih pumpi odrediti snagu turbine ako kotao proizvodi 2!lh0t pare i skicirati procese na Ut
dijagramu.
U
2
B
3
4
C
6
5
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 30
q>2!cbsy>2
lK
lK
- t3!>!8/47!
i3!>!3786!
lhL
lh
ta~ka 2:
ta~ka 3:
q!>1/4!cbs-
y4!>!1/:5-
i4!>!3595/:!!!
ta~ka 4:
q>1/2!cbs-
y5!>!1/9:-
i5!>!3431/:!!!
ta~ka 5:
q>1/2!cbs-
(suva para)
t>8/47!
lK
lhL
(vla`na para)
lK
lhL
(vla`na para)
lK
lh
t>8/47!
lK
lh
y>1
(kqu~ala te~nost)
lK
i6!>!2:2/:!
lh
ta~ka A = ta~ka 5!
(jer se zanemaruje rad pumpe)
q>!1/4!cbs-
ta~ka B:
iC!>!39:/4!
u>7:/23pD
(kqu~ala te~nost)
lK
lh
ta~ka 1 = ta~ka B!
(jer se zanemaruje rad pumpe)
prvi zakon termodinamike za proces u me{nom zagreja~u vode:
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⇒
⋅
⋅
I2 = I3
⋅
⋅
⋅ ⋅
n− n 4 ⋅ i B + n 4 ⋅ i 4 = n⋅ iC
⋅
39:/4 − 2:2/3
lh
n4 > 2 ⋅
>1/154!
3595/: − 2:2/3
t
⇒
⋅
⋅
n 4 = n⋅
iC − i B
i4 − i B
⋅
⋅
⋅
prvi zakon termodinamike za proces u parnoj turbini: R 23 = ∆ I23 + X 23
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅ ⋅
X 23 = − ∆ I23 > I2 − I3 > n⋅ i 3 − n 4 ⋅ i 4 + n− n 4 ⋅ i 5 >!453!lX!> X uvs
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 31
6/34/ Parno turbinsko postrojewe (slika), radi po Rankinovom kru`nom procesu. U parnoj turbini
nekvazistati~ki adijabatski {iri se pregrejana vodena para stawa 3)q>36!cbs-!u>571pD* do pritiska q5
>1/15!cbs. Deo pare pri pritisku q4>4!cbs!se oduzima iz turbine radi regenerativnog zagrevawa napojne
vode (u zagreja~u vode povr{inskog tipa) od temperature!)uLMK*Q5!do temperature )uLK*Q4. Ako prvi deo
turbine (do oduzimawa pare) radi sa stepenom dobrote ηefy>1/:!i masenim protokom
⋅
⋅
n >23/6!lh0t!i ako je korisna snaga turbine X uvs >22!NX, zanemaruju}i rad napojnih pumpi odrediti:
a) toplotni protok koji para predaje okolini u kondenzatoru
b) stepen dobrote adijabatske ekspanzije u drugom delu turbine (nakon oduzimawa pare)
c) termodinami~ki stepen korisnog dejstva ciklusa
d) skicirati procese sa vodenom parom na Ut dijagramu
Repw
!2
!3
!C
⋅
⋅
n
!4
n4
Xuvs
⋅
⋅
n− n 4
!B
!6
!5
Rpew
U
!3
2
B
4
C
6
dipl.ing. @eqko Ciganovi}
!4l
5l
!5
!t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 32
u>571pD
lK
t3!>!8/313!
lhL
q>36!cbs-
ta~ka 2:
lK
i3!>!4484!
lh
q!>4!cbs-
ta~ka 3k:
i4l!>!3927/:!!!
lK
lhL
(pregrejana para)
lK
lh
q!>4!cbs-
ta~ka 3:
i4!>!3982/8!!!
t>8/313!
(pregrejana para)
lK
lh
t4!>8/433!
q>1/15!cbs-
ta~ka 4k:
y5l!>!1/97-
ηfy
e =
i3 − i 4
>1/:
i3 − i4l
lK
lhL
t>8/433!
i5l!>!3324/5!!!
q>1/15!cbs-
ta~ka 5:
lK
lhL
(vla`na para)
lK
lh
y>1
(kqu~ala te~nost)
lK
i6!>!232/53!
lh
ta~ka A = ta~ka 5!
(jer se zanemaruje rad pumpe)
q>!1/15!cbs-
ta~ka B:
y>1
(kqu~ala te~nost)
lK
iC!>!672/5!
lh
ta~ka 1 = ta~ka B!
(jer se zanemaruje rad pumpe)
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
ograni~enom isprekidanom konturom:
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⇒
⋅
⋅
⋅
⋅ ⋅
n− n 4 ⋅ i B + n 4 ⋅ i 4 = n⋅ iC
⋅
672/5 − 232/53
lh
>3!
n 4 > 23/6 ⋅
3982/8 − 232/53
t
dipl.ing. @eqko Ciganovi}
⋅
I2 = I3
⇒
⋅
⋅
n 4 = n⋅
iC − i B
i4 − i B
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 33
⋅
⋅
⋅
prvi zakon termodinamike za proces u parnoj turbini: R 23 = ∆ I23 + X 23
⋅
⋅
⋅
⋅
⋅
⋅ ⋅
⋅
⋅
X 23 = − ∆ I23 > I2 − I3 > n⋅ i 3 − n 4 ⋅ i 4 + n− n 4 ⋅ i 5 >! X uvs
⋅
i5!>!
⋅
⋅
n⋅ i 3 − X uvs − n 4 ⋅ i 4
⋅
⋅
n− n4
i′!=!i5!!=!i″
>
23/6 ⋅ 4484 − 22 ⋅ 21 4 − 3 ⋅ 3982/8
lK
>3531/:!
23/6 − 3
lh
ta~ka 4 je u vla`noj pari
b*
⋅
⋅
⋅ ⋅
R lpoe!>! R 56> n− n4 ⋅ (i 6 − i 5 ) !>! 2(23/6 − 3 ) ⋅ (232/53 − 3531/: ) >−35/2!NX
c*
ηfy
e =
i4 − i 5
3982/8 − 3531/:
>
>1/79
i4 − i5l 3982/8 − 3324/5
c)
⋅
⋅
R epw + R pew
η=
⋅
>!///>!
R epw
⋅
⋅
46/2 − 35/2
>1/42
46/2
⋅
R epw = R 23 = n⋅ (i 3 − i2 ) > 23/6 ⋅ (4484 − 672/5 ) >46/2!NX
⋅
⋅
⋅ ⋅
R epw = R 56 = n− n 4 ⋅ (i 6 − i 5 ) > (23/6 − 3 ) ⋅ (232/53 − 3531/: ) >−!35/2!NX
6/35/ Sa vodenom parom kao radnim telom, izvr{ava se Rankin−Klauzijus−ov kru`ni proces sa
maksimalnom regeneracijom toplote. Regeneracija toplote, koja se odvija sa beskona~no mnogo predajnika
toplote povr{inskog tipa, naizmeni~no povezanih sa beskona~no mnog toplotno izolovanih turbina,
vr{i se sa ciqem predgrevawa napojne vode pre ulaza u parni kotao. Kru`ni proces se odvija izme|u
pritisaka qnjo>1/16!cbs i qnby>61!cbs i najve}om temperaturom u tokou procesa od unby>511pD. Kotao
⋅
proizvodi n >1/2!lh0t!pare, a procesi u turbinama su ravnote`ni (kvazistati~ki). Skicirati proces na
Ts dijagramu i zanemaruju}i rad napojne pumpe, odrediti:
a) termodinami~ki stepen korisnosti kru`nog procesa
b) snagu turbine visokog pritiska, UWQ
c) snagu turbine niskog pritiska, UOQ, (sve turbine osim prve)
d) relativno pove}awe stepena korisnog dejstva (%) u odnosu na ciklus bez regeneracije toplote(sa samo
jednom turbinom)
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 34
U
U
3 Unby
2
4
B
B
4
2
qnby
5
qnjo
6
5
t
t
napomena:
{rafiran povr{ina (ispod linije A−1) i povr{ina ispod
stepenaste linije 3−4 je jednaka i predstavqa maksimalno
mogu}u regenerisanu (rekuperisanu) toplotu u ovom ciklusu
q>1/16!cbsy>1
lK
lK
-!
t6>!1/5872
i6!>!248/94!
lhL
lh
ta~ka 5:
ta~ka A = ta~ka 5!
(jer se zanemaruje rad pumpe)
q>61!cbsy>1
lK
lK
t2>3/:32!
i2!>!2265/5!
lhL
lh
ta~ka 1:
u>511pD
lK
t3!>!7/75!
lhL
q>61!cbs-
ta~ka 2:
lK
i3!>!42:4!
lh
(kqu~ala te~nost)
!u2>374/:2pD
(pregrejana para)
u4>!u2>374/:2pD-!!!!!t4>t3!>!7/75!
ta~ka 3:
i4!>!3:51!
(kqu~ala te~nost)
lK
lh
lK
lhL
(pregrejana para)
(ova vrednost se ~ita sa it dijagrama za vodenu paru)
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 35
∆t45!>!−∆tB2!(uslov ekvidistantnosti)
q>1/16!cbs-
ta~ka 4:
t5!−!t4!>!tB!−!t2!!!!!⇒!!!!!t5!>!tB!−!t2!,!t4!>1/5872!−!3/:32!,7/75!>5/2:6!
y5>1/58
!i5>2387/8
lK
lhL
lK
lh
a)
η!>!
repw + rpew
3349/7 − 2249/:
= ///!>
= !1/55
3349/7
repw
lK
lh
lK
rpew!>!r56!>i6!−!i5!>!248/94!−!2387/8>−2249/:!
lh
repw!>!r23!>!i3!!−!i2!>!42:4!−!2265/5!>!3149/7!
c*
prvi zakon termodinamike za proces u UWQ:
⋅
⋅
⋅
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⋅
⋅
X 23 = − ∆ I23 > I2 − I3 > n⋅ (i3 − i 4 ) >! 1/2⋅ (42:4 − 3:51) >36/4!lX!> X UWQ
d*
prvi zakon termodinamike za proces u UOQ:
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⋅
⋅
X 23 = R 23 − ∆ I23 > R sfl − ∆ I23 > n⋅ (i B − i2 ) − n⋅ (i 5 − i 4 ) >! X UOQ
⋅
⋅
X UOQ > 1/2 ⋅ (248/94 − 2265/5 ) − 1/2 ⋅ (2387/8 − 3:51) >75/8!lX!> X UOQ
e*
U
3
B
6
C
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
η′!>!
strana 36
r epw (+r pew (
4166/28 − 29:1/18
= ///!>
= !1/49
r epw (
4366/28
lK
lh
lK
rpew′>!rC6!>i6!!−!iC!>!248/94!−!3138/:>−29:1/18!
lh
repw′!>!rB3!>!i3!!−!iB!>!42:4!−!248/94!>!4166/28!
ta~ka C
qC>!1/16!cbs-!!!!!tC!>t3!>!7/75!
y4!>!1/89-
i4!>!3138/:!
η′!;!211!>!)!η!−!η′!*!;!y
lK
lhL
(pregrejana para)
lK
lh
⇒
y=
η − η( 1/55 − 1/49
⋅ 211 >26/9&
>
η(
1/49
zadatak za ve`bawe (1.25.)
6/36/ Parni kotao proizvodi paru stawa 3)q>31!cbs-!u>471pD*/!Para se po izlasku iz kotla deli: jedan
deo ide u turbinu, a drugi deo se prigu{uje. Prigu{ena para se zatim me{a sa onom koja je
kvazistati~ki adijabatski ekspandirala u turbini, a dobijena me{avina odvodi u kondenzator u kojoj
se kondezuje na 231pD. Dobijeni kondenzat se pumpom vra}a u kotao. Snaga turbine iznosi 2!NX, a
toplota predana okolini u kondenzatoru iznosi 6/:!NX. Skicirati procese sa paroma na Ut
koordinatnom sistemu i odrediti:
a) koliko pare proizvodi kotao, koliko se prigu{uje a koliko ide u turbinu
b) termodinami~ki stepen korisnosti ovog postrojewa
,R23
2
3
XQ
XUVS
4
7
5
6
−R67
⋅
re{ewe:
nlpubp >3/7
lh ⋅
lh ⋅
lh
-! n uvscjob >2/:5
- n wfoujm >1/77
-!η>1/25
t
t
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 37
6/37/ Vodena para stawa 2)u>511pD-!q>91!cbs*!kvazistati~ki izentropski ekspandira u turbini visokog
pritiska do stawa!4)q>5!cbs*-!posle ~ega joj se izobarski dovodi r45>599!lK0lh toplote. Nakon dovo|ewa
toplote para kvazistati~ki izentropski ekspandira u turbini niskog pritiska do stawa )q>1/19!cbs*/
Proces se daqe nastavqa po idealnom Rankinovom ciklusu (slika). Skicirati ciklus na Ut dijagramu i
odrediti termodinami~ki stepen korisnosti ovog kru`nog procesa.
2
3
,R23
XUWQ
5
4
XQ
,R45
XUOQ
7
6
−R67
U
3
5
2
4
7
6
t
ta~ka 2:
u>511pD
lK
t3!>!7/469!
lhL
q>91!cbs-
lK
i3!>!4246!
lh
dipl.ing. @eqko Ciganovi}
(pregrejana para)
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
ta~ka 3:
q!>5!cbs-
y4!>!1/9:-
i4!>!3625!!!
ta~ka 4:
q!>5!cbs-
i5!>!4113!!!
lK
lh
y6!>!1/:1-
q>1/19!cbs-
lK
-!
i7!>!284/:!
lh
r45>599!
lK
lh
(vla`na para)
lK
lh
lK
lhL
t6!>t5>8/558!
(pregrejana para)
lK
lhL
(vla`na para)
lK
lh
y>1
(kqu~ala te~nost)
lK
t7>!1/6:38
lhL
q!>91!cbs-
ta~ka 1:
η!>!
lK
lhL
i6!>!3446/9!!!
ta~ka 6:
i2!>!293/7!!!
t>7/469!
t5>8/558!
q>1/19!cbs-
ta~ka 5:
strana 38
t2>!t7>!1/6:38!
lK
lhL
(te~nost)
lK
lh
repw + rpew
4551/5 − 3268/2
= ///!>
= !1/48
4551/5
repw
repw!>!r23!,!r45!>!i3!!−!i2!,!r45!>!4246!−!293/7!,!599!>!4551/5!
rpew!>!r67!>!i7!−!i6!>!284/:!−!3446/9>−3268/2!
dipl.ing. @eqko Ciganovi}
lK
lh
lK
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 39
6/38/ Parno turbinsko postrojewe radi po Rankin-Klauzijus-ovom kru`nom procesu sa dvostepenim
adijabatskim {irewem vodene pare. Pregrejana vodena para stawa 2)q2>211!cbs-!u2>551pD* {iri se u
turbini visokog pritiska nekvazistati~ki, sa stepenom dobrote ηEUWQ>1/:, do pritiska!q3>6!cbs.
Potom se para izobarski zagreva do temperature u4>411pD, nakon ~ega se, u turbini niskog pritiska,
nekvazistati~ki {iri, sa stepenom dobrote ηEUOQ>1/9, do pritiska q5>1/16!cbs, koji vlada u
kondenzatoru.
a) da li je termodinami~ki stepen korisnosti ovog kru`nog procesa mogu}e dosti}i u RankinKlauzijus-ovom kru`nom procesu sa jednostepenim adijabatskim {irewem vodene pare stawa 2 do
pritiska q5, uz maksimalno pove}awe stepena dobrote procesa u turbini
b) koliko mimimalno mora da iznosi stepen dobrote jednostepene adijabatske ekspanzije da bi
dostigli stepen korisnog dejstva koji ima navedeni ciklus sa dvostepenom adijabatskom
ekspanzijom
U svim slu~ajevima zanemariti rad napojne pumpe.
2
!i
4
3l
3
5l
5
1
6
!t
ta~ka 1:
u>551pD
q>211!cbs-
(pregrejana para)
lK
lK
- t2!>!7/488!
i2!>!4322!
lhL
lh
ta~ka 2k:
q!>6!cbs-
y3l>1/:2-
i3l!>!366:/3!
dipl.ing. @eqko Ciganovi}
t3l>t2>7/488!
lK
lhL
(vla`na para)
lK
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
ta~ka 2:
strana 40
ηeuwq =
q!>6!cbs-
i3!>! i2 − η euwq ⋅ (i2 − i 3l ) >3735/5!!!
ta~ka 3:
i2 − i3
>1/:
i2 − i3l
lK
lh
(vla`na para)
u>411pD
q>6!cbs-
(pregrejana para)
lK
lK
- t4!>!8/574!
i4!>!4173!
lhL
lh
ta~ka 4k:
q>1/16!cbs-
y5l!>!1/99-
i5l!>!3381/3!
ta~ka 4:
q!>1/16!cbs-
t>8/574!
q>1/16!cbs-
(vla`na para)
lK
lh
ηeuoq =
i5!>! i 4 − η euoq ⋅ (i 4 − i 5l ) >3544/6!
ta~ka 5:
lK
lhL
i2 − i3
>1/9
i2 − i3l
lK
lh
y>1
(vla`na para)
(kqu~ala te~nost)
lK
i6!>!248/94!
lh
ta~ka 0 = ta~ka 5:
η!>!
(jer se zanemaruje rad pumpi)
repw + rpew
4621/9 − 33:6/67
= ///!>
= 1/46
repw
4621/9
repw!>!r12!,!r34>!i2!−!i1!,!i4!−!i3!>!4322−248/94!,!4173!−3735/5>4621/9!
rpew!>!r56!>!i6!−!i5!>!248/94!−!3544/6!>−!33:6/67!
dipl.ing. @eqko Ciganovi}
lK
lh
lK
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 41
ciklus sa jednostepenom ekspanzijom sa maksimalnim stepenom dobrote!ηe>2
2
!i
B
1
6
!t
ta~ka A:!
q!>1/16!cbs-
yB>1/85-
iB!>!2:41/:!!!
η′!>!
(
(
+ rpew
repw
(
repw
= ///!>
t>7/488!
kJ
(vla`na para)
kgK
lK
lh
4184/3 − 28:4/26
= !1/53
4184/3
(
repw
!>!r12!>!i2!−!i1!>!4322−248/94!!>!4184/3!
lK
lh
(
rpew
!>!rB6!>i6!−!iB!>!248/94!−!2:41/:!>−!28:4/26!
lK
lh
kako je η′!?!η- u ciklusu sa jednostepenom adijabatskom ekspanzijom sa maksimalnim pove}awem
stepena dobrote ekspanzije ( ηfy
e = 1) mo`e se dosti}i stepen korisnog dejstva navedenog ciklusa
sa dvostepenom adijabatskom ekspanzijom.
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 42
c*!
ciklus sa jednostepenom adijabatskom ekspanzijom sa stepenom
dobrote ekspanzije ηnjo
e
2
!i
B
C
1
6
!t
η>1/42
((
r epw
!>!r12!>!i2!−!i1
((
r pew
>!rC6!>i6!−!iC
!η!>!
((
((
r epw
+ r pew
((
r epw
=
i2 − i1 + i6 − iC
i2 − i1
⇒
iC> i2 − i 1 + i 6 − η ⋅ (i2 − i 1 )
iC!> 4322 − 248/94 + 248/94 − 1/42 ⋅ (4322 − 248/94 ) >!3369/4!
ηnjo
=
e
lK
lh
i2 − iC
4322 − 3369/4
>
>1/85
4322 − 2:41/:
i2 − i B
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 43
6/39/ Parno turbinsko postrojewe radi sa dvostepenim {irewem i me|uzagrevawem pare uz jednostepeno
regenerativno zagrevawe napojne vode od temperature koja vlada u kondenzatoru do temperature od
uC>323pD (slika). Zanemaruju}i radove napojnih pumpi i ako je:
− pritisak pare u kondenzatoru 7!lQb
− pritisak pare u kotlu 23!NQb
− pritisak pare na izlazu iz turbine visokog pritiska q>5!NQb
− temperatura pare na ulazu u turbinu visokog pritiska u>641pD
− temperatura pare na ulazu u turbinu niskog pritiska u>641pD
⋅
− protok pare kroz turbinu visokog pritiska n =1/5!lh0t
⋅
⋅
− protok pare kroz turbinu niskog pritiska n− n4 =1/4!lh0t
− stepen dobrote adijabatske ekspanzije u turbini niskog pritiska ηeuoq>1/93
a) odrediti stepen dobrote adijabatske ekspanzije u turbini visogog pritiska, ηeuwq
b) odrditi termodinami~ki stepen korisnog dejstva kru`nog procesa
c) skicirati promene stawa vodene pare na hs dijagramu
2
3
⋅
n
R23
XUWQ
C
⋅
n4
4
5
R45
B
7
XUOQ
6
R67
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
a)
η euwq =
strana 44
i3 − i4
4537 − 4286/6
= ///>
= 1/87
4537 − 41:8/7
i 3 − i 4l
lK
i3!>!4537!
lh
u>!641pD
lK
t3!>!7/6996!
lhL
ta~ka 3k:
q!>51!cbs-
q>231!cbs-
ta~ka 2:
i4l!>!41:8/7!!!
t>7/6996!
(pregrejana para)
lK
(pregrejana para)
lhL
lK
lh
q!>51!cbs
ta~ka 3:
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
ograni~enom isprekidanom konturom:
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⇒
⋅
⋅
⋅
⋅ ⋅
n− n 4 ⋅ i B + n 4 ⋅ i 4 = n⋅ iC
ta~ka 6:
q>!1/17!cbs-
⋅
I2 = I3
⇒
i4 =
⋅
⋅ ⋅
n⋅ iC − n− n 4 ⋅ i B
⋅
>///
n4
y>1
(kqu~ala te~nost)
lK
i7!>!262/6!
lh
ta~ka A = ta~ka 6
ta~ka B:
q!>!51!cbs-
(jer se zanemaruje rad pumpi)
u>!323pD
(voda)
lK
iC!>!:18/6!
lh
i4 =
1/5 ⋅ :18/6 − 1/4 ⋅ 262/6
lK
>!4286/6!
1/2
lh
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 45
b)
⋅
η=
⋅
R epw + R pew
⋅
>!///>!
R epw
2219/46 − 797/8
>1/49
2219/46
⋅
⋅
⋅
⋅ ⋅
R epw = R 23 + R 45 = n⋅ (i 3 − i2 ) + n− n 4 ⋅ (i 5 − i 4 ) >
⋅
⋅
R epw > 1/5 ⋅ (4537 − :18/6 ) + 1/4 ⋅ (4625 − 4288/6 ) >2219/46!lX
⋅
⋅
⋅ ⋅
R pew = R 67 = n− n 4 ⋅ (i 7 − i 6 ) > 1/4 ⋅ (262/6 − 3551/5 ) >−!797/8!lX
u>641pD
ta~ka 4:
q>51!cbs-
lK
i5!>!4625!
lh
lK
t5!>!8/2856!
lhL
ta~ka 5k:
q!>1/17!cbs-
y6l!>!1/96-
i6l!>!3315/8!!!
ta~ka 5:
q>1/17!cbs-
t>8/2856!
(pregrejana para)
lK
lhL
(vla`na para)
lK
lh
ηeuoq =
i 5 − i6
i5 − i6l
i6!>!i5!−! η euoq ⋅ (i 5 − i 6l ) = 4625!−! 1/93 ⋅ (4625 − 3315/8 ) = 3551/5!
ta~ka 1 = ta~ka B
lK
lh
(jer se zanemaruje rad pumpi)
5
3
i
4
4l
6
6l
2
C
B
7
dipl.ing. @eqko Ciganovi}
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 46
⋅
6/3:/ Parni kotao proizvodi n =31!u0i pare stawa 2)q3>27!cbs-!u3>511pD*/ Para u turbini ekspandira
⋅
u dva stepena. Nakon prvog stepena, deo pare ( n4 )se odvodi za potrebe nekog spoljnog predajnika
toplote u kojem se vr{i potpuna kondenzacija pare na u>291pD pri ~emu se od pare odvodi 3!NX
toplote. Tako nastalo kondenzat se ne vra}a u kotao, nego ispu{ta u okolinu, a umesto wega se u kotao
dodaje ista koli~ina vode iz okoline stawa )q>2!cbs-!u>26pD*. Ostatak pare ekspandira u drugom
stepenu turbine a zatim odvodi u kondenzator, u kome vlada temperatura od 41pD. Ekspanzije u
turbinama su ravnote`ne (kvazistati~ke) i adijabatske. Zanemaruju}i snage napojnih pumpi, skicirati
proces na Ts dijagramu i odrediti:
⋅
a) maseni protok sve`e vode, n4
b) ukupnu snagu koja se dobije u turbinama
c) termodinami~ki stepen korisnog dejstva ciklusa
3
5
4
,R23
−R47
−R56
2
7
1
6
U
3
4
2
6
7
5
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 47
u>!511pD
ta~ka 2:
q>27!cbs-
lK
i3!>!4364!
lh
lK
t3!>!8/344!
lhL
ta~ka 3:
q!> (q)u =291p D >21!cbs- t4>!t3!>!8/344!
i4!>!4228/4!!!
lK
lhL
(pregrej. para)
lK
lh
ta~ka 4:
u>!41pD
y5!>1/96
i5!>!32:2/5!
ta~ka 5:
u>!41pD-
lK
i6!>!236/82!
lh
ta~ka 1:
i2!>!239/2!!!
(pregrejana para)
t5!>!t4>!t3!>!8/344!
lK
(vla`na para)
lhL
lK
lh
y>1
(kqu~ala te~nost)
lK
t6!>1/5477!
lhL
q!>27!cbs-
t2>!t6!>!1/5477!
u>!291pD-
y>1
lK
lhL
(te~nost)
lK
lh
ta~ka 6:
(kqu~ala te~nost)
lK
i7!>!874/2!
lh
b*
prvi zakon termodinamike za proces u spoqnom predajniku toplote:
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⇒
⋅
⋅
R qsfebkojlb = n4 ⋅ (i 7 − i 4 )
⋅
lh
− 3 ⋅ 21 4
R qsfebkojlb
>1/96
>
n4 =
874/2 − 4228/4
t
i7 − i4
⋅
b)
prvi zakon termodinamike za proces u turbini visokog pritiska:
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⇒
⋅
⋅
X uwq = n⋅ (i 3 − i 4 ) >
31 ⋅ 21 4
⋅ (4364 − 4228/4 )
4711
⋅
X uwq >1/86!NX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 48
prvi zakon termodinamike za proces u turbini niskog pritiska:
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⇒
⋅
31 ⋅ 21 4
⋅ ⋅
X uoq = n− n4 ⋅ (i 4 − i 5 ) >
− 1/96 ⋅ (4228/4 − 32:2/5 ) >
4711
⋅
X uwq >5/47!NX
c)
prvi zakon termodinamike za proces u parnom kotlu:
⋅
⋅
⋅
⇒
R 23 = ∆ I23 + X 23
⋅
⋅
R lpumb = n⋅ (i3 − i2 ) >
31 ⋅ 21 4
⋅ (4364 − 239/2)
4711
⋅
R lpumb >28/47!NX
⋅
η=
⋅
X uwq + X uoq
⋅
>
R lpumb
1/86 + 5/47
>1/3:
28/47
zadatak za ve`bawe (1.30.)
6/41/ Parno turbinsko postrojewe radi po Rankin−Klauzijus−ovom kru`nom procesu sa dvostepenim
adijabatskim {irewem vodene pare(slika kao u zadatku 1.26) . Pregrejana vodena para stawa 3)q>21
NQb-!u>551pD* {iri se u turbini visokog pritiska nekvazistati~ki, sa stepenom dobrote η euwq >1/:, do
pritiska od q4>1/6!NQb. Potom se izobarski zagreva do temperature od u5>411pD, nakon ~ega se, u
turbini niskog pritiska, {iri nekvazistati~ki, sa stepenom dobrote η euoq >1/9 do pritiska od
q6>1/116!NQb, koji vlada u kondenzatoru. Skicirati proces na it dijagramu i zanemaruju}i snage
napojnih pumpi odredit stepen korisnosti posmatranog kru`nog procesa.
re{ewe:
η>1/476
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 1
6. LEVOKRETNI KRU@NI PROCESI
7/2/ Odrediti minimalan rad koji treba ulo`iti da se od nekog tela, konstantne temperature u>!−24pD,
oduzme 21!lK!toplote i preda okolnom vazduhu, konstantne temperature od 48pD. Koliko se toplote u tom
slu~aju predaje okolnom vazduhu.
Najmawe rada se mora ulo`iti ako levokretna tolotna ma{ina radi po
Karnoovom levokretnom ciklusu (sve promene stawa radne materije su
povratne).
U
4
3
5
2
UUQ
UUJ
!t
UUJ>−24pD!>!371!LR epw
Xofup
=
UUQ>48pD>421!L-
Repw!>21!lK
Uuq − Uuj
Uuj
421 − 371
!> 21 ⋅
!!!!!! ⇒ !!!!!! Xofup = R epw ⋅
>2/:3!lK
Uuq − Uuj
371
Uuj
R pew >!Repw!,! Xofup !>!21!,!2/:3!>!22/:3!lK
7/3/ Rashladni ure|aj (slika) koristi kao radnu materiju vazduh (idealan gas) i radi po levokretnom
kru`nom Xulovom procesu. Stawe vazduha na ulazu u izentropski kompresor je 2)u2>−21pD-!q2>2!cbs*- a na
izlazu iz kompresora 3)q3>5!cbs*/!Temperatura vazduha na ulazu u izentropsku turbinu je u4>31pD. Maseni
protok vazduha kroz rashladni ure|aj 2311!lh0i a sve promene stawa radne materije su ravnote`ne
(kvazistati~ke). Skicirati promene stawa vazduha na Ut dijagramu i odrediti:
a) rashladni efekat instalacije )lX*
b) koeficijent hla|ewa instalacije, εi
c) ako je svrha rashladnog ure|aja proizvodwa leda temperature um>−4pD od vode temperature ux>21pD,
odrediti masu proizvedenog leda za vreme od τ>2!i
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 2
U
4
3
3
3
2
5
5
2
!!!ledomat
t
a)
U3!>!U2 ⋅ q 3
q
2
U5!>!U4 ⋅ q 5
q
4
⋅
⋅
κ −2
κ
κ −2
κ
> 374 ⋅ 5
2
> 3:4 ⋅ 2
5
⋅
2/5 −2
2/5
2/5 −2
2/5
R epw = R 52 = n⋅ d q ⋅ (U2 − U5 ) >
>!4:1/9!L
>2:8/3!L
2311
⋅ 2 ⋅ (374 − 2:8/3 ) >32/:4!lX
4711
b)
⋅
εi!>!
R epw
⋅
X ofup
⋅
= ///!>
32/:4
>!3/16
21/78
⋅
⋅
⋅
X ofup > X lpnqsftps!, X uvscjob!> n⋅ d q ⋅ (U2 − U3 + U4 − U5 )
⋅
X ofup >
2311
⋅ 2 ⋅ (374 − 4:1/9 + 3:4 − 2:8/3 ) !>!−21/78!lX
4711
c)
⋅
32/:4 ⋅ 4711
R epw ⋅ τ
nmfe =
= /// =
>318/6!lh
53 + 449/5
i x − im
lK
)q>!2!cbs-!u>21pD*
ix!>!53!
lh
im = d m ⋅ (Um − 384 ) − sm > 3 ⋅ (−4 ) − 443/5 >−449/5!
dipl.ing. @eqko Ciganovi}
lK
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 3
7/4/ Helijum (idalan gas) obavqa realan levokretni Xulov proces sa potpunim rekuperativnim
(regenegrativnim) zagrevawem radne materije. Rashladna snaga ovog postrojewa je 33!lX. Temperatura
u rashladnoj komori je stala i jednaka je temperaturi na ulazu u gasnu turbinu UUJ>U4!>356!L>dpotu.
Temperatura okoline je stalna i jednaka temperaturi na ulazu u kompresor UUQ>U2>431!L. Odnos
q
pritiska na ulazu i izlazu iz gasne turbine iznosi 4 >3/2. Stepeni dobrote u adijabatskom
q5
fy
kompresoru i adijabatskoj turbini su jednaki i iznose ηlq
e = η e >1/93. Prikazati ovaj proces u Ut
koordinatnom sistemu i odrediti:
a) neto snagu potrebnu za pogon ovog postrojewa
b) faktor hla|ewa ovog postrojewa
4
!C
5
R
E
K
U
P
E
R
A
T
O
R
!B
3
Rpew
2
Repw
3
U
3l
B
2
!Rsfl
C
4
5l
!UQ
!UJ
5
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 4
a)
κ −2
2/78 −2
κ
>! 431 ⋅ (3/2) 2/78 >541/:!L
U3L!>!U2 ⋅ q 3L
q
2
U −U
541/:6 − 431
U3!>!U2!,! 3L lq 2 !>! 431 +
>566/4!L
1/93
ηe
κ −2
2/78 −2
κ
> 356 ⋅ 2 2/78 >293/2!L
U5L!>!U4 ⋅ q 5L
q
3/2
4
U
U5!>!U4!, η e ⋅ (U5l − U4 ) !>! 356 + 1/93 ⋅ (293/2 − 356 ) >2:4/5!L
⋅
⋅
⋅
⋅
R epw = R 5C = n⋅ d q ⋅ (UC − U5 ) ⇒
⋅
n=
⋅
R epw
n=
>
d q ⋅ (UC − U5 )
33
lh
>9/3!/21.3!
6/3 ⋅ (356 − 2:4/5 )
t
⋅
⋅
⋅
⋅
X ofup > X lpnqsftps!, X uvscjob!> n⋅ d q ⋅ (U2 − U3 + U4 − U5 )
⋅
X ofup > 9/3 ⋅ 21 −3 ⋅ 6/3 ⋅ (431 − 566/4 + 356 − 2:4/5 ) !>!−46/8!lX
UC>U3, uslov potpune (maksimalne) regeneracije
(rekuperacije) toplote za Xulov ciklus
napomena:
b)
⋅
εi!>!
R epw
⋅
>
⋅
R pew − R epw
⋅
⋅
33
>!1/73
68/8 − 33
⋅
R pew = R 3 B = n⋅ d q ⋅ (UB − U3 ) > 9/3 ⋅ 21 −3 ⋅ 6/3 ⋅ (431 − 566/4 ) >−68/8!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 5
7/5. Rashladno postrojewe (slika) koristi kao radni fluid freon 12 )S23*/ Temperatura isparavawa je
354!L, a teperatura kondenzacije 426!L. Snaga kompresora u kojem se vr{i kvazistati~ko adijabatsko
sabijawe freona iznosi 1/94!lX/ Skicirati promene stawa S23 u Ut i it koordiantnom sistemu i
odrediti:
⋅
a) rashladni kapacitet ) R epw* i koeficijent hla|ewa ovog postrojewa )εi)
b) ako bi se kqu~ala te~nost S23 pre prigu{ivawa podhladila za ∆U>21!L koliko bi tada iznosio
⋅ -
⋅-
rashladni kapacitet ) R epw * i koeficijent hla|ewa ovog postrojewa ) ε i ) koeficijent hla|ewa
c) na Ut dijagramu {rafirati povr{inu koja predstavqa pove}awe rashladnog kapaciteta postrojewa
usled pothla|ivawem kondenzata pre prigu{ivawa
4
3
⋅
⋅
R pew
n
⋅
X lq
⋅
5
2
n
⋅
n⋅ y 5
⋅
n ⋅ (2 − y 5
)
⋅
R epw
3
i
U
3
4
2
4
5
2
t
dipl.ing. @eqko Ciganovi}
5
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
!
strana 6
U>354!L>!−!41pD
ta~ka 1:
lK
i2!>!752/92!
lh
y>2
lK
t2!>!2/6993!
lhL
q!>!)q*Ul>426L!>!21/28!cbs
ta~ka 2:
i3!>!794/1:!
t!>!t2!>!2/6993!
lK
lhL
lK
lh
U>426!L>53pD
ta~ka 3:
y>1
lK
i4!>!652/58!
lh
i5!>!i4!>!652/58!
ta~ka 4:
lK
lh
a)
⋅
⋅
⋅
⋅
⋅
R epw = R 52 = n⋅ (i2 − i 5 ) >///> 3 ⋅ 21 −3 ⋅ (752/92 − 652/58 ) >3!lX
X lq = X u23
⋅
εi!>!
R epw
⋅
X lq
=
⋅
−1/94
lh
X lq
> 3 ⋅ 21 −3
= n⋅ (i2 − i 3 ) !!!!!!⇒!!!!!! n =
>
725/92 − 794/1:
t
i2 − i3
⋅
⋅
3
>3/52
1/94
b)
3
i
U
3
4
2
4
B
C
5
B
2
t
dipl.ing. @eqko Ciganovi}
C
5
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 7
ta~ka A:
iB! ≅ !)i′*Ub>416!L!>!642/22!
ta~ka B:
iC!>!iB!>!642/22!
⋅ -
⋅
lK
lh
lK
lh
⋅
R epw = R C2 = n⋅ (i2 − iC ) > 3 ⋅ 21 −3 ⋅ (752/92 − 642/22) >3/32!lX
⋅ -
εi!>!
R epw
⋅
X lq
=
3/32
>3/78
1/94
U
3
4
B
2
5
C
t
⋅
∆ R epw
⋅
6.5. Levokretni kru`ni proces obavqa se sa n =711!lh0i amonijaka )OI4* , izme|u Unjo>−24pD i
qnby>2!NQb. U toplotno izolovan kompresor ulazi suva para koja se nekvazistati~ki sabija do stawa
3)U3>211pD*/ Po izlasku iz kondenzatora vr{i se pothla|ivawe do temperature od 26pD. Skicirati
proces na Ut dijagramu i odrediti:
a) koeficijent hla|ewa
b) za koliko bi se pove}ala vrednost koeficijenta hla|ewa )&*, ako bi sabijawe u kompresoru bilo
kvazistati~ko
c) u{tedi u snazi za pogon kompresora u slu~aju kvazistai~kog sabijawa u odnosu na stvarno
nekvazistati~ko sabijawe )lX* i u Ut koordinatnom sistemu {rafirati povr{inu ekvivalentnu
toj u{tedi
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 8
U
3
3l
4
2
5
t
a)
u!>!−24pD
ta~ka 1:
lK
i2!>!3318!
lh
t2!>!21/5:3!
q!>!21!cbs
ta~ka 2:
i3!>!3552/6!
q>21!cbs
lK
i4! ≅ !)i′*u>26pD!>!2141/2!
lh
u!>211pD
u>26pD
i5!>!i4!>!2141/2!
ta~ka 4:
⋅ -
R epw
⋅
= ///!>
X lq
⋅
⋅
⋅
⋅
lK
lhL
lK
lh
ta~ka 3:
εi!>!
y>2
lK
lh
2:7/26
>6
4:/19
⋅
711
⋅ (3318 − 2141/2) >2:7/26!lX
4711
⋅
711
= n⋅ (i2 − i 3 ) >
⋅ (3318 − 3552/6 ) >−4:/19!lX
4711
R epw = R 52 = n⋅ (i2 − i 5 ) >
X lq = X u23
U
3
4
5
dipl.ing. @eqko Ciganovi}
2
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 9
b)
i3l>3513/:!
⋅ -
R epw
= ///!>
⋅ X lq
⋅
lK
lhL
lK
lh
⋅
εi!>!
t!>!t2>21/5:3!
q>21!cbs
ta~ka 2k:
2:7/26
>7
43/76
⋅
X lq = X u23l = n⋅ (i2 − i 3l ) >
711
⋅ (3318 − 3513/: ) >−43/76!lX
4711
ε
7
∆ε i (&) = i − 2 ⋅ 211& !> − 2 ⋅ 211& >31&
εi
6
d*
⋅
⋅ -
⋅
∆ X lq !>! X lq !.! X lq !>!7/54!lX
U
3
3l
∆Xlq
4
5
2
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 10
7/7/ Amonija~ni kompresorski rashladni ure|aj sa prigu{nim ventilom radi izme|u pritisaka
qnjo>4/92:!cbs!i!qnby>26!cbs. Kompresor usisava suvozasi}enu paru amonijaka i kvazistati~ki
adijabatski je sabija. Ure|aj je projektovan tako da iz prostorije koju hladi oduzima 61!lX toplote.
Odrediti:
a) snagu kompresora
c* koliko bi trebalo da iznosi stepen suvo}e vla`ne pare koja napu{ta kondenzator da bi
koeficijent hla|ewa iznosio εh=0
a)
q>4/92:!cbs
ta~ka 1:
lK
i2!>!332:!
lh
ta~ka 2:
i3!>!3528/6!
y>2
t2!>!21/465!
lK
lhL
q!>26!cbs
t!>!t2!>!21/465!
q>26!cbs
y>1
lK
lhL
lK
lh
ta~ka 3:
lK
i4!>!2255/2!
lh
i5!>!i4!>!2255/2!
ta~ka 4:
⋅
⋅
⋅
⋅
R epw = n⋅ (i2 − i 5 )
⇒
lK
lh
⋅
⋅
61
lh
R epw
n=
>
> 5/76 ⋅ 21 −3
t
i2 − i 5 332: − 2255/2
⋅
X lq = X u23 = n⋅ (i2 − i 3 ) > 5/67 ⋅ 21 −3 ⋅ (332: − 3528/6 ) >−:/16!lX
b)
i 4( = i2
⇒
i
332: − 2255/2
i − i(
y 4 = 2
>1/:8
>
i( (−i( q=26cbs 3363 − 2255/2
3
4′
2
4
5
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 11
7/8/!Levokretni kru`ni proces sa pothla|ivawem, prigu{ivawem, isparavawem i pregrevawem pare pre
ulaska u kompresor obavqa se sa freonom 12 )S23*- kao radnim telom (slika). Rashladni kapacitet
postrojewa je 6/9!lX, a snaga kompresora, koji vr{i nekvazistati~ko sabijawe pare freona, je 3!lX. Radna
materija obavqa ciklus izme|u pritisaka qnjo>1/2!NQb i qnby>1/7!NQb i pri tom dosti`e maksimalnu
temperaturu od 71pD. Temperatura pothla|ivawa je 27pD. Skicirati promene stawa radnog tela na qw i Ut
dijagramu i odrediti:
a) stepen dobrote adijabatske kompresije
b) {rafirati na Ut dijagramu potrebnu snagu kompresora
LE!,!QI
4
3
5
JT!,!QH
2
3
U
3l
4
5
2
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
ta~ka 2:
strana 12
q!>!7!cbs
u!>!71pD
q>7!cbs
u>27pD
lK
i3!>!7:1/6!
lh
ta~ka 3:
lK
i4! ≅ !)i′*u>27pD!>!626/3:!
lh
ta~ka 4:
i5!>!i4!>!626/3:!
ta~ka 1:
q2!>!2!cbs
⋅
⋅
R epw >6/9!lX
⋅
⋅
⋅
⋅
lK
lh
X lq >−3!lX
⋅
R epw = R 52 = n⋅ (i2 − i 5 )
⋅
X lq = X u23 = n⋅ (i2 − i 3 )
)2*
)3*
Kombinovawem jedna~ina (1) i (2) dobija se:!i2!>!756/7!
t2!>!2/716!
lK
-!
lh
n>55/6!
h
t
lK
lhL
q!>!7!cbs
ta~ka 2k:
i3l!>!789/8!
t!>!t2!>!2/716!
lK
lhL
lK
lh
b*
ηlq
e =
i2 − i3l
756/7 − 789/8
!>!
>1/85
i2 − i3
756/7 − 7:1/6
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 13
b)
3
U
3
U
3l
3l
4
4
2
2
5
5
t
t
⋅
⋅
R pew
R epw
⋅
⋅
⋅
X lq = R pew − R epw
3
U
3l
4
2
2
5
t
⋅
X lq
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 14
7/9/!Levokretna instalacija radi sa amonijakom kao radnim telom. Rashladni kapacitet postrojewa je
51!lX. Temperatura isparavawa je −44pD, pritisak u kondenzatoru 6!cbs-!a temperatura prehla|ivawa
−4pD. Toplota oslobo|ena prehla|ivawem kondenzata koristi se za pregrevawe suve pare amonijaka,
tako da u kompresor ulazi pregrejana para koja se sabija nekvazistati~ki adijabatski sa stepenom
dobrote ηLQ
e >1/9. Predstaviti kru`ni proces na Ut dijagramu i odrediti:
a) maseni protok amonijaka kroz instalaciju )lh0t*
b) snagu kompresora )lX*
5
4
3
6
7
2
U
3
3L
4
5
6
2
7
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
ta~ka 3:
i4>:88/:6!
strana 15
q>6!cbs-!
y>1
q>!6!cbs-!
U>381!L
lK
lh
ta~ka 4:
lK
i5>:56/8!!
lh
ta~ka 5:
i6!>!i5>:56/8!!
ta~ka 6:
U>351!L-
lK
lh
y>2
lK
i7!>!3288!
lh
i4!−!i5!>!i2!−!i7!
i2>!i4!−!i5!,!i7
i2>!:88/:6!−!:56/8!,!3288!>!331:/36
q3L>6!cbs-!
ta~ka 2K:
i3l>3537!
lK
lh
t3>!t2>21/:!
t2>21/:!
lK
lhL
lK
lhL
lK
lh
ηLQ
e >1/9
q3>!6!cbs-!
ta~ka 2:
ηLQ
e >
−∆i45!>!∆i72
q2!>!q7!>2/7647!cbs
ta~ka 1:
i2 − i 3L
i −i
331:/36 − 3537
lK
i 3 = i2 − 2 LQ 3L > 331:/36 −
>3591/2:!
i2 − i 3
1
/
9
lh
ηe
a)
⋅
⋅
R epw = n⋅ (i 7 − i 6 )
⇒
⋅
⋅
lh
R epw
51
> 4/36 ⋅ 21 −3
n=
=
t
i7 − i6
3288 − :56/8
b)
⋅
⋅
M lpnq = n⋅ (i2 − i3 ) = 4/36 ⋅ 21 −3 ⋅ (331:/36 − 3591/2: ) >!−9/9!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 16
7/:/!Rashladno postrojewe (slika) radi, sa freonom 12 kao radnim telom, izme|u pritisaka q2>366!lQb i
q3>2!NQb. U kompresoru snage 2/4!lX nekvazistati~ki adijabatski sabija se pare freona pri ~emu
specifi~na entropija freona (usled mehani~ke neravnote`e) poraste za ∆t23>4/3!K0lhL. Odrediti:
a) rashladnu snagu postrojewa (lX)
b) koeficijent hla|ewa postrojewa
c) {rafirati na Ut dijagramu povr{inu koja koja je evivalentna rashladnoj snazi postrojewa
!kondenzator
4
3
2
5
ispariva~
6
7
b*
q>3/66!cbs
y>2
lK
lK
t2!>!2/6829!
i2!>!764/12!
lhL
lh
ta~ka 1:
ta~ka 2:
i3!>!789/6!
q!>!21!cbs
t>t2!,!∆t23!>!2/686!
q>21!cbs
y>1
lK
lhL
lK
lh
ta~ka 3:
lK
i4!>!651/84!
lh
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 17
Prvi zakon termodinamike za proces u odvaja~u te~nosti:
⋅
⋅
⋅
/
⋅
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⋅
⋅
n⋅ i 4 + n⋅ i 7 = n⋅ i 5 + n⋅ i2
X lq = X u23
⋅
⋅
⇒
I2 = I3
⇒
i7!−!i5!>!i2!−!i4
⋅
X lq
lh
−2/4
= n⋅ (i2 − i 3 ) !!!!!!⇒!!!!!! n =
>
> 6/2 ⋅ 21 −3
764/12 − 789/6
t
i2 − i3
⋅
⋅
⋅
⋅
⋅
R epw = R 67 = n⋅ (i 7 − i 6 ) = n⋅ (i 7 − i 5 ) = n⋅ (i2 − i 4 )
⇒
⋅
R epw > 6/2 ⋅ 21 −3 ⋅ (764/12 − 651/84 ) >6/84!lX
b)
⋅
εi!>!
R epw
⋅
X lq
=!
6/84
>!5/5
2/4
c)
3
3l
U
4
5
⋅
7
R sfl
!!2
6
t
⋅
R epw
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 18
⋅
7/21/!Levokretni kru`ni proces sa amonijakom kao radnim telom ) n >1/12!lh0t* odvija se izme|u qnjo>2
cbs i qnby>36!cbs. Kompresija je ravnote`na izentropska i dvostepena. Stepen povi{ewa pritiska u oba
q
q
stepena je jednak ( 3 = 5 ) . Na ulazu u kompresor niskog pritiska (stawe 1) para amonijaka je
q2 q 4
suvozasi}ena. Nakon prvog stepena kompresije para amonijaka se hladi do temperature od T>431!L/
Skicirati proces na Ut dijagramu i odrediti:
a) rashladni efekat instalacije koja radi po ovom ciklusu (lX)
b) snagu kompresora niskog pritiska i snagu kompresora visokog pritiska )lX*
c) koeficijent hla|ewa )εi*
d) procentualno pove}awe koeficijenta hla|ewa koje je ostvareno dvostepenom kompresijom (u
odnosu na jednostepenu kompresiju izme|u istih pritisaka i istog stawa na ulazu u kompresor)
e) na Ut dijagramu {rafirati povr{inu koja predstavqa u{tedu u snazi kompresora usled dvostepene
kompresijeϕ
6
5
⋅
X lwq
⋅
R 56
⋅
n
3
4
⋅
n
⋅
⋅
⋅
X loq
R 34
2
n⋅ y 7
7
⋅
n⋅ (2 − y 7 )
⋅
Repw
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 19
U
5
6
3
4
2
7
t
q>2!cbs
ta~ka 1:
lK
i2!>!3272!
lh
t2!>!22/14!
t>t2>22/14!
lK
lhL
lK
lh
q!>6!cbs
ta~ka 3:
U>431!L
lK
t4!>!21/74!
lhL
lK
i4!>!3444!
lh
ta~ka 4:
i5!>!3739!
lK
lhL
q!>! q2 ⋅ q 5 >6!cbs
ta~ka 2:
i3!>!3584!
y>2
lK
lhL
q!>36!cbs
t!>21/74!
q>36!cbs
y>1
q>2!cbs
i>!i6!>!2353!
lK
lh
ta~ka 5:
lK
i6!>!2353!
lh
ta~ka 6:
dipl.ing. @eqko Ciganovi}
lK
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 20
b*
⋅
⋅
⋅
R epw = R 72 = n⋅ (i2 − i 7 ) = 1/12 ⋅ (3272 − 2353) >:/2:!lX
c*
⋅
⋅
⋅
⋅
⋅
⋅
X loq = X u23 = n⋅ (i2 − i3 ) > 1/12 ⋅ (3272 − 3584) >−4/23!lX
X lwq = X u45 = n⋅ (i 4 − i 5 ) > 1/12 ⋅ (3444 − 3739 ) >−3/:6!lX
d*
⋅
R epw
εi!>!
⋅
=!
⋅
X loq + X lwq
:/2:
>!2/6
4/23 + 3/:6
e*
U
B
6
2
7
t
q!>36!cbs
ta~ka A:
iB!>!3915/3!
⋅
t!>22/14!
lK
lhL
lK
lh
⋅
⋅
X lq = X u2B = n⋅ (i2 − i B ) > 1/12 ⋅ (3272 − 3915 ) >−7/54!lX
⋅
ε i-
!>!
R epw
⋅
X lq
=!
:/2:
>!2/54
7/54
ε
2/6
∆ε i (&) = i- − 2 ⋅ 211& !>
− 2 ⋅ 211& >8/25&
ε
2/54
i
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 21
f*
U
B
5
⋅
∆X
6
3
4
2
7
t
zadaci za ve`bawe:
(2.11. − 2.12.)
7/22/!U komori za hla|ewe potrebno je odr`avati stalnu temperaturu od −26pD, pri ~emu temperatura
spoqa{weg (okolnog) vazduha iznosi 41/4pD. Toplotni dobici kroz zidove komore iznose 91!lK0t. Za
hla|ewe komore primeweno je kompresiono rashladno postrojewe bez pothla|ivawa kondenzata i sa
wegovim prigu{ivawem. Pri tome kompresor usisava suvu paru freona 22 )S33*!i sabija je adijabatski.
Odrediti minimalnu snagu za pogon rashladnog postrojewa kao i faktor hla|ewa. Skicirati promene
stawa freona na Ut i it dijagramu.
⋅
re{ewe:
X l >28/5!lX-
εi>5/7
7/23/ U postrojewe koje radi po levokretnom kru`nom procesu, kondenzuje se i pothla|uje amonijak pri
pritisku od q>2!NQb. Te~ni rashladni fluid ulazi u prigu{ni ventil pri temperaturi od 28pD, gde se
prigu{uje do temperature isparavawa u>−34pD. Kompresor, u kojem se obavqa adijabatsko sabijawe radi
lq
sa stepenom dobrote η e >1/9, usisava suvu paru, koja od stawa vla`ne pare prelazi u stawe suve pare na
ra~un pothla|ivawa te~ne faze. Snaga kompresora je 67!lX. Odrediti rashladnu snagu ovog postrojewa i
u Ut koordinatnom sistemu {rafirati povr{inu ekvivalentnu snazi za pogon kompresora.
⋅
re{ewe:
R epw!>!29:/7!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 22
7/24/ Kaskadna rashladna instalacija (slika), sastoji se iz me|usobno spregnutog “kola visoke
temperature” i “kola niske temperature”. “Kola” su spregnuta preko toplotno izolovanog predajnika
toplote, u kome rashladni fluid kola niske temperature (preko kondenzatora kola niske temperature) u
potpunosti predaje toplotu rashladnom fluidu kola visoke temperature (preko ispariva~a kola visoke
temperature). Kolo visoke temperature radi sa freonom 11 )S22*, izme|u pritisaka qnjo>q{)−45pD* i
⋅
qnby>1/3!NQb i masenim protokom n 2>1/45!lh0t. Kolo niske temperature radi sa freonom 22 )S33*izme|u pritisak qnjo>q{)−:1pD* i qnby>1/3!NQb. Izra~unati stepen (koeficijent) hla|ewa ovog
postrojejwa, ako oba kola rade bez pothla|ivawa kondenzata i sa kvazistati~kom adijabatskom
kompresijom suvozasi}ene pare.
3
4
kolo visoke temperature
⋅
X lwu
S22
5
4′
2
3′
kolo niske temperature
⋅
X lou
S33
5′
2′
⋅
lh
n S22!>!1/45!
kolo visoke temperature:
t
y>2
ta~ka 1:
u!>!−45pD
lK
lK
t2!>!2/8415!
i2!>!783/86!
lhL
lh
ta~ka 2:
i3!>!841!
q!>!3!cbs
t!>!t2!>2/8415!
q!>!3!cbs
y>1
lK
lhL
lK
lh
ta~ka 3:
lK
i4!>!651!
lh
ta~ka 4:
i5!>!i4!>!651!
dipl.ing. @eqko Ciganovi}
lK
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 23
kolo niske temperature:
u!>!.!:1pD
ta~ka 1′:
lK
i2′!>!774/:7!
lh
t2′!>!2/::74!
q>3!cbs
ta~ka 2′:
i3′!>!863!
y>2
lK
lhL
t!>!t2′!>!2/::74!
lK
lhL
lK
lh
q!>!3!cbs
ta~ka 3′:
y>1
lK
i4′!>!582/3!
lh
i5′!>!i4′!>!582/3!
ta~ka 4′:
lK
lh
prvi zakon termodinamike za toplotno izolovani predajnik toplote
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⋅
⋅
⇒
⋅
/
⋅
nS33 ⋅ i 3( + nS22 ⋅ i 5 = nS33 ⋅ i 4( + nS22 ⋅ i2
⋅
nS33 = 1/45 ⋅
⋅
⋅
I2 = I3
⇒
⋅
⋅
nS33 = nS22⋅
i2 − i 5
i 3( − i 4 (
kg
783/6 − 651
>1/27!
863 − 582/3
s
⋅
⋅
R epw = R 5 (2( = nS33 ⋅ (i2( − i 5 ( ) = 1/27 ⋅ (774/:7 − 582/3 ) = 41/9!lX
⋅
⋅
⋅
X loq = X u2(3( = nS33 ⋅ (i2( − i3( ) > 1/27 ⋅ (774/:7 − 863) >−25/2!lX
⋅
⋅
⋅
X lwu = X u23 = nS22 ⋅ (i2 − i3 ) > 1/45 ⋅ (783/86 − 841) >−2:/6!lX
⋅
εi!>!
R epw
⋅
⋅
=!
X lou + X lwu
dipl.ing. @eqko Ciganovi}
41/9
>!1/:3
25/2 + 2:/6
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 24
7/25/!Rashladno postrojewe sa dva prigu{na ventila, dva odvaja~a te~nosti, dva kompresora i jednim
⋅
ispariva~em prikazano je na slici. Ako iz kondenzatora rashladnog postrojewa (stawe 1) izlazi n =1/2
lh0t kqu~alog freona 12 temperature 41pD, i ako se prvim prigu{nim ventilom sni`ava pritisak freona
na q>281!lQb, a drugim na q>31!lQb, skicirati proces u it koordinatnom sistemu i odrediti rashladni
kapacitet postrojewa.
1
7
5
n
6
4
2
3
ispariva~
5
7
i
4
6
2
1
3
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
ta~ka 0:
strana 25
u>41pD
y>1
q!>!2/8!cbs
i!>!i1!>!63:/19!
q!>1/3!cbs
i3!>!)!i′!*q>2/8!cbs!>!595/6!
lK
i1!>!63:/19!
lh
ta~ka 1:
lK
lh
y2!>!1/38
ta~ka 2:
y3!>!1/32
s!>!291/44!
q!>1/3!cbs
ta~ka 3:
lK
lh
lK
lh
y>1
lK
i4!>!737/6!
lh
1. na~in:
⋅
⋅
R epw = n⋅ (2 − y 2 ) ⋅ (2 − y 3 ) ⋅ (s )q=1/3cbs > 1/2 ⋅ (2 − 1/38) ⋅ (2 − 1/32) ⋅ 291/34 >21/5!lX
2. na~in:
⋅
⋅
R epw = n⋅ (2 − y2 ) ⋅ (i 4 − i 3 ) > 1/2 ⋅ (2 − 1/38) ⋅ (737/6 − 595/6 ) >21/5!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 26
7/26/!Freon 12 (R12) kao rashladni fluid obavqa levokretni ciklus sa dvostepenim isparavawem (slika).
U nisko−temperaturskom ispariva~u vlada pritisak 2!cbs, u visoko-temperaturskom ispariva~u 4!cbs, a u
kondenzatoru 9!cbs. Kondenzovani fluid (stawa 6) se razdvaja na dve struje i svaka od wih se adijabatski
prigu{uje u odgovaraju}em prigu{nom ventilu (do stawa 7 odnosno stawa 8). Suva para (stawa 2) iz
visoko−temperaturskog ispariva~a (VTI) se adijabatski prigu{uje do pritiska 1 bar
(stawe 3) i zatim izobarski me{a sa suvom parom (stawa 1) iz nisko−temperaturskog ispariva~a (NTI).
Dobijena me{avina (stawa 4) se kvazistati~ki izentropski sabija u kompresoru do stawa 5. Ako je
rashladni kapacitet visokotemperaturskog ispariva~a 25!lX, a niskotemperaturskog 8!lX, skicirati
promene stawa freona 12 na Ut dijagramu i odrediti:
a) masene protoke rashladnog fluida kroz oba ispariva~a
b) snagu kompresora
c) faktor hla|ewa rashladnog postrojewa
7
6
VTI
8
3
NTI
9
4
2
5
6
U
7
3
8
4
9
2
5
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 27
a)
ta~ka 6:
q!>!9!cbs
y>1
ta~ka 7:
i8!>!i7!>!642/6!
ta~ka 8:
i9!>!i8!>!i7!>!642/6!
ta~ka 2:
q!>!4!cbs
y>2
ta~ka 1:
q!>!2!cbs
y>2
i>642/6!
lK
lh
lK
lh
lK
lh
lK
lh
lK
i2!>!752/2!
lh
i3!>!766/69!
⋅
⋅
8
R ouj
lh
n ouj!>!
>!
>1/17!
752/2 − 642/6
t
i2 − i 9
⋅
⋅
25
R wuj
lh
n wuj!>!
>
>!1/22!
t
i 3 − i 8 766/69 − 642/6
b)
lK
lh
i!>@
ta~ka 3:
i4!>!i3!>!766/69!
ta~ka 4:
q>!4!cbs
prvi zakon termodinamike za me{awe fluidnih struja:
⋅
⋅
⋅
R 23 = ∆ I23 + X 23
⋅
⋅
n wuj ⋅ i 4 + nouj ⋅ i2 = n wuj + nouj ⋅ i 5
⋅
i5 =
⋅
⇒
⋅
⋅
⇒
1/22 ⋅ 766/69 + 1/17 ⋅ 752/2
lK
>761/5!
1/22 + 1/17
lh
q!>!9!cbs
ta~ka 5:
⋅
I2 = I3
t!>!t5!>!2/734!
i5 =
⋅
n wuj ⋅ i 4 + nouj ⋅ i2
⋅
⋅
n wuj + nouj
lK
t5!>!2/734!
lhL
lK
lhL
i6!>!799/2!
lK
lh
⋅
⋅
⋅
X lq = nouj + nouj ⋅ (i 5 − i 6 ) > (1/17 + 1/22) ⋅ (761/5 − 799/2) >−7/5!lX
c)
⋅
εi!>!
R epw
⋅
X lq
⋅
=!
⋅
R ouj + R wuj
⋅
X lq
dipl.ing. @eqko Ciganovi}
!!>
8 + 25
>!4/4
7/5
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
zadatak za ve`bawe:
strana 28
(2.16.)
7/27/ Za potrebe hla|ewa dve odvojene rashladne komore koristi se levokretni kru`ni proces sa
zajedni~kim kondenzatorom (KD) ,pri ~emu je rashladni fluid freon 12. U nisko temperaturskom
ispariva~u (NTI) vlada temperatura od −41PD, u visoko temperaturskom ispariva~u (VTI) temperartura
−2PD, dok je pritisak u kondenzatoru 1/8:42!NQb. Kondezovani fluid (stawa 6) razdvaja se na dve struje i
svaka od wih se adijabatski prigu{uje u odgovaraju}em ventilu (do stawa 7 odnosno stawa 8). Suva para
(stawa 1) iz nisko temperaturskog ispariva~a se kvazistati~ki adijabatski sabija u prvom kompresoru do
pritiska koji vlada u visoko temperaturskom ispariva~u
(stawe 3) i zatim izobarski me{a sa suvom parom (stawa 2) iz visoko temperaturskog ispariva~a.
Dobijena me{avina se kvazistati~ki adijabatski sabija u drugom kompresoru do temperature od 51PD
(stawe 5). Ako je maseni protok rashladnog fluida kroz nisko temperaturski ispariva~ 1/174!lh0t, a kroz
visoko temperaturski ispariva~ 1/224!lh0t, odrediti:
a) rashladne snage oba ispariva~a
b) snage oba kompresora
c) koeficijent hla|ewa rashladnog postrojewa
KD
7
6
VTI
8
3
NTI
9
⋅
a)
R ouj!>8!lX⋅
b) X u24 >!−2/29!lX
c) εi!>!6
2
4
5
⋅
! R wuj>25!lX
⋅
X u 56 >!−4!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 29
7/28. U toplotnoj pumpi, radna materija obavqa levokretni kru`ni proces koji se sastoji od ravnote`nog
(kvazistati~kog) adijabatskog sabijawa, izotermskog ravnote`nog (kvazistati~og) sabijawa, ravnote`nog
(kvazistati~kog) adijabatskog {irewa i ravnote`nog (kvazistai~kog) izotermnog {irewa. Maksimalna
odnosno minimalna temperatura radne materije iznose: Unby>!431!L i Unjo>!391!L, a temperature
toplotnog izvora odnosno toplotnog ponora su stalne i iznose
Uuj>!3:1!L i Uuq>!421!L. Nepovratnost predaje toplote radnoj materiji iznosi
∆TJ!>!6!K0L. Predstaviti proces u Ut koordinatnom sistemu i odrediti:
a) nepovratnost predaje toplote toplotnom ponoru
b) koeficijent grejawa toplotne pumpe
U
Unby
4
3
!Uuq
!Uuj
Unjo
5
2
t
a)
∆TJ!>! (∆T tj )52 = (∆Tsu )52 −
(∆T tj )52
Repw!>!
2
Unjo
−
2
Uuj
R epw
R
R
!>! epw − epw !!!!!!!
Uuj
Unjo
Uuj
6
!>
2
2
−
391 3:1
⇒
>!51/7!lK
Rpew!>!R34!> Unby ⋅ (∆T SU )34 >///!> 431 ⋅ (−1/256 ) >!−57/5!lK
(∆TSU )34
= −(∆T SU )52 = −
(∆Ttj )34 = (∆Tsu )34 − R pew
Uuq
R epw
51/7
K
>−256
=−
L
Unjo
391
!>! − 256 −
.57511
K
>!5/79!
421
L
b)
εh!>!
R pew
R pew − R epw
>
57/5
>!9
57/5 − 51/7
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 30
7/29/!Dve toplotne pumpe me|usobno spojene redno (slika) rade po idealnom Karnoovom kru`nom procesu.
Toplotna pumpa 2 uzima 511!lK toplote od toplotnog izvora stalne temperature UUJ>411!L. Toplotu
odvedenu od toplotne pumpe 2 preuzima toplotna pumpa 3, koja predaje toplotu toplotnom ponoru stalne
temperature UUQ>2311!L. Ako obe toplotne pumpe rade sa istim faktorom grejawa odrediti:
a) temperaturu radne materije pri kojoj se vr{i razmena toplote izme|u dve toplotne pumpe, UY
b) neto mehani~ke radove koji se dovode radnoj materiji u toplotnoj pumpi, 2(UQ2) i toplotnoj pumpi,
3(UQ3)
TOPLOTNI Repw
IZVOR
Ry
Rpew
UQ2
UQ3
X2
X3
TOPLOTNI
PONOR
a)
ε h2 =
UY
UY − UUJ
ε h2 = ε h3
UY =
ε h3 =
⇒
UUQ
UUQ − UY
UY
UUQ
>
UY − UUJ UUQ − UY
⇒
UUJ ⋅ UUQ > 411 ⋅ 2311 >711!L
b)
ε h2 =
εh2!>!
UY
711
>
>3
711 − 411
UY − UUJ
RY
R Y − R epw
⇒
ε h3 =
RY =
ε h2
ε h2 − 2
UUQ
2311
>
>3
2311 − 711
UUQ − UY
⋅ R epw >911!lK
X2> R Y − R epw >911!−!511!>!511!lK
εh3!>!
R pew
R pew − R Y
⇒
R pew =
ε h3
ε h3 − 2
⋅ R Y >2711!lK
X3> R pew − R Y >2711!−!911!>!911!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 31
7/2:/!Izobarskim odvo|ewem toplote od 4!lh vodene pare stawa (y>1/8-!q>31!lQb) sve dok se ne postigne
stawe kqu~ale te~nosti, pomo}u levokretnog kru`nog procesa u postrojewu sa toplotnom pumpom,
toplota se predaje vodoniku (idealan gas). Masa vodonika je 29!lh, a po~etna temperatura :6pD. Vodonik
se nalazi u zatvorenom sudu. Koeficijent grejawa toplotne pumpe je εh>2/9. Odrediti krajwu temperaturu
vodonika u sudu kao i snagu kompresora ako toplotna pumpa radi jedan sat.
Koli~ina toplote koja se oduzme od vodene pare u procesu kondenzacije )Rqbsb*
istovremeno predstavqa dovedenu toplotu za toplotnu pumpu )Repw*
Repw!>!Rqbsb!>!nqbsb!/!)iy!−!i′*!>!///!> 4 ⋅ (2:12/83 − 362/5 ) >!5:62!lK
iy!>!i′!,!y/!)i′′!−!i′*!>!!///!> 362/5 + 1/8 ⋅ (371: − 362/5 ) >2:12/83!
i′!>!362/5!
lK
-!
lh
i′′!>!371:!
lK
-!
lh
lK
lh
)q!>!1/3!cbs*
Koli~ina toplote koju primi vodonik )RW*!istovremeno predstavqa odvedenu toplotu za
toplotnu pumpu!)Rpew*
εh!>!
R pew
R pew − R epw
R pew =
⇒
R pew =
εh
εh − 2
⋅ R epw
2/9
⋅ 5:62>!2224:/86!lK!>!RW
2/9 − 2
RW!>!nw!/!dw!)!Uw3!.!Uw2* !!!!!!
UW3> 479 +
⇒
!!!!!!Uw3!> UW2 +
RW
>
nw ⋅ d W
2224:/86
>!538/62!L
29 ⋅ 21/5
XL!>! R pew − R epw = 7299/86!lK
•
XL =
XL
7299/86
= 2/83!lX
>
4711
τ
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 32
7/31. Instalacija toplotne pumpe (slika) radi sa ugqendioksidom (idealan gas) kao radnim fluidom po
Xulovom kru`nom procesu izme|u qnjo>1/2!NQb i qnby>1/5!NQb. Stepen dobrote adijabatske kompresije
je ηlq>!1/:7, a stepen dobrote adijabatske ekspanzije ηfy>!1/:3. Svrha toplotne pumpe je da se u prostoriji
odr`ava temperatura UUQ>28pD pri temperaturi okolnog vazduha UUJ>1pD. Pri tome se iz okoline
radnom telu dovodi 311!lK0t toplote. Usvojiti da se ugqendioksid ispred kompresora zagreva do
temperature koja vlada u okolini, a ispred ekspanzionog ure|aja hladi do temperature koja vlada u
prostoriji. Odrediti faktor grejawa toplotne pumpe kao i snagu kompresora i snagu ekspanzionog
ure|aja (turbine).
U
grejana
prostorija
3
3l
4
3
4
2
5
5l
t
κ −2
2
5
2/39 −2
κ
>!384 ⋅ 5 2/39 >!47:/82!L
U3L!>!U2 ⋅ q 3L
q
2
2
U −U
47:/82 − 384
!>484/85!L
U3!>!U2!,! 3L lq 2 !>!384!,!
1/:7
η
U5L!>!U4 ⋅ q 5L
q
4
κ −2
κ
>!3:1 ⋅ 2
5
2/39 −2
2/39
>!325/25!L
U5!>!U4!,!ηfy!/)U5l!.!U4*!>!3:1!,!1/:3 ⋅(325/25 − 3:1) !>331/32!L
⋅
⋅
⋅
R epw = R 52 = n⋅ d q ⋅ (U2 − U5 )
⋅
n=
⋅
⇒
⋅
R epw
311
lh
>
>5/57!
d q ⋅ (U2 − U5 ) 1/96 ⋅ (384 − 331/32)
t
⋅
⋅
R pew = R 34 = n⋅ d q ⋅ (U4 − U3 ) > 5/57 ⋅ 1/96 ⋅ (3:1 − 484/85 ) >−428/57!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 33
⋅
R pew
εh!>!
⋅
>!
⋅
R pew − R epw
⋅
⋅
428/57
>3/8
428/57 − 311
⋅
X lpnqsftps!>! X u23 > n ⋅ d q ⋅ (U2 − U3 ) > 5/57 ⋅ 1/96 ⋅ (384 − 484/85 ) >−492/:!lX
⋅
⋅
⋅
X uvscjob!>! X u45 > n ⋅ d q ⋅ (U4 − U5 ) > 5/57 ⋅ 1/96 ⋅ (3:1 − 331/32) >!375/68!lX
7/32/!Toplotna pumpa koja se koristi za zagrevawe vazduha (idealan gas) od Uw2>66pD!do Uw3>71pD! na
ra~un hla|ewa vode od (q>2!cbs- Ux2>29pD*!do!)q>2!cbs-!Ux3>25pD*, radi izme|u pritisaka qnjo>5/66!cbs
i!qnby>27/83!cbs, sa freonom 12 )S23* kao radnim telom, po idealnom Rankin−Klauzijusovom kru`nom
procesu sa prigu{ivawem te~ne faze,. Protok vode kroz ispariva~ toplotne pumpe je 2/6!lh0t/ Skicirati
promene stawa freona 12 na Ut!i!it dijagramu i odrediti faktor grejawa toplotne pumpe, snagu
kompresora kao i maseni protok vazduha kroz kondenzator toplotne pumpe.
vazduh
4
3
S23
5
2
3
voda
3
U
i
4
2
4
5
5
2
t
dipl.ing. @eqko Ciganovi}
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 34
q>5/66!cbs
ta~ka 1:
lK
i2!>!771/95!
lh
t2!>!2/675!
ta~ka 2:
i3!>!795/13!
y>2
lK
lhL
q>27/83!cbs
t!>!t2!>2/675!
q>27/83!cbs
y>1
lK
lhL
lK
lh
ta~ka 3:
lK
i4!>!677/2!
lh
i5!>!i4!>!677/2!
ta~ka 4:
⋅
⋅
⋅
⋅
⋅
⋅
lK
lh
R epw = R 52 = n g ⋅ (i2 − i 5 ) >!///
R pew = R 34 = n g ⋅ (i 4 − i3 ) >!///
⋅
R pew
εh =
⋅
>
⋅
R pew − R epw
i4 − i3
i 4 − i 3 − i2 − i 5
>
677/2 − 795/13
677/2 − 795/13 − 771/95 − 677/2
>6/19
prvi zakon termodinamike za proces u ispariva~u toplotne pumpe:
⋅
⋅
⋅
R 23 = ∆ I23 + X U23
⋅
⋅
⋅
⇒
⋅
⋅
n x ⋅ i x2 + n g ⋅ i 5 = n x ⋅ i x3 + n g ⋅ i2
⋅
⋅
ng = nx ⋅
⋅
I2 = I3
⇒
i x2 − i x3
86/46 − 69/71
lh
>1/376!
> 2/6 ⋅
771/95 − 677/2
t
i2 − i 5
napomena:
lK
lh
lK
ix3!>!69/71!
lh
ix2!>!86/46!
dipl.ing. @eqko Ciganovi}
)q!>!2!cbs-!u>29pD*
)q!>!2!cbs-!u>25pD*
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 35
prvi zakon termodinamike za proces u kompresoru toplotne pumpe:
⋅
⋅
⋅
R 23 = ∆ I23 + X U23
⇒
⋅
⋅
⋅
X U23 = −∆ I23 = − n g ⋅ (i3 − i2 )
⋅
X U23 = −1/376 ⋅ (795/13 − 771/95 ) >−7/25!lX
prvi zakon termodinamike za proces u kondenzatoru toplotne pumpe:
⋅
⋅
⋅
R 23 = ∆ I23 + X U23
⋅
⇒
⋅
⋅
⋅
I2 = I3
⋅
⋅
n w ⋅ d q ⋅ Uw2 + n g ⋅ i 3 = n w ⋅ d q ⋅ Uw3 + n g ⋅ i 4
⋅
⋅
nw = ng ⋅
⇒
i3 − i 4
795/13 − 677/2
lh
>7/36!
> 1/376 ⋅
d q ⋅ (Uw3 − Uw2 )
2 ⋅ (71 − 66 )
t
7/33/!Termodinami~ki stepen korisnosti Xulovog kru`nog procesa (2−3−4−5−2), koji obavqa vazduh
(idealan gas) iznosi η =0.3. Koliki bi bio faktor grejawa toplotne pumpe, kada bi metan (idealan gas)
obavqao levokretni Xulov kru`ni procec izme|u istih stawa )2−5−4−3−2*.
desnokretni
levokretni
U
U
4
4
5
5
3
3
2
2
t
dipl.ing. @eqko Ciganovi}
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 36
desnokretni kru`ni proces:
R epw = R 34 = n ⋅ d q ⋅ (U4 − U3 )
R pew = R 52 = n ⋅ d q ⋅ (U2 − U5 )
η!>!
U − U3 + U2 − U5
R epw + R pew
!
>! 4
R epw
U4 − U3
)2*
levokretni kru`ni proces:
R epw = R 25 = n ⋅ d q ⋅ (U5 − U2 )
R pwe = R 43 = n ⋅ d q ⋅ (U3 − U4 )
!!!εh!>!
R pew
R pew − R epw
>
U3 − U4
U3 − U4 − U5 + U2
pore|em izraza (1) i (2) uo~ava se:
zadatak za ve`bawe:
>
U4 − U3
U4 − U3 − U5 + U2
εh =
)3*
2
2
>4/44
>
η 1/4
(2.23.)
7/34. Toplotna pumpa radi, sa vodenom parom kao radnim fluidom, po realnom levokretnom
Rankin−Klauzijusovom kru`nom procesu bez podhla|ivawa kondenzata izme|u pritisaka qnjo>9!lQb!i
qnby>1/7!NQb. U kompresor ulazi suvozasi}ena vodena para, a na izlazu iz kompresora temperatura pare
je 811pD. Kao izvor toplote koristi se otpadna voda nekog industrijskog procesa, temperature
61pD>dpotu, koja predaje vodenoj pari 2311!lX toplote. Potro{a~ toplote (toplotni ponor) nalazi se na
temperaturi 261pD>dpotu. Odrediti:
a) koli~inu toplote koja se predaje potro{a~u (lX)
b) koeficijent grejawa toplotne pumpe )εh*
c) promenu entorpije sistema koji sa~iwavaju radna materija, toplotni izvor i toplotni ponor
⋅
a)
R pew >3161/4!lX
b)
εh!>!3/5
c)
∆ T tj!>!2/24!
⋅
lX
L
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 1
VLA@AN VAZDUH
8/2/!U sudu zapremine!W>1/96!n4!nalazi se!nww>2/12!lh!nezasi}enog vla`nog vazduha stawa!)q>2!cbsu>31pD*/!Odrediti:
a) masu suvog vazduha u sudu i masu vodene pare u sudu
b) pritisak suvog vazduha i pritisak vodene pare u sudu
c) gustinu suvog vazduha, gustinu vodene pare i gustinu vla`nog vazduha u sudu
lhI3 P
d) sadr`aj vlage (apsolutnu vla`nost) vla`nog vazduha,
lhTW
e) relativnu vla`nost vla`nog vazduha
f) specifi~nu entalpiju vla`nog vazduha
a)
nww!>!ntw!,!nI3P
)2*
(
)
q ⋅ W = nTW ⋅ S hTW + nI3PS hI3P ⋅ U
)3*
re{avawem sistema jedna~ina (1) i (2) dobija se;
2⋅ 216 ⋅ 1/96
2
2
q⋅W
n tw =
>
>2!lh
− 2/12⋅ 573 ⋅
− n ww ⋅ S hI3P ⋅
−
U
S
S
3:4
398
573
−
htw
hI3P
nI3P!>!nww!−!ntw!>!2/12!−!2!>1/12!lh
b)
q TW =
nTW ⋅ S hTW U
q I3P =
=
W
nI3P ⋅ S hI3P U
W
2 ⋅ 398 ⋅ 3:4
= :9:41!Qb
1/96
=
1/12 ⋅ 573 ⋅ 3:4
= 26:3!Qb
1/96
c)
ρtw>
!!
nTW
q TW
:9:41
lhTW
=
=
>2/287!
W
S hTW U 398 ⋅ 3:4
n4
!ρI3P>
nI3P
W
=
q I3P
S hI3P U
=
lhI3 P
26:3
>1/119!
573 ⋅ 3:4
n4
ρww!>!ρtw!,!ρI3P!>2/287!,!1/119!>2/195!!
lhWW
n4
d)
y>
nI3P
nTW
=
NI3P
NTW
⋅
q I3P
q − q I3P
dipl.ing. @eqko Ciganovi}
=
lhI3 P
29
26:3
⋅
>1/12!
6
3: 2 ⋅ 21 − 26:3
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 2
e)
ϕ>
q I3P
(qqt )U =31p D
=
26:3
>1/79
3448
f)
i = d qTW ⋅ u + y ⋅ )2/97 ⋅ u + 3611* = 2 ⋅ 31 + 1/12 ⋅ (2/97 ⋅ 31 + 3611) >56/48!
lK
lhTW
8/3/!Odrediti temperaturu!vla`nog vazduha ~ije je stawe pri!q>2!cbs!zadato na na~in:
lhI3 P
a) y>1/13!
!)apsolutna vla`nost*-!ϕ>1/9!(relativna vla`nost)
lhTW
b) Uwu>31pD!)temperatura vla`nog termometra*-!Us>21pD!)ta~ka rose*
a)
q I3P =
y
NI3P
NTW
q qt =
q I3P
ϕ
u4!>!39/6pD
b)
=
⋅q =
+y
1/13
⋅ 2 ⋅ 21 6 >4232/75!Qb
29
+ 1/13
3:
4232/75
>4:13!Qb
1/9
)temperatura kqu~awa vode na!q>4:13!Qb*
(qI3P )S = ϕS ⋅ (qqt )US =21 > 2⋅ 2338 >2338!Qb
NI3P
yS>
NTW
⋅
q I3P
q − q I3P
=
lhI3 P
29
2338
⋅
>1/1188!
6
3: 2 ⋅ 21 − 2338
lhTW
y!>!yS
(qI3P )wu = ϕ wu ⋅ (qqt )Uwu =31 > 2⋅ 3448 >3448!Qb
ywu>
NI3P
NTW
⋅
q I3P
q − q I3P
=
lhI3 P
29
3448
⋅
>1/1259!
6
3: 2 ⋅ 21 − 3448
lhTW
iwu = dq ⋅ u + y wu ⋅ )2/97 ⋅ u wu + 3611* = 2 ⋅ 31 + 1/1259 ⋅ (2/97 ⋅ 31 + 3611) >56/48
lK
lhTW
i!>!iwu
u!>!
i − y ⋅ 3611
68/66 − 1/1188 ⋅ 3611
!>
>48/87pD
2 + 1/1188 ⋅ 2/97
dq + y ⋅ 2/97
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 3
8/4/!Vla`nom vazduhu stawa!2)q2>2!cbs-!u2>31pD-!ϕ2>1/9-!nww>31!lh0i) dovodi se toplota u zagreja~u
vazduha dok vazduh ne dostigne stawe!3)q3>2!cbs-!u3>91pD), a zatim se tako zagrejan vazduh u adijabatski
izolovanoj komori vla`i pregrejanom vodenom parom stawa!Q)q>2!cbs-!u>231pD-!nqq>2!lh0i*!do stawa
4)q>2!cbs). Skicirati promene stawa vla`nog vazduha na Molijerovom i!−y!dijagramu i odrediti:
a) toplotnu snagu zagreja~a vazduha!)lX*
b) entalpiju!)i*-!apsolutnu vla`nost!)y*!i temperaturu!)u*!vla`nog vazduha stawa!4
i
4
3
3828
u3
ϕ>2
u2
2
ϕ2
y
ta~ka 1:
q qt >3448!Qb
)napon pare ~iste vode na!u>31pD*
q I3P = ϕ ⋅ q qt > 1/9 ⋅ 3448 >297:/7!Qb
y2!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
297:/7
⋅
>1/1229!
6
lhTW
3: 2 ⋅ 21 − 297:/7
i2> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1229 ⋅ )2/97 ⋅ 31 + 3611* >61/13!
ntw!>!
lK
lhTW
n ww
31
lhTW
>
>2:/88!
2 + y 2 2 + 1/1229
i
ta~ka 2:
y3!>!y2>1/1229!
lhI3 P
lhTW
i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 91 + 1/1229 ⋅ )2/97 ⋅ 91 + 3611* >222/44!
dipl.ing. @eqko Ciganovi}
lK
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 4
⋅
⋅
⋅
prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
2:/88
⋅ (222/44 − 61/13) >1/45!lX
4711
R 23 = n tw ⋅ (i3 − i2 ) =
ta~ka 3:
⋅
⋅
⋅
prvi zakon termodinamike za proces vla`ewa vazduha:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
n tw ⋅ i3 + nqq ⋅ iqq = n tw ⋅ i 4
⋅
⇒!
i4 =
⋅
ntw ⋅ i3 + nqq ⋅ iqq
⋅
ntw
2
2:/88
⋅ 222/44 +
⋅ 3828
lK
4711
4711
i4 =
>359/87!
2:/88
lhTW
4711
materijalni bilans vlage za proces vla`ewa vazduha:
⋅
⋅
⋅
n tw ⋅ y 3 + nqq = n tw ⋅ y 4
⋅
⇒!
y4 =
⋅
n tw ⋅ y 3 + nqq
⋅
n tw
y4 =
lhI3 P
2:/88 ⋅ 1/1229 + 2
>1/1735!
2:/88
lhTW
u4!>!
i 4 − y 4 ⋅ 3611
359/87 − 1/1735 ⋅ 3611
>94/22pD
!>
d q + y 4 ⋅ 2/97
2 + 1/1735 ⋅ 2/97
napomena:
iqq>!3828!
lK
-!entalpija pregrejane vodene pare stawa!Q!)q>2!cbs-!u>231pD*
lh
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 5
8/5/!Za pripremu vla`nog vazduha stawa!5)q>2!cbs-!u>47pD-!ϕ>1/4) koristi se sve` vazduh stawa!2)q>2
cbs-!u>21pD-!ϕ>1/9). Sve` vazduh se najpre zagreva u zagreja~u do stawa!3)q>2!cbs*-!a onda adijabatski
vla`i ubrizgavawem vode!X)q>2!cbs-!ux>61pD*!dok ne postane zasi}en!)q>2!cbs-!ϕ>2*/!Na kraju se vazduh
dogreva u dogreja~u. Potro{wa vode u fazi vla`ewa vazduha iznosi!71!lh0i. Skicirati promene stawa
vla`nog vazduha na Molijerovom!i!−y!dijagramu i odrediti:
a) veli~ine stawe vla`nog vazduha na ulazu u dogreja~!!4)i-!y-!u*
b) toplotne snage zagreja~a i dogreja~a!)lX*
i
5
u5
ϕ5
3
ϕ>2
u2
4
2
ϕ2
y
31:
ta~ka 1:
q qt >2338!Qb
)napon pare ~iste vode na!u>21pD*
q I3P = ϕ ⋅ q qt > 1/9 ⋅ 2338 >:92/7!Qb
y2!>
NI3P
NTW
⋅
q I3P
q − q I3P
lhI3 P
29
:92/7
⋅
>1/1173!
6
3: 2 ⋅ 21 − :92/7
lhTW
>
i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 21 + 1/1173 ⋅ )2/97 ⋅ 21 + 3611* >36/73!
lK
lhTW
ta~ka 4;
q qt >6:51!Qb
)napon pare ~iste vode na!u>47pD*
q I3P = ϕ ⋅ q qt > 1/4 ⋅ 6:51 >2893!Qb
y5!>
NI3P
NTW
⋅
q I3P
q − q I3P
lhI3 P
29
2893
⋅
>1/1224!
6
3: 2 ⋅ 21 − 2893
lhTW
>
i5> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 47 + 1/1224 ⋅ )2/97 ⋅ 47 + 3611* >76/12!
lK
lhTW
ta~ka 3:
y4!>y5!>1/1224!
q I3P =
y
NI3P
NTW
q qt =
q I3P
ϕ
p
u4!>!27 D
=
lhI3 P
lhTW
⋅q =
+y
1/1224
⋅ 2 ⋅ 21 6 >2899!Qb
29
+ 1/1224
3:
2899
>2899!Qb
2
)temperatura kqu~awa vode na!q>2899!Qb*
i4> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 27 + 1/1224 ⋅ )2/97 ⋅ 27 + 3611* >55/6:!
dipl.ing. @eqko Ciganovi}
lK
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 6
ta~ka 2:
y3!>y2!>1/1173!
lhI3 P
lhTW
materijalni bilans vlage za proces vla`ewa vazduha!)3−4*;
⋅
⋅
⋅
⋅
ntw ⋅ y3 + nX = ntw ⋅ y4 !!!!!!!!⇒!!!!!!!! n tw
71
⋅
nX
lh
4711
>4/38!
=
=
t
1/1224 − 1/1173
y4 − y3
⋅
⋅
⋅
prvi zakon termodinamike za proces vla`ewa vazduha:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
n tw ⋅ i3 + nqq ⋅ iqq = n tw ⋅ i 4
⋅
⇒
i3 =
⋅
ntw ⋅ i4 − nX ⋅ iX
⋅
ntw
i3 =
71
⋅ 31:
lK
4711
>54/63!
4/38
lhTW
4/38 ⋅ 55/6: −
⋅
⋅
⋅
⋅
⋅
⋅
prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
R 23 = n tw ⋅ (i3 − i2 ) > 4/38 ⋅ (54/63 − 36/73) >69/64!lX
prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!! R 45 = ∆ I45 + X u 45
⋅
⋅
R 45 = n tw ⋅ (i 5 − i 4 ) > 4/38 ⋅ (76/12 − 55/6:) >77/88!lX
8/6/!Vla`an vazduh, pri konstantnom pritisku!)q>2/37!cbs*-!struji kroz adijabatski izolovan kanal i pri
tome se najpre zagreva a potom i vla`i suvozasi}enom vodenom parom!)q>2/4!cbs*!)slika*/!Jedan deo vodene
parekoristi se za zagrevawe vazduha (ulazi u cevnu zmiju i iz we izlazi potpuno kondenzovan tj. kao
kqu~ala te~nost), a drugi deo pare (istog po~etnog stawa) koristi se za vla`ewe vla`nog vazduha (isti~e
kroz mlaznicu i me{a se sa vla`nim vazduhom stawa 2). Zapreminski protok vla`nog vazduha na ulazu u
kanal iznosi!1/65!n40t-!a wegovo stawe je definisano temperaturom suvog termometra i temperaturom
vla`nog termometra!2)utu>33pD-!uwu>23pD*/!Odrediti potrebne masene protoke vodene pare posebno kroz
cevnu zmiju i posebno kroz mlaznicu, da bi se ostvarilo stawe!4)u>71pD-!ϕ>1/4*!vla`nog vazduha na izlazu
iz kanala. Skicirati promene stawa vla`nog vazduha na!Molijerovom!i−y!dijagramu.
nB
nC
vla`an
vazduh
2
dipl.ing. @eqko Ciganovi}
3
4
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 7
ta~ka WU:
(qI3P )wu = ϕ wu ⋅ (qqt )Uwu =23 > 2⋅ 2512 >2512!Qb
ywu>
NI3P
⋅
NTW
q I3P
lhI3 P
29
2512
⋅
>1/1181!
6
3: 2/37 ⋅ 21 − 2512
lhTW
=
q − q I3P
iwu = dq ⋅ u + y wu ⋅ )2/97 ⋅ u wu + 3611* = 2⋅ 23 + 1/1181 ⋅ (2/97 ⋅ 23 + 3611) >3:/77
lK
lhTW
ta~ka 1:
i2>iwu!>3:/72!
y2 =
lK
lhTW
i2 − d q ⋅ u 2
=
2/97 ⋅ u 2 + 3611
(qI3P )2 = N
y2
I3P
lhI3 P
3:/77 − 2 ⋅ 33
>1/1141!
lhTW
2/97 ⋅ 33 + 3611
⋅q =
+ y2
NTW
1/114
⋅ 2/37 ⋅ 21 6 >717!Qb
29
+ 1/114
3:
(q tw )2 = q − (q I3P )2 = 2/37 ⋅ 21 6
(ρ tw )2 =
(q TW )2
S hTW ⋅ U2
⋅
=
− 717 > 2/36 ⋅ 21 6 !Qb
2/36 ⋅ 21 6
lhTW
>2/59!
398 ⋅ 3:6
n4
⋅
n tw = (ρ tw )2 ⋅ W 2 = 2/59 ⋅ 1/65 >1/9!
lhTW
t
ta~ka 3:
q qt >2::21!Qb )napon pare ~iste vode na!u>71pD*
q I3P = ϕ ⋅ q qt > 1/4 ⋅ 2::21 >6:84!Qb
y4!>
NI3P
⋅
NTW
q I3P
q − q I3P
>
lhI3 P
29
6:84
⋅
>1/141:!
6
lhTW
3: 2/37 ⋅ 21 − 6:84
i4> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 71 + 1/141: ⋅ )2/97 ⋅ 71 + 3611* >251/8
lK
lhTW
ta~ka 2:
y3>y2>1/1141!
lhI3 P
lhTW
i3>@
materijalni bilans vlage za proces vla`ewa vazduha!)3−4*;
⋅
⋅
⋅
ntw ⋅ y3 + nB = ntw ⋅ y4 !!!!!!!!⇒
⋅
n B = 1/9 ⋅ (1/141: − 1/1141) > 3/34 ⋅ 21 −3
dipl.ing. @eqko Ciganovi}
⋅
⋅
n B = ntw ⋅ (y 4 − y 3 )
lh
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 8
prvi zakon termodinamike za proces u otvorenom sistemu ograni~enom isprekidanom konturom:
⋅
⋅
⋅
R 23 = ∆ I23 + X u23
⋅
⋅
⋅
⋅
⋅
n tw ⋅ i2 + n B ⋅ i B2 + nC ⋅ iC = n tw ⋅ i 4 + nC ⋅ i B3
⋅
⋅
n tw ⋅ (i 4 − i2 ) − nC ⋅ iC
1/9 ⋅ (251/8 − 3:/77 ) − 3/34 ⋅ 21 −3 ⋅ 3798
>
nC =
=
i B2 − i B3
3798 − 55:/3
⋅
⋅
lh
t
lK
iB2>iC2!>!3798!
lh
iB3>55:/3
nC > 2/3: ⋅ 21 −3
)suva para!q>2/4!cbs*
)kqu~ala voda!q>2/4!cbs*
i
4
3
ϕ>2
3798
2
WU
y
)8/7/*
zadatak za ve`bawe:
8/7/!21!)2,y*!lh0t!vla`nog vazduha stawa!2)q>2!cbs-!u>71pD-!qI3P>1/12!cbs) vla`i se vodenom parom
stawa!Q)q>2!cbs-!u>271!pD). Parcijalni pritisak vodene pare u vla`nom vazduhu nakon vla`ewa iznosi
3)qI3P>1/16!cbs*/!Dobijeni vla`an vazduh stawa 2 hladi se do zasi}ewa (stawe 3). Svi procesi sa vla`nim
vazduhom su izobarski. Skicirati procese sa vla`nim vazduhom na Molijerovom iy!dijagramu i odrediti:
a) temperaturu )u* i apsolutnu vla`nost!)y*!vla`nog vazduha stawa!3!i stawa!4
b) koliko se toplote odvede od vla`nog vazduha u procesu hla|ewa!)3−4*-!)lX*
re{ewe:
a) u3>75/79pD-!y3>1/1438!
lhI3 P
lhI3 P
-!u4>43/:pD-!y4>1/1438!
lhTW
lhTW
⋅
b)
R 34!>!−447!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 9
8/8/!U adijabatski izolovanom rashladnom torwu, za potrebe hla|ewa neke prostorije, hladi se voda
X2)q>2!cbs-!ux2>68pD) isparavawem u struji vazduha, ~ije je stawe na ulazu u toraw 2)q>2!cbs-!u>33pDϕ>1/3) a na izlazu iz torwa 3)q>2!cbs-!u>38pD-!ϕ!>1/:*/!Protok suvog vazduha kroz toraw iznosi!9/6!lh0t.
Ohla|ena voda iz torwa!X3)q>2!cbs-!ux3>33!D*-!se me{a sa sve`om vodom!Xp)q>2cbs-!uxp>27pD*!da bi se
nadoknadila isparena koli~ina vode i ponovo odvodi u prostoriju koju treba ohladiti. Odrediti:
a) potro{wu sve`e vode!)X1*
b) razmewenu toplotu u torwu!)lX*
c) protoke tople )X2* i ohla|ene vode!)X3*
d) koli~inu toplote koju prostorija koja se hladi predaje vodi-!R′!!)lX*
X2
2)u3-!ϕ3*
Rups
R′
X1
X3
2)u2-!ϕ2*
Xp
vla`an vazduh:
ta~ka 1:
q qt >3754!Qb
)napon pare ~iste vode na!u>33pD*
q I3P = ϕ ⋅ q qt > 1/3 ⋅ 3754 >639/7!Qb
y2!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
639/7
⋅
>1/1144!
6
lhTW
3: 2 ⋅ 21 − 639/7
i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 33 + 1/1144 ⋅ )2/97 ⋅ 33 + 3611* >41/48!
dipl.ing. @eqko Ciganovi}
lK
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 10
ub•lb!3;
)napon pare ~iste vode na!u>38pD*
q qt >4675!Qb
q I3P = ϕ ⋅ q qt > 1/: ⋅ 4675 >4318/7!Qb
y3!>
NI3P
NTW
⋅
q I3P
>
q − q I3P
lhI3 P
29
4318/7
⋅
>1/1317!
6
lhTW
3: 2 ⋅ 21 − 4318/7
i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 38 + 1/1317 ⋅ )2/97 ⋅ 38 + 3611* >8:/64
lK
lhTW
voda:
lK
lh
lK
ix3>:2/:7!
lh
lK
ix1>77/99!
lh
entalpija vode!q>2!cbs-!u>68pD
ix2>349/37!
entalpija vode!q>2!cbs-!u>33pD
entalpija vode!q>2!cbs-!u>27pD
materijalni bilans vlage za proces vla`ewa vazduha u torwu:
⋅
⋅
X2 + ntw ⋅ y2 = X3 + ntw ⋅ y3
⇒
⋅
X2 − X3 = ntw ⋅ (y 3 − y2 ) > Xp
⋅
Xp > n tw ⋅ (y 3 − y 2 ) > 9/6 ⋅ (1/1317 − 1/1144 ) >1/258!
lh
t
⋅
⋅
⋅
prvi zakon termodinamike za proces u torwu:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
X2 ⋅ i x2 + n tw ⋅ i2 = X3 ⋅ i x3 + n tw ⋅ i3
⋅
R ups > X2 ⋅ i x2 − X3 ⋅ i x3 = n tw ⋅ (i 3 − i2 )
⋅
R ups > n tw ⋅ (i 3 − i2 ) > 9/6 ⋅ (8:/64 − 41/68) >528/97!lX
Xp > X2 − X3
R ups > X2 ⋅ i x2 − X3 ⋅ i x3
)2*
)3*
Kombinovawem jedna~ina!)2*!i!)3*!dobija se:
X 2>3/869!
lh
lh
-! X 3>3/722!
t
t
prvi zakon termodinamike za proces u hladwaku prostorije koju treba hladiti:
⋅
⋅
⋅
R 23 = ∆ I23 + X u23
R ( = X2 ⋅ ix2 − X3 ⋅ ix3 − Xp ⋅ ixp
R ( = 3/869 ⋅ 349/37 − 3/722 ⋅ :2/:7 − 1/258 ⋅ 77/99 >518/29!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 11
8/9/!U nekom procesu izobarski se hladi vla`an vazduh, po~etnog stawa!)q2>2!cbs-!u2>41pD-!ϕ2>1/9nww>211!)2,y*!lh0i*/!Odrediti:
a) koli~inu toplote koja se odvodi od vla`nog vazduha kao i koli~inu izdvojenog kondenzata ako se
hla|ewe vazduha vr{i do!u3>21pD
b) koli~inu toplote koja se odvodi od vla`nog vazduha kao i koli~inu izdvojenog leda ako se hla|ewe
vazduha vr{i do!u4>−21pD
c) koli~inu toplote koja se odvodi od vla`nog vazduha kao i koli~inu izdvojenog leda i kondenzata ako
se hla|ewe vazduha vr{i do!u4>1pD!i pri tome nastaje jednaka koli~ina leda i kondenzata
Sve procese predstaviti na Molijerovom i−y dijagramu za vla`an vazduh
⋅
− R 23
2
3′
3
izdvojeni
kondenzat
i/ili led
ta~ka 1:
q qt >5352!Qb
)napon pare ~iste vode na!u>41pD*
q I3P = ϕ ⋅ q qt > 1/9 ⋅ 5352>44:3/9!Qb
y2!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
44:3/9
⋅
>1/1329!
6
lhTW
3: 2 ⋅ 21 − 44:3/9
i2> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 41 + 1/1329 ⋅ )2/97 ⋅ 41 + 3611* >96/83!
dipl.ing. @eqko Ciganovi}
lK
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 12
a)
i
2
ϕ2
u2
ϕ>2
3
u3
3′
y
ta~ka 2:
q qt >2338!Qb
)napon pare ~iste vode na!u>21pD*
q I3P = ϕ ⋅ q qt > 2⋅ 2338 >2338!Qb
y3!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
2338
>1/1188!
⋅
6
3: 2 ⋅ 21 − 2338
lhTW
i3> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 21 + 1/1188 ⋅ )2/97 ⋅ 21 + 3611* >3:/4:!
lK
lhTW
koli~ina izdvojenog kondenzata:
⋅
⋅
nlpoe = n tw ⋅ (y2 − y 3 ) = 211 ⋅ (1/1329 − 1/1188) >2/52!
lh
i
⋅
⋅
⋅
prvi zakon termodinamike za proces u hladwaku vazduha:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
R23 = ntw ⋅ (i3 − i2) + nlpoe ⋅ ix >
ix!>!53!
lK
lh
dipl.ing. @eqko Ciganovi}
211
2/52
⋅ (3:/4: − 96/83) +
⋅ 53 >−2/66!lX
4711
4711
entalpija kondenzata (vode)!q>2!cbs-!u>21pD
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 13
b)
i
2
ϕ2
u2
ϕ>2
y
3
u3
3′
ta~ka 2:
q qt >36:/5!Qb )napon pare ~iste vode na!u>−21pD*
q I3P = ϕ ⋅ q qt > 2⋅ 36:/5 >36:/5!Qb
y3!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
36:/5
⋅
>1/1127!
6
lhTW
3: 2 ⋅ 21 − 36:/5
i3> dqtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ (−21) + 1/1127 ⋅ )2/97 ⋅ (−21) + 3611* >
!!!>−7/14
lK
lhTW
koli~ina izdvojenog leda:
⋅
⋅
nmfe = ntw ⋅ (y2 − y3 ) = 211 ⋅ (1/1329 − 1/1127) >3/13!
lh
i
⋅
⋅
⋅
prvi zakon termodinamike za proces u hladwaku vazduha:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
R23 = ntw ⋅ (i3 − i2) + nlpoe ⋅ im >
211
3/13
⋅ (− 7/14 − 96/83) +
⋅ (− 463/5 ) >−3/86!lX
4711
4711
im!>! d m ⋅ (Um − 384) − sm = 3 ⋅ (−21) − 443/5 >−463/5!
dipl.ing. @eqko Ciganovi}
lK
!!!!!!entalpija leda-!u>−21pD
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 14
c)
i
2
ϕ2
u2
ϕ>2
3
u3
y
3′
ta~ka 2:
q qt >721/9!Qb )napon pare ~iste vode na!u>1pD*
q I3P = ϕ ⋅ q qt > 2⋅ 721/9 >721/9!Qb
y3!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
721/9
⋅
>1/1149!
6
3: 2 ⋅ 21 − 721/9
lhTW
i3> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 1 + 1/1149 ⋅ )2/97 ⋅ 1 + 3611* >:/6
lK
lhTW
koli~ina izdvojenog kondenzata i leda:
⋅
⋅
⋅
nlpoe + nmfe = n tw ⋅ (y2 − y 3 ) = 211 ⋅ (1/1329 − 1/1149) >2/9!
⋅
⋅
nlpoe = nmfe >1/:!
lh
i
lh
i
⋅
⋅
⋅
prvi zakon termodinamike za proces u hladwaku vazduha:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
⋅
R 23 = ntw ⋅ (i3 − i2 ) + nlpoe ⋅ i x + nm ⋅ im >
⋅
211
1/:
⋅ (:/6 − 96/83) +
⋅ (− 443/5 ) >−3/3!lX
4711
4711
lK
im!>! d m ⋅ (Um − 384) − sm = 3 ⋅ 1 − 443/5 >−443/5!
!!!!!!entalpija leda-!u>1pD
lh
lK
ix!>1!
entalpija kondenzata (vode)!q>2!cbs-!u>1pD
lh
R 23 =
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 15
8/:/!Iz!21!)2,y*!lh0t!vla`nog vazduha stawa!2)q>2!cbs-!u>31pD-!y>1/13!lhI3P0!lhTW*!izdvaja se vlaga u
te~nom stawu, a zatim se preostali vazduh zagreva izobarski dok se ne postigne relativna vla`nost od
ϕ>1/4/!Odrediti maseni protok izdvojene vlage )lh0t* kao i temperaturu vla`nog vazduha nakon
zagrevawa. Prikazati procese sa vla`nim vazduhom na Molijerovom!i−y!dijagramu.
ta~ka 1:
i2>g)u2-!y2*!>!69!
ta~ka 2:
lK
!)pro~itano sa Molijerovog!i−y!dijagrama*
lhTW
(qI3P )3 = ϕ3 ⋅ (qqt )U3 =31 > 2⋅ 3448 >3448!Qb
y3>
NI3P
q I3P
lhI3 P
29
3448
>1/125:!
⋅
=
⋅
6
lhTW
NTW q − q I3P 3: 2 ⋅ 21 − 3448
ta~ka 3:
lhI3 P
lhTW
y4
1/125:
=
⋅q =
⋅ 2 ⋅ 21 6 >1/1345!cbs
NTW
29
+ 1/125:
+ y4
3:
NI3P
y4>y3>1/125:!
(qI3P )4
(qqt )4 =
(qI3P )4
ϕ4
=
1/1345
>1/189!cbs
1/4
⇒
u4!>!)ul*Q>1/189!cbs>52/6pD
koli~ina odstrawene vlage:
⋅
⋅
n x = n tw ⋅ (y 2 − y 3 ) = 21 ⋅ (1/13 − 1/125:) >1/162!
lh
t
i
ϕ4
i2
ϕ>2
3
u2
2
y
y2
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 16
8/21/!Vla`an vazduh stawa!2)q>286!lQb-!utu>39pD-!uwu>33pD*!i zapreminskog protoka!W>1/6!n40t
izobarski se vla`i u adijabatski izolovanoj komori sa!1/13!lh0s pregrejane vodene pare stawa!)q>286
lQb-!u>511pD*!do stawa 2. Odrediti:
a) temperaturu vla`nog vazduha stawa!3
b) koli~inu toplote koju bi trebalo odvesti od vla`nog vazduha stawa!2 da bi ga izobarski ohladili do
temperature od −21pD!)stawe!4*-!kao i masu leda u jedinici vremena!)lh0i*!koja se tom prilikom
izdvoji iz vla`nog vazduha
c) skicirati sve procese sa vla`nim vazduhom na Molijerovom i−y dijagramu
ta~ka WU:
q qt >3784!Qb
)napon pare ~iste vode na!u>33pD*
q I3P = ϕ ⋅ q qt > 2⋅ 3784 >3784!Qb
ywu!>
NI3P
NTW
⋅
q I3P
>
q − q I3P
lhI3 P
29
3784
⋅
>1/11:7!
6
3: 2/86 ⋅ 21 − 3784
lhTW
iwu> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 33 + 1/11:7 ⋅ )2/97 ⋅ 33 + 3611* >57/4:
lK
lhTW
ta~ka 1:
lK
lhTW
i2 − d q ⋅ u 2
i2>iwu!>57/25!
y2 =
2/97 ⋅ u 2 + 3611
(qI3P )2 = N
y2
I3P
NTW
=
lhI3 P
57/25 − 2 ⋅ 39
>1/1182!
lhTW
2/97 ⋅ 39 + 3611
⋅q =
+ y2
1/1182
⋅ 2/86 ⋅ 21 6 >2:8:!Qb
29
+ 1/1182
3:
(q tw )2 = q − (q I3P )2 = 2/86 ⋅ 21 6 − 2:8: > 2/84 ⋅ 21 6 !Qb
(ρ tw )2 =
⋅
(q TW )2
S hTW ⋅ U2
=
2/84 ⋅ 21 6
lhTW
>3!
398 ⋅ 412
n4
⋅
n tw = ρ tw ⋅ W = 3 ⋅ 1/6 >2!
dipl.ing. @eqko Ciganovi}
lhTW
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 17
ta~ka 2:
materijalni bilans vlage za proces vla`ewa vazduha:
⋅
⋅
⋅
⋅
n tw ⋅ y2 + nqq = ntw ⋅ y 3
⇒!
y3 =
⋅
n tw ⋅ y2 + nqq
⋅
n tw
y3 =
lhI3 P
2 ⋅ 1/1182 + 1/13
>1/1382!
2
lhTW
⋅
⋅
⋅
prvi zakon termodinamike za proces vla`ewa vazduha:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
⋅
n tw ⋅ i2 + nqq ⋅ iqq = n tw ⋅ i 3
⇒!
i3 =
⋅
ntw ⋅ i2 + nqq ⋅ iqq
⋅
ntw
lK
2 ⋅ 57/25 + 1/13 ⋅ 4387/6
>222/78!
lhTW
2
i 3 − y 3 ⋅ 3611
222/78 − 1/1382 ⋅ 3611
u3!>!
!>
>52/9pD
2 + 1/1382 ⋅ 2/97
d q + y 3 ⋅ 2/97
i3 =
iqq>!4387/6!
lK
-!
lh
entalpija pregrejane vodene pare, q>2/86!cbs-!u>511pD
ta~ka 3:
q qt >36:/5!Qb )napon pare ~iste vode na!u>−21pD*
q I3P = ϕ ⋅ q qt > 2⋅ 36:/5 >36:/5!Qb
y4!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
36:/5
⋅
>1/111:!
6
lhTW
3: 2/86 ⋅ 21 − 36:/5
i3> dqtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ (−21) + 1/111: ⋅ )2/97 ⋅ (−21) + 3611* >
!!!>−8/88
lK
lhTW
koli~ina izdvojenog leda:
⋅
⋅
nmfe = n tw ⋅ (y 3 − y 4 ) = 2 ⋅ (1/1382 − 1/111:) >1/1373!
lh
lh
>!:5/43!
t
i
⋅
⋅
⋅
prvi zakon termodinamike za proces u hladwaku vazduha:!!!!! R 34 = ∆ I34 + X u34
⋅
⋅
⋅
R 34 = n tw ⋅ (i 4 − i 3 ) + nm ⋅ im > 2 ⋅ (− 8/88 − 222/78) +
im!>! d m ⋅ (Um − 384) − sm = 3 ⋅ (−21) − 443/5 >−463/5!
dipl.ing. @eqko Ciganovi}
:5/43
⋅ (− 463/5 ) >−239/8!lX
4711
lK
!!!!!!entalpija leda-!u>−21pD
lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 18
i
4387/6
3
2
u2
ϕ>2
uwu
WU
y
4
u4
4′
zadatak za ve`bawe:
)8/22/*
⋅
8/22/!Pri izobarskom hla|ewu W >91!n40i vla`nog vazduha stawa!2)q>2!cbs-!u>31pD-!ϕ>1/7* do stawa
3)u>11D) od vla`nog vazduha odvede se 9:1!X toplote. Rashladna povr{ina sastoji se iz 23 plo~a
dimenzija 31!Y!41!dn zanemarqive debqine. Odrediti vreme potrebno da se na rashladnim plo~ama
stvori sloj leda debqine δ=5!dn. Pretpostaviti ravnomernost debqine leda. )ρM>:11!lh0n4*
re{ewe:
τ>351111!t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 19
⋅
8/23/!Struja vla`nog vazduha stawa 2)q>2!cbs-!u>41!pD-!ϕ>31&- W >26
n4
) me{a se sa strujom vla`nog
njo
n4
*/!Skicirati proces me{awa na Molijerovom i−y
njo
dijagramu i odrediti temperaturu!)u*!-!apsolutnu vla`nost!)y) i entalpiju!)i*!novonastale me{avine ako se
me{awe vr{i:
a) adijabatski
b) neadijabatski, pri semu se okolini predaje!!4!lX!toplote
⋅
vazduha stawa!3)q>2!cbs-!u>51!pD-!ϕ>91&- W >31
ϕ3
i
3
N
u3
O
u2
ϕ>2
2
4
ϕ2
y
ta~ka 1:
q qt >5352!Qb
)napon pare ~iste vode na!u>41pD*
q I3P = ϕ ⋅ q qt > 1/3 ⋅ 5352>959/3!Qb
y2!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
959/3
⋅
>1/1164!
6
lhTW
3: 2 ⋅ 21 − 959/3
i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 41 + 1/1164 ⋅ )2/97 ⋅ 41 + 3611* >54/66!
lK
lhTW
q tw = q − q I3P > 2 ⋅ 21 6 − 959/3 >::262/9!Qb
ρtw>
⋅
⋅
q TW
::262/9
26
lhTW
lh
= 2/25!
- !!! n tw2> ρ tw ⋅ W 2 > 2/25 ⋅
>
>1/396!
4
S hTW ⋅ U 398 ⋅ 414
71
t
n
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 20
ta~ka 2:
q qt >8486!Qb
)napon pare ~iste vode na!u>51pD*
q I3P = ϕ ⋅ q qt > 1/9 ⋅ 8486 >6:11!Qb
y3!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
6:11
⋅
>1/149:!
6
lhTW
3: 2 ⋅ 21 − 6:11
i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 51 + 1/149: ⋅ )2/97 ⋅ 51 + 3611* >251/25
lK
lhTW
q tw = q − q I3P > 2 ⋅ 21 6 − 6:11 >:5211!Qb
ρtw>
⋅
⋅
q TW
:5211
31
lh
lhTW
n
>
W
!!!
= 2/16!
ρ
⋅
>1/46!
>
3 > 2/16 ⋅
tw3
tw
4
S hTW ⋅ U 398 ⋅ 424
71
t
n
ta~ka!N;
materijalni bilans vlage za proces me{awa dva vla`na vazduha:
⋅
⋅
n tw2 ⋅ y 2 + n tw3 ⋅ y 3 = ntw2 + n tw3 ⋅ y n ⇒!
⋅
⋅
yn =
⋅
yn =
⋅
n tw2 ⋅ y 2 + ntw3 ⋅ y 3
⋅
⋅
n tw2 + n tw3
lhI3 P
1/396 ⋅ 1/1164 + 1/46 ⋅ 1/149:
>1/1349!
1/396 + 1/46
lhTW
⋅
⋅
⋅
n tw2 ⋅ i2 + ntw3 ⋅ i 3 = n tw2 + n tw3 ⋅ in ⇒!
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X u23
prvi zakon termodinamike za proces me{awa:
⋅
in =
⋅
ntw2 ⋅ i2 + n tw3 ⋅ i 3
⋅
⋅
n tw2 + n tw3
1/396 ⋅ 54/66 + 1/46 ⋅ 251/25
lK
>:7/8:!
1/396 + 1/46
lhTW
in − y n ⋅ 3611
:7/8: − 1/1349 ⋅ 3611
un!>!
>46/82pD
!>
d q + y n ⋅ 2/97
2 + 1/1349 ⋅ 2/97
in =
ta~ka!O;
⋅
⋅
R 23
⋅
⋅
R 23 = ∆ I23 + X u23
prvi zakon termodinamike za proces me{awa:
⋅
⋅
⋅
⋅
⋅
⋅
n tw2 ⋅ i2 + n tw3 ⋅ i 3 − R 23
⋅
= ntw2 + n tw3 ⋅ io − ntw2 ⋅ i2 − n tw3 ⋅ i3 -!! io =
⋅
⋅
n tw2 + ntw3
1/396 ⋅ 54/66 + 1/46 ⋅ 251/25 − 4
lK
>:3/18!
1/396 + 1/46
lhTW
lhI3 P
:3/18 − 1/1349 ⋅ 3611
!
uo!>!
yo!>yn!>1/1349!
>!42/26pD
2 + 1/1349 ⋅ 2/97
lhTW
io =
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 21
⋅
8/24/!Za prostoriju u kojoj se gaje {ampiwoni (slika) priprema se! n 4>6111!lh0i!vla`nog vazduha na
slede}i na~in: sve` vazduh stawa!2)q>2!cbs-!u>−21pD-!ϕ>1/9*!!adijabatski se me{a se sa delom
iskori{}enog vazduha stawa!5)q>2cbs-!u>33!pD-!ϕ>1/:*!u odnosu!2;3/!Dobijeni vla`an vazduh stawa
N)q>2!cbs) se zagreva u zagreja~u do stawa!3)q>2cbs-!u>36pD) a zatim adijabatski vla`i uvo|ewem
suvozasi}ene vodene pare stawa!Q)u>211pD*!do stawa!4)q>2!cbs*!kada vazduh dosti`e apsolutnu
vla`nost otpadnog vazduha. Tako dobijen vazduh se u komori sa {ampiwonima hladi. Skicirati
promene stawa vla`nog vazduha na Molijerovom!i!−y!dijagramu i odrediti:
a) temperaturu vla`nog vazduha stawa!N
b) temperaturu vla`nog vazduha stawa!4
c) toplotnu snagu zagreja~a vazduha!)lX*
d) potro{wu vodene pare u fazi vla`ewa!)lh0t*
n′′
4
komora za
vla`ewe
3
5
prostorija sa
{ampiwonima
N
5
5
recirkulacioni
vazduh
2
sve`
vazduh
otpadni
vazduh
i
4
3
3786
5
ϕ>2
N
2
y
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 22
ta~ka 1:
q qt >36:/5!Qb )napon pare ~iste vode na!u>−21pD*
q I3P = ϕ ⋅ q qt > 1/9 ⋅ 36:/5 >318/6!Qb
y2!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
318/6
⋅
>1/1124!
6
lhTW
3: 2 ⋅ 21 − 318/6
i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2⋅ (−21) + 1/1124 ⋅ )2/97 ⋅ (−21) + 3611* >−7/88!
lK
lhTW
ta~ka 4:
q qt >3754!Qb
)napon pare ~iste vode na!u>33pD*
q I3P = ϕ ⋅ q qt > 1/: ⋅ 3754 >3489/8!Qb
y5!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
3489/8
>1/1262!
⋅
3: 2⋅ 216 − 3489/8
lhTW
i5> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 33 + 1/1262 ⋅ )2/97 ⋅ 33 + 3611* >71/48!
lK
lhTW
ta~ka!N;
materijalni bilans vlage za proces me{awa dva vla`na vazduha:
⋅
n tw2
⋅
⋅
n tw2 ⋅ y2 + n tw 5 ⋅ y 5 = n tw2 + n tw 5 ⋅ y n
⋅
⋅
⋅
⇒!
yn =
⋅ y2 + y 5
ntw 5
⋅
n tw2
⋅
+2
n tw 5
yn
2
⋅ 1/1124 + 1/1262
lhI3 P
>1/1216
= 3
2
lhTW
+2
3
⋅
⋅
⋅
R 23 = ∆ I23 + X u23
prvi zakon termodinamike za proces me{awa:
⋅
n tw2
⋅
⋅
n tw2 ⋅ i2 + n tw 5 ⋅ i 5 = n tw2 + n tw 5 ⋅ in
⋅
⋅
⋅
⇒!
in =
⋅ i2 + i 5
n tw 5
⋅
ntw2
⋅
+2
n tw 5
in
2
⋅ (− 7/88) + 71/48
lK
3
=
>48/::!
2
lhTW
+2
3
un!>!
in − y n ⋅ 3611
48/:: − 1/1216 ⋅ 3611
!>
>22/63pD
d q + y n ⋅ 2/97
2 + 1/1216 ⋅ 2/97
⋅
napomena:
n tw5!je oznaka za maseni protok samo recirkulacionog vazduha !!
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 23
ta~ka 2:
y3!>!yn>1/1216!
lhI3 P
lhTW
i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 36 + 1/1216 ⋅ )2/97 ⋅ 36 + 3611* >62/85
lK
lhTW
ta~ka 3:
lhI3 P
lhTW
6111
⋅
H4
lhTW
n tw4!>!
> 4711 >2/49!
2 + y 4 2 + 1/1262
t
y4!>!y5>1/1262!
⋅
⋅
⋅
⋅
⋅
n tw4!>! n tw3!>! n twn> n tw2!,! n tw5
⋅
n tw2 =
)2*
⋅
ntw 5
3
)3*
⋅
Kombinovawem jedna~ina!)2*!i!)3*!dobija se;! n tw2>1/57!
lh ⋅
lh
-! n tw5>1/:3!
t
t
materijalni bilans vlage za proces vla`ewa vazduha;
⋅
⋅
⋅
⋅
⋅
⋅
n tw2 + ntw 5 ⋅ y 3 + n( ( = ntw2 + ntw 5 ⋅ y 4 !!!!⇒ n( ( = n tw2 + n tw 5 ⋅ (y 4 − y 3 )
! n( ( = 2/49 ⋅ (1/1262 − 1/1216) >7/46!/21−4!
dipl.ing. @eqko Ciganovi}
lh
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 24
⋅
⋅
⋅
prvi zakon termodinamike za proces vla`ewa vazduha:! R 23 = ∆ I23 + X u23
⋅
⋅
ntw2+ ntw 5 ⋅ i3 + n( (⋅i#
⋅
⋅
⋅
⋅
ntw2+ ntw 5 ⋅ i3 + n#⋅i# = ntw2+ ntw 5 ⋅ i4 !!!!⇒!!! i4 =
⋅
⋅
ntw2+ ntw 5
i4 =
2/49 ⋅ 62/85 + 7/46 ⋅ 21 −4 ⋅ 3786
lK
>75/16!
2/49
lhTW
u4!>!
i 4 − y 4 ⋅ 3611
75/16 − 1/1262 ⋅ 3611
>36/69pD
!>
d q + y 4 ⋅ 2/97
2 + 1/1262 ⋅ 2/97
napomena:
i′′>!3786!
lK
-!entalpija suvozasi}ene vodene pare stawa!Q)u>211pD*
lh
⋅
⋅
⋅
prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!!!! R 23 = ∆ I23 + X u23
R {bh = (Htw2 + Htw 5 ) ⋅ (i3 − in ) > 2/49 ⋅ (62/85 − 48/:: ) >29/:9!lX
zadatak za ve`bawe:
)8/25/*
8/25/ n2>3!)2,y*!lh0t!vla`nog vazduha stawa!2)q>2!cbs-!y>1/116!lh0lhTW) adijabatski se me{a sa n3>4
)2,y*!lh0t!vla`nog vazduha stawa!3)q>2!cbs-!y>1/17!lh0lhTW-!u>61!pD*/!Ne koriste}i i−y dijagrama
odrediti:
a) temperaturu vla`nog vazduha 1 tako da vazduh dobijen me{awem vazduha 1 i 2 bude zasi}en
b) temperaturu dobijenog zasi}enog vla`nog vazduha
c) temperaturu vla`nog vazduha 1 tako da vazduh dobijen me{awem vazduha 1 i 2 bude zasi}en za slu~aj da
je me{awe neadijabatsko uz toplotne gubitke u okolinu od Rp>5!lX
d) skicirati sve procese na i−y dijagramu
a) u2>23/6pD
b) uN>46/5pD
c) u2′>25/8pD
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 25
8/26/ Za klimatizaciju nekog objekta potrebno je obezbediti vla`an vazduh stawa
⋅
4)q>1/23!NQb-!u>33pD-!ϕ>61&-! W >1/5!n40t*/!U tu svrhu koristi se ure|aj koji se sastoji iz filtera,
hladwaka, zagreja~a vazduha i ventilatora-duvaqke, (slika). Snaga ventilatora koji adijabatski sabija
vazduh sa pritiska!q3)>q2>qp*!na pritisak!q4!je 2/5!lX/!Stawe okolnog nezasi}enog vla`nog vazduha je
P)qp>1/2!NQb-!up>41pD-!ϕ>61&-*/!Prikazati proces pripreme vla`nog vazduha na Molijerovom!i!−y
dijagramu i odrediti:
b* koli~inu izdvojenog kondenzata!)lh0i*
c* toplotnu snagu hladwaka vazduha,!Rimb!)lX*
d* toplotnu snagu zagreja~a vazduha,!R{bh!)lX*
4
W4
Rimb
h
l
a
d
w
a
k
f
i
l
t
e
r
1
,R{bh
2
z
a
g
r
e
j
a
~
v
e
n
t
i
l
a
t
o
r
3
X
kondenzat
i
ϕ>2-!q>2/3!cbs
4
i
1
y
3
ϕ>2-!q>2!cbs
2
y
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 26
ta~ka 3:
)napon pare ~iste vode na!u>33pD*
q qt >3754!Qb
q I3P = ϕ ⋅ q qt > 1/6 ⋅ 3754 >2432/6!Qb
y4!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
2432/6
⋅
>1/117:!
6
lhTW
3: 2/3 ⋅ 21 − 2432/6
i4> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 33 + 1/117: ⋅ )2/97 ⋅ 33 + 3611* >4:/64!
lK
lhTW
q tw = q − q I3P > 2/3 ⋅ 21 6 − 2432/6 >229789/6!Qb
ρtw>
⋅
q TW
229789/6
lh
lhTW
>
> 2/51 ⋅ 1/5 >1/67!
!!!H
= 2/51!
ρ
⋅
W
>
tw
tw
4
S hTW ⋅ U 398 ⋅ 3:6
t
n
ta~ka 2:
y3>y4>1/117:!
lhI3 P
lhTW
⋅
⋅
⋅
prvi zakon termodinamike za proces u ventilatoru:!!!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
X u23 = ntw ⋅ (i 3 − i 4 )
⇒
i3 = i4 +
X u23
⋅
> 4:/64 −
n tw
2/5
lK
>48/14!
1/67
lhTW
ta~ka 1:
y2>y3>1/117:!
q I3P =
q qt =
lhI3 P
lhTW
y
NTW
+y
NI3P
q I3P
=
ϕ
p
u2!>9/6 D
⋅ q2 >
1/117:
⋅ 2 ⋅ 21 6 = 21::/5!Qb
29
+ 1/117:
3:
21::/5
= 21::/5!Qb
2
)temperatura kqu~awa vode na!q>21::/5!cbs*
i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 9/6 + 1/117: ⋅ )2/97 ⋅ 9/6 + 3611* >36/97
lK
lhTW
ta~ka 0:
q qt >5352!Qb
)napon pare ~iste vode na!u>41pD*
q I3P = ϕ ⋅ q qt > 1/6 ⋅ 5352>3231/6!Qb
yp!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
3231/6
>1/1245!
⋅
3: 2 ⋅ 21 6 − 3231/6
lhTW
ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 41 + 1/1245 ⋅ )2/97 ⋅ 41 + 3611* >75/36!
dipl.ing. @eqko Ciganovi}
lK
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 27
⋅
koli~ina izdvojenog kondenzata:
⋅
X l = n tw ⋅ (y 2 − y 3 )
⋅
X l = 1/67 ⋅ (1/1245 − 1/117:) ⋅ 4711 >24/21!
lh
i
⋅
⋅
⋅
prvi zakon termodinamike za proces u hladwaku vazduha:!!!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
R imb = n tw ⋅ (i2 − i 1 ) + X l ⋅ il > 1/67 ⋅ (36/97 − 75/36 ) +
napomena:
24/21
⋅ 46/64 >−32/48!lX
4711
il!−!entalpija kondenzata (voda!q>2!cbs-!u>9/6pD*
⋅
⋅
⋅
prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
R {bh = n tw ⋅ (i 3 − i2 ) > 1/67 ⋅ (48/14 − 36/97 ) >7/37!lX
8/27/ Postrojewe za delimi~no su{ewe vazduha sastoji se od vodom hla|enog klipnog kompresora i
⋅
hladwaka za vla`an vazduh (slika). U klipnom kompresoru se sabija!! n ww>1/38!)2,y*!lh0t!vla`nog
vazduha stawa!2)q2>1/2!NQb-!u2>31pD-!ϕ2>1/9*!do stawa!3)q3?q2-!u3>56pD-!ϕ3>2), a potom se uz
izdvajawe te~ne faze vla`an vazduh stawa!3!izobarski hladi do stawa!4)u4>u2). Ukupan toplotni fluks
sa vla`nog vazduha na rashladnu vodu u toku procesa sabijawa i izobarskog hla|ewa vla`nog vazduha
iznosi!R>RI2,RI3>24!lX/!Odrediti pritisak vla`nog vazduha na kraju procesa sabijawa, koli~inu
izdvojenog kondenzata kao i pogonsku snagu za pogon klipnog kompresora.
RI3
4
3
X
2
vla`an
vazduh
vla`an
vazduh
kondenzat
RI2
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 28
ta~ka 1:
)napon pare ~iste vode na!u>31pD*
q qt >3448!Qb
q I3P = ϕ ⋅ q qt > 1/9 ⋅ 3448 >297:/7!Qb
y2!>
NI3P
NTW
⋅
q I3P
q2 − q I3P
>
lhI3 P
29
297:/7
⋅
>1/1229!
lhTW
3: 2 ⋅ 21 6 − 297:/7
i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1229 ⋅ )2/97 ⋅ 31 + 3611* >61/13!
lK
lhTW
ta~ka 2:
lhI3 P
lhTW
q qt >:695!Qb )napon pare ~iste vode na!u>56pD*
y3>y2>1/1229!
q I3P = ϕ ⋅ q qt > 2⋅ :695 >:695!Qb
NI3P
N tw
q3!>!
y3
+ y3
⋅ q I3P
29
+ 1/1229
= 3:
⋅ :695 >!624821!Qb!>!6/248!cbs
1/1229
i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 56 + 1/1229 ⋅ )2/97 ⋅ 56 + 3611* >86/5:!
lK
lhTW
ta~ka 3:
q qt >3448!Qb
)napon pare ~iste vode na!u>31pD*
q I3P = ϕ ⋅ q qt > 2⋅ 3448 >3448!Qb
y4!>
NI3P
NTW
⋅
q I3P
q 4 − q I3P
>
lhI3 P
29
3448
⋅
>1/1139!
lhTW
3: 6/248 ⋅ 21 6 − 3448
i4> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1139 ⋅ )2/97 ⋅ 31 + 3611* >38/21!
lK
lhTW
koli~ina izdvojenog kondenzata:
⋅
⋅
X = ntw ⋅ (y 3 − y 4 ) = 1/38 ⋅ (1/1229 − 1/1139) > 3/54 ⋅ 21 −4 !
dipl.ing. @eqko Ciganovi}
lh
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 29
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
⋅
⋅
⋅
⋅
⋅
⋅
! R 23 = ∆ I23 + X u23
ograni~enom isprekidanom konturom:
⋅
⋅
X !>! − ntw ⋅ i 4 − X x ⋅ i x + n tw ⋅ i2 − R I2 − R I3
⋅
X >! − 1/38 ⋅ 38/21 − 3/54 ⋅ 21 −4 ⋅ 299/5 + 1/38 ⋅ 61/13 − 24 >−8/38!lX
ix>299/5!
napomena:
lK
lh
entalpija vode!!q>2!cbs-!u>56pD
)8/28/*
zadatak za ve`bawe:
8/28/ 611!lh0i!vla`nog vazduha stawa!2)q>2!cbs-!u>3pD!ϕ>1/9*!me{a se izobarski sa!611!lh0i!vla`nog
vazduha stawa!3)q>2!cbs-!u>57pD-!ϕ>1/8). Zatim se kondenzat koji je nastao me{awem izdvaja, a
preostali vazduh zagreva do!81pD. Nakon zagrevawa vazduhu se dodaje vodena para ~ija entalpija iznosi
3111!lK0lh!i vla`ewe se obavqa do postizawa stawa zasi}ewa. Skicirati procese sa vla`nim
vazduhom na Molijerovom i!−y! dijagramu i odrediti:
a) apsolutnu vla`nost me{avine )y* kada kondenzat jo{ nije izdvojen (ra~unskim putem)
b) maseni protok odvedenog kondenzata!)lh0i*
c) toplotnu snagu greja~a!)lX*
d) maseni protok vodene pare koja se dodaje u ciqu vla`ewa!)lh0i*
za stavke b), c) i d)!mo`e se koristiti Molijerov dijagram za vla`an vazduh
lhI3 P
a) yn!>!1/1355!
lhTW
lh
b) nlpoefo{bu!>!2/5!
i
c) R45!>!23/2!lX
lh
d) nwpefob!qbsb!>!54/8!
i
i
5
3
6
ϕ>2
4
N
2
y
ix>3111!lK
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 30
DRUGI VLA@NI GASOVI
8/29/!Me{avina vodonika (idealan gas) i vodene pare (idealan gas) ima temperaturu!u>41pD-!relativnu
vla`nost!ϕ>:1&!i pritisak!q>311!lQb/!Za navedenu gasnu me{avinu odrediti:
a) apsolutnu vla`nost!)y*!i specifi~nu entalpiju!)i*!vla`nog vodonika
b) masene udele vodonika i vodene pare u vla`nom vodoniku
a)
)napon pare ~iste vode na!u>41pD*
q qt >5352!Qb
q I3P = ϕ ⋅ q qt > 1/: ⋅ 5352>4927/:!Qb
y2!>
NI3P
NI3
⋅
q I3P
q2 − q I3P
>
lhI3 P
29
4927/:
⋅
>1/2862!
6
lhTW
3 3 ⋅ 21 − 4927/:
i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 25/66 ⋅ 41 + 1/2862 ⋅ )2/97 ⋅ 41 + 3611* >995/13
lK
lhI3
b)
nI3P
hI3P =
nI3P
nI3P + nI3
=
nI3
nI3P
nI3
+
nI3
=
y
1/2862
=
>1/26
y + 2 1/2862 + 2
nI3
nI3
hI3 =
nI3
nI3P + nI3
=
nI3
nI3P
nI3
dipl.ing. @eqko Ciganovi}
+
nI3
=
2
2
=
>1/96
y + 2 1/2862 + 2
nI3
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 31
8/2:/!U vertikalnom cilindru sa klipom, po~etne zapremine!W>1/2!n4 nalazi se, pri stalnom pritisku
q>3!cbs!sme{a ugqen−dioksida (idealan gas) i pregrejane vodene pare. Maseni udeo vodene pare u sme{i
je! hI3P >1/2-!a po~etna temperatura!sme{e!:1pD/!Odrediti koli~inu toplote koju treba odvesti od
vla`nog ugqen-dioksida da bi zapo~ela kondenzacija vodene pare.
ta~ka 1:
y2!>!
hI3P
2 − hI3P
=
lhI3 P
1/2
>1/2222!
2 − 1/2
lhDP3
i2> dqDP3 u + y ⋅ )2/97 ⋅ u + 3611* > 1/96 ⋅ :1 + 1/2222 ⋅ )2/97 ⋅ :1 + 3611* >483/96
qI3P =
y2
NI3P
NDP3
!
⋅q =
+ y2
lK
lhI3
1/2222
⋅ 3 ⋅ 216 > 1/4 ⋅ 216 Qb!>1/41!cbs
29
+ 1/2222
55
qDP3 = q − qI3P >3!−!1/4!>2/8!cbs
ρDP3 =
qDP3
ShDP3 ⋅U
=
lhDP3
2/8 ⋅ 216
>3/59
29: ⋅ 474
n4
nDP3 = ρDP3 ⋅ W = 3/59 ⋅ 1/2 >1/359!lh
ta~ka 2:
y3!>!y2!>!1/2222!
qqt3 =
qI3P
ϕ3
=
lhI3 P
!!
lhDP3
qI3P = dpotu > 1/4 ⋅ 216 Qb!>1/41!cbs
1/4
>1/4!cbs
2
u3!>!)ulr*q>1/4!cbs!≈!7:pD
i3> dqDP3 u + y ⋅ )2/97 ⋅ u + 3611* > 1/96 ⋅ 7: + 1/2222 ⋅ )2/97 ⋅ 7: + 3611* >461/77
lK
lhI3
prvi zakon termodinamike za proces u cilindru:
R23!>!∆V23!,!X23
⇒
R23!>!V3!−!V2!,!q!/)W3!.−!W2*!
R23!>!I3!−!I2!
⇒
R23!>! nDP3 ⋅ (i3 − i2)
R23!>! 1/359 ⋅ (461/77 − 483/96 ) >−6/6!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 32
8/31/!U toplotno izolovanoj komori me{aju se dva toka razli~itih vla`nih gasova zadatih
⋅
termodinami~kih stawa : zasi}en vla`an kiseonik!)P3*!stawa!2)q>1/6!NQb-!U>464!L-! n2 >3!)2,y*!lh0t*!i
⋅
vla`an metan!)DI5) stawa!3)q>1/4!NQb-!U>3:4!L-!ϕ>1/5-! n3 >4!)2,y*!lh0t*/!Promene kineti~ke i
potencijalne energije gasnih tokova su zanemarqive. Odrediti temperaturu vla`ne gasne sme{e koja
izlazi iz komore.
1. vla`an kiseonik
M. me{avina vla`nog
kiseonika i vla`nog metana
2. vla`an metan
ta~ka 1:
q qt >58471!Qb )napon pare ~iste vode na!u>91pD*
q I3P = ϕ ⋅ q qt > 2⋅ 58471 >58471!Qb
y2!>
NI3P
NP3
⋅
qI3P
q2 − qI3P
>
lhI3P
29
58471
⋅
>1/169:!
43 6 ⋅ 216 − 58471
lhP3
i2> dqP3 ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 1/:2 ⋅ 91 + 1/169: ⋅ )2/97 ⋅ 91 + 3611* >339/92!
⋅
nP3 >3
lK
lhP3
lh
t
ta~ka 2:
q qt >3448!Qb
)napon pare ~iste vode na!u>31pD*
q I3P = ϕ ⋅ q qt > 1/5 ⋅ 3448 >:45/9!Qb
y3!>
NI3P
NDI5
⋅
q I3P
q 3 − q I3P
>
lhI3 P
29
:45/9
⋅
>1/1146!
lhDI5
27 4 ⋅ 21 6 − :45/9
i3> dqDI5 ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 3/45 ⋅ 31 + 1/1146 ⋅ )2/97 ⋅ 31 + 3611* >66/79!
⋅
nDI5 >4
lK
lhDI5
lh
t
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 33
ta~ka!N;
materijalni bilans vlage za proces me{awa dva vla`na gasa:
⋅
⋅
⋅
⋅
nP3 ⋅ y2 + nDI5 ⋅ y3 = nP3 + nDI5 ⋅ yn
⋅
⇒!
yn =
⋅
nP3 ⋅ y2 + nDI5 ⋅ y3
⋅
⋅
nP3 + nDI5
lhI3 P
3 ⋅ 1/169: + 4 ⋅ 1/1146
>1/1368!
3+4
lh(P 3 + DI 5 )
yn =
⋅
⋅
⋅
nP3 ⋅ i2 + nDI5 ⋅ i 3 = nP3 + nDI5
⋅
in =
⋅
⋅
⋅
⋅
nP3 ⋅ i2 + nDI5 ⋅ i 3
⋅ in !!!!!!⇒!!!!! in =
⋅
⋅
nP3 + nDI5
3 ⋅ 339/92 + 4 ⋅ 66/79
lK
>235/:4!
3+4
lh(P 3 + DI 5 )
⋅
hP3 =
nP3
⋅
⋅
nP3 + nDI5
⋅
3
>1/5
=
3+4
hDI5 =
nDI5
⋅
⋅
nP3 + nDI5
d qn = h P3 ⋅ d qP3 + h DI5 ⋅ d qDI5 = 1/5 ⋅ 1/:2 + 1/7 ⋅ 3/45 >2/88!
un!>!
⋅
R 23 = ∆ I23 + X u23
prvi zakon termodinamike za proces me{awa:
=
4
>1/7
3+4
lK
lh(P 3 + DI 5 )
in − y n ⋅ 3611
235/:4 − 1/1368 ⋅ 3611
!>
>44/5pD>417/5!L
d qn + y n ⋅ 2/97
2/88 + 1/1368 ⋅ 2/97
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 34
8/32/!U toplotno izolovanom kanalu izobarski se me{aju tok kiseonika stawa!2)q>1/3!NQb-!U>391!L⋅
⋅
W >1/657!n40t*!i tok pregrejane vodene pare stawa!Q)q>1/3!NQb-!U>664!L-! nq >1/17!lh0t*/!Nastali
vla`an kiseonik stawa!3, biva potom u vodom hla|enom klipnom kompresoru, pogonske snage!Q>76!lXsabijan do stawa!4)q>1/4!NQb-!ϕ>1/83*/!Odrediti toplotni protok sa vla`nog kiseonika na vodu za
hla|ewe kompresora i prikazati sve procese u!!i−y!koordinatnom sistemu.
nqq
2
3
4
ta~ka!Q;
iqq!>!4141!!
lK
lh
)q>!3!cbs-!u>391pD*
ta~ka 1:
y2!>1!
lhI3 P
lhP3
i2!>! d qP3 ⋅ u 2 + y2 ⋅ )2/97 ⋅ u 2 + 3611* > 1/:2 ⋅ 8 + 1 ⋅ 2/97 ⋅ 8 + 3611* >!7/48!
ρ P3 =
lK
lhP3
⋅
⋅
kg
q P3
lhP3
3 ⋅ 21 6
=
>3/86!
!!!!!
n
P3 = ρ P3 ⋅ W = 3/86 ⋅ 1/657 >2/6
4
S hP3 U2 371 ⋅ 391
n
s
ta~ka 2:
materijalni bilans vlage za proces vla`ewa kiseonika!)2−3*;
⋅
⋅
⋅
⋅
nP3 ⋅ y2 + nqq = nP3 ⋅ y3 !!!!⇒
y3 =
⋅
nP3 ⋅ y2 + nqq
⋅
=
nP3
lhI3 P
1/17
>1/15!
lhP3
2/6
⋅
⋅
⋅
prvi zakon termodinamike za proces vla`ewa kiseonika:!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
nP3 ⋅ i2 + nqq ⋅ iqq = nP3 ⋅ i3
⋅
⇒!
i3 =
⋅
nP3 ⋅ i2 + nqq ⋅ iqq
⋅
nP3
i3 =
2/6 ⋅ 7/48 + 1/17 ⋅ 4141
lK
>238/68!
2/6
lhP 3
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 35
ta~ka 3:
y4!>!y3!>1/15!
y4
1/15
⋅ q4 =
⋅ 4 ⋅ 21 6 !>!2::28!Qb
NI3P
29
+ 1/15
+ y4
43
NP3
q I3P =
q qt =
lhI3 P
lhP3
q I3P
ϕ
=
2::28
>38774Qb!≈1/39!cbs
1/83
u4!>!)ulr*Q>1/39!cbs>78/6pD
i4!>! d qP3 ⋅ u 4 + y 4 ⋅ )2/97 ⋅ u 4 + 3611* > 1/:2 ⋅ 78/6 + 1/15 ⋅ )2/97 ⋅ 78/6 + 3611* !>
>277/56!
⋅
prvi zakon termodinamike za proces u kompresoru:!
⋅
⋅
⋅
lK
lhP3
⋅
R 34 = ∆ I34 + X u34
⋅
R 34 = nP3 ⋅ (i 4 − i 3 ) + X u34 = 2/6 ⋅ (277/56 − 238/68) − 76 >−7/79!lX
i
i
ϕ>2-!q>4!cbs
4
y
3
4141
ϕ>2-!q>3!cbs
2
y
zadatak za ve`bawe:
)8/33/*
8/33/!Vla`an azot, masenog protoka!1/5!lh0t-!stawa!2)q2>4!cbs-!u2>55pD-!ϕ2>1/:*!izobarski se ohladi do
temperature od!1pD!)stawe 2*-!pri ~emu se od azota odvede!47!lX!toplote. Odrediti masuformiranog
kondenzata i masu formiranog leda ako je proces trajao!2!sat.
re{ewe:!
nlpoefo{bu!>!23/:7!lh
dipl.ing. @eqko Ciganovi}
nmfe!>!21/9!lh
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 36
8/34/!^asovni kapacitet teorijske tunelske teorijske su{are iznosi!261!lh suvih banana. Vla`nost
sirovih banana (maseni udeo vlage) je!z2>81!nbt&!a suvih!z3>23!nbt&/!Temperatura vazduha na
izlazu iz su{are kf!51pD a maksimalna temperatura vazduha u su{ari!96pD/!Atmosferski vazduh ima
temperaturu od!29pD!i ta~ku rose!23pD/!Skicirati promene stawa vla`nog vazduha na Molijerovom!i
−y!dijagramu i odrediti potro{wu grejne pare u zagreja~u vazduha (suvozasi}ena vodena para) ako joj je
temperatura za!31!L!vi{a od maksimalne temperature vazduha u su{ari (smatrati da je kondenzat
grejne pare na izlazu iz zagreja~a vazduha neprehla|en). Sve promene stawa vla`nog vazduha su
izobarske na q>2!cbs/
nwn
1
vazduh
2
zagreja~
vazduha
npn
komora za
su{ewe
materijala
3
grejna para
i
2
u2
3
u3
ϕ>2
1
up
us
S
y
napomena:
Teorijski uslovi su{ewa (adijabatska su{ara) podrazumevaju:
1−2;
y>dpotu
2−3;
i>dpotu
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 37
ub•lb!S;
yS>!g)uS-!ϕS>2*!>!1/1199!
lhI3 P
lhTW
ub•lb!1;
yp!>yS!>!1/1199!
lhI3 P
lhTW
ip!>!g)up-!yp*!>!51/4!
lK
lhTW
ub•lb!2;
y2!>!y1!>!yS!>!1/1199!
lhI3 P
lhTW
i2!>!g)u2-!y2*!>!219/5!
lK
lhTW
ub•lb!3;
lhI3 P
lhTW
i3!>!i2>219/5!
lK
lhTW
napomena:
Sve vrednosti pro~itane sa Molijerovog!i!−!y!dijagrama
y3!>!g)u3-!i3*!>!1/1376!
materijalni bilans vlage za proces su{ewa banana:
⋅
n tw ⋅ (y 3 − y 2 ) > n pn ⋅
⋅
n tw =
z2 − z 3
2 − z2
⇒
⋅
n tw = npn ⋅
z2 − z 3
2
⋅
2 − z 2 y 3 − y2
261 1/8 − 1/23
2
lh
⋅
⋅
>5/66
4711
2 − 1/8
1/1376 − 1/1199
t
⋅
⋅
⋅
prvi zakon termodinamike za proces u zagreja~u vazduha:!! R 12 = ∆ I12 + X u12
⋅
⋅
⋅
⋅
n tw ⋅ i p + nq ⋅ i( ( = n tw ⋅ i2 + nq ⋅ i(
⋅
nq =
⇒
⋅
n tw ⋅ (i2 − i p )
nq =
i( (−i(
⋅
5/66 ⋅ (219/5 − 51/4 )
lh
>1/25
3354
t
i′′!−!i′!>!s!>3354!
lK
lh
dipl.ing. @eqko Ciganovi}
toplota kondenzacije vodene pare na!u>216pD
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 38
8/35/!U teorijskoj su{ari su{i se, pri!q>2!cbs-!61!lh0i!kva{~eve biomase koja sadr`i!z2>81!nbt&
vlage, pri ~emu se dobija suvi kvasac sa!z3>8!nbt&!vlage. Na ulazu u zagreja~ stawe vazduha odre|eno
je temperaturom suvog termometra i temperaturom vla`nog termometra!1)utu>27pD-!uwu>21pD*/!Stawe
otpadnog vazduha odre|eno je entalpijom i relativnom vla`nosti vazduha!3)i>:1!lK0lhTW-!ϕ>1/7).
Skicirati promene stawa vla`nog vazduha na Molijerovom!i!−y!dijagramu i odrediti:
b* potro{wu suvog vazduha u su{ari!) no4 0t*
c* toplotnu snagu zagreja~a vazduha!)lX*
d* koliko bi se toplote moglo u{tedeti hla|ewem otpadnog vazduha do stawa zasi}ewa i
rekuperativnim kori{}ewem oslobo|ene toplote za zagrevawe sve`eg vazduha u predgreja~u!)lX*
i
i3
2
ϕ3
ϕ>2
3
up
uwu
1
WU
y
ta~ka 0:
ip>3:/6
lK
lhTW
yp>1/1168
lhI3 P
lhTW
y3>1/1326
lhI3 P
lhTW
ta~ka 2:
i3>:1
lK
lhTW
ta~ka 1:
i2>i3>:1
napomena:
lK
lhTW
y2>yp>1/1168
lhI3 P
lhTW
Sve vrednosti pro~itane sa Molijerovog!i!−!y!dijagrama
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 39
b*
materijalni bilans vlage za proces su{ewa kva{~eve biomase:
⋅
n tw ⋅ (y 3 − y 2 ) > n wn ⋅
⋅
ntw =
z2 − z 3
2− z3
z2 − z 3
2
⋅
2 − z 3 y 3 − y2
⋅
⇒
n tw = n wn ⋅
61 1/8 − 1/18
2
lh
>1/7
⋅
⋅
4711 2 − 1/18 1/1326 − 1/1168
t
⋅
⋅
W o tw = n tw ⋅
n4
33/5
33/5
>1/57! o
> 1/7 ⋅
3:
t
N tw
c*
⋅
⋅
⋅
prvi zakon termodinamike za proces u zagreja~u vazduha:!! R 12 = ∆ I12 + X u12
⋅
⋅
R {bh = n tw ⋅ (i2 − i p ) = 1/7 ⋅ (:1 − 3:/6 ) >47/3:!lX
d*
n wn
n pn
X
Htw
1
C
2
3
Rsfl
B
prvi zakon termodinamike za proces u otvorenom sistemu ograni~enom
⋅
⋅
⋅
isprekidanom konturom:!! R 23 = ∆ I23 + X u23
⋅
⋅
R sfl > − ntw ⋅ (i3 − i B ) > −1/7 ⋅ (:1 − 92/3) >−6/39!lX
napomena:
iB>g)yB>y3-!ϕ>2*>92/3!
dipl.ing. @eqko Ciganovi}
lK
)!Molijerov!i−y!dijagram)
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 40
8/36/!Jednostepena, teorijska su{ara, radi sa vazduhom kao agensom za su{ewe po zatvorenom ciklusu
(slika) na pritisku!q>:1!lQb>jefn/!Nakon zagrevawa vazduha )2−3*- wegovog prolaska kroz komoru za
su{ewe )3−4*-!te hla|ewa )4−5*- u predajniku toplote, u kome se kondenzuje vodena para, ulazi zasi}en
vla`an vazduh stawa!5)U>424!L*-!a napu{ta ga ohla|eni zasi}en vla`an vazduh i izdvojeni kondenzat
temperature!U2>3:4!L. Maseni protok odvedenog kondenzata je!X>1/14!lh0t. Toplotna snaga zagreja~a
vazduha je!R{bh>:6!lX/!Skicirati promene stawa vla`nog vazduha na Molijerovom i!−!y!dijagramu i
odrediti potreban maseni protok suvog vazduha i relativnu vla`nost!)ϕ4*!do koje se, su{ewem vla`nog
materijala, ovla`i vazduh.
vla`an
materijal
zagreja~
2
3
komora
za su{ewe
4
osu{en
materijal
5
!!!L
hladwak
kondenzat
i
3
4
5
ϕ>2
2
y
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 41
ta~ka 1:
)napon pare ~iste vode na!u>31pD*
q qt >3448!Qb
q I3P = ϕ ⋅ q qt > 2⋅ 3448 >3448!Qb
y2!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
3448
⋅
>1/1276!
6
lhTW
3: 1/: ⋅ 21 − 3448
i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1276 ⋅ )2/97 ⋅ 31 + 3611* >72/97!
lK
lhTW
ta~ka 4:
)napon pare ~iste vode na!u>51pD*
q qt >8486!Qb
q I3P = ϕ ⋅ q qt > 2⋅ 8486 >8486!Qb
y5!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
8486
⋅
>1/1665!
lhTW
3: 1/: ⋅ 21 6 − 8486
i5> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 51 + 1/1665 ⋅ )2/97 ⋅ 51 + 3611* >293/73!
⋅
X
⋅
⋅
!>! n tw/!)y4!−!y3*!>! n tw/!)y5!−!y2*!
⋅
⇒!
⋅
lK
lhTW
⋅
X
n tw!>
y 5 − y2
1/14
lh
>!1/88!
1/1665 − 1/1276
t
n tw!>
ta~ka 2:
y3!>!y2!>!1/1276!
lhI3 P
lhTW
i3!>!@
⋅
⋅
⋅
prvi zakon termodinamike za proces u zagreja~u vazduha:!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
R {bh = n tw ⋅ (i 3 − i2 )
i 3 = i2 +
R {bh
⋅
> 72/97 +
n tw
:6
lK
>296/35!
lhTW
1/88
ta~ka 3:
lhI3 P
lK
i4>i3>296/35!
lhTW
lhTW
i 4 − y 4 ⋅ 3611
296/35 − 1/1665 ⋅ 3611
>53/48pD
u4!>!
!>
d q + y 4 ⋅ 2/97
2 + 1/1665 ⋅ 2/97
y4>y5>!1/1665!
q I3P =
y4
NI3P
N tw
ϕ4!>!
q I3P
(qqt )U4
⋅q>
+ y4
!>
1/1665
⋅ 1/: ⋅ 21 6 >8485/9!Qb
29
+ 1/1665
3:
8485/9
>!1/99
9472
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 42
8/37/!U dvostepenu teorijsku su{nicu uvodi se vla`an vazduh zapreminskog protoka!Wp>1/94!n40t!i stawa
⋅
1)q>1/2!NQb-!u>25pD-!ϕ>1/5). Nakon zagrevawa vazduha u zagreja~u toplotne snage! R J>62/:!lX!)do stawa
2) vazduh se uvodi u prvi stepen su{are odakle izlazi sa temperaturom!u>41pD!)stawe!3). Ovaj vazduh se
⋅
zatim zagreva u drugom zagreja~u toplotne snage! R JJ>35!lX!)do stawa!4), te uvodi u drugi stepen su{are
koji napu{ta sa relativnom vla`no{}u!ϕ>1/9!)stawe!5*/!Ako se zanemare padovi pritiska odrediti masu
vlage uklowenu iz vla`nog materijala u prvom i drugom stepenu su{ewa (posebno za svaki stepen) za
vreme od τ=1 sat. Skicirati promene stawa vla`nog vazduha na!i!−y!dijagramu.
4
i
2
5
3
ϕ>2
1
y
ta~ka 0:
q qt >26:8!Qb
)napon pare ~iste vode na!u>25pD*
q I3P = ϕ ⋅ q qt > 1/5 ⋅ 26:8 >749/9!Qb
yp!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
749/9
⋅
>1/1151!
6
lhTW
3: 2 ⋅ 21 − 749/9
ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 25 + 1/1151 ⋅ )2/97 ⋅ 25 + 3611* >35/2!
lK
lhTW
q tw = q − q I3P > 2 ⋅ 21 6 − 749/9 >::472/3!Qb
ρtw>
⋅
⋅
q TW
::472/3
lh
lhTW
= 2/32!
!!!
n
>
tw = ρ tw ⋅ W > 2/32 ⋅ 1/94 >2!
4
S hTW ⋅ U 398 ⋅ 398
t
n
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 43
ta~ka 1:
y2!>!yp>1/1151!
lhI3 P
lhTW
i3!>!@
⋅
⋅
⋅
⋅
⋅
prvi zakon termodinamike za proces u 1. zagreja~u vazduha:! R 12 = ∆ I12 + X u12 R J = n tw ⋅ (i2 − i 1 )
⋅
i2 = i p +
RJ
⋅
> 35/2 +
n tw
62/:
lK
>87!
lhTW
2
ta~ka 2:
lK
lhTW
i . dq ⋅ u
i3!>!i2!>!87!
y3!>!
2/97 ⋅ u + 3611
>!
lhI3 P
87 . 2 ⋅ 41
>!1/129!
2/97 ⋅ 41 + 3611
lhTW
ta~ka 3:
y4!>!y3!>!1/129!
lhI3 P
lhTW
i4!>!@
⋅
⋅
⋅
prvi zakon termodinamike za proces u 2. zagreja~u vazduha: R 34 = ∆ I34 + X u34
⋅
⋅
⋅
R JJ = n tw ⋅ (i 4 − i 3 )
i 4 = i3 +
R JJ
⋅
n tw
> 87 +
35
lK
>211!
lhTW
2
ta~ka 4:
i5!>!i4!>!211!
lK
lhTW
y5!>!g)ϕ5-!i5*!>!1/1374!
lhI3 P
! )i!−!y!dijagram*
lhTW
⋅
X2!>! n tw!)y3!−!y2* ⋅ τ >! 2 ⋅ (1/129 − 1/115 ) ⋅ 4711 >61/5!lh
⋅
X3!>! n tw!)y5!−y4* ⋅ τ >! 2 ⋅ (1/1374 − 1/129 ) ⋅ 4711 >3:/:!lh
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 44
8/38/!U teorijskoj konvektivnoj su{ari su{i se neki materijal koji ne sme biti izlo`en temperaturi
vi{oj od!91pD/!Maksimalna relativna vla`nost, koju dosti`e vazduh pri svakom prolasku preko vla`nog
materijala, iznosi!ϕnby>:1%. Odrediti koli~inu vlage, koja se u toku jednog sata odstrani iz materijala,
⋅
ako je stawe vla`nog vazduha na ulazu u su{aru odre|eno sa!P)q>2cbs-!u>31pD-!ϕ>1/6-! n ww>1/6!lh0t*;
a) u slu~aju dvostepene teorijske su{are
b) u slu~aju teorijske su{are sa beskona~no mnogo stepeni su{ewa (naizmeni~no povezanih komora za
su{ewe i zagreja~a vazduha)
Smatrati da se tokom svih proces pritisak vazduha u su{ari ne mewa.
a)
i
4
2
u2>u4
ϕ3>ϕ5>ϕnby
5
3
ϕ>2
1
y
ta~ka 0:
q qt >3448!Qb
)napon pare ~iste vode na u>31pD*
q I3P = ϕ ⋅ q qt > 1/6 ⋅ 3448 >2279/6!Qb
yp!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
2279/6
>!1/1184!
⋅
lhTW
3: 2 ⋅ 21 6 − 2279/6
ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1184 ⋅ )2/97 ⋅ 31 + 3611* >49/63!
⋅
n tw =
lK
lhTW
⋅
1/6
n ww
lh
>
>1/5:7!
t
2 + y 1 2 + 1/1184
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 45
ta~ka 1:
y2>yp>!1/1184!
lhI3 P
lhTW
i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 91 + 1/1184 ⋅ )2/97 ⋅ 91 + 3611* >::/45!
lK
lhTW
ta~ka 2:
i3>i2!>::/45
lK
lhTW
y3>!g)ϕ3-!i3*>1/1376!
lhI3 P
!
lhTW
)i!−!y!dijagram*
ta~ka 3:
y4>y3>!1/1376!
lhI3 P
lhTW
i4> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 91 + 1/1376 ⋅ )2/97 ⋅ 91 + 3611* >261/2:!
lK
lhTW
ta~ka 4:
lK
lhTW
lhI3 P
y5>!g)ϕ5-!i5*>1/154!
!
lhTW
i5>i4!>261/2:
)i!−!y!dijagram*
⋅
X!>! n tw!)y5!−!yp* ⋅ τ >! 1/5:7 ⋅ (1/154 − 1/1184) ⋅ 4711 >74/86!lh
b)
ta~ka!3o;
q qt >58471!Qb )napon pare ~iste vode na!u>91pD*
q I3P = ϕ nby ⋅ q qt > 1/: ⋅ 58471 >53735!Qb
y3o!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
53735
⋅
>!1/5722!
6
3: 2 ⋅ 21 − 53735
lhTW
⋅
X′!>! n tw!)y3o!−!yp* ⋅ τ >! 1/5:7 ⋅ (1/5722 − 1/1184) ⋅ 4711 >921/4!lh
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 46
8/39/!U teorijskoj su{ari sa recirkulacijom, jednog dela iskori{}enog vazduha, protok atmosferskog
⋅
vla`nog vazduha, stawa!1)i>61!lK0lhTW-!y>1/12!lhI3P0lhTW*-!iznosi! n p>7!u0i. Stawe me{avine sve`eg
i opticajnog vazduha na ulazu zagreja~ vazduha je N)u>51pD-!y>1/145!lhI3P0lhTW*/!Me{avina se u
kaloriferu zagreva do stawa!2)u>99pD*/!Po~etna vla`nost materijala je! Z2 >81&!ra~unato na suvu
materiju, a krajwa! Z3 >9&!tako|e ra~unato na suvu materiju. Skicirati promene stawa vla`nog vazduha
na!i!−!y!dijagramu i odrediti:
a) masene protoke: odstrawene vlage i osu{enog materijala!)lh0i*
b) maseni udeo sve`eg i opticajnog vazduha u me{avini
c) potrebnu koli~inu toplote za zagrevawe vla`nog vazduha!)lK0t*
d) kolika bi bila potro{wa toplote da se su{ewe izvodi samo sve`im vazduhom tj. da nema
recirkulacije i kolika bi bila temperaturu vla`nog vazduha na ulazu u komoru za su{ewe u tom
slu~aju
i
2
u2
3
ϕ>2
un
ip
N
1
y
y2
yN
ta~ka 0:
y1>1/12!
⋅
n twp!>!
lhI3 P
lhTW
⋅
n1
2+ y1
ip!>!!61!
lK
lhTW
21 4
4711 = 2/76! lh
=
2 + 1/12
t
7⋅
ta~ka M:
yn>1/145!
lhI3 P
lhTW
in!> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 51 + 1/145 ⋅ )2/97 ⋅ 51 + 3611* >238/64
dipl.ing. @eqko Ciganovi}
lK
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 47
ta~ka 1:
y2>yn>1/145!
lhI3 P
lhTW
i2!> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 99 + 1/145 ⋅ )2/97 ⋅ 99 + 3611* >289/67
lK
lhTW
ta~ka 2:
i3>i2>289/67!
lK
lhTW
y3!>!@
prvi zakon termodinamike za proces me{awa dva vla`na vazduha:
⋅
⋅
⋅
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X u23
n twp ⋅ i p + n tw3 ⋅ i 3 = ntwp + n tw3 ⋅ in
⋅
n twp ⋅ (in − i p ) 2/76 ⋅ (238/64 − 61)
lh
>
=
>3/62
i 3 − in
289/67 − 238/64
t
⋅
n tw3
materijalni bilans vlage za proces me{awa dva vla`na vazduha:
⋅
⋅
⋅
n twp + n tw3 ⋅ y n − n twp ⋅ y p
⋅
⋅
⋅
⋅
n twp ⋅ y p + n tw3 ⋅ y 3 = n twp + n tw3 ⋅ y n !!! y 3 =
⋅
n tw3
(2/76 + 3/62) ⋅ 1/145 − 2/76 ⋅ 1/12 >1/15:9! lhI3 P
y3 =
3/62
lhTW
a)
⋅
⋅
⋅
lh
X = ntwp + n tw3 ⋅ (y 3 − y 2 ) > (2/76 + 3/62) ⋅ 4711 ⋅ (1/15:9 − 1/145 ) >347/73!
i
Z2
Z
1/8
1/19
3
z2 =
>
z3 =
>
>1/52>1/18
2 + Z2 2 + 1/8
2 + Z3 2 + 1/19
⋅
X = npn ⋅
z2 − z 3
2 − z2
⋅
n pn = X⋅
2 − z2
2 − 1/52
lh
>521/72!
> 347/73 ⋅
z2 − z 3
1/52 − 1/18
i
b)
⋅
hp>
n TWp
⋅
⋅
nTWp + nTW3
2/76
>
>1/52/76 + 3/62
⋅
h3>
nTW3
⋅
⋅
nTWp + nTW3
>
3/62
>1/7
2/76 + 3/62
c)
⋅
⋅
⋅
prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
R {bh = !n twp + n tw3 ⋅ (i2 − in ) (2/76 + 3/62) ⋅ (289/67 − 238/64 ) >323/39!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 48
d)
i
2′
2
u2
3
ϕ>2
un
ip
N
1
y
y1
yN
lhI3 P
lK
i2′>!i2>!i3!>289/67!
lhTW
lhTW
289/67 − 1/12 ⋅ 3611
i − y ⋅ 3611
u2′!>!
!>
>261/87pD
2 + 1/12 ⋅ 2/97
d q + y ⋅ 2/97
y2′!>!yp>1/12!
8/3:/!U teorijskoj su{ari se obavqa proces izdvajawa vlage iz koncentrata paradajza. Maseni protok
koncentrata paradajza na ulazu u su{aru je!1/237!lh0t. Na ulazu u su{aru koncentrat paradajza sadr`i
z 2 =31!nbt&!vode, a prah na izlazu! z 3 =6!nbt%. Parcijalni pritisak vodene pare u okolnom (sve`em)
vazduhu je! q I3P 1 >2/44!lQb, dok na izlazu iz su{are ne sme biti vi{i od! q I3P 3 >37/8!lQb/!Da bi se taj
(
)
(
)
uslov ispunio potrebno je me{awe dela iskori{}enog i okolnog sve`eg vazduha tako da parcijalni
pritisak vodene pare u vla`nom vazduh na ulazu u zagreja~ iznosi! q I3P >7/8!lQb. Pritisak vazduha za
(
)n
vreme su{ewa je konstantan i iznosi!q>212/4!lQb/!Odrediti:
a) maseni protok sve`eg i recirkulacionog vazduha (ra~unato na suv vazduh)
c* specifi~nu potro{wu toplote u su{ari!)lK0lh!odstrawene vlage) ako se u fazi zagrevawa vazduh
zagreje za u 2 − u n >41pD
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 49
ta~ka 0:
y1!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
2/44
⋅
>1/1194!
3: 212/4 − 2/44
lhTW
>
lhI3 P
29
37/8
⋅
>1/3333!
3: 212/4 − 37/8
lhTW
ta~ka 2:
y3!>
NI3P
NTW
⋅
q I3P
q − q I3P
ta~ka M:
yn!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
7/8
⋅
>1/155!
3: 212/4 − 7/8
lhTW
ta~ka 1:
y2>yn!>1/155!
lhI3 P
lhTW
a)
materijalni bilans vlage za proces su{ewa:
⋅
z − z3
⋅
n tw p + n tw3 ⋅ (y 3 − y 2 ) > n wn ⋅ 2
2− z3
⋅
⋅
n tw p + ntw3 = 1/237 ⋅
⇒
⋅
z2 − z 3
2
⋅
2 − z 3 y 3 − y2
⋅
n tw p + ntw3 = n wn ⋅
1/3 − 1/16
2
lh
⋅
>1/22
2 − 1/16 1/3333 − 1/155
t
materijalni bilans vlage za proces me{awa dva vla`na vazduha:
⋅
⋅
⋅
⋅
n twp ⋅ y p + n tw3 ⋅ y 3 = n twp + n tw3 ⋅ y n
⋅
)2*
⋅
n tw p + n tw3 >1/22
)3*
⋅
Kombinovawem jedna~ina!)2*!i!)3*!dobija se;! n tw p >1/1:3
lh ⋅
lh
-! ntw 3 >1/129
t
t
c*
i2!> d q ⋅ u 2 + y2 ⋅ )2/97 ⋅ u 2 + 3611*
)2*
in!> d q ⋅ u n + y n ⋅ )2/97 ⋅ u n + 3611*
)3*
(
Oduzimawem jedna~ina!)2*!i!)3*!dobija se;!!!!!!i2!−in!> (u 2 − u n ) ⋅ d q + 2/97 ⋅ y2
⋅
⋅
n twp + n tw3 ⋅ (i2 − in )
(u 2 − u n ) ⋅ d q + 2/97 ⋅ y2
R n2
lK
>293/2!
= ⋅ =
=
⋅
⋅
lhX
y 3 − y2
X
n twp + n tw3 ⋅ (y 3 − y2 )
⋅
rx
)
dipl.ing. @eqko Ciganovi}
(
)
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 50
8/41/!Jabuke koje ne podnose temperaturu vi{u od!81pD su{e se u teorijskoj su{ari sa recirkulacijom
dela iskori{}enog vazduha. Stawe sve`eg vazduha odre|eno je sa!1)u>7pD-!y>6/42!hI3P0lhTW*/
Apsolutna vla`nost iskori{}enog vazduha je
y3>45!hI3P0lhTW-a specifi~na potro{wa toplote u su{ari iznosi!rx>4761!lK0lh!odstrawene vlage.
Skicirati promene stawa vla`nog vazduha na!i!−!y!dijagramu i odrediti:
b* masene udele sve`eg i recirkulacionog vazduha u me{avini
b) minimalnu temperaturu do koje se mora zagrejati sve` vazduh pre me{awa da bi se izbeglo
stvarawe magle za vreme procesa me{awa
a)
ta~ka 0:
yp>1/11642!
lhI3 P
lhTW
ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 7 + 1/11642 ⋅ )2/97 ⋅ 7 + 3611* >2:/44!
lK
lhTW
ta~ka 2:
y3>1/145!
lhI3 P
lhTW
⋅
rx =
R n2
⋅
=
i − h p ⋅ i p − h3 ⋅ i3
i2 − in
i − in
= 3
= 3
y 3 − y2 y 3 − y n
y 3 − h p ⋅ y p − h3 ⋅ y 3
X
i3 ⋅ (2 − h3 ) − h p ⋅ i p
i − ip
rx>
> 3
y 3 ⋅ (2 − h3 ) − y p ⋅ i p
y3 − yp
⇒
i3>2:/44!, 4761 ⋅ (1/145 − 1/11642) >235/16
⇒
i3>ip!, r x ⋅ (y 3 − y p ) ⇒
lK
lhTW
ta~ka 1:
lK
lhTW
i2 − d q ⋅ u 2
lhI3 P
235/16 − 2 ⋅ 81
y2 =
>
>1/1316!
2/97 ⋅ u 2 + 3611 2/97 ⋅ 81 + 3611
lhTW
ta~ka M:
lhI3 P
yn>y2>1/1316!
lhTW
y − yn
1/145 − 1/1316
>
hp> 3
>1/58
1/145 − 1/11642
y3 − yp
i2>i3!>235/16!
h3>
yn − yp
1/1316 − 1/11642
>
>1/64
1/145 − 1/11642
y3 − yp
in = h p ⋅ i p + h3 ⋅ i3 > 1/58 ⋅ 2:/44 + 1/64 ⋅ 235/16 >85/94
dipl.ing. @eqko Ciganovi}
lK
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 51
b)
i
2
u2
4761
3
N′
N
1′
up
ϕ>2
1
y
y3
y2
P−N−3;
P−N′−3;
pravac me{awa pre zagrevawa okolnog vazduha
pravac me{awa nakon zagrevawa okolnog vazduha
ta~ka 0′:
grafi~ki postupak:
Konstrui{e se prava kroz ta~ke!3!i!N′)!yN′>yN). Presek ove prave sa
linijom!yp>dpotu!!defini{e polo`aj ta~ke O′. Iz dijagrama se o~itava!uP′!
/
ra~unski postupak:
lhI3 P
y p( >yp>1/11642!
lhTW
lK
in′!>88/68
lhTW
in = h p ⋅ i p( + h 3 ⋅ i 3
i p( >@
⇒
i p( =
in − h3 ⋅ i 3
hp
88/68 − 1/58 ⋅ 235/16
lK
>44/46
lhTW
1/64
47/46 − 1/11642 ⋅ 3611
i − y ⋅ 3611
up′!>!
>33/96pD
!>
d q + y ⋅ 2/97
2 + 1/11642 ⋅ 2/97
i p( >
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 52
8/42/!U teorijskoj su{ari sa recirkulacijom jednog dela iskori{}enog vazduha su{i se vla`an
lhX
materijal po~etne vla`nosti!!411&!ra~unato na suvu materiju!)Z2>4!
*/!U su{ari se odstrani
lhTN
91% od vlage koju sa sobom u su{aru unosi vla`an materijal i pri tom dobijamo!43!lh0i!osu{enog
materijala. Stawe sve`eg vazduha odre|eno je sa!)u>31pD-!ϕ>1/7*!a stawe otpadnog vazduha odre|eno je
sa!)u>51pD-!ϕ>1/9*/!Temperatura vazduha nakon faze zagrevawa iznosi!u>87pD/!Odrediti:
b* toplotnu snagu zagreja~a vazduha!R{bh!)lX*
b) koliko bi se toplote moglo u{tedeti (u zagreja~u) hla|ewem otpadnog vazduha do stawa zasi}ewa i
rekuperativnim kori{}ewem tako oslobo|ene toplote za zagrevawe vazduha nastalog me{awem
sve`eg i recirkulacionog vazduha (slika)
ta~ka 0:
)napon pare ~iste vode na!u>31pD*
q qt >3448!Qb
q I3P = ϕ ⋅ q qt > 1/7 ⋅ 3448 >2513/3!Qb
yp!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
2513/3
⋅
>!1/1199!
6
lhTW
3: 2 ⋅ 21 − 2513/3
ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1199 ⋅ )2/97 ⋅ 31 + 3611* >53/44!
lK
lhTW
ta~ka 2:
)napon pare ~iste vode na!u>51pD*
q qt >8486!Qb
q I3P = ϕ ⋅ q qt > 1/9 ⋅ 8486 >6:11!Qb
y3!>
NI3P
NTW
⋅
q I3P
q − q I3P
>
lhI3 P
29
6:11
⋅
>!1/149:!
3: 2 ⋅ 21 6 − 6:11
lhTW
i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 51 + 1/149: ⋅ )2/97 ⋅ 51 + 3611* >251/25!
lK
lhTW
ta~ka 1:
i2>!i3>!251/25!
y2>!
lK
lhTW
i . dq ⋅ u
2/97 ⋅ u + 3611
>!
lhI3 P
251/25 . 2 ⋅ 87
>1/1354!
2/97 ⋅ 87 + 3611
lhTW
ta~ka M:
lhI3 P
lhTW
1/149: − 1/1354
>
>1/596
1/149: − 1/1199
yn>y2>1/1354!
hp>
y3 − yn
y3 − yp
h3>2−h2>1/626
in = h p ⋅ i p + h3 ⋅ i3 > 1/596 ⋅ 53/44 + 1/626 ⋅ 251/25 >:3/81
dipl.ing. @eqko Ciganovi}
lK
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 53
a)
z2 =
Z2
4
lhX
>
>1/86!
lh)X + TN*
2 + Z2 4 + 2
materijalni bilans komore za su{ewe materijla: nwn>npn,!X
!!!!)2*
⋅
bilans vlage komore za su{ewe materijala:
n wn ⋅ z 2 = n pn ⋅ z 3 + X !!)3*
uslov zadatka:
1/9 ⋅ n wn ⋅ z 2 = X !!
⋅
!!!!)4*
⋅
kada se odstrawena vlaga!) X *!iz jedna~ine )4*!uvrsti u jedna~ine!)2*!i!)3*!⇒
n wn ⋅ z 2 = n pn ⋅ z 3 + 1/9 ⋅ n wn ⋅ z 2
tj.
n wn = n pn + 1/9 ⋅ n wn ⋅ z 2
tj.
1/3 ⋅ n wn ⋅ z 2 = npn ⋅ z 3 !!!)5*
npn
n wn =
!!!!)6*
2 − 1/9 ⋅ z 2
kada se jedna~ina!)6) uvrsti u jedna~inu!)5*!dobija se:
z3 >
1/3 ⋅ z 2
1/3 ⋅ 1/86
lhX
>1/486
>
lh)X + TN*
2 − 1/9 ⋅ z 2 2 − 1/9 ⋅ 1/86
n pn ⋅ z 3 43 ⋅ 1/486
lh
>
>91
1/3 ⋅ 1/86
1/3 ⋅ z 2
i
)5*
⇒
n wn =
)2*
⇒
X = n wn − n pn !>91!−!43!>59
⋅
⋅
⋅
lh
i
⋅
59
2
lh
X
⋅
n tw1!,! n tw3!>
>
>1/:2
t
y 3 − y2 1/149: − 1/1354 4711
⋅
⋅
⋅
prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
R {bh = !n twp + n tw3 ⋅ (i2 − in ) > 1/:2 ⋅ (251/25 − :3/8) >54/28!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 54
b)
nwn
zagreja~
vazduha
C
predgreja~
vazduha
B
komora za
su{ewe
materijala
2
npn
3
otpadni vazduh
3
recirkulacioni
vazduh
N
3
1
sve` vazduh
i
2
u2
3
C
u3
ϕ>2
N
up
ϕ3
B
1
ϕp
⋅
⋅
⋅
lh
n tw1!>! h p ⋅ n tw1 !+!n tw3 > 1/596 ⋅ 1/:2 >1/55
t
⋅
⋅
⋅
⋅
lh
n tw3!>! h3 ⋅ n tw1 !+!n n tw3 > 1/626 ⋅ 1/:2 >1/58
t
ta~ka A:
lK
y B = y3 iB>g (y B - ϕ = 2) >247/28
lhTW
⋅
⋅
R qsfe!>! n twp ⋅ (i3 − i B ) >!1/55 ⋅(251/25 − 247/28) >2/86!lX
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 55
8/43/!U dvostepenoj teorijskoj su{ari za!7!sati osu{i se!2111!lh!vla`nog materijala.!Maseni odnos vlage
lhX
prema suvoj materiji u materijalu koji ulazi u prvi stepen su{ewa je 1/54!
!a maseni odnos vlage
lhTN
lhX
prema suvoj materiji u materijalu koji napu{ta drugi!stepen su{ewa je!1/25!
. Sve` ulazni vazduh
lhTN
stawa!1)q>2!cbs-!u>27pD-!ϕ>1/6*!me{a se sa recirkulacionim vazduhom stawa!6)q>2!cbs-!u>57pD-!ϕ>1/7*!u
odnosu!3;2, a zatim se predgreja~u vazduha (razmewiva~ toplote) pomo}u dela vla`nog vazduha oduzetog iz
prvog stepena su{are. U greja~ima vazduha!H2!i!H3 vla`an vazduh se zagreva do temperature od!91pD/
Temperatura vla`nog vazduha na izlazu iz postrojewa je!41pD/!Skicirati promene stawa vla`nog vazduha
na Molijerovom!i!−!y!dijagramu i odrediti:
a) veli~ine stawa vla`nog vazduha!)i-!y-!u*!u karakteristi~nim ta~kama
b) toplotne snage greja~a vazduha-!H2!i!H3
c) vla`nost materijala (maseni udeo vlage) na kraju prvog stepena su{ewa
sve`
vazduh
1
nwn
N
p
r
e
d
g
r
e
j
a
~
6
7
2
npn
prvi
stepen
su{ewa
3
drugi
stepen
su{ewa
4
H2
H3
6
5
otpadni vazduh
recirkulacioni
vazduh
i
5
3
6
4
2
7
ϕ>2
N
1
y
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 56
b*
ta~ka 0:
lhI3 P
lhTW
yp!>!g)up-!ϕp*>1/1168!
ip>!g)up-!yp*>41/53!
lK
lhTW
ta~ka 5:
y6!>!g)u6-!ϕ6*!>1/14::7!
lhI3 P
lhTW
i6!>!g)u6-!y6*!>25:/43!
lK
lhTW
ta~ka M:
prvi zakon termodinamike za proces me{awa dva vla`na vazduha:
⋅
⋅
⋅
⋅
⋅
⋅
⋅
R 23 = ∆ I23 + X u23
n twp ⋅ i p + n tw 6 ⋅ i 6 = n twp + n tw 6 ⋅ in
⋅
n twp
⋅
in =
⋅ ip + i6
n tw 6
⋅
n twp
⋅
+2
3
⋅ (41/53) + 25:/43
lK
>81/16!
>2
3
lhTW
+2
2
n tw6
materijalni bilans vlage za proces me{awa dva vla`na vazduha:
⋅
ntwp
⋅
⋅
n tw 1 ⋅ y p + n tw 6 ⋅ y 6 = n twp + n tw 6 ⋅ y n
⋅
⋅
⋅
yn =
⇒!
⋅ yp + y6
ntw 6
⋅
ntwp
⋅
+2
ntw 6
3
⋅ 1/1168 + 1/15
lhI3 P
yn = 2
>1/1282
3
lhTW
+2
2
i − y n ⋅ 3611
81/13 − 1/1282 ⋅ 3611
!>
un!>! n
>37/54pD
d q + y n ⋅ 2/97
2 + 1/1282 ⋅ 2/97
ta~ka 2:
y3>yn!>1/1282!
lhI3 P
lhTW
i5!>i6!>25:/43!
lK
lhTW
y5>!g)u5-!i5*>1/1373!
lK
lhTW
y4>!y5>1/1373!
i3!>!g)u3-!y3*>236/46!
lK
lhTW
ta~ka 4:
lhI3 P
lhTW
ta~ka 3:
i4!>i3>236/46!
dipl.ing. @eqko Ciganovi}
lhI3 P
lhTW
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 57
ta~ka 6:
y7!>!y4!>!y5>1/1373!
lhI3 P
lhTW
i7!>!g)u7-!y7*!>!:7/9:!
lK
lhTW
ta~ka 1:
y2!>!yn!>!1/1282!
lhI3 P
lhTW
i2!>!@
⋅
⋅
⋅
prvi zakon termodinamike za proces u predgreja~u vazduha:!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
n twp ⋅ (i 4 − i 7 ) = ntwp + ntw 7 ⋅ (i2 − iN )
⇒
⋅
n twp
⋅
i2 = in −
⋅ (i 7 − i 4 )
n tw 7
⋅
n twp
3
⋅ (:7/9: − 236/46 )
lK
>9:/13
> 81/16 − 2
3
lhTW
+2
2
+2
⋅
n tw 7
i − y2 ⋅ 3611
9:/13 − 1/1282 ⋅ 3611
!>
u2!>! 2
>88/95pD
d q + y2 ⋅ 2/97
2 + 1/1282 ⋅ 2/97
c*
z2 =
Z2
1/54
lhX
>
>1/41!
2 + Z2 1/54 + 2
lh)X + TN*
z3 =
Z3
1/25
lhX
>
>1/23!
1/25 + 2
2 + Z3
lh)X + TN*
materijalni bilans vlage za oba stepena su{ewa zajedno:
z 2 − z 3 2111 1/41 − 1/23
lh
>
⋅
>45/1:
7
2 − 1/23
2− z3
i
⋅
X = n wn ⋅
⋅
⋅
⋅
⋅
X = ntwp + n tw 6 ⋅ (y 4 − y 3 ) + n tw 7 ⋅ (y 6 − y 5 )
⋅
n tw 6
⋅
⇒
45/1:
⋅
lh
X
4711
=
>
>1/34
t
4 ⋅ (y 4 − y 3 ) + (y 6 − y 5 ) 4 ⋅ (1/1373 − 1/1282) + 1/15 − 1/1373
⋅
n twp = 3 ⋅ n tw 6 > 3 ⋅ 1/34 >1/57
dipl.ing. @eqko Ciganovi}
lh
t
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 58
⋅
⋅
⋅
⋅
⋅
prvi zakon termodinamike za proces u greja~u vazduha H2 :!!! R 23 = ∆ I23 + X u23
⋅
⋅
⋅
R 23 = ntwp + ntw3 ⋅ (i3 − i2 ) > 1/7: ⋅ (236/46 − 9:/13) >36/18!lX
⋅
prvi zakon termodinamike za proces u greja~u vazduha H3 :!!! R 45 = ∆ I45 + X u 45
⋅
⋅
R 45 = n tw6 ⋅ (i 5 − i 4 ) > 1/34 ⋅ (25:/43 − 236/46 ) >6/62!lX
d*
⋅
X 2!>! n wn ⋅
z 2 − z(
2 − z(
⋅
⋅
⋅
X 2 = n twp + n tw 6 ⋅ (y 4 − y 3 )
)2*
)3*
⋅
Kombinovawem jedna~ina!)2*!i!)3*!dobija se! X 2>!33/7!
zadatak za ve`bawe:!
lh
!!j!z′>1/3
i
)8/44/*
8/44/!U dvostepenoj teorijskoj su{ari su{i se!2911!lh0i nekog proizvoda koji sadr`i!4:!nbt&!vlage.
Nakon su{ewa proizvod sadr`i!:3!nbt% suve materije. Vazduh izlazi iz su{are na temperaturi od
56pD/!Temperatura okoline je!31pD/!Vazduh se pred svakim stepenom zagreva do!91pD!a na izlazu iz
svakog stepena ima relativnu vla`nost!81&/!Sve promene stawa vla`nog vazduah u su{ari se doga|aju
pri!q>2!cbs>dpotu/!Skicirati promene stawa vla`nog vazduha na!i!−y!!dijagramu i odrediti:
a) ukupnu potro{wu toplote u su{ari!)lX*
b) izra~unati vla`nost materijala (maseni udeo vlage) na izlazu iz prvog stepena su{ewa
re{ewe:
a) R>684!lX
b) z′>1/36
dipl.ing. @eqko Ciganovi}
{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 1
KRETAWE TOPLOTE
9/2/!Sa jedne strane ravnog zida povr{ine B>4!n3 nalazi se suva vodena para U2>247pD, a sa druge
ravnog zida nalazi se vazduh U6>36pD. Zid je sastavqen od dva sloja: ~eli~nog lima (1) )λ2>61-!X0nLδ2>21!nn* i izolacionog materijala
(2) )λ3>1/17!X0nL-!δ3>31!nn*/ Koeficijent prelaza toplote sa pare na zid iznosi α2>6111!X0n3L, a
sa zida na okolni vazduh α3>2111!X0n3L. Odrediti:
b* toplotni protok sa pare na vazduh, kroz zid )X*
b) temperaturu izolacije (do vazduha) i temperaturu lima (do pare)
δ3
δ2
α2
para
λ2
λ3
vazduh
α3
U6
U2
U4
U3
U5
a)
⋅
R=
U2 . U6
δ
δ
2
2
+ 2 + 3 +
α2 λ2 λ 3 α 3
B=
247 . 36
⋅ 4 >2111!X
2
1/12 1/13
2
+
+
+
6111
61
1/17 2111
b)
U − U3
R= 2
⋅B ⇒
2
α2
⋅
⋅
U .U
R= 5 6
2
α3
⋅
2111
R
>246/:4pD
U3!>!U2!−!
>247!−!
6111 4
α2 ⋅ B
⋅
B
⇒
2111
R
>36/69pD
U5!>!U6!,
>36,!
2111 4
α3 ⋅ B
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 2
9/3/!Za izgradwu privremenog skloni{ta u polarnim oblastima istra`iva~i mogu upotrebiti ponetu
{per plo~u, debqine!δ2>6!nn!)λ2>1/218!X0)nL**-!vla`nu zemqu!)λ3>1/767!X0)nL**-!i nabijen sneg
)λ4>1/218!X0)nL**/!Na unutra{woj povr{i zida skloni{ta ustali se temperatura U2>3:4!L-!a
koeficijent prelaza toplote, sa spoqa{we povr{i zida na okolni vazduh temperature!U6>341!L,
iznosi!α>:/7!X0)n3L*/!′′Povr{inski toplotni protok′′ (toplotni fluks) pri tome treba da iznosi
r>69!X0n3. Odrediti:
a) najmawu debqinu sloja vla`ne zemqe u konstrukciji zida, tako da ne do|e do topqewa snega
b) potrebnu debqinu sloja nabijenog snega u konstrukciji zida
δ2
{per
plo~a
U2
δ3
δ4
vla`na
zemqa
U3
U4
nabijen
sneg
U5
U6
a)
napomena:
!r!>!
Uo~iti da je U4!>384!L (uslov ne topqewa snega na grani~noj
povr{ini vla`na zemqa sneg).
U2 − U4
δ2 δ3
+
λ2 λ3
⇒
U −U
δ
δ3!>!λ3! 2 4 − 2
λ2
r
3:4 − 384 6 ⋅ 21 −4
−
δ3!>! 1/767 ⋅
69
1/218
b)
r!>!
U2 − U6
δ2 δ3 δ4 2
+
+
+
λ2 λ3 λ 4 α
⇒
⇒
>2:7!nn
U −U
δ
δ
2
δ4!>!λ4! 2 6 − 2 − 3 −
λ2 λ3 α
r
3:4 − 341 6 ⋅ 21 −4 79 ⋅ 21 −4
2
>!79!nn
δ4!>!λ4! ⋅
−
−
−
69
1/218
1/767
:/7
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 3
9/4/![upqi cilindar od stiropora!)λ>1/138!X0nL) unutra{weg pre~nika!ev>1/3!n, spoqa{weg
pre~nika!et>1/4!n!i visine!I>2/6!n!napuwen je ledom sredwe!temperature!u2>1pD/!Temperatura
okolnog vazduha je!u4>41pD, a koeficijent prelaza toplote na stiropor sa okolnog vazduha iznosi!α>9
X0n3L/!Temperatura unutra{we povr{ine cilindra je!1pD/!Zanemaruju}i razmenu toplote kroz baze
cilindra, odrediti:
a) toplotne dobitke cilindra )lX*
b) temperaturu spoqa{we povr{ine zida cilindra
d* vreme za koje }e se sav led otopiti ako toplota topqewa leda iznosi sm>443/5!lK0lh-!a gustina leda
ρm>:11!lh0n4
α
led
vazduh
u2
u3
u4
a)
⋅
!R =
u2 − u 4
36 − 1
⋅I =
⋅ 2/6 = 28/95!X
et
2
2
1/4
2
2
mo
+
mo
+
1/4π ⋅ 9 3π ⋅ 1/138 1/3
e t π ⋅ α 3π ⋅ λ e v
b)
u − u3
R= 2
⋅I ⇒!
2
et π ⋅ α
⋅
⋅
28/95
R
u3!>!u2!−!
>!36!−
>39/5pD
1/4π ⋅ 9 ⋅ 2/6
ev π ⋅ α ⋅ I
c)
τ!>!
Rm
⋅
R
= ///!>
251:4/87
28/95 ⋅ 21 −4
= 8:1119!t!
(9 dana 3 sata 27 min)
Rm!>!nm!/!sm!>!///!>53/5/!443/5!>251:4/87!lK
nm!>!ρm!/!Wm!>!ρm!/
1/3 3 π
ev3 π
⋅ M >!:11 ⋅
⋅ 2/6 >!53/5!lh
5
5
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 4
9/5/!Odrediti kolika se maksimalna debqina leda!)λmfe!>3/67!X0)nL**!mo`e obrazovati na spoqnoj
povr{ini aluminijumske cevi!)λBm>33:/2!X0)nL**-!pre~nika!∅>:6094!nn-!du`ine!M>2!n-!koju obliva
voda, ako je temperatura na wenoj unutra{woj povr{ini U4>374!L-!pri ~emu je toplotni protok sa
vode na cev!2661!X/
e4
e3
e2
U4
napomena:
!U2!>!384!L
U3
U2
(uslov stvarawa leda)
e4 − e3
1/216 − 1/1:6
>1/116!n>!6!nn
>!///>!
3
3
⋅
U2 − U4
R=
⋅M
⇒
e
2
e4
2
mo
mo 3
+
3π ⋅ λmfe e3 3π ⋅ λ Bm e2
δ mfe !>!
3π ⋅ λ ⋅ M
λ mfe e 3
mfe
(
)
e4!> e 3 ⋅ fyq
U
U
mo
⋅
−
−
2
4
⋅
λ Bm
e2
R
3/67 :6
3π ⋅ 3/67 ⋅ 2
⋅ (384 − 374 ) −
mo
>1/216!n
e4!> 1/1:6 ⋅ fyq
33:/2 94
2661
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
zadaci za ve`bawe:
strana 5
)9/6/!−!9/7/*
9/6/!Sa jedne strane staklene!)λ>1/9!X0nL*!plo~e, ukupne povr{ine!B>21!n3-!nalazi se vla`an vazduh
temperature!71pD!dok je sa druge strane voda temperature!31pD/!Pad temperature kroz staklenu plo~u
iznosi!6pD. Koeficijent prelaza toplote sa vla`nog vaduha na plo~u iznosi!α2>31!X0n3L-!a sa plo~e
na vodu
α3>211!X0n3L/!Odrediti:
a) temperature staklene povr{ine u dodiru sa vla`nim vazduhom i vodom
b) debqinu staklene plo~e
c) toplotni protok sa vla`nog vazduha na vodu
d) toplotni protok sa vla`nog vazduha na vodu ako bi se sa strane vode formirao sloj kamenca
toplotnog otpora S>1/2!n3L0X
re{ewe:
b* u3!>!41/94pD-!u4>36/94pD
b) δ>7/97!nn
⋅
c) R >6945!X
⋅
d)
R( >3484!X
9/7/!U ~eli~noj cevi!)λ>57/6!X0)n3L**-!pre~nika!26:y5/6!nn, po celoj du`ini deonice ime|u dva
ventila, usled du`eg prekida rada u ma{inskoj hali, u zimskom periodu, obrzaovao se ledeni ~ep
sredwe temperature!1pD/!^eli~na cev je toplotno izolovana slojem stiropora debqine!δ>61!!nn/
Naglim zagrevawem, temperatura vazduha u ma{inskoj hali povisi se do!41pD i potom ostaje
nepromewena. Koeficijent prelaza toplote sa okolnog vazduha na spoqa{wu povr{ stiropora tako|e
je stalan i iznosi!α>26!X0)n3L*/!Uz predpostavku da je temperatura na unutra{woj povr{i cevi
stalna i da iznosi!1pD-!odrediti vreme (u danima) za koje }e se ledeni ~ep potpuno otopiti.
re{ewe:
τ!≈!7 dana
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 6
9/8/!U zidanom kanalu od hrapave crvene opeke!)ε3>1/:4*-!du`ine!M>2!n, kvadratnog popre~nog preseka
stranice!b>511!nn!postavqena je ~eli~na cev!)ε2>1/9*!spoqa{weg pre~nika!e>211!nn/
Temperatura spoqa{we povr{i cevi je!U2>684!L-!a unutra{wih povr{i zidova kanala!U3>434!L/
Prostor izme|u cevi i kanala je vakumiran. Odrediti:
a) toplotni protok koji zra~ewem razmene cev i zidani kanal
b) toplotni protok koji zra~ewem razmene cev i zidani kanal (pri istim temperaturama U2 i U3 ) ako
se izme|u cevi i zidova kanala postavi cilindri~ni toplotni ekran koeficijenta emisije εF>1/96
i pre~nika eF>311!nn
c) temperaturu tako postavqenog ekrana (zanemariti debqinu ekrana )
ε3
ε2
U2
U3
a)
5
⋅
R {
23
5
5
5
U2
U
684
434
− 3
−
211
211
211
211
=
⋅ M >///>
⋅ 2 >2496!X
2
2
1/2 ⋅ π ⋅ 5/56
eπ ⋅ D23
D23!>!Dd/!ε23!>//!!/!>!6/78/!1/8:!>!5/56!
X
n3L 5
2
2
ε23!>!
>///>!
>!1/8:
2
1/425 2
2 B2 2
− 2
+
− 2
+
1/9
2/7 1/:4
ε2 B3 ε3
B2!>!e π /!M!> 1/2 ⋅ π ⋅ 2 >1/425!n3
B3!>! 5 ⋅ b ⋅ M > 5 ⋅ 1/5 ⋅ 2 >2/7!n3
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 7
b)
ε3
ε2
U
εF
U2
5
⋅
R {
2F3
U3
5
5
5
U
U2
684
434
− 3
−
211
211
211
211
⋅ M >///>
⋅ 2>!99:/77!X
=
2
2
2
2
+
+
eπ ⋅ D2F eF π ⋅ D F3
1/2 ⋅ π ⋅ 5/35 1/3 ⋅ π ⋅ 5/8
D2F!>!Dd!ε2F!>5/35!
X
3 5
nL
BF!>!eF π /!M!> 1/3 ⋅ π ⋅ 2 >!1/739!n3
X
DF3!>!Dd/!εF3!>5/8!
n3L 5
ε2F!>!
2
>!1/86
2 B2 2
+
− 2
ε2 BF εF
εF3!>!
2
>!1/94
2 BF 2
− 2
+
εF B3 ε3
c)
U trenutku uspostavqawa stacionarnog re`ima kretawa toplote
⋅
⋅
postavqamo toplotni bilans za toplotni ekran: R [ = R [ odakle
2F
F3
sledi da je:
UF!>! 5
D2F ⋅ e ⋅ U25 + D F3 ⋅ eF ⋅ U35
D2F ⋅ e + D F3⋅ eF
>
5
5/35 ⋅ 1/2 ⋅ 684 5 + 5/8 ⋅ 1/3 ⋅ 434 5
5/35 ⋅ 1/2 + 5/8 ⋅ 1/3
= >561!L
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 8
9/9/!U industrijskoj hali nalazi se pe} od vaqanog ~eli~nog lima!)ε2>1/68*!ukupne povr{ine!B>3/6
n3/!Temperatura spoqa{we povr{i pe}i je!u2>271pD, a okolnog vazduha i unutra{wih povr{i zidova
hale!u3>u4>26pD. Koeficijent prelaza toplote sa spoqa{we povr{i pe}i na vazduh u hali je!α>24/24
X0)n3L*/!Odrediti:
a) ukupan toplotni protok (zra~ewe + prelaz) koji odaje spoqa{wa povr{ina pe}i
b) temperaturu unutra{we povr{i pe}i ako je debqina zida pe}i!δ>31!nn-!a koeficijent toplotne
provodqivost zida pe}i!λ>61!X0)n3L*
α
ε2
U1
ε3
U2
U3
U4
a)
⋅
⋅
⋅
R >! R QSFMB[ ,! R [SB•FOKF >!///>!5871!,!3351!>!8111!X
∑
23
23
⋅
U −U
271 − 26
R QSFMB[ >! 2 3 ⋅ B2 >!
⋅ 3/6 >5871!X
2
2
23
α
24/24
5
5
5
5
544
399
U2
U
−
− 4
⋅
211
211
211
211 ⋅ B >!
R [SB•FOKF >
⋅ 3/6 >3351!X
2
2
2
23
D24
4/34
ε24!>!ε2!>!1/68!)B2!==!B4*
D24!>!Dd/!ε24!>! 6/78 ⋅ 1/68 >!4/34!
X
n3L 5
b)
⋅
⋅
U − U2
R >! R QSPWPEKFOKF >! 1
⋅ B !!! ⇒
δ
∑
12
λ
−4
8111 31 ⋅ 21
⋅
!>!271/4pD
U1!>!271!,
3/6
61
⋅
R
Σ δ
⋅ !>!271/4pD
U1!>!U2!,
B
λ
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 9
9/:/!Temperatura vrelih gasova, koji se kre}u kroz kanal, meri se temperaturskom sondom!)ε2>1/9*/
Pri stacionarnim uslovima sonda pokazuje temperaturu!u2>411pD/!Temperatura povr{i zidova kanala
je!u4>311pD/!Koeficijent prelaza toplote sa vrelih gasova na povr{ sonde iznosi!α>69!X0)n3L).
Odrediti stvarnu temperaturu vrelih gasova u kanalu!)!u3>@*/
u4
u3
u2
ε
α
(r{sb•fokf )24 = (rqsfmb{ )32
toplotni bilans temperaturske sonde:
5
5
5
U2
U
− 4
211
211 = U3 − U2
2
2
D24
α
5
684
584
−
211
211
U3!>!684!,!
2
5/65
ε24!>!ε2!>!1/9!)B2!==!B4*
⇒
5
U2
U
− 4
211
211 !!⇒
U3!>!U2!,!
2
D24
5
>729!L!)!456pD!*
D24!>!Dd/!ε24!> 6/78 ⋅ 1/9 >!5/65!
X
n3L 5
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 10
9/21/!U prostoru izme|u dve, koncentri~no postavqene posrebrene cevi ostvaren je potpuni vakum.
Temperatura na spoqa{woj povr{i unutra{we cevi, spoqa{weg pre~nika!e2>261!nn-!iznosi
u2>711pD-!a temperatura na unutra{woj povr{i spoqa{we cevi, unutra{weg pre~nika!e3>311!nniznosi u3>311pD/!Emisivnost svake posrebrene cevi je!ε>1/16/!Odrediti “ekvivalentnu” toplotnu
provodqivost materijala!)λ*-!~ijim bi se postavqewem u prostor izme|u cevi, pri nepromewenim
temperaturama i pre~nicima cevi ostvarila ista linijska gustina toplotnog protoka!)!toplotni
fluks-!X0n*
e2
e3
e3
e2
(r{sb•fokf )23 = (rqspwp} fokb )23
5
5
U2
U
− 3
211
211 = U2 − U3
2
e
2
mo 3
3π ⋅ λ e2
e2π ⋅ D23
5
⇒
5
e
U2
U
mo 3
− 3
e2
211
211 ⋅
!>///
λ!>!
2
3π ⋅ (U2 − U3 )
e2π ⋅ D23
D23!>!Dd!ε23!>///!>! 6/78 ⋅ 1/14 >1/276!
ε23!>!
X
n3L 5
2
2
2
>!1/14
>
>!
2
1/26 2
2 e2 2
2 B2 2
+
− 2
− 2
+
ε + e ε − 2
1/16 1/3 1/16
ε2 B3 ε3
2
3 3
5
5
984
584
1/3
−
mo
211
211
X
1/26
λ!>
⋅
>!1/158!
2
3π ⋅ (984 − 584 )
nL
1/26 ⋅ π ⋅ 1/276
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 11
9/22/!Cilindri~ni kolektor za vodenu paru, spoqa{weg pre~nika!e2>386!nn-!nalazi se u velikoj
prostoriji. Koeficijent emisije kolektora iznosi!ε2>1/:2/!Radi smawewa toplotnih gubitaka
zra~ewem postavqa se osno postavqen toplotni {tit (ekran), zanemarqive debqine, koeficijenta
emisije!εF>1/66/!Predpostavqaju}i da se postavqawem toplotnog {tita ne mewa temperatura na
spoqa{woj povr{i kolektora i unutra{woj povr{i zidova prostorije odrediti pre~nik toplotnog
{tita!)eF*-!tako da je u odnosu na neza{ti}eni kolektor smawewe toplotnih gubitaka zra~ewem 61&!/
(r{sb•fokf )23 = (r{sb•fokf )2F3
5
5
U2
U
− 3
211
211 = 3 ⋅
2
e2π ⋅ D23
2
3
=
2
2
2
+
e2ε23
e2ε2F eFεF3
5
5
U2
U
− 3
211
211
2
2
+
e2π ⋅ D2F eFπ ⋅ DF3
⇒
2
3
=
2
2 e2 2
− 2
+
e2ε23
ε2 eF εF
e2
3
2
2 2
2
=
+
− 2 +
⇒
e2ε2 e2ε2 eF εF
eFεF
⇒
+
2
eFεF3
3
eF!>! e2 ⋅ ε2 ⋅ − 2
εF
3
− 2 !>!1/76:!n!>!76:!nn
eF!>! 1/386 ⋅ 1/:2 ⋅
1/66
napomena:
ε23!>!ε2 )!B2!==!B3!*
ε2F!>!
εF3!>!εF)!BF!==!B3!*
2
2
>
2 e2 2
2 B2 2
− 2
− 2
+
+
ε2 BF εF
ε2 eF εF
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 12
9/23/!Oko duga~kog cilindra, pre~nika!e2>361!nn-!koncentri~no je postavqen ekran pre~nika
eF>461!nn-!zanemarqive debqine. Ukupan koeficijent emisije povr{i cilindra i povr{i ekrana su
jednaki i iznose!!ε!>1/9/!U stacionarnim uslovima, temperatura ekrana je!2:1pD-!temperatura okolnog
vazduha i okolnih povr{i iznosi!61pD-!a sredwi koeficijent prelaza toplote sa spoqa{we povr{i
ekrana na okolni vazduh!α>46!X0)n3L). Zanemaruju}i konvektivnu predaju toplote izme|u cilindra i
ekrana odrediti temperaturu povr{i cilindra.
α
ε
ε
U2
5
5
Uw
U3
(r{sb•fokf )23 = (r{sb•fokf )F3 + (rqsfmb{ )FW
toplotni bilans ekrana:
U2
U
− F
211
211 =
2
e2π ⋅ D2F
UF
5
5
UF
U
− 3
211
211 + UF − UW
2
2
eFπ ⋅ α
eFπ ⋅ DF3
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 13
D2F!>!Dd!ε2F!>///> 6/78 1/8 >4/:8!
X
n3L 5
2
2
>
>!>!1/8
2 B2 2 2 e2 2
+
+
− 2
− 2
ε BF ε ε eF ε
X
DF3!>!Dd!εF3!>//!!/!>!5/65!
n3L 5
εF3!>!εF!>!1/8
ε2F!>!
5
2
U
U2 = 211 ⋅ 5 3 +
e2 ⋅ D2F
211
5
U 5
3 − U2
211 U − U
211
+ 3
4 >!835!L
2
2
eF ⋅ α
eF ⋅ DF3
9/24/!U kanalu kvadratnog popre~nog preseka!)b>711!nn*!nalazi se ~eli~na cev!∅>3310311!nnλ>57!X0nL/!Kroz kanal proti~e suv vazduh. Temperatura spoqa{we povr{i cevi je!U2>711!L-!a
unutra{we povr{i zida je!U4>411!L/!Koeficijenti emisije zra~ewa su!ε2>1/92!(za cev) i!ε4>1/97!(za
zidove kanala). Koeficijent prelaza toplote sa cevi na vazduh je!α2>41!X0n3L/!Odrediti:
a) temperaturu vazduha u kanalu )U3>@* ako su toplotni gubici spoqa{we povr{i cevi, radijacijom i
konvekcijom jednaki
c* temperaturu unutra{we povr{i cevi )Up>@*
c) koeficijent prelaza toplote )α3*!sa vazduha na zidove kanala
α3
U4
α2
U3
U2
Up
ε2
ε4
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
a)
strana 14
(rqsfmb{b )23 = (r{sb•fokf )24
uslov zadatka:
5
U
U2
− 4
U2 − U3
211
211
=
2
2
et π ⋅ α
et π ⋅ D24
5
5
⇒
D24!>!Dd/!ε24!>///> 6/78 ⋅ 1/9 >!5/53!
5
U
U2
− 4
211
211 >!///
U3!>!U2! −
α
D24
X
n3L 5
2
2
2
>1/89
>
>!!
2
1/33 ⋅ π 2
2 et π 2
2 B2 2
+
2
−
+
− 2
+
− 2
1/92 5 ⋅ 1/7 1/97
ε2 B 4 ε4
ε2 5 ⋅ b ε3
B2!>! et π ⋅ M
B4!>!5b!/!M
ε24!>!
5
5
711
411
−
211
211
U3!>!711! −
>!532!L
41
5/53
b)
toplotni bilans spoqa{we povr{ine cevi:
(rqspwp} fokf )12 = (rqsfmb{b )23 + (r{sb•fokf )24 !!⇒ (rqspwp} fokf )12 = 3 ⋅ (rqsfmb{b )23
U − U3
3⋅ 2
=
2
et π ⋅ α
⇒
e
mo t
ev
U2 − U3
⋅
U1!>!U2!.! 3 ⋅
2
3π ⋅ λ
et π ⋅ α
U1 − U2
!>
e
2
mo t
3π ⋅ λ ev
⇒
1/33
mo
711 − 532
1/3 !>!713/5!L
Up!>!711!−! 3 ⋅
⋅
2
3π ⋅ 57
1/33 ⋅ π ⋅ 41
c)
toplotni bilans vazduha u kanalu:
U − U4
U2 − U3
⋅M = 3
⋅b⋅M ⋅ 5
2
2
et π ⋅ α2
α3
α3!>!
⇒
•
•
Rqsfmb{ = Rqsfmb{
23
34
U − U3
2
⋅
α3!>! 2
2
5b ⋅ (U3 − U4 )
et π ⋅ α2
⇒
X
711 − 532
2
>23/9!
⋅
2
5 ⋅ 1/7 ⋅ (532 − 411)
n3L
1/33π ⋅ 41
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 15
9/25/!!Unutar metalnog!cilindri~nog rezervoara, unutra{weg pre~nika!e>1/6!n! i visine i>2!n,
ostvaren je potpuni vakum. Temperatura na unutra{woj povr{i doweg dna je stalna i iznosi U2>511!L,
dok su temperature na preostalim unutra{wim povr{ima tako|e stalne i iznose U3>411!L/
Koeficijent emisije svih unutra{wih povr{i je jednak i iznosi ε>1/9/ Odrediti debwinu
izolacionog materijala )δj{* toplotne provodnosti λj>1/4!X0)nL* koeficijenta emisije εj{>1/:6, kojim
treba izolovati dowe dno, da bi pri stacionarnim uslovima i pri nepromewenim temperaturama (na
dodirnoj povr{i doweg dna i izolacionog sloja U2 i na ostalim unutra{wim povr{inama U3) toplotni
protok sa doweg dna bio smawen za 31&/
e
U3
vakum
i
δj{
Uj{
U2
5
⋅
R {
23
5
5
5
U2
U
511
411
− 3
−
211
211
211
211
=
⋅ B 2 >///>
⋅ 1/2:7 >262/72!X
2
2
D23
5/53
D23!>!Dd/!ε23!>//!!/!>!6/78/!1/89!>!5/53!
X
n3L 5
2
2
>!1/89
ε23!>!
>///>!
2
1/2:7 2
2 B2 2
+
−
2
− 2
+
1/9 2/878 1/9
ε2 B3 ε3
e 3 π 1/6 3 π
!>1/2:7!n3
=
5
5
e3 π
1/6 3 π
> 1/6 ⋅ π ⋅ 2 +
>2/878!n3
B3!>! e ⋅ π ⋅ i +
5
5
B2!>
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 16
⋅
⋅
R( = 1/9 ⋅ R { >232/3:!X
23
5
5
Uj{
U
− 3
⋅
211
211
⋅ B2
R( =
2
⋅
⇒
R(
U
Uj{ = 211 ⋅ 5
+ 3
D J{3 ⋅ B 2 211
5
D J{3
DJ{3>!Dd/!εJ{3′!>//!!/!>!6/78/!1/:4!>!6/38!
X
n3L 5
2
2
>!1/:4
εJ{3!>!
>///>!
2
1/2:7 2
B2 2
2
+
−
2
+
− 2
1/:6 2/878 1/9
ε j{ B 3 ε 3
5
Uj{ = 211 ⋅ 5
232/3:
411
+
>486/43!L
6/38 ⋅ 1/2:7 211
toplotni bilans gorwe povr{i izolacije:
U − Uj{
⋅
R qspwp} fokf = 2
⋅ B2
δ {j
2J{
λ j{
δ j{ =
⇒
δ j{ =
(511 − 486/43) ⋅ 1/4 ⋅ 1/2:7 >22/:7!nn
napomena:
⋅
⋅
R qspwp} fokf = R(
2J{
(U2 − Uj{ ) ⋅ λ j{ ⋅ B 2
⋅
R qspwp} fokf
2J{
>
232/3:
Pri izra~unavawu A2 za slu~aju sa izolaciju zanemaruje se
smawewe povr{ine A2 zbog male debqine izolacije
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 17
9/26/!Ravan zid debqine!6!dn!sa jedne svoje strane izlo`en je dejstvu toplotnog zra~ewa, ~iji
intenzitet u pravcu normale na zid iznosi!r{>2111!X0)n3L*/!Usled na ovaj na~in prenete koli~ine
toplote ozra~ena povr{ zida odr`ava se na temperaturi od!u2>62/2pD/!Ukupan koeficijent emisije
povr{i zida je!ε>1/8-!a toplotna provodqivost materijala od kojeg je zid na~iwen je!λ>1/86!X0)nL*/
Temperatura okolnog vazduha (sa obe strane zida) je 31pD/!Zanemaruju}i sopstveno zra~ewe zida i
smatraju}i da je koeficijent prelaza toplote sa obe strane zida na okolni vazduh (α) isti odrediti
temperaturu neozra~ene povr{i zida!)u3*
r{
rsfg
rqsfmb{2
rqspw
rqsfmb{3
Uw
U2
U3
Uw
toplotni bilans ozra~ene povr{i zida (povr{ 1)
r{!>!rsfg!,!rqsfmb{2!,!rqspw
r{!>!)2.!ε!*!r{!,!
U2 − Uw
U −U
!,! 2 3 !!!)2*
2
δ
α
λ
toplotni bilans ne ozra~ene povr{i zida (povr{ 2)
U2 − U3
U − Uw
> 3
δ
2
λ
α
rqspw!>!rqsfmb{3
!!!
!!!!!!!!)3*
re{avawem sistema dve jedna~ine )2*!i )3* sa dve nepoznate )U3!i!α* dobija se:
U3!>419/7!L-!!α!>26!
X
n3L
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 18
9/27/!Ravan zid od mermera!)λ>3/9!X0)nL*-!ε>1/66*-!debqine!1/167!n!izlo`en je sa obe strane dejstvu
toplotnog zra~ewa ~iji intenziteti u pravcu normala na povr{i iznose!r{2>253:!X0n3!i!r{3>2:4
X0n3/!Hla|ewe mermera sa obe strane zida obavqa se iskqu~ivo konvektivnim putem (zanemaruje se
sopstveno zra~ewe mermera). Temperatura vazduha sa jedne strane zida je!UW2>61pD-!a sa druge
UW3>51pD-!a odgovaraju}i koeficijenti prelaza toplote!α2>9!X0)n3L*!i!α3>31!X0)n3L*/!Odrediti
temperature obe povr{i mermera!)U2!j!U3*/
r{2
r{3
rsfg3
rsfg
rqsfmb{2
rqspw
rqsfmb{3
Uw2
U2
U3
Uw3
toplotni bilans ozra~ene povr{i zida (povr{ 1)
r{2>!rsfg!,!rqsfmb{2!,!rqspw
r{2>)2−ε!*!r{2,
U2 − Uw2
U −U
!,! 2 3 !!)2*
δ
2
λ
α
toplotni bilans ozra~ene povr{i zida 2 (povr{ 2)
r{3!,!rqspw!>!rsfg3!,!rqsfmb{
r{3,
U − Uw3
U2 − U3
!)3*
>!)2−ε!*!r{3!,! 3
δ
2
λ
α3
re{avawem sistema dve jedna~ine )2* i )3* sa dve nepoznate )U3!i U2*!dobija se:
U2>91pD-!U3>81pD
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 19
9/28/!Toplotna provodnost materijala od kojeg je na~iwen ravan zid, mo`e da se izrazi u funkciji
temperature zida u obliku:! λ = 1/5 + 7 ⋅ 21 −4 u -!!pri ~emu je toplotna provodnost, λ, izra`ena!v!X0)nL*a temperatura, u, u!pD/!Debqina zida je!δ>41!cm a temperature sa jedne i druge strane zida su!u2>231pD
i!u3>41pD/!Pdrediti toplotni fluks (povr{insku gustinu toplotnog protoka) kroz zid i predstaviti
grafi~ki raspored temperatura u zidu.
1. na~in:
diferencijalna jedna~ina provo|ewa toplote kroz ravan zid pri λ>g)u*:
r = −λ(u ) ⋅
δ
eu
ey
u3
∫(
∫
(
)
r ⋅ δ = −1/5 ⋅ (u 3 − u 2 ) − 7 ⋅ 21 −4
r ⋅ ey = − 1/5 + 7 ⋅ 21 −4 u ⋅ eu ⇒
1
u2
(
)
− 1/5 ⋅ (41 − 231) − 4 ⋅ 21
− 1/5 ⋅ (u 3 − u 2 ) − 4 ⋅ 21 −4 u 33 − u 23
>
δ
41 ⋅ 21 −3
X
r>366! 3
n
r=
)
r ⋅ ey = − 1/5 + 7 ⋅ 21 −4 u ⋅ eu
⇒
−4
(41
3
u 33 − u 23
3
− 231 3
)
2. na~in:
Jedna~ina za toplotni fluks kroz ravan zid pri λ>g)u*;
2
r!>! −
δ
r!>! −
U3
∫
U2
2
λ)u* ⋅ eu = −
δ
U3
∫ (1/5 + 1/117 ⋅ u ) ⋅ eu
⇒
U2
2
u3 − u3
1/5 ⋅ (u3 − u2 ) + 1/117 ⋅ 3 2
δ
3
231 3 − 41 3
2
>!366! X
r=−
1/5 ⋅ (231 − 41) + 1/117 ⋅
1/14
3
n3
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
e3 y
1. na~in:
eu 3
eu
r
=−
ey
1/5 + 7 ⋅ 21 −4 u
e3 y
=−
eu 3
strana 20
>@
⇒
ey
1/5 + 7 ⋅ 21 −4 u
=−
eu
r
7 ⋅ 21 −4
=1
r
Kako je drugi izvod funkcije y>g)u* negativan to zna~i da je y>g)u*!konkavna
(ispup~ena na gore).
δ>g)!u!*>@
2. na~in:
Do istog zakqu~ka se mo`e do}i posmatrawem funkcije δ>g)u*, kada
koristimo 2. na~in za izra~unavawe toplotnog fluksa.
δ!> −
2
u3 − u23
1/5 ⋅ (u − u2 ) + 1/117 ⋅
r
3
∂δ
2
= − ⋅ [1/5 + 1/117 ⋅ u ]
∂u
r
3
∂ δ
∂u3
!=!1
⇒
⇒
∂ 3δ
∂u
3
=−
2
⋅ 1/117
r
funkcija!δ!>!g)!u!*!je konkavna (ispup~ena na gore)
u
u2
isprekidana linija predstavqa
temperaturni profil pri λ>dpotu
puna linija predstavqa
temperaturni profil pri λ≠dpotu,
tj. pri! λ = 1/5 + 7 ⋅ 21 −4 u
u3
δ
y
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 21
9/29/!Cev od {amotne opeke!∅>4310391!nn!)λ>1/5,1/113/!u) oblo`ena je sa spoqa{we strane slojem
izolacije!)δj>6!nn-!λj>1/16,1/1112/!u*-!gde je!u!temperatura zida u!pD-!b!λ!v!X0)n/L). Temperatura
unutra{we povr{i cevi od {amotne opeke je!241pD-!a spoqa{we povr{i izolacionog materijala
41pD/!Odrediti topotni fluks (gustinu toplotnog protoka) po du`nom metru cevi.
u4
u3
u2
3π
r23!!>−
e
mo 3
e2
U3
∫ (1/5 + 1/113 ⋅ u ) ⋅eu !>! mo e3 [1/5(u3 − u2) + 1/112⋅ (u3 − u2 )]
3π
U2
3
3
e2
U4
∫(
)
[
(
)]
[
(
)]
3π
3π
1/15 + 21 ⋅ 21−6 u ⋅eu !>
1/15 ⋅ (u 4 − u3 ) + 6 ⋅ 21−6 u34 − u33
e4
e4
mo
mo
U
e3 3
e3
toplotni bilans spoqa{we povr{i {amotne opeke (unutra{we povr{i
izolacije):
r23!>!r34
r34!>−
[
(
)]
3π
3π
1/5 ⋅ (u3 − u2) + 1/112⋅ u33 − u23 !>!
1/15 ⋅ (u 4 − u3 ) + 6 ⋅ 21−6 u34 − u33
e3
e4
mo
mo
e2
e3
4/856 ⋅ 21 −6 ⋅ u 33 + 2/876 ⋅ 21 −3 ⋅ u 3 − 3/397 >1
u3 =
− 2/876 ⋅ 21 −3 ±
(2/876 ⋅ 21 )
−3 3
− 5 ⋅ 4/856 ⋅ 21 −6 ⋅ (− 3/397)
3 ⋅ 4/856 ⋅ 21 −6
>216/9pD
u3!>!216/9pD-!(pozitivno re{ewe)
r24!>!r23!>!r34!>!−
r24!>−
[
[
(
3π
1/5 ⋅ (u3 − u2) + 1/112⋅ u33 − u23
e3
mo
e2
(
)]
)]
3π
X
1/5 ⋅ (216/9 − 241) + 1/112 ⋅ 216/9 3 − 241 3 !>!835!
1/43
n
mo
1/39
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 22
9/2:/!Preko gorwe povr{i horizontalne ravne plo~e postavqena je toplotna izolacija!debqine!δ>39
nn-!toplotne provodqivosti!λ>1/4!,21−5!/!u!,!3/21−8/!u3!-!gde je!λ!)X0nL*!b!u!)pD*-!i emisivnosti
hrapave povr{ine u pravcu normale!εo>1/:/!Na dodirnoj povr{i plo~e i izolacije je stalna
temperatura!u2>511pD. Temperatura gorwe povr{i izolacije je tako|e stalna i iznosi!u3>211pD.
Temperatura zidova velike prostorije u kojoj se nalazi izolovana plo~a iznosi!u5>31pD/!Koeficijent
prelaza toplote sa gorwe povr{i izolacije na vazduh iznosi α>48/:!X0)n3L*/!Odrediti temperaturu
vazduha u prostoriji )u4*/
u5
u4
ε3
α
u3
u2
izolacija
plo~a
toplotni bilans gorwe povr{i izolacije:
(rqspwp} fokf )23 = (rqsfmb{ )34 + (r{sb•fokf )35
5
−
2
δ
u3
U
U3
− 5
u
−
u
211
211
1/4 + 2⋅ 21− 5 ⋅ u + 3 ⋅ 21−8 ⋅ u3 ⋅ eu !>! 3 4 ,
2
2
D35
α
∫(
u2
5
)
⇒
5
5
U3
U
u3
− 5
2
2
211
211
u4!>!u3! − −
1/4 + 2⋅ 21− 5 ⋅ u + 3 ⋅ 21−8 ⋅ u3 ⋅ eu −
2
α
δ
u2
D35
∫(
u4>u3!−
[(
)
)
(
2
⋅ rqspwp } fokb 23 − r{sb•fokb
α
)35 ] >///>211!− 482/: ⋅ [4743 − 711] >31pD
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 23
napomena:
u3
(rqspwp} fokf )23 !>! − 2δ ∫ (1/4 + 2⋅ 21−5 ⋅ u + 3 ⋅ 21−8 ⋅ u3 )⋅ eu !
u2
3
3
4
4
X
n3
(rqspwp} fokf )23 !>! − 2δ 1/4 ⋅ (u3 −u2) + 2⋅ 21−5 u3 3− u2 + 3 ⋅ 21−8 u3 4− u2 !>!4743!
5
5
5
5
U3
U
484 3:4
− 5
−
(r{sb•fokf )35 > 211 2 211 > 211 2 211 >711! X3
n
6
D35
D35!>!Dd/!ε35!>!///>6/78!/1/993>6
X
n3L 5
ε35!>!ε3!)!kfs!kf!B3!==!B5!*
ε3!>!L/εo!>!///>!1/:91/:>!1/993
L!>!1/:9-!(hrapava povr{ izolacije)
)9/31/*
zadatak za ve`bawe:
9/31/!Izme|u homogenog i izotropnog izolacionog materijala debqine δj>51!nn!i cevi, spoqa{weg
pre~nika e>91!nn, ostvaren je idealan dodir. Zavisnost toplotne provodqivosti materijala od
temperature data je izrazom:
2 U
λ
= 1/59 + 1/27 ⋅ mo
− 384 . Termoelementima, pri ustaqenim uslovima, izmerene su
[X 0 (nL )]
31 [L ]
slede}e temperature:
−
na spoqa{woj povr{i cevi
U2!>!534!L
U3!>!429!L
−
na spoqa{woj povr{i izolacionog materijala
U4!>!384!L
−
za okolni fluid.
Odrediti koeficijent prelaza toplote sa spoqa{we povr{i izolacionog materijala na okolni
fluid.
re{ewe: α!>!41/6!
X
n3L
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 24
9/32.!Tanka plo~a visine!i>1/3!n-!{irine b>1/6!n-!potopqena je, vertikalno, u veliki rezervoar sa
vodom temperature ug!>!31pD. Odrediti snagu greja~a, ugra|enog u plo~u, potrebnu za odr`avawe
temperature povr{i na!u{>!71pD/
⋅
u − ug
R= {
⋅ b ⋅ i ⋅ 3 !>!///
2
α
1. korak:
fizi~ki parametri za vodu na temperaturi!ug!>!31pD
X
n3
2
λg!>!6:/:!/21.3!
-!βg!>!2/93!/21.5!
-!νg!>!2/117!/21.7!
t
L
nL
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>!i!>!1/3!n
3. korak:
potrebni kriterijumi sli~nosti
Hsg!>!
β g ⋅ h ⋅ mL4 ⋅ )U{ − Ug *
υ 3g
Qsg!>!8/13
!>!
2/93 ⋅ 21 −5 ⋅ :/92 ⋅ 1/3 4 ⋅ )71 − 31*
(2/117 ⋅ 21 )
−7 3
>6/76!/219
Qs{!>!3/:9
1/36
Qsg
4. korak:
konstante u kriterijalnoj jedna~ini
Qs{
Hsg!/Qsg!>!4/:8!/21:
turbulentno strujawe fluida u grani~nom sloju
Ovg!>!D!)!Hsg!/Qsg!*o!
D>!1/26
5. korak:
o>1/44
izra~unavawe Nuseltovog broja
8/13
Ovg!>!1/26!)!4/:8!/21:!*1/44!
3/:9
1/36
>384/46
6. korak:
izra~unavawe koeficijenta prelaza toplote!)α*
λ
6:/: ⋅ 21 −3
X
α!>!Ovg! ⋅ g >!384/46 ⋅
>929/8!
1/3
mL
n3L
⋅
R=
71 − 31
⋅ 1/6 ⋅ 1/3 ⋅ 3 >765:/7!X
2
929/8
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 25
9/33/!Odrediti povr{insku gustinu toplotnog protoka (toplotni fluks) konvekcijom, sa spoqa{we
povr{i vertikalnog zida neke pe}i na okolni prividno miran vazduh stalne temperature!ug>31pD/
Temperatura spoqa{we povr{ine pe}i je!u{>!91pD/!Smatrati!da je strujawe vazduha u grani~nom sloju
turbulentno po celoj visini zida.
U − Ug
r!>! {
>///
2
α
α!>!Ovg!
Qs
λg
>!D!/!)!Hsg!/!Qsg!*o!/ G
mL
Qs{
1/36
λg
>///
mL
/
fizi~ki parametri za vazduh na temperaturi tf = 20oC
λg!>!3/6:!/21.3!
n3
X
!!!!! νg!>!26/17!/21.7!
nL
t
βg!>!
2
2
2
>
>!4/52!/21.4!
UG 3:4
L
potrebni kritrijumi sli~nosti:
Qsg!>!1/814
Qs{!>!1/7:3
konstante u kriterijalnoj jedna~ina za prirodnu konvekciju
D>1/26
o>!1/44!!
(turbulentno strujawe u grani~nom sloju)
β g ⋅ h ⋅ ml4 ⋅ (u { − u g )
⋅ Qsg
3
υg
1/44
α!>!1/26/
β g ⋅ h ⋅ (u { − u g )
⋅ Qsg
3
υg
1/44
α!>!1/26/
4/52⋅ 21
α!>!1/26/
r!>!
−4
⋅ :/92⋅ (91 − 31)
(26/17 ⋅21 )
−7 3
Qsg
Qs
{
/!
Qsg
Qs
{
/!
⋅ 1/814
1/36
/
λg
mL
⇒
1/36
1/44
/λ
g>
/ 1/814
1/7:3
1/36
⋅ 3/6: ⋅ 21−3 !>7/74!
X
n3 L
X
91 − 31
>4:8/9!
2
n3
7/74
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 26
9/34/!Ukupan toplotni protok koji odaje horizontalna ~eli~na cev!)ε>1/86) spoqa{weg pre~nika!et>91
nn!iznosi 611!X. Ako sredwa temperatura spoqa{we povr{i cevi iznosi!U2>91pD-!temperatura mirnog
okolnog vazduha!U3>31pD!i temperatura unutra{we povr{i zidova velike prostorije u kojoj se cev nalazi
U4>26pD!-!odrediti:
a) du`inu cevi
b) ukupan toplotni protok koji odaje ista ova cev kada bi je postavili vertikalno (pretpostaviti da je
strujawe u grani~nom sloju turbulentno po celoj visini cevi)
a)
5
5
U
U2
− 4
⋅
⋅
⋅
211
U − U3
211
R >! R qsfmb{ ,! R {sb•fokf > 2
⋅ M !,!!
⋅M
2
2
23
23
∑
e t πα
e t π ⋅ D24
M!>!
⋅
R
Σ
e t πα ⋅ (U2 − U3 ) + D24
U 5 U 5
⋅ 2 − 4
211
211
ε24!>!ε2!>!1/86!)B2!==!B4*
= ///
D24!>!Dd!ε24!>!6/78!/1/86>!5/37!
X
n3 L 5
fizi~ki parametri za vazduh na temperaturi!ug!>!31pD
2
n3
2
2
X
>
λg!>!3/6:!/21.3!
!!!!!!βg!>!
>!4/52!/21.4!
νg!>!26/17!/21.7!
nL
t
UG 3:4
L
1. korak:
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>!et!>!91!nn
3. korak:
Hsg!>!
potrebni kriterijumi sli~nosti
β g ⋅ h ⋅ mL4 ⋅ )U{ − Ug *
υ 3g
Qsg!>1/814
4. korak:
4/52 ⋅ 21 −4 ⋅ :/92 ⋅ 1/19 4 ⋅ )91 − 31*
(26/17 ⋅ 21 )
−7 3
>!5/65!/217
Qs{!>!1/7:3
Qs
Ovg!>!D!)!Hsg!/Qsg!*o! g
Qs{
laminarno strujawe fluida u grani~nom sloju
konstante u kriterijalnoj jedna~ini
Hsg!/Qsg!>!4/2:!/217
D>!1/6
!>
1/36
o>1/36
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
5. korak:
strana 27
izra~unavawe Nuseltovog broja
1/814
Ovg!>!1/6!)!4/2:!/217!*1/36!
1/7:3
1/36
>32/32
6. korak:
izra~unavawe koeficijenta prelaza toplote!)α*
λ
X
3/6: ⋅ 21 −3
α!>!Ovg! g >32/32!
>!7/98!
.
4
mL
91 ⋅ 21
n3L
M!>!
611
464 5 399 5
1/19π7/98 ⋅ (91 − 31) + 5/37 ⋅
−
211
211
>!3/66!n
b)
5
5
U
U2
− 4
⋅
211 ⋅ M >///>591!X
R ( > U2 − U3 ⋅ M !,!! 211
2
2
∑
eT π ⋅ α (
eT π ⋅ D24
α′>@
2. korak:
ml( >M!>!3/66!n
3. korak:
Hsg′!>!
4.korak:
Hsg′!/!Qsg!>!2/14!/2122
4/52 ⋅ 21 −4 ⋅ :/92 ⋅ 3/66 4 ⋅ )91 − 31*
(3/6: ⋅ 21 )
−3 3
⇒
>!2/58!/2122
D>!1/26
o>1/44
)turbulentno strujawe fluida u grani~nom sloju du` cele visine cevi)
5. korak:
1/814
Ovg′!>!1/26!)!2/14!/2122!*1/44!
1/7:3
6. korak:
α′!>!75:/32!
1/36
>75:/32
3/6: ⋅ 21 −3
X
>!7/6:!
3/66
n3L
5
5
464
399
−
211
R ( > 91 − 31 ⋅ 3/66 !,!! 211
⋅ 3/66 >59:!X
2
2
∑
1/19π ⋅ 7/6:
1/19π ⋅ 5/37
⋅
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 28
9/35/!Vertikalna cev, visine!M>1/9!n, nalazi se u prividno mirnom vazduhu stalne temperature
Ug>31pD!j!pritiska!q>2!cbs/!Temperatura na grani~noj povr{i cevi je stalna i iznosi U{>91pD/
Odrediti toplotni protok sa cevi na okolni vazduh u laminarnom delu strujawa, turbulentnom delu
strujawa i du` cele cevi.
fizi~ki parametri za vazduh na temperaturi!ug!>!31pD
2
X
n3
2
2
>!4/52!/21−4!
!!!!!!βg!>!
νg!>!26/17!/21−7!
>
λg!>!3/6:!/21−3!
nL
t
UG 3:4
L
1. korak:
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>!M!>!1/9!n
3. korak:
potrebni kriterijumi sli~nosti
β ⋅ h ⋅ ml4 ⋅ )U{ − Ug *
4/52 ⋅ 21 −4 ⋅ :/92 ⋅ 1/9 4 ⋅ )91 − 31*
>!5/64!/21:
!>
Hsg!>! g
−7 3
υ 3g
26/17 ⋅ 21
Qsg!>1/814
Qs{!>!1/7:3
(
4. korak:
)
Qs
Ovg!>!D!)!Hsg!/Qsg!*o! g
Qs{
)kriti~na vrednost proizvoda!Hsg!/Qsg!*
konstante u kriterijalnoj jedna~ini
(Hsg ⋅ Qsg )ls >2/21:
Hsg!/Qsg!>!4/2:!/21:!?! (Hsg ⋅ Qsg )ls
1/36
prirodno strujawe fluida u grani~nom
sloju do visine M>mls je laminarno a
nakon toga turbulentno
D>1/87-!
o>1/36
mls
5. korak:
2
4
⋅ υ 3g
)
izra~unavawe Nuseltovog broja za laminarnu oblast strujawa
(Ov g )mbn >!1/87!)!2!/21:!*1/36! 1/814
1/7:3
6. korak:
(
2
3
4
2 ⋅ 21
2 ⋅ 21 : ⋅ 26/17 ⋅ 21 −7
>1/65!n
>
=
4/52 ⋅ 21 −4 ⋅ :/92 ⋅ )91 − 31* ⋅ 1/814
β g ⋅ h ⋅ )U{ − Ug * ⋅ Qsg
:
1/36
>246/26
izra~unavawe koeficijenta prelaza toplote!)αmbn) u laminarnom delu
strujawa.
αmbn!>! (Ov g )mbn ⋅
λg
3/6: ⋅ 21 −3
X
>!135.15 ⋅
>7/59!
1/65
mls
n3L
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
⋅
R mbn =
strana 29
91 − 31
⋅ 1/65 >76/5!X
2
1/2 ⋅ π ⋅ 7/5
oblast turbulentnog strujawa u grani~nom sloju:
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>!M!−!mls!>!1/:!−!1/65!>1/47!n
3. korak:
potrebni kriterijumi sli~nosti
(Hsg ⋅ Qsg )M >4/2:!/21:- (Hsg ⋅ Qsg )ls >2!/21:
4. korak:
konstante u kriterijalnoj jedna~ini
Hsg!/Qsg!>!4/2:!/21:!?! (Hsg ⋅ Qsg )ls
5. korak:
⇒
Qsg
Qs{
1/36
Ovg!>!D!)!Hsg!/Qsg!*o!
D>1/26-!
o>1/44
izra~unavawe Nuseltovog broja za turbulentnu oblast strujawa
(Ov g )uvs >! D ⋅ [(Hsg ⋅ Qsg )oM − (Hsg ⋅ Qsg )omls ]⋅
Qsg
Qs{
(Ov g )uvs > 1/26 ⋅ (4/2: ⋅ 21 : )
1/44
6. korak:
(
− 2 ⋅ 21 :
)
1/36
1/814
⋅
1/7:3
1/44
1/36
>76/3:
izra~unavawe koeficijenta prelaza toplote!)αmbn) u turbulentnom
delu strujawa.
αuvs!>! (Ov g )uvs ⋅
⋅
R uvs =
λg
3/6: ⋅ 21 −3
X
>!76/3: ⋅
>5/81!
ml
1/47
n3L
91 − 31
⋅ 1/47 >42/:!X
2
1/2 ⋅ π ⋅ 5/8
⋅
⋅
⋅
R = R mbn !,! R uvs >76/5!,!42/:!>:8/4!X
Σ
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
zadaci za ve`bawe:
strana 30
)9/36/!−!9/37/*
9/36/!Toplotni gubici prostorije, u kojoj je potrebno odr`avati temperaturu od!29pD-!iznose!2!lX/
Grejawe vazduha u prostoriji se ostvaruje grejnim telima (konvektorima). Ova tela imaju oblik
kvadra!)2111!y!291!y!711!nn*-!i toplotno su izolovana na gorwoj i dowoj osnovi ({rafirani deo na
slici). Temperatura na spoqa{wim povr{ima konvektora je!55pD/!Odrediti potreban broj
konvektora za nadokna|ivawe toplotnih gubitaka prostorije. Zanemariti razmenu toplote zra~ewem.
re{ewe:
o>7
291!nn
!711!nn
2111!nn
⋅
9/37/!U horizontalnoj cevi spoqa{weg pre~nika e>231!nn!vr{i se potpuna kondenzacija n >36!lh0i
suvozasi}ene vodene pare pritiska q>211!lQb. Cev se nalazi u velikoj prostoriji i okru`ena je
mirnim okolnim vazduhom stalne temperature 31pD. Sredwa temperatura na povr{i zidova prostorije
iznosi 28pD, a sredwa temperatura na spoqa{woj povr{i cevi :8pD. Koeficijent emisije zra~ewa sa
cevi iznosi ε>1/:. Odrediti du`inu cevi.
re{ewe:
M>47/23!n
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 31
9/38/!Kroz cev unutra{weg pre~nika!e>91!nn!struji transformatorsko uqe sredwe temperature
ug>61pD-!sredwom brzinom!x>1/3!n0t. Temperatura unutra{we povr{i zidova cevi je!u{>36pD/!Ukupno
⋅
razmewen toplotni protok izme|u uqa i unutra{we povr{i cevi iznosi! R >896!X/!Odrediti du`inu
cevi.
M
ug
ug
u{
⋅
⋅
u − u{
R= g
⋅M
2
eπ ⋅ α
R
M!>!
= ///
eπ ⋅ α ⋅ (u g − u { )
⇒
fizi~ki parametri za transformatorsko uqe temperaturi ug!>!61pD
2
lh
X
-!
βg>7:/6/21−6! λg>1/233!
-!
ρg>956!
4
nL
L
n
lK
µg>!:/:!/21.4! Qb ⋅ t -!
dqg>3/154!
lhL
1. korak:
2. korak:
karakteristi~na du`ina ~vrste povr{i
e3 π
B
ml!>! 5 ⋅ = 5 ⋅ 5 >e>!91!nn
P
eπ
3. korak:
potrebni kriterijumi sli~nosti
Sfg!>!
Hsg>
dqg ⋅ µ g
λg
ρ g ⋅ x ⋅ mL
956 ⋅ 1/3 ⋅ 1/19
>
>2476/7
µg
:/: ⋅ 21 −4
β g ⋅ h ⋅ mL4 ⋅ )Ug − U{ * ⋅ ρ 3g
µ 3g
4
>
3/154 ⋅ 21 ⋅ :/: ⋅ 21
1/233
d q{ ⋅ µ {
>
)Sf=3411-!laminarno strujawe*
7:/6 ⋅ 21 −6 ⋅ :/92 ⋅ 1/19 4 ⋅ )61 − 36* ⋅ 956 3
(:/: ⋅ 21 )
−4 3
>7/47!/216 Qsg!>
−4
>276/9
2/:29 ⋅ 21 4 ⋅ 35 ⋅ 21 −4
>!485/3
1/234
λ{
fizi~ki parametri za transformatorsko uqe temperaturi u{!>!36pD
lK
X
λ{>1/234!
dq{>2/:29!
-!
µ{>!35!/21.4! Qb ⋅ t -!
lhL
nL
Qs{!>!
>///>
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
4. korak:
o
q
konstante u kriterijalnoj jedna~ini:!!!!!!Ovg> D ⋅ Sfn
g ⋅ Qsg ⋅ Hsg ⋅ ε U
Qs
n>1/44-!!!o>1/54-!!!q>1/2-!!!εU> g
Qs{
M
pretpostavimo:
> 61
ml
5. korak:
strana 32
1/2:
276/9
>
485/3
⇒
εM>2
>1/97-!!!D> 1/26 ⋅ ε M
⇒!
D>1/26
izra~unavawe Nuseltovog broja
(
Ov g = 1/26 ⋅ (2476/7 )1/44 ⋅ (276/9 )1/54 ⋅ 7/47 ⋅ 21 6
6. korak:
1/2:
)
1/2
⋅ 1/97 1/36 >56/5
izra~unavawe koeficijenta prelaza toplote!)α*
α!>!Ovg! ⋅
λg
X
1/233
>56/5!
>!7:/3!
.
4
mL
91 ⋅ 21
n3L
⋅
R
896
M!>!
>
>2/9!n
eπ ⋅ α ⋅ (u g − u { ) 1/19π ⋅ 7:/3 ⋅ (61 − 36 )
provera pretpostavke:
2/9
M
>
>33/6!=!61!
mL 1/19
pretpostavimo:
M(
>33/6
ml
M( = M ⋅
εM
ε M(
> 2/9 ⋅
⇒
pretpostavka nije ta~na
ε M( >2/22
2
>2/73!n
2/22
provera pretpostavke:
M( 2/73
>
>31/36!≠!33/6!
1/19
ml
pretpostavimo:
M( (
>31/36
ml
M(( = M ⋅
εM
ε M((
> 2/9 ⋅
⇒
pretpostavka nije ta~na
ε M(( >2/24
2
>2/6:!n
2/24
provera pretpostavke:
M( ( 2/6:
>
>2:/99!≈!31/36!
ml
1/19
pretpostavka ta~na !!
stvarna du`ina cevi iznosi M>2/6:!n/
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 33
9/39/ Kroz prav kanal, prstenastog popre~nog preseka dimenzija ∅2>910:1!nn, ∅3>3110321!nn,
proti~e voda sredwom brzinom od x>2/3!n0t. Ulazna temperatura vode je Ux2>91pD, a sredwa
temperatura zidova kanala iznosi U{>31pD. Odrediti du`inu cevi na kojoj }e temperatura vode pasti
na Ux3>71pD. Zanemariti prelaz
toplote sa vode na spoqa{wu cev.
Ux2
Ux3
U{
M
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu:
⋅
⋅
⋅
⇒
R 23!>!∆ I 23!,! X U23
lK
lh
lK
ix3!>!362/2!
lh
⋅
⋅
R 23!>! n x /!)!ix3!−!ix2!*!>!///
)q>2!cbs-!u>91pD*
!ix2!>!445/:!
)q>2!cbs-!u>71pD*
⋅
⋅
1/3 3 ⋅ π 1/1: 3 ⋅ π
e3 ⋅ π e33 ⋅ π
>!3:/5!lh0t
!> :88/9 ⋅ 2/3 ⋅
nx = ρ x ⋅ x ⋅ 4
−
−
5
5
5
5
lh
napomena:!
ρx>:88/9! 4 -!je gustina vode odre|ena za sredwu
n
U + Ux 3
91 + 71
temperaturu vode u cevi;! Uxts = x2
>81pD
=
3
3
⋅
R 23 = 3:/5 ⋅ (362/2 − 445/: ) >−3574/8!lX
∆Uts
R=
⋅M
2
e3 π ⋅ α
⋅
⇒
⋅
⋅
R !>! R23 >3574/8!lX
⋅
⇒
R
M!>!
= ///
e3 π ⋅ α ⋅ ∆Uts
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 34
U
∆Unby!>!91!−!31!>71pD
∆Unjo!>!71!−!31!>51pD
91pD
wpeb
71pD
∆Uts!>!
71 − 51
= 5:/4pD
71
mo
51
dfw
31pD
31pD
M
1. korak:
fizi~ki parametri za vodu odre|eni za sredwu temperaturu vode u
U + Ux 3
91 + 71
=
>81pD
cevi: Uxts = x2
3
3
n3
X
λg>77/9!/21−3!!
!
!
υg>1/526/21−7!
t
nL
2. korak:
karakteristi~na du`ina ~vrste povr{i
e34 ⋅ π e33 ⋅ π
−
B
5 = e − e = 311 − :1 >221!nn
ml!>! 5 ⋅ = 5 ⋅ 5
4
3
P
e 4 ⋅ π − e3 ⋅ π
3. korak:
potrebni kriterijumi sli~nosti
Qsg!> (Qs )Ug =81p D >3/66Sfg!>!
4. korak:
Qs{!> (Qs )U{ =31p D >8/13
x ⋅ mL
2/3 ⋅ 1/22
!>!4/29!/216! (turbulentno strujawe)
=
νg
1/526 ⋅ 21 −7
konstante u kriterijalnoj jedna~ini
1/36
1/36
Qs
3/66
>
>1/89-!!!!D>1/132!/!εM
n>1/9-!!!o>1/54-!!!q>1-!!!εU!>! g
Qs
8/13
{
M
D>1/132
> 61
⇒
εM>2 ⇒!
pretpostavimo:
ml
5. korak:
izra~unavawe Nuseltovog broja
Ovg!>!1/132!/!)!4/29!/216!*1/9!/!)!3/66!*1/54!/!1/89!>!726/5
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
6. korak:
strana 35
izra~unavawe koeficijenta prelaza toplote!)α*
λg
X
77/9 ⋅ 21 .3
>!4848! 3
> 726/5 ⋅
.4
mL
221 ⋅ 21
nL
3574/8
M!>!
>!58/4!n
:1 ⋅ 21 −4 π ⋅ 4/848 ⋅ 5:/4
α!> Ov g ⋅
provera pretpostavke:
58/4
M
>541!?!61!
>
mL
1/22
pretpostavka ta~na
stvarna du`ina cevi iznosi M>58/4!n/
9/3:/!Predajnik toplote se sastoji od cilindri~nog, toplotno izolovanog omota~a, unutra{weg
pre~nika!E>1/5!n!i snopa od!o>66!pravih cevi, spoqa{weg pre~nika!e>41!nn/!Podu`no, kroz
prostor izme|u omota~a i cevi, struji suv vazduh i pritom se izobarno, pri!q>6!NQb!-!hladi od
temperature!Ug2>447!L!ep!Ug3>424!L/ Maseni protok vazduha iznosi n>1/5!lh0t/!Uemperatura na
spoqa{woj povr{i cevi pre~nika e je stalna i iznosi!U{>3:4!L. Odrediti du`inu predajnika
toplote.
E
e
M
⋅
∆Uts
R=
⋅o⋅M
2
e⋅ π⋅α
⋅
R
M=
>!///
e ⋅ π ⋅ α ⋅ ∆Uts ⋅ o
⇒
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu:
⋅
⋅
⋅
R 23!>!∆ I 23!,! X U23
!
⇒
⋅
⋅
R 23!>! n w ⋅ d qg ⋅ (Ug 3 − Ug2 )
⋅
R 23!>! 1/5 ⋅ 2/196 ⋅ (424 − 447) >−:/:9!lX
⇒
⋅
⋅
R !>! R23 >:/:9!lX
U
∆Unby!>!447!−!3:4!>!54!L
447!L
∆Unjo!>!424!−!3:4!>!31!L
∆Uts =
54 − 31
!>!41!L
54
mo
31
wb{evi
424!L
3:4!L
dfw
3:4!L
M
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 36
fizi~ki parametri za vazduh (q>61!cbs- Ugts =
1. korak
λg>4/14/21−3!
X
nL
ρg>64/84!
lh
4
n
Ug2 + Ug 3
>435/6!L*
3
µg!>31/74/21−7!Qb/t
-
2. korak
karakteristi~na du`ina ~vrste povr{i
E3 π
e3 π
−o⋅
3
3
3
3
B
5 > E − o ⋅ e = 1/5 − 66 ⋅ 1/14 >1/165!n
ml = 5 ⋅ = 5 ⋅ 5
E +o⋅e
1/5 + 66 ⋅ 1/14
P
Eπ + o ⋅ e π
3. korak
potrebni kriterijumi sli~nosti
Qsg!> (Qs )Ug =435/6L -q=61 cbs >1/849Sfg!>!
ρ g ⋅ x ⋅ ml
64/84 ⋅ 1/197 ⋅ 1/165
>!///>!
>2/32/215!!!!(turbulentno strujawe)
µg
31/74 ⋅ 21 −7
⋅
E3 π
e 3 π
ng = ρ g ⋅ x ⋅
o⋅
−
5
5
⋅
x=
Qs{!> (Qs )U{ =3:4L -q=61 cbs >1/887
⇒
1/5
1/5 π
1/14 π
64/84 ⋅
− 66 ⋅
5
5
3
3
x=
>1/197!
ng
E3 π
e 3 π
ρg ⋅
−o⋅
5
5
n
t
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
4. korak:
strana 37
konstante u kriterijalnoj jedna~ini
Qs
n>1/9-!!!o>1/54-!!!q>1-!!!εU!>! G
Qs{
M
pretpostavimo:
> 61
ml
5. korak:
1/36
1/849
>
1/887
⇒
1/36
εM>2
≈2-!!!!D>1/132!/!εM
⇒!
D>1/132
izra~unavawe Nuseltovog broja
(
Ov g = 1/132 ⋅ 2/32 ⋅ 21 5
)
1/9
⋅ (1/849)1/54 >45
6. korak:
izra~unavawe koeficijenta prelaza toplote!)α*
λ
X
4/14 ⋅ 21 −3
>2:/2!! 3
α = Ov g ⋅ g !>! 45 ⋅
1/165
ml
nL
M=
R
:/:9
>4/47!n
>! M =
e ⋅ π ⋅ α ⋅ ∆Uts ⋅ o
1/14 ⋅ π ⋅ 2:/2 ⋅ 21 −4 ⋅ 41 ⋅ 66
provera pretpostavke:
4/47
M
>73!?!61!
>
mL 1/165
pretpostavka ta~na !!
stvarna du`ina cevi iznosi M>4/47!n/
9/41/ Kroz prav kanal pravougaonog popre~nog preseka proti~e voda brzinom x>2n0t/!Dimenzije
unutra{wih stranica!pravougaonika iznose b>21!nn!i!c>31!nn/!Temperatura vode na ulazu u kanal je
ux2>21pD-!a na izlazu!ux3>81pD/!Temperatura zidova kanala je!u{>211pD>dpotu/!Odrediti:
a) toplotni protok sa zidova kanala na vodu )lX*
b) du`inu kanala
c
M
α
U{
voda
b
Ux>g)!M!*
voda
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 38
b*
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu:
⋅
⋅
⋅
⇒
R 23!>!∆ I 23!,! X U23
lK
lh
lK
ix3!>3:4!
lh
⋅
⋅
R 23!>! n x /!)!ix3!−!ix2!*!>!///
)q>2!cbs-!u>21pD*
!ix2!>53!
)q>2!cbs-!u>81pD*
⋅
⋅
n x = ρ x ⋅ x ⋅ b ⋅ c !> ::3/3 ⋅ 2 ⋅ 21 ⋅ 21 −4 ⋅ 31 ⋅ 21 −4 >1/3!
napomena:!
ρx>::3/3!
lh
n4
lh
t
-!je gustina vode odre|ena za sredwu
temperaturu vode u cevi;! Uxts =
⋅
Ux2 + Ux3 21 + 81
=
>51pD
3
3
⋅
⋅
R 23 = 1/3 ⋅ (3:4 − 53) >61/3!lX
⇒
∆U
R >! ts ⋅ (3b + 3c) ⋅ M
2
α
R
M!>!
>!///
α ⋅ ∆Uts ⋅ 3 ⋅ (b + c)
R !>! R23 >61/3!lX
b)
⋅
⋅
⇒
U
∆Unby!>!:1pD
∆Unjo!>!41pD
∆Uts!>!
:1 − 41
= 65/7pD
:1
mo
41
211pD
{je
211pD
81pD
wpeb
21pD
M
1. korak:
fizi~ki parametri za vodu odre|eni za sredwu temperaturu vode u
U + Ux3 21 + 81
=
>51pD
cevi: Uxts = x2
3
3
n3
X
lh
υg>1/76:/21−7!
λg>74/6!/21−3!
!-!
ρg!>::3/3!
t
nL
n4
2. korak:
ml!>! 5 ⋅
3. korak:
karakteristi~na du`ina ~vrste povr{i
1/12 ⋅ 1/13
B
b ⋅c
= 5⋅
!> 5 ⋅
>24/44!nn
P
3 ⋅ (b + c)
3 ⋅ (1/12 + 1/13)
potrebni kriterijumi sli~nosti
x ⋅ mL 2 ⋅ 24/24 ⋅ 21 −4
>
>3!/215
(turbulentno strujawe)
νg
1/76: ⋅ 21 −7
Qs{!> (Qs )U{ =211p D >2/86
Qsg!> (Qs )Ug =51p D >5/42Sfg!>!
4. korak:
konstante u kriterijalnoj jedna~ini:
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
Qs
n>1/9-!!!o>1/54-!!!q>1-!!!εU!> G
Qs{
M
> 61
pretpostavimo:
ml
5. korak:
(
1/36
5/42
>
2/86
⇒
1/36
>2/36-!!!!D>1/132!/!εM
εM>2
⇒!
D>1/132
)
1/9
⋅ (5/42)1/54 ⋅ 2/36 >246/9
izra~unavawe koeficijenta prelaza toplote!)α*
α!>!Ovg! ⋅
M!>!
izra~unavawe Nuseltovog broja
Ov g = 1/132 ⋅ 3 ⋅ 21 5
6. korak:
strana 39
λg
74/6 ⋅ 21 .3
X
>246/9! ⋅
>!757:/2!
.
4
mL
24/44 ⋅ 21
n3L
61/3
757:/2 ⋅ 21
−4
⋅ 65/7 ⋅ 3 ⋅ (1/12 + 1/13)
provera pretpostavke:
>!3/48!n
3/48
M
>288/9?!61! pretpostavka ta~na !!
>
mL 24/44 ⋅ 21 −4
stvarna du`ina kanala iznosi M>3/48!n/
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 40
9/42/ Dve cevi spoqa{wih pre~nika!e>1/229!n-!od kojih je jedna od wih toplotno izolovana nalaze se
u toplotno izolovanom kanalu kvadratnog popre~nog preseka, unutra{we stranice!b>1/4!m (slika).
Kroz slobodan prostor, pri stalnom pritisku!q>212/436!lQb-!proti~e voda, sredwom brzinom!x>1/3
n0t/!Temperatura vode na ulazu u kanal je!Ug2>3::!L, a iz kanala izlazi voda sredwne temperature
Ug3>418!L/
Temperatura povr{i neizolovane cevi je stalna i iznosi U{>474!L/!Odrediti toplotni protok koji
razmeni voda ca neizolovanom cevi kao i du`inu kanala,
α
e
b
e
voda
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu:
⋅
⋅
⋅
R 23!>!∆ I 23!,! X U23
lK
lh
lK
!ix3!>!253/5!
lh
!ix2!>!21:/1
⇒
⋅
⋅
R 23!>! n x /!)!ix3!−!ix2!*!>!///
)q>2!cbs-!u>37pD*
)q>2!cbs-!u>45pD*
⋅
⋅
lh
e 3 π
1/229 3 π
n x = ρ x ⋅ x ⋅ b3 − 3 ⋅
!> ::6/8 ⋅ 1/3 ⋅ 1/4 3 − 3 ⋅
>24/68
5
5
t
lh
napomena:!
ρx>::6/8! 4 -!je gustina vode odre|ena za sredwu
n
U + Ux3 37 + 45
temperaturu vode u cevi;! Uxts = x2
=
>41pD
3
3
⋅
R 23 = 24/68 ⋅ (253/5 − 21:/1) >564/3!lX
⇒
⋅
⋅
R !>! R23 >564/3!lX
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
∆Uts
R=
⋅M
2
eπ ⋅ α
⋅
strana 41
⋅
⇒
R
!>!///
M!>!
eπ ⋅ α ⋅ ∆Uts
U
∆Unby!>474!−!3::!>75pD
∆Unjo!>474!−!418!>!67pD
cev
474!L
75 − 67
= 6:/:pD
∆Uts!>
75
mo
67
474!L
418!L
voda
3::!L
M
1. korak:
fizi~ki parametri za vodu odre|eni za sredwu temperaturu vode u
U + Ux3 37 + 45
=
>41pD
cevi: Uxts = x2
3
3
n3
X
lh
υg>1/916/21−7!
λg>72/9!/21−3!
!-!
ρg!>::6/8!
t
nL
n4
2. korak:
karakteristi~na du`ina ~vrste povr{i
B
1/179
>1/25!n
ml!>! 5 ⋅ >///> 5 ⋅
P
2/:52
e3π
1/229 3 π
>1/179!n3
!> 1/4 3 − 3 ⋅
5
5
P!>! 5 ⋅ b + 3 ⋅ eπ !>! 5 ⋅ 1/4 + 3 ⋅ 1/229 ⋅ π >2/:52!n
B = b3 − 3 ⋅
3. korak:
potrebni kriterijumi sli~nosti
x ⋅ mL
1/3 ⋅ 1/25
>
>4/59/215 (turbulentno strujawe)
νg
1/916 ⋅ 21 −7
Qs{!> (Qs )U{ =:1p D >2/:6
Qsg!> (Qs )Ug =41p D >6/53Sfg!>!
4. korak:
konstante u kriterijalnoj jedna~ini:
Qs
n>1/9-!!!o>1/54-!!!q>1-!!!εU!> G
Qs{
M
> 61
pretpostavimo:
ml
1/36
6/53
>
2/:6
⇒
1/36
>2/3:-!!!!D>1/132!/!εM
εM>2
⇒!
D>1/132
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
5. korak:
izra~unavawe Nuseltovog broja
(
Ov g = 1/132 ⋅ 4/59 ⋅ 21 5
6. korak:
strana 42
)
1/9
⋅ (6/53)1/54 ⋅ 2/3: >351/9
izra~unavawe koeficijenta prelaza toplote!)α*
α!>!Ovg! ⋅
λg
72/9 ⋅ 21 .3
X
>351/9! ⋅
>!2174!
mL
1/25
n3L
⋅
R
564/3
M!>
>
>!2:/3!n
eπ ⋅ α ⋅ ∆Uts
1/229 ⋅ π ⋅ 2174 ⋅ 21 −4 ⋅ 6:/:
provera pretpostavke:
M 2:/3
>248/2?!61!
>
mL 1/25
pretpostavka ta~na !!
stvarna du`ina cevi iznosi M>2:/3!n/
9/43/!U tolplotno izolovanom kanalu kvadratnog popre~og preseka stranice!b>1/6!n!postavqena je
cev spoqa{weg pre~nika!e>1/3!n/!Kroz kanal struji suv vazduh temperature!ug>41pD-!brzinom!x>9
n0t/!Temperatura zidova kanala iznosi U3>321pD-!a temperatura spoqa{we povr{i cev!U2?U3>@/
Koeficijent emisije zra~ewa cevi iznosi!ε2>1/:6-!a zidova kanala!ε3>1/9/!Smatraju}i da
koeficijenti prelaza toplote )α*!sa obe povr{i (sa cevi na vazduh i sa kanala na vazduh) imaju istu
vrednost, odrediti:
b* temperaturu spoqa{we povr{i cevi
b) ukupan toplotni fluks koji odaje spoqa{wa povr{ cevi
Pri odre|ivawu Nuseltovog broja smatrati da εU!i!εM!iznose!εU>εM>2
α
ε2
ε3
α
U3
Ug
U2
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 43
a)
⋅
⋅
toplotni bilans unutra{wih povr{i zidova kanala: R {sb•okf = R qsfmb{
23
3g
5
5
U2
U
− 3
U − Ug
211
211
⋅M = 3
⋅b⋅M ⋅ 5
2
2
α
eπ ⋅ D23
⇒
5
5b ⋅ α
U
U2 = 211 ⋅ 5 3 + (U3 − Ug ) ⋅
>!///
eπ ⋅ D23
211
D23!>!Dd!ε23!>//!!/!>!6/78/!1/995!>!6!
ε23!>!
X
n3L 5
2
2
2
>!1/995
>
>!
2
1/3π 2
2 B2 2
2 eπ ⋅ M 2
+
2
−
+
− 2
+
− 2
1/:6 5 ⋅ 1/6 1/9
ε2 B3 ε3
ε 2 5b ⋅ M ε 3
fizi~ki parametri za vazduh na temperaturi!ug!>41pD
lh
X
νg!>!27!/21−7! Qb ⋅ t
λg!>!3/78!/21−3!!
-! !ρg!>!::3/3!
4
nL
n
1. korak:
2. korak:
karakteristi~na du`ina ~vrste povr{i
1/3 3 π
e3π
1/6 3 −
b3 −
B
5 >1/444!n
5 !> 5 ⋅
ml!>! 5 ⋅ = 5 ⋅
5 ⋅ 1/6 + 1/3π
P
5 ⋅ b + eπ
3. korak:
potrebni kriterijumi sli~nosti
x ⋅ mL
9 ⋅ 1/444
>2/77!/216
>
νg
27 ⋅ 21 −7
Qsg!> (Qs )Ug =41p D >1/812
Sfg!>!
4. korak:
(turbulentno strujawe)
konstante u kriterijalnoj jedna~ini:
n>1/9-!!!o>1/54-!!!q>1-!!!εU!>2-!!!!D>1/132!/!εM>1/132
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
5. korak:
strana 44
izra~unavawe Nuseltovog broja
(
Ov g = 1/132 ⋅ 2/77 ⋅ 21 6
6. korak:
)
1/9
⋅ (1/812)1/54 >381/5
izra~unavawe koeficijenta prelaza toplote!)α*
α!>Ovg! ⋅
λg
3/78 ⋅ 21 .3
X
>32/8!
>381/5! ⋅
1/444
mL
n3L
5
5 ⋅ 1/6 ⋅ 32/8
594
U2 = 211 ⋅ 5
>853!L
+ (321 − 41) ⋅
211
1/3 ⋅ π ⋅ 6
b)
5
(
r Σ = r {sb•okf
5
U2
U
− 3
U − Ug
211
211
+ 2
>
2g
2
2
eπ ⋅ α
eπ ⋅ D23
)23 + (rqsfmb{ )
5
5
853
594
−
211
X
853 − 414
211
+
>8924/15!,!6:96/66!>!248:9/7!
rΣ =
2
2
n
1/3π ⋅ 32/8
1/3π ⋅ 6
9/44/!Horizontalna cev, spoqa{weg pre~nika!e>41!nn!i du`ine!M>6!n, se hladi popre~nom strujom
vode sredwe temperature!ug>21pD/!Voda struji brzinom!x>3!n0t-!pod napadnim uglom od!β>71p/
Temperatura spoqa{we povr{i cevi iznosi!u{>91pD/!Odrediti toplotni protok konvekcijom sa cevi
na vodu.
u{
voda
ug
α
β
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
⋅
R=
strana 45
u{ − ug
⋅ M >!///
2
e⋅π⋅α
fizi~ki parametri za vodu na temperaturi!ug>21pD
X
n3
λg!>!68/5!/21.3!!
!
!υg!>!2/417!/21−7!
t
nL
1. korak:
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>!et!>!41!nn
3. korak:
potrebni kriterijumi sli~nosti
x ⋅ mL
3 ⋅ 1/14
>
>5/6:!/215
(prelazni re`im strujawa)
νg
2/417 ⋅ 21 −7
Qs{!> (Qs )U{ =91p D >3/32
Qsg!> (Qs )Ug =21p D >:/63-
Sfg!>!
4. korak:
konstante u kriterijalnoj jedna~ini:
Qs
n>1/7-!!!o>1/48-!!!q>1-!!!εU!> G
Qs{
/!
/!
D>1/37! εβ>1/37! 1/:4>1/35
5. korak:
1/36
1/36
:/63
>
>2/55
3/32
)β>71pD! ⇒
εβ>!1/:4*
izra~unavawe Nuseltovog broja
Ovg!>!1/35/!)!5/6:!/215!*1/7!/!)!:/63!*1/48!/!2/55!>!5:9/7
6. korak:
izra~unavawe koeficijenta prelaza toplote!)α*
α!>!Ovg! ⋅
⋅
R=
λg
68/5 ⋅ 21 .3
X
>5:9/7! ⋅
>!:46:/:!
mL
1/14
n3L
91 − 21
2
⋅ 6 >419/86!lX
1/14 ⋅ π ⋅ :46:/: ⋅ 21 −4
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 46
9/45/!Suvozasi}ena vodena para!)u>291pD*!transportuje se kroz parovod na rastojawe od!M>5!ln/!Parovod
je napravqen od ~eli~nih cevi!)λ2>61!X0nL*-!pre~nika!)∅>211091!nn*!i izolovan je slojem staklene
vune!)λ3>1/15!X0nL*!debqine!δ>:1!nn/!Pra}ewe atmosferskih uslova pokazalo je da:
- maksimalna brzina vetra koji duva normalno na parovod je x>21!n0t
- minimalna temperatura okolnog vazduha je!−21pD
U parovod treba ugraditi kondenzacione lonce na drugom!)3/*!i ~etvrtom!)5/*!kilometru. Ukoliko gubici
zra~ewem iznose!71&!od gubitaka konvekcijom, a koeficijent prelaza toplote sa strane pare koja se
kondenzuje du` celog cevovoda iznosi α2>:111!X0n3L!-!odrediti potreban kapacitet kondenzacionih
lonaca!)lh0t*/!Pri izra~unavawu koeficijenta prelaza toplote sa strane vazduha zanemariti popravku!εUtj. smatrati da je!εU>2
razmewen toplotni protok na prva dva kilometra!)M2>3111!n*;
⋅
⋅
⋅
⋅
R 2 = R
+ R
> 2/7 ⋅ R
qsfmb{ {sb•fokf
qsfmb{
Uqbsb − Uwb{evi
⋅
R 2 = 2/7 ⋅
e
e
2
2
2
2
+
mo 3 +
mo 4 +
e2π ⋅ α 2 3π ⋅ λ 2 e2 3π ⋅ λ 3 e 3 e 4 π ⋅ α 3
⋅ M2 >!///
fizi~ki parametri za vazduh!ob!ufnqfsbuvsj!ug>−21pD
n3
X
λg!>!3/47!/21−3!
νg!>!23/54/21−7!
nL
t
1. korak:
2. korak:
karakteristi~na du`ina ~vrste povr{i
mfl>e4>!1/39!n
3. korak:
izra~unavawe potrebnih kriterijuma sli~nosti
x ⋅ mL
21 ⋅ 1/39
>3/36!/216
!>
υg
23/54 ⋅ 21 −7
Qsg!> (Qs )Ug = −21p D >1/823
Sfg!>!
4. korak:
(turbulentno strujawe)
konstante u kriterijalnoj jedna~ini
D>1/134-
n>1/9-
o>1/5-
q>1
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
5. korak:
strana 47
izra~unavawe Nuseltovog broja
(
Ov g = 1/132 ⋅ 3/36 ⋅ 21 6
6. korak
)
1/9
⋅ (1/823)1/5 ⋅ (Hsg )1 ⋅ 2 >495/2
izra~unavawe koeficijenta prelaza toplote!)α*
α = Ov g ⋅
λg
3/47 ⋅ 21 −3
X
>43/5!!
!>! 495/2 ⋅
mfl
1/39
n3L
⋅
R 2 = 2/7 ⋅
⋅
⋅
R 2 = n2 ⋅ s
napomena:
291 + 21
⋅ 3111 >!
2
2
211
2
391
2
mo
mo
+
+
+
1/19π ⋅ :111 3π ⋅ 61 91 3π ⋅ 1/15 211 1/39π ⋅ 58/6
⇒
⋅
⋅
R 2 >!258!lX
⋅
258
lh
R2
>1/184!
>
n2 =
3126
t
s
Razmewena toplota na drugom delu cevovoda!)du`ine!M3>3!ln) je
identi~na kao na prvom delu cevovoda!)du`ine!M2>3!ln*-!pa je i
kapacitet drugog kondenzazcionog lonca jednak kapacitetu prvog
⋅
lh
kondenzacionog lonca-! n3 >1/184!
t
9/46/!Upravno na cev, spoqa{weg pre~nika!e>311!nn!i du`ine!M>9!n-!struji suv vazduh temperature
ug>−31pD-!pri!q>212/4!lQb/!Temperatura na spoqa{woj povr{i cevi je konstantna i iznosi u{>291pD/
Odrediti brzinu strujawa vazduha pri kojoj toplotni protok sa cevi na vazduh iznosi 31!lX/
⋅
u − ug
X
R
R= {
!>!31! 3
⋅M
⇒
α!>!
2
e ⋅ π ⋅ M ⋅ (u { − u g )
nL
e⋅π⋅α
X
31
α!>!
!>!31! 3
1/3 ⋅ π ⋅ 9 ⋅ (291 + 31)
nL
⋅
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 48
fizi~ki parametri za vazduh na!ug>−31pD
λg!>!3/39/21−3!!
X
!
nL
ml>1/3!n-!
!Qsg!>!1/827-!
α!>! Ovg ⋅
νg!>!23/8:/21−7!
!
Qs{!>!1/792-
λ
λ
= D ⋅ Sfn ⋅ Qso ⋅ Hsq ⋅ ε U ⋅
ml
ml
m
2
Sf!>! α ⋅ l ⋅
o
λ
D ⋅ Qsg ⋅ Hsgq ⋅ ε U
lh
n4
Qs
εU> G
QsB
1/36
>2
⇒
2
n
>!///
predpostavimo da je strujawe vazduha oko cevi turbulentno tj. da va`i:
3/216!=!Sfg!=!2/218
⇒
D!>1/134/!εβ!>1/134-
n!>1/9-!!!o!>1/5-!!!q>1
2
1/9
1/3
2
>9/53/215
⋅
Sf> 31 ⋅
−3
1/5
3
/
39
⋅
21
1
/
134
⋅
1
/
827
⋅
2
pretpostavka neta~na!"
predpostavimo da je strujawe vazduha oko cevi preobra`ajno tj. da va`i:
2/214!=!Sfg!=!3/216
⇒
D!>1/37/!εβ!>1/37-!!!n!>1/7-!!!o!>1/48-!!!q>1
2
1/7
1/3
2
>7/49/215
⋅
Sf> 31 ⋅
−3
1/48
3
/
39
⋅
21
1
/
37
⋅
1
/
827
⋅
2
x!>!
pretpostavka ta~na!"
Sf g ⋅ υ g
n
7/49 ⋅ 21 5 ⋅ 23/8: ⋅ 21 −7
>5/19!
>!
1/3
ml
t
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
zadatak za ve`bawe:
strana 49
)9/47/*
9/47/!Kroz cev od ner|aju}eg ~elika!)!λ>28!X0nL!*-!pre~nika!∅>62y3/:!nn!i du`ine!M>3/6!n-!struji
voda!ux!>91pD>dpotu-!!sredwom brzinom!xx>2!n0t. Upravno na cev struji vazduh sredwe temperature
ug>31pD>dpotu-!sredwom brzinom!xg>3!n0t/!Odrediti toplotni protok sa vode na vazduh kao i
temperaturu spoqa{we povr{i cevi ( smatrati da je!εUv!>!εUt!>!2*/
⋅
re{ewe:
u!>!8:/6pD
R >655/3!X-
9/48/!Dve kvadratne plo~e stranica du`ine!b>2!n obrzauju ravnu povr{ (zanemarqive debqine) du`
koje brzinom!x>2!n0t!struji suv vazduh!)u>21pD-!q>2!cbs*/!Odrediti koliko se toplote preda vazduhu
za slede}a tri slu~aja:
a) obe plo~e su stalne temperature u{>231pD
b) prva plo~a je stalne temperature u{>231pD, a druga je adijabatski izolovana
d* prva plo~a je adijabatski izolovana, a druga plo~a je stalne temperature u{>231pD
b
b
b
b*
ml!>!
Sfls ⋅ ν g
6 ⋅ 21 6 ⋅ 25/27 ⋅ 21 −7
>
>7/:!n
x
2
kako je M>b,b>!3!n! ≤ !mls strujawe vazduha du` cele plo~e je laminarno
napomena:
υg!>25/27!/21−7
n3
t
(vazduh na temperaturi!ug!>21pD)
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 50
b*
⋅
R 1−3b >!
u{ − ug
⋅ b ⋅ b ⋅ 3 >///
2
α 1−3b
fizi~ki parametri za vazduh na temperaturi!ug!>21pD
n3
X
λg!>!3/62!/21−3!
υg!>25/27!/21−7
t
nL
1. korak:
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>3b!>3!n
3. korak:
potrebni kriterijumi sli~nosti
x ⋅ ml
2⋅ 3
>2/52!/216
=
υg
25/27 ⋅ 21 −7
Qsg!> (Qs )Ug =21p D >1/816Qs{!> (Qs )U{ =231p D >1/797
Sfg!>
4. korak:
konstante u kriterijalnoj jedna~ini
Qs
n>1/6-!!!o>1/44-!!!q>1 -!!!εU!>! G
Qs{
5. korak:
1/36
1/816
=
1/797
1/36
>2-!!!D>1/775
izra~unavawe Nuseltovog broja
Ovg!>!1/775/!)!2/52!/216!*1/6!/!)!1/816!*1/44!/!2!>333/28
6. korak:
izra~unavawe koeficijenta prelaza toplote!)α*
α1−3b!>Ovg!/
⋅
R 1−3b >!
λg
X
3/62 ⋅ 21 .3
>3/8:! 3
>333/28/!
3
mL
nL
231 − 21
⋅ 2 ⋅ 2 ⋅ 3 >!724/9!X
2
3/8:
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 51
b)
!
⋅
R 1−b >!
u{ − ug
⋅ b ⋅ b >///
2
α 1 −b
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>b!>2!n
3. korak:
potrebni kriterijumi sli~nosti
x ⋅ ml
2⋅ 2
>1/8!/216
=
Sfg!>
−7
υg
25/27 ⋅ 21
5. korak:
izra~unavawe Nuseltovog broja
Ovg!>!1/775/!)!1/8!/216!*1/6!/!)!1/816!*1/44!/!2!>!267/65
6. korak:
izra~unavawe koeficijenta prelaza toplote!)α*
λg
3/62⋅ 21.3
X
>!4/:4! 3
>267/65/!
mL
2
nL
⋅
231 − 21
⋅ 2 ⋅ 2>!543/4!X
R 1−b >!
2
4/:4
α1−b!>!Ovg!/
napomena:
prikazani su samo oni koraci koji nisu isti kao pod a)
c)
⋅
⋅
⋅
R b−3b !>! R 1−3b !−! R 1−b !>!724/9!−!534/4!>292/6!X
napomena:
⋅
R b−3b >!
zadatak pod c) se mo`e re{iti i na slede}i na~in:
u{ − ug
231 − 21
⋅ 2 ⋅ 2>292/6!X
⋅ b ⋅ (M − b) >///>
2
2
α b−M
2/76
(Ov g )b−M > D ⋅ [(Sf g )n1−M − (Sf g )n1−b ]⋅ Qsgo ⋅ ε u
(Ov g )b−M >!1/775/! (2/52 ⋅ 21 6 )
1/6
αb−M!>! (Ov g )b−M /
(
− 1/8 ⋅ 21 6
)
1/6
/!)!1/816!*1/44!/!2>76/74
λg
X
3/62⋅ 21.3
>!2/76! 3
>76/74/!
2
mL
nL
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 52
9/49/ Vazduh temperature!ug>31pD-!struji sredwom brzinom!3/6!n0t preko ravne plo~e du`ine!b>6!n!i
{irine!c>2/6!n. Povr{i plo~e se odr`avaju na stalnoj temperaturi od u{>:1pD/!Odrediti toplotni
protok sa plo~e na vazduh u laminarnom delu strujawa, turbulentnom delu strujawa kao i ukupni du`
cele plo~e.
mls
vazduh
c
b
Sfls ⋅ ν g
6 ⋅ 21 6 ⋅ 26/17 ⋅ 21 −7
>4!n
>
3/6
x
kako je M>b>!6!n!?!mls strujawe vazduha na du`ini mls je laminarno a na
du`ini M−mls turbulentno
n3
napomena:
υg!>26/17!/21−7
(vazduh na temperaturi!ug!>31pD)
t
mls!>!
u{ − ug
⋅ mls ⋅ c >///
2
α mbn
1. korak:
fizi~ki parametri zavazduh na temperaturi!ug!>!31pD
n3
X
λg!>!3/6:!/21−3!
-!!υg!>26/17!/21−7!
t
nL
⋅
R mbn!>!
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>!mls!>4!n
3. korak:
potrebni kriterijumi sli~nosti
Sfg!>Sfls>6!/216
Qsg!> (Qs )Ug =31p D >1/814-
Qs{!> (Qs )U{ =:1p D >1/7:
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
4. korak:
strana 53
konstante u kriterijalnoj jedna~ini
Qs
n>1/6-!!!o>1/44-!!!q>1 -!!!εU!>! G
Qs{
5. korak:
1/36
1/814
=
1/7:
1/36
>2-!!!D>1/775
izra~unavawe Nuseltovog broja
Ovg!>!1/775/!)!6!/216!*1/6!/!)!1/814!*1/44!/!2!>!528/:4
izra~unavawe koeficijenta prelaza toplote!)αmbn*
6. korak:
3/6: ⋅ 21.3
λg
X
>!4/72! 3
>528/:4!
4
mL
nL
⋅
:1 − 31
R mbn!>!
⋅ 4 ⋅ 2/6 >2248/3!X
2
4/72
αmbn!>!Ovg!/
⋅
R uvs!>!
u{ − ug
⋅ (b − mls ) ⋅ c >///
2
α uvs
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>!b!−!mls!>3!n
3. korak:
potrebni kriterijumi sli~nosti
(Sf g )mls >Sfls>6!/216-! (Sf g ) M > x ⋅ M =
υg
4. korak:
26/17 ⋅ 21 −7
>9/4!/216
konstante u kriterijalnoj jedna~ini
Qs
n>1/9-!!!o>1/54-!!!q>1 -!!!εU!>! G
Qs{
5. korak:
3/6 ⋅ 6
1/36
1/814
=
1/7:
1/36
>2-!!!D>1/148
izra~unavawe Nuseltovog broja
(Ov g )mls −M > D ⋅ [(Sf g )nM − (Sf g )nls ]⋅ Qsgo ⋅ ε u
(Ov g )mls −M >!1/148/! (9/4 ⋅ 21 6 )
1/9
(
− 6 ⋅ 21 6
)
1/9
/!)!1/814!*1/54!/!2>687/24
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 54
izra~unavawe koeficijenta prelaza toplote!)αuvs*
6. korak:
αuvs> (Ov g )m
ls −M
⋅
R uvs!>!
/
λg
X
3/6: ⋅ 21 .3
>!8/57! 3
>687/24/!
3
mL
nL
:1 − 31
⋅ (6 − 4 ) ⋅ 2/6 >2677/9!X
2
8/57
⋅
⋅
⋅
+ R uvs
>2248/3!,!2677/9!>3815!X
R 1−m = R mbn
1−mls
mls −M
9/4:/!Vertikalnu plo~u (zanemarqive debqine), visine!i>3/5!n!i {irine!b>1/9!n!sa obe strane (u
pravcu kra}e strane) opstrujava vazduh temperature!ug>31pD. Toplotni protok konvekcijom sa plo~e na
vazduh iznosi!6!lX/!Odrediti sredwu brzinu strujawa vazduha, tako da se temperatura na povr{ima
plo~e odr`ava konstantnom i iznosi u{>81pD, smatraju}i da je strujawe vazduha turbulentno po celoj
plo~i.
b
i
⋅
u − ug
R >! {
⋅b⋅i⋅3
⇒
2
α
X
6111
α!>!
!>37! 3
(81 − 31) ⋅ 1/9 ⋅ 3/5 ⋅ 3
nL
vazduh
⋅
R
α!>!
!>
(u { − u g ) ⋅ b ⋅ i ⋅ 3
fizi~ki parametri zavazduh na temperaturi!ug!>!31pD
n3
X
λg!>!3/6:!/21−3!
-!!υg!>26/17!/21−7!
t
nL
1. korak:
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 55
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>!b!>1/9!n
3. korak:
potrebni kriterijumi sli~nosti
Qsg!> (Qs )Ug =31p D >1/8144. korak:
Qs{!> (Qs )U{ =81p D >1/7:5
konstante u kriterijalnoj jedna~ini
Qs
n>1/9-!!!o>1/54-!!!q>1 -!!!εU!>! G
Qs{
6. korak:
1/36
1/814
=
1/7:5
1/36
>2-!!!D>1/148
izra~unavawe Nuseltovog broja
Ov g = α ⋅
5. korak:
ml
1/9
= 37 ⋅
>914/1:
λg
3/6: ⋅ 21 −3
izra~unavawe Rejnoldsovog broja
o
q
Ovg> D ⋅ Sfn
g ⋅ Qsg ⋅ Hsg ⋅ ε U
⇒
Ov g
Sfg!>!
D ⋅ Qs o ⋅ Hs q ⋅ ε
U
g
g
2
n
2
1/9
914/1:
> 4/295 ⋅ 21 6
Sfg!>!
1/148 ⋅ 1/814 1/54 ⋅ 2 ⋅ 2
x!>!
Sf g ⋅ υ g
n
4/295 ⋅ 21 6 ⋅ 26/17 ⋅ 21 −7
!>!7!
=
ml
1/9
t
zadatak za ve`bawe:
)9/51/*
9/51/!Vaqanu, vertikalnu bakarnu plo~u, visine!3/3!n!i {irine!1/:!m sa obe strane opstrujava vazduh
sredwe temperature!31pD-!sredwom brzinom!7!n0t. Sredwe temperature obe povr{i plo~e iznose
81pD-!a zidova velike prostorije u kojoj se plo~a nalazi!31pD/!Odrediti toplotni protok koji se
odvodi sa plo~e (debqinu plo~e zanemariti) ako se strujawe vr{i u pravcu kra}e strane.
⋅
⋅
⋅
re{ewe:!!!!!!! R > R qsfmb{ , R {sb•fokf >!3126/7!,!:23/4!>!3:83/:!X
∑
23
24
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 56
9/52/!Iznad horizontalne ravne betonske plo~e, du`ine M>3!n, toplotno izolovane sa dowe strane,
suv vazduh stawa (q>2!cbs-!Ug>394!L) proti~e brzinom x>4!n0t. Ako se pod dejstvom toplotnog
zra~ewa, na gorwoj povr{i plo~e ustali temperatura U{>434!L, odrediti povr{inski toplotni
protok (toplotni fluks) tog zra~ewa i grafi~ki predstaviti raspored temperatura u betonskoj plo~i
i okolnom vazduhu.
rep{sb•fop
Ug
U{
vazduh, x>4!n0t
rtpqtuw/{sb•fokf
α
rsfgmflupwbop
U{>dpotu
M
toplotni bilans ozra~ene povr{i:
r ep{sb•fop = rsfgmflupwbop + r bqtpscpwbop
r ep{sb•fop = r ep{sb•fop ⋅ (2 − ε ) +
5
U{
2 U − Ug 211
r ep{sb•fop > ⋅ {
+
2
ε 2
α
ε ⋅ Dd
U{
U{ − U3 U2 − Ug 211
+
+
2
2
δ
ε ⋅ Dd
λ
α
5
>///
ε!>!L!/!εo!>!1/:9!/!1/:5!>!1/:3
α!>!@
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 57
fizi~ki parametri za vazduh na temperaturi!ug!>21pD
n3
X
λg!>!3/62!/21−3!
υg!>25/27!/21−7
t
nL
1. korak:
2. korak:
karakteristi~na du`ina ~vrste povr{i
ml!>M!>3!n
3. korak:
Sfg!>
potrebni kriterijumi sli~nosti
x ⋅ ml
4⋅3
>5/35!/216
=
υg
25/27 ⋅ 21 −7
Qsg!> (Qs )Ug =21p D >1/8164. korak:
Qs{!> (Qs )U{ =231p D >1/7:9
konstante u kriterijalnoj jedna~ini
Qs
n>1/6-!!!o>1/44-!!!q>1 -!!!εU!>! G
Qs{
5. korak:
1/36
1/816
=
1/7:9
1/36
>2-!!!D>1/775
izra~unavawe Nuseltovog broja
Ovg!>!1/775/!)!5/35!/216!*1/6!/!)!1/816!*1/44!/!2!>496/37
6. korak:
izra~unavawe koeficijenta prelaza toplote!)α*
α!>Ovg!/
λg
X
3/62 ⋅ 21 .3
>5/95! 3
>496/37/!
3
mL
nL
5
U{
2 U − Ug 211
r ep{sb•fop > ⋅ {
+
2
ε 2
α
ε ⋅ Dd
5
434
2 434 − 394
X
211
⋅
+
>
>938/7! 3
2
2
n
1/:3
5/95
1/:3 ⋅ 6/78
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 58
⋅
9/53/!U istosmernom razmewiva~u toplote tipa cev u cevi!zagreva se! n f>2611!lh0i!etanola!od
U2>21pD!to!U3>41pD/!Grejni fluid je suvozasi}ena vodena para!q>1/23!cbs!koja se u procesu razmene
toplte sa etanolom potpuno kondenzuje. Etanol proti~e kroz cev a para se kondenzuje u anularnom
prostoru. Unutra{wi pre~nik unutra{we cevi iznosi!e>211!nn/!Zanemaruju}i toplotni otpor
prelaza sa pare na cev, toplotni otpor provo|ewa kroz cev kao i toplotne gubitke u okolinu odrediti:
a) maseni protok grejne pare
b) du`inu cevi
b*
prvi zakon termodinamike za proces u razmewiva~u toplote:
⋅
⋅
⋅
⋅
⋅
⋅
⇒
R 23!>!∆ I 23!,! X U23
⋅
⋅
I 2!>! I 3
⋅
nq ⋅ i2 + nf ⋅ d qf ⋅ U2 = nq ⋅ i3 + nf ⋅ d qf ⋅ U3
⋅
⋅
nq =
nf ⋅ d qf ⋅ (U3 − U2 )
i2 − i3
lK
lh
lK
!i3>318!
lh
!i2>36:2
dqf>3/56!
lK
lhL
2611
⋅ 3/56 ⋅ (41 − 21)
lh
>9/67!/21−3!
> 4711
3495
t
)i′′-!!q>1/23!cbs*
)i′-!!!q>1/23!cbs*
specifi~ni toplotni kapacitet etanola odre|en za sredwu
temperaturu etanola: Uf!>!
21 + 41
= 31pD
3
b)
⋅
R sb{
∆UTS
=
⋅M
2
l
⋅
⇒
R sb{
M!>!
!>///
l ⋅ ∆Uts
⋅
R sb{!−!interno razmewena toplota u razmewiva~u izme|u pare i etanola
l!−!koeficijent prolaza toplote sa pare na etanol
∆Uts!−!sredwa logaritamska razlika temperatura izme|u pare i etanola
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 59
L
para-!i2
para-!i3
⋅
R sb{
etanom-!U2
etanol-!U3
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ograni~enom
⋅
konturom K:
⋅
⋅
⋅
R sb{!>!∆ I 23!,! X U23
⋅
R sb{!> nq ⋅ (i2 − i3 ) = 9/67 ⋅ 21 −3 ⋅ (36:2 − 318) >315/18!lX
U
∆Unby!>5:/26!−!21>!4:/26pD
∆Unjo!>5:/26!−!41!>!2:/26pD
4:/26 − 2:/26
= 39/4pD
∆Uts!>
4:/26
mo
2:/26
para
5:/26
5:/26
41
etanom
21
M
e
2
2
2
2
mo 3 +
=
+
l e3 π ⋅ α Q 3π ⋅ λ e3 e2π ⋅ α f
⇒
l> e2 ⋅ π ⋅ α >///
2
!!!!−!!toplotni otpor prelaza sa strane pare, zanemaren uzadatku
e3 π ⋅ α Q
e
2
⋅ mo 3 !−!toplptni otpor provo|ewa kroz cev, zanemaren u zadatku
3π ⋅ λ
e3
2
!!!!−!!toplotni otpor prelaza sa strane etanola
e2π ⋅ α f
αf!>!@
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 60
1. korak:
fizi~ki parametri za etanol odre|eni za sredwu temperaturu
U + Uf3 21 + 41
=
>31pD
etanola u cevi: Ufts = f2
3
3
lh
X
λg!>!1/294!!
!
!
ρg!>!89:! 4
nL
n
lK
dqg!>!3/56!
µg!>!2/2:!/21−4! Qb ⋅ t
lhL
2. korak:
karakteristi~na du`ina ~vrste povr{i
e23 ⋅ π
B
ml!>! 5 ⋅ = 5 ⋅ 5 >e2>211!nn
P
e2 ⋅ π
3. korak:
Qsg!>
Qs{!>
potrebni kriterijumi sli~nosti
dqg ⋅ µ g
λg
d q{ ⋅ µ {
λ{
>
3/56 ⋅ 21 4 ⋅ 2/2: ⋅ 21 −4
>26/:4
1/294
>///>
3/92 ⋅ 21 4 ⋅ 1/7:6 ⋅ 21 −4
>21/:8
1/289
µ{-!λ{-!dq{
fizi~ki parametri etanola na temperaturi!U{>5:/26pD
lK
X
µ{!>!1/7:6!/21−4! Qb ⋅ t !
λ{!>!1/289!
!
dq{!>!3/92!
lhL
nL
Sfg!>
ρ g ⋅ x ⋅ mL
89: ⋅ 7/8 ⋅ 21 −3 ⋅ 1/2
>!5553/4!!!!!prelazni re`im strujawa
>!///>
µg
2/2: ⋅ 21 −4
2611
4711 >7/8!/21−3! n
x!>!
>
3
t
ρ g ⋅ e2 ⋅ π 89: ⋅ 1/23 ⋅ π
⋅
5 ⋅ nF
4. korak:
5⋅
konstante u kriterijalnoj jedna~ini
Qs
n>1-!!!o>1/54-!!!q>1-!!!εU!>! g
Qs{
L1!>g!)!Sf!*>!24/9
5. korak:
1/36
26/:4
>
21/:8
1/36
>2/2-!!!!D>L1>//!/>24/9
izra~unavawe Nuseltovog broja
Ovg!>24/9!/)!26/:4!*1/54!/!2/2!>!61
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
6. korak:
strana 61
izra~unavawe koeficijenta prelaza toplote!)α*
αf!>!Ovg! ⋅
λg
X
1/294
>61! ⋅
>!:2/6 3
.
4
mL
211 ⋅ 21
nL
l> e2 ⋅ π ⋅ α f> 1/2 ⋅ π ⋅ :2/4 >39/86!
X
nL
⋅
R sb{
315/18
>!362!n
M!>!
!>
l ⋅ ∆Uts
39/86 ⋅ 21 .4 ⋅ 39/4
9/54/!U kqu~alu vodu pritiska!q>21!cbs-!koja pri konstantnom pritisku isparava u proto~nom kotlu,
potopqeno je 31 pravih cevi, unutra{weg pre~nika!e>:6!nn/!Kroz cev !)pri konstantnom pritisku,
q>2!cbs*-!brzinom!xg>7!n0t-!struji dimni gas!)sDP3>1/24-!sI3P>1/22-!sO3>1/87*/!Temperatura gasa na
ulazu u cevi je!571pD-!a na izlazu iz wih!371pD/!Ako se zanemari toplotni otpor provo|ewa kroz
zidove cevi kao i toplotni otpor prelaza sa cevi na kqu~alu vodu odrediti:
a) koli~inu vode koja ispari u kotlu )lh0t*
b) !potrebnu du`inu cevi
b*
prvi zakon termodinamike za proces u razmewiva~u toplote:
⋅
⋅
⋅
⇒
R 23!>!∆ I 23!,! X U23
⋅
⋅
⋅
⋅
⋅
I 2!>! I 3
⋅
n x ⋅ i2 + neh ⋅ d qeh ⋅ U2 = n x ⋅ i3 + neh ⋅ d qeh ⋅ U3
⋅
nx =
⋅
neh ⋅ d qeh ⋅ (U2 − U3 )
⋅
i3 − i2
n eh = ρ eh ⋅ x ⋅
>///>
lh
1/59 ⋅ 2/25 ⋅ (571 − 371)
>6/54!/21−3!
3126/4
t
lh
e3 π
1/1:6 3 ⋅ π
⋅ o = 1/673 ⋅ 7 ⋅
⋅ 31 >1/59!
5
5
t
lh
lK
!
specifi~ni toplotni kapacitet i gustina dimnnog ρeh>1/673! 4
lhL
n
gasa odre|eni za sredwu temperaturu dimnog gasa:
Uf!>
571 + 371
= 471pD
3
lK
!i2>!873/8
)i′-!!q>21!cbs*
lh
lK
!i3>3889!
)i′′-!!!q>21!cbs*
lh
dqeh>2/25!
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 62
b)
⋅
R sb{ =
∆UTS
⋅M
2
l
⋅
⇒
M!>!
R sb{
!>///
l ⋅ ∆Uts
⋅
R sb{!−!interno razmewena toplota u razmewiva~u izme|u dimnog gasa i vode
l!−!koeficijent prolaza toplote sa dimnog gasa na vodu
∆Uts!−!sredwa logaritamska razlika temperatura izme|u dimnog gasa i vode
L
dimni gas gas-!U2
dimni gas-!U3
⋅
R sb{
voda-!i2
para-!i3
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
ograni~enom konturom K:
⋅
⋅
⋅
R sb{!>!∆ I 23!,! X U23
⋅
R sb{!> n eh ⋅ d qeh (U2 − U3 ) = 1/59 ⋅ 2/25 ⋅ (571 − 371) >21:/55!lX
∆Unby!>571!−291>!391pD
∆Unjo!>371!−!291!>!91pD
U
571
dimni gas
391 − 91
= 26:/76pD
∆Uts!>
391
mo
91
291
voda
371
291
M
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 63
e
2
2
2
2
=
+
mo 3 +
l e3 π ⋅ α x 3π ⋅ λ e3 e2π ⋅ α eh
⇒
l> e2 ⋅ π ⋅ α eh>///
2
!!!!−!!toplotni otpor prelaza sa strane vode, zanemaren uzadatku
e3 π ⋅ α x
e
2
⋅ mo 3 !−!toplptni otpor provo|ewa kroz cev, zanemaren u zadatku
3π ⋅ λ
e3
2
!!!!−!!toplotni otpor prelaza sa strane dimnog gasa
e2π ⋅ α eh
αeh!>!@
1. korak:
fizi~ki parametri za dimni gas odre|eni za sredwu temperaturu
Ueh2 + Ueh3 571 + 371
=
>471pD
dimnog gasa: Uehts =
3
3
lh
X
λg!>!6/47!/21−3!!
! !
ρg!>!1/673! 4
nL
n
lK
dqg!>!2/24:!
µg!>!41/4!/21−7! Qb ⋅ t
lhL
2. korak:
karakteristi~na du`ina ~vrste povr{i
e23 ⋅ π
B
ml!>! 5 ⋅ = 5 ⋅ 5 >e2>:6!nn
P
e2 ⋅ π
3. korak:
potrebni kriterijumi sli~nosti
Qsg!> (Qs )Ug =471p D >1/75Sfg!>
Qs{!> (Qs )U{ =291p D >1/78
ρ g ⋅ x ⋅ mL
1/673 ⋅ 7 ⋅ 1/1:6
>21683
!>
µg
41/4 ⋅ 21 −7
4. korak:
konstante u kriterijalnoj jedna~ini
U
564
n>1-!!!o>1/54-!!!q>1-!εU!>2/38!−!1/38 ⋅ { >2/38!−!1/38 ⋅
>2/188Ug
744
predpostavimo !
M
?61!
mL
⇒
εM>!2
⇒
D>1/132
D>1/132!/!εM!>1/132
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
5. korak:
strana 64
izra~unavawe Nuseltovog broja
Ovg!>1/132/!)21683!*1/9!/)!1/75!*1/54!/2/188>41/:4
6. korak:
izra~unavawe koeficijenta prelaza toplote!)α*
αeh!>!Ovg! ⋅
λg
X
6/47 ⋅ 21 .3
>41/:4 ⋅
>28/56! 3
.4
mL
:6 ⋅ 21
nL
l> e2 ⋅ π ⋅ α eh> 1/1:6 ⋅ π ⋅ 28/56 >!6/32!
X
nL
⋅
R
21:/55
=
M!>!
>!7/7!n
l ⋅ ∆Uts ⋅ o 6/32 ⋅ 21 −4 ⋅ 26:/76 ⋅ 31
provera pretpostavke iz 4. koraka:!
M
?61
mL
M
7/7
>7:/6!?61!!
>
1/1:6
mL
pretpostavka je ta~na
⇒
⋅
9/55/!U razmewiva~u toplote sa suprotnosmerim tokom fluida zagreva W >7111!n40i!se )!pri!q>212/4
lQb-!u>1pD*!vazduha (ideala gas) od po~ete temperature!U2>!51pD!do krajwe temperature!U3>91pDpomo}u vode temperature!Ux2>:1pD. Procewena vrednost koeficijenta prolaza toplote iznosi!l>611
X0)n3L). Ukupna povr{ina za razmenu toplote iznosi!B>29!n3/!Odrediti maseni protok vode!)lh0t*/
voda-!Ux2
voda-!Ux3
⋅
R sb{
vazduh-!U2
vazduh-!U3
L
⋅
n wb{evi
⋅
q⋅W
=
>
Sh ⋅ U
7111
4711 >3/27! lh
398 ⋅ 384
t
212/4 ⋅ 214 ⋅
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
zbirka zadataka iz termodinamike
strana 65
prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu
⋅
⋅
⋅
⋅
R sb{!>!∆ I 23!,! X U23
ograni~enom konturom K:
⋅
R sb{!> n w ⋅ d qw (U3 − U2 ) = 3/27 ⋅ 2 ⋅ (91 − 51) >97/5!lX
⋅
R sb{
⋅
∆UTS
=
⋅B
2
l
∆Uts =
97/5
R sb{
∆Uts !>!
>
!>!:/7pD
l ⋅ B 1/6 ⋅ 29
⇒
(:1 − 91) − (Ux3 − 51)
:1 − 91
mo
Ux3 − 51
:1pD
voda
91pD
Ux3!>!@
@
vazduh
51pD
pretpostavimo!ux3>71pD
⇒
∆Uts >25/5pD
(nije ta~no!)
pretpostavimo!ux3>59pD
⇒
∆Uts >9/:pD
(nije ta~no!)
pretpostavimo!ux3>5:pD
⇒
∆Uts >:/6pD
(ta~no!)
prvi zakon termodinamike za proces u razmewiva~u toplote:
⋅
⋅
⋅
⋅
⋅
⋅
⇒
R 23!>!∆ I 23!,! X U23
⋅
⋅
I 2!>! I 3
⋅
n x ⋅ i x2 + n w ⋅ d qw ⋅ U2 = n x ⋅ i x3 + n w ⋅ d qw ⋅ U3
⋅
nx =
⋅
n w ⋅ d qw ⋅ (U3 − U2 )
i x2 − i x3
lK
lh
lK
ix3>31:/4!
lh
ix2>488/1!
>
lh
3/27 ⋅ 2 ⋅ (91 − 51)
>1/6!
488 − 31:/4
t
)q>2!cbs-!Ux2>:1pD*
)q>2!cbs-!Ux3>5:pD*
dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv
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