MAHALAKSHMI
ENGINEERING COLLEGE
TIRUCHIRAPALLI-621213.
Department: Mechanical
Semester: III
Subject Code: ME2202
Subject Name: ENGG. THERMODYNAMICS
UNIT – II
SECOND LAW, ENTROPY &AVAILABILITY
1) A reversible heat pump is used to maintain a temperature of 0oC is a refrigerator
when it rejects the heat to surroundings at 27oC If the heat removal rate from the
Refrigerator is 1500 KJ/min. determine the C.O.P of the machine and work input
required. If the required input to run the pump is developed by a reversible
engine which receives heat at 400oC and reject heat to atmosphere, then
determine the overall C.O.P of the system
(AUC DEC 2011)
GIVEN:
REVERSIBLE HEAT PUMP
Temp high (T1) = 27oC +273
=300K
Temp low (T2) = 0oC +273
= 273K
Heat removal rate (Q) = 1500/60 KJ/min
= 25 KJ/ sec
REVERSIBLE ENGINE
Temp (T3)
= 400oC + 273
= 673K
TO FIND:
(i) C.O.P
(ii) work input
(iii) over all C.O.P of the system
SOLUTION:
STEP: 1 Reversible heat pump
C.O.P machine =
THigh
=
THigh - TLow
300
300 – 273
L.VIJAYAKUMAR /A.P-MECH
Page 1
C.O.P(machine) = 11.1111
Thigh - Tlow
Work input(machine) = Q1
Thigh
300 – 273
= 25
300
Work input(machine) = 2.25 KW
STEP:2
Reversible heat engine
C.O.P engine
= Q1/Q2
Where
Q3/T3 = Q4/T4
Q4 + W = Q4/T4
T3
Q4 +2.25 = Q4/300
673
(Q 4 + 2.25)300 = Q 4 × 673
300Q4 + 675 = Q 4 × 673
675 = 673Q 4 – 300Q4
675 = 373 Q 4
Q4 = 675/373
Q4 = 1.8096 KJ/S
Q3 = Q4 + W
Q3 = 1.8096 + 2.25
L.VIJAYAKUMAR /A.P-MECH
Page 2
Q3 = 4.0596 KJ/S
COP engine = Q1/Q3
= 25 / 4.0596
COP engine = 6.1582
STEP:3
To find over all COP
C.O.Pov erall
= Q2 + Q4
Q3
Where
Q2 = Q1 + W
Q2
= 25 + 2.25
Q2 = 27.25 KJ/S
C.O.P ov erall = Q2 + Q4
Q3
= 27.25 + 1.8096
C.O.P ov erall
L.VIJAYAKUMAR /A.P-MECH
4.0596
= 7.1582
Page 3
2. A reversible engine is supplied with heat from two constant temp source 900K and 600K
and reject heat at const temperature sink at 300K if the engine execute a number of
complete cycles while developing 70 KW, and rejecting 3200 KJ Of heat per min.
determine heat supplied by each source per mine and efficiency of the engine. (AUC
DEC 2011)
GIVEN:
Reversible heat engine
TempHigh (T1)
Temp (T2)
Temp (T3)
Work (W)
Heat rejected
= 900K
= 600K
= 300K
= 70 KJ/ Sec (or) KW
= 3200 KJ/min = 63.33KJ/s
60
TO FIND:
a)
b)
Heat supplied by each source per min
Efficiency of the engine
SOLUTION:`
STEP:1
Total heat supplied (Q) = work + heat rejected
= 123.333 KJ/sec
STEP:2
Q1 × ŋ1
+
(Q – Q1) ᵑ
2
=w
Where
ŋ1 = T
H
- Tlow
× 100
TH
ŋ1= 900 – 300 × 100
900
L.VIJAYAKUMAR /A.P-MECH
Page 4
ŋ1= 0.6666 × 100
ŋ1= 66.6666%
Where
Ŋ2 = T
2
– T3
× 100
T2
Ŋ2 = 600 – 300
600
Ŋ2 = 0.5 × 100
Ŋ2 = 50%
Q1 × Ŋ 1
Q1
+
(Q – Q1) Ŋ2 = w
× 0.6666 + ( 123.3333 –Q ) 0.5
1
=70
0.6666 Q1 + 61.666 – 0.5 Q1 = 70
0.6666 Q1 – 0.5 Q1
0.1666 Q1
= 70 – 61.6666
= 8.3333
Q1 = 8.3333
0.1666
Q1 = 50.020 KJ/S
STEP:3
Q2 = total heat supplied – heat rejected
Q2 = 123.3333 – 50.020
Q2 = 73.3131 KJ/S
L.VIJAYAKUMAR /A.P-MECH
Page 5
2) Two reversible heat engine A and B are arranged in series. Engine ‘A’ rejecting
heat directly to engine ‘B’ receives 200 KJ at a temp of 421oC from a hot source,
while engine ‘B’ is in communication with a cold sink at a temp 5oC. if the work
out A is twice that B
(AUC DEC 2012)
Find
a) The intermediate temp b/w A and B
b) The efficiency of each engine
c) The heat rejected to the cold sink
GIVEN:
Reversible heat engine series arrange
Temp (T1) = 4210C +273 = 694K
Temp ( T2) = 50C + 273 = 278K
Work out ‘A’ is the twice that ‘B’ (W A) = 2 × W B
(Q1) = 200 KJ
TO FIND:
a) The intermediate temp b/w A and B (T2)=?
b) The efficiency of each engine (ᵑ A) = ? , (ᵑ B)
c) The heat rejected to the cold sink (Q 2) = ? , Q3
SOLUTION:
STEP:1
T2 = T1 + 2 T3
3
T2 = 694 + (2 ×278)
3
T2 = 416.666K
STEP:2
Ŋa= T – T
1
2
× 100
T1
L.VIJAYAKUMAR /A.P-MECH
Page 6
Ŋa= 694 – 416.666
× 100
694
Ŋa = 0.3996 × 100
Ŋa = 39.96%
Ŋb = T
2
– T3
T2
= 416.6666 – 278
416.6666
Ŋb = 0.3328 × 100
Ŋb = 33.28%
STEP:3
Ŋa = 1 Ŋa =
1 - Q2 / Q4
0.3996 = 1 -
Q2 / 200
Q2 / Q4
Q2 / 200
= 1- 0.3996
Q2 / 200 = 0.6004
Q2 = 200 × 0.6004
L.VIJAYAKUMAR /A.P-MECH
Page 7
Q2 = 120.08 KJ
STEP:4
Ŋb = 10.3328 = 1 -
Q3 – Q2
Q3 / 120.08
Q3 / 120.08 = 1 – 0.3328
Q3 / 120.08 = 0.6672
Q3 = 0.6672 × 120.08
Q3 = 80.1173KJ
L.VIJAYAKUMAR /A.P-MECH
Page 8
3. 2 kg of water at 900C is mixed with 3 kg of water at 100C is an isolated system. Calculate
the change of entropy due to the mixing process (AUC DEC 2012)
GIVEN:
Isolated system problem
Mass (m 1) = 2 kg
Temp (T1) = 00C + 273 =273K
(T2) = 900C + 273 = 363K
(T3) = 100C +273 = 283K
(m 2) = 3 kg
Mass = m 1 + m 2
= 2 + 3 =5 kg
TO FIND:
Change in entropy due to mixing
SOLUTION:
STEP:1
Total energy before mixing = total energy after mixing
m 1 cpw (T2 – T1) + m 2 cpw (T3 – T1) = mass × cpw (Tm – T1)
2 (90 – 0) + 3 (10 – 0) = 5 (Tm -0)
180 + 30 = 5 Tm
210 = 5 Tm
210/5 = Tm
420C = Tm
Tm = 420C + 273
Tm = 315 K
STEP:2
Initial entropy of the system
= m 1 cpw × ln T2 + 273
+ m 2 cpw × ln T3 + 273
T1
T1
= 2 × ln 363 + 3 × ln 283
273
273
= 0.5698 + 0.1079
S1 = 0.6777 KJ /K
Final entropy of the system
=m × cpw × ln
Tm
T1
= 5 × ln 315
273
S2
= 0.7155 KJ/KG
L.VIJAYAKUMAR /A.P-MECH
Page 9
Change in entropy (
S) =
S2
-
S1
= 0.7155 – 0.6777
S = 0.0378 KJ / Kg.K
L.VIJAYAKUMAR /A.P-MECH
Page 10
4. A heat engines operates carnot cycles with a efficiency of 75 percent. What COP would
a refrigerator operating an the same cycle have? The low temp is 00C.(AUC MAY 2013)
GIVEN:
Carnot cycle
ᵑ
H.E
= 75 / 100 =0.75%
Templow = 00C + 273 = 273K
TO FIND:
a) COP of refrigerator
SOLUTION:
Carnot efficiency (
ᵑ
c) = Th – TL
TH
(
ᵑ
TH – 273
c) =
TH
0.75 =
TH – 273
TH
0.75 × TH = TH – 273
273 = TH – 0.75 TH
273 = 0.25 TH
273 / 0.25 = TH
1092 K = TH
(COP)ref rigerator =
TL
TH - TL
= 273
1092 – 273
(COP)ref
L.VIJAYAKUMAR /A.P-MECH
= 0.3333
Page 11
5. One kg of ice at -50C is exposed to the atmosphere which is at 200C. the ice melts and
come into thermal equilibrium with the atmosphere
(i) determine the entropy increase to the turbine
(ii) what is the minimum amount of work necessary to the ice -50C
Assume CP for ice as 2.093 KJ /Kg.K and the latent heat of fusion of ice as 333.3
KJ/KG
(AUC MAY 2013)
GIVEN:
Temp (Tice) = -50C +273 = 268K
Temp (Tatm ) = 200C +273 =293K
(Cpice) = 2.093 KJ/KG.K
Latent heat of fusion (l) =333.3 KJ/Kg.K
TO FIND:
a)
b)
Entropy increase of universe ( s)univ =?
Minimum amount of work (W) =?
SOLUTION:
STEP:1
(
S)univ = (
s)sy s + (
s)atm
Where
(
s)atm = - (SQ)
Tatm
Where
Heat absorbed by air from atm (SQ) = m Cpice (To -Ti) + m l + m CpW ( Tatm – 0)
= 1 × 2.093 ( 0 – (-5)) + 1 × 333.3 + 0.4 .187 (20 -0)
= 10.465 + 333.3 + 83.74
= 427.505 KJ
(
s)atm = - (SQ)
Tatm
= -
427.505
293
L.VIJAYAKUMAR /A.P-MECH
Page 12
(
s)atm = -1.45906 KJ/K
Where
(
s)sy s = m×Cpice × ln (TO/Ti) + m × L + m CpW × ln (Ta/T0)
TO
= 1 × 2.093 × ln (273/268) + 1×333.3 + 1 × 4.187 × ln (293/273)
= 0.0386 + 1.2208 + 0.2960
(
s)sy s = 1.5554 KJ/K
STEP:2
(
S)univ = (
s)sy s + (
s)atm
= -1.4590 + 1.5554
= 0.0964 KJ/K
STEP:3
(
S)univ = (
s)sy s + (
s)ref + (
s)atm
Where
(
s)sy s = - 1.5554 KJ/K
(
s)ref = 0
(
s)atm =
Q+W
0 = -1.5554 + 0 + Q + W
0 = -(1.5554 × 293) + Q + W
0 = -455.7322 + ( Q + W)
455.7322 = Q + W
455.7322 - Q = W
455.7322 – 427.505 = W
W min =28.2272 KJ
L.VIJAYAKUMAR /A.P-MECH
Page 13
6. A reversible engine operates between a source at 972oC and two sinks, one at127oC
and another at 27oC. The energy rejected is same at both the sinks. What is the ratio
ofheat supplied to the heat rejected? Also calculated the efficiency.(AUC DEC ’05)
GIVEN:
Reversible engine
Temp (T1) = 9720C + 273 = 1245K
Temp (T2) =127oC + 273 =400 K
Temp (T3) = 27oC + 273 = 300K
Energy rejected is same at both sinks
(Q2) = (Q3)
(Q1) = (Q2) + (Q3)
(Q1) = 2 Q2
TO FIND:
a)
What is the ratio ofheat supplied to the heat rejected?
b)
efficiency.
SOLUTION:
STEP:1
Q1
=
Q2 + Q3
T1
=
T2 + T3
1245
400 + 300
= 1.78
STEP :2
Efficiency (
ᵑ
) = 1 - Q2 – Q3
× 100
Q1
= 1-
1
× 100
1.78
= 0.4382 × 100
Efficiency (
L.VIJAYAKUMAR /A.P-MECH
Ŋ)
= 43.82%
Page 14
7. 50 kg of water is at 313 K and enough ice at -5oC is mixed with water in
an adiabatic Vesselsuch that at the end of the process all the ice melts and water at
0oC isobtained. Find the mass of ice required and the entropy change of water and
ice. Given c pofWater = 4.2kJ/kg-K. c pof ice = 2.1 kJ/kg-K and latent heat of ice = 335
(AUC DEC ’05)
kJ/kg.
GIVEN:
Mass of water (m w) = 50 Kg
Temp water
(T1) = 313K
Temp ice (T2) = -5oC + 273 = 268K
(T3) = 0oC + 273 = 273K
(Cp)water = 4.2 KJ / Kg.K
(Cp)ice = 2.1 KJ/Kg.K
Latent heat of ice (L)ice = 335 KJ/Kg
TO FIND:
a) Mass of ice (m ice)
b) Change in entropy of water and ice
SOLUTION:
STEP:1
Energy balance equation
m w Cpw T1 + m ice Cpice T2 + m ice L = (m w + m ice) Cpw T3
= (50 × 4.187 × 313) + ( m ice × 2.1 × 268) + (m ice × 335)
= (50 + m ice)4.187×273
=65526.55 + 562.8 m ice + 335 m ice = (50 + m ice) × 1143.09
=65526.55 + 897.8 m ice = 57152.55 + 1143.051 m ice
1143.05 m ice - 897.8 m ice = 65526.55 – 57152.55
245.25 m ice = 8374
m ice = 8374 / 245.25
m ice = 34.1447 Kg
STEP:2
Change in entropy of water
( s)w = m w Cpw ln (T2/T3)
L.VIJAYAKUMAR /A.P-MECH
Page 15
( s)w = 50 × 4.187 × ln (313/273)
( s)w = 28.6247 KJ/K
Change in entropy of ice
( s)ice = m ice × Cpice × ln (T2 /T3) + m ice (l/T3)
(
s)ice = 34.1447 × 2.1 × ln (268/273) + 34.1447 × ln (335/273)
= -1.3254 + 6.9880
( s)ice = 5.6625 KJ/K
L.VIJAYAKUMAR /A.P-MECH
Page 16
8. A heat engine operating between two reservoirs at 100 K and 300 K is used to
drive heat pump which extracts heat from the reservoir at 300 Kat a rate twice that at
which engine rejects heat to it. If the efficiency of the engine is 40Vo of the maximum
possible and the co-efficient of performance of the heat pump is 50% of the maximum
possible, make calculations for the temperature of the reservoir to which the heat pump
rejects heat. Also work out the rate of heat rejection from the heat pump if the rate of
supply of heat to the engine is 50 kW.
(AUC DEC ’06)
GIVEN:
Temp (T1) = 1000K
(T2) = 300K
Efficiency of heat engine = 40% of max possible efficiency
C.O.P heat pump = 50% of max possible COP rate of supply heat engine
Qs1 = 50 KW
TO FIND:
The temp of the reservoir (T4)
SOLUTION:
STEP:1
Carnot efficiency (
STEP:2
Actual efficiency
Ŋc ) = T
1 – T2 × 100
T1
= 1000 – 300 × 100
1000
= 0.7 × 100
= 70 %
= 40% carnot efficiency
= 0.4 × 0.7
= 0.28
STEP:3
Ŋa)
= Qs1 – Qr1
Qr1
0.28 = 50 – Qr1
50
(0.28 × 50) = 50 – Qr1
14 = 50 – Qr1
Actual efficiency (
L.VIJAYAKUMAR /A.P-MECH
Page 17
Qr1 = 50 -14
Qr1 = 36KW
Qs2 = 2 × Qr2
Qs2 = 2 × 36
Qs2 = 72KW
Work output of heat engine (W E) = Qs1 – Qr1
= 50 – 36
= 14 KW
Qr2 = Qs2 + work output of H.E
Qr2 = 72 + 14
Qr2 = 86 KW
(C.O.P) Heat pump = Heat supplied
Work done
=
Qr2
Qr2 – Qs2
=
86
86 – 72
= 6.14
Actual co-efficient of
Performance of heat pump
= 50 % COP carnot
T4
6.14 = 0.5
T4 – 300
6.14 =
T4
T4 – 300
0.5
12.28 =
T4
T4 – 300
12.28 T4 – 3684 = T4
12.28 T4 – T4 = 3684
T4 = (3684 / 11.28)
T4 = 326.5957 K
L.VIJAYAKUMAR /A.P-MECH
Page 18
(10)
One kg of air is contained in a piston cylinder assembly at 10barpressure and
500Ktemperature. The piston moves outwards and the air expands to 2 bar
pressure and 350 K temperature. Determine the maximum work obtainable.
Assume the environmental conditions to be 1 bar and 290K.Also make
calculations for the availability in the initial and final states.
(AUC MAY ’06)
GIVEN:
Mass ( m) = 1 Kg
Pressure (P1) = 10 bar × 100 =1000 KN / m 2
Temp (T1) = 500K
Pressure (P2) = 2 bar × 100 =200 KN / m 2
Temp (T2) = 350K
Environmental condition
Po = 1 bar × 100 = 100 KN / m 2
To = 290 K
TO FIND:
a)
Availability of initial state
b) Availability of final state
c) Max . work
SOLUTION:
STEP:1
Availability of initial state
ψ1 = m Cp T1 - To Cp ln T2
- R ln P1
To
Po
L.VIJAYAKUMAR /A.P-MECH
Page 19
=1
1.005 × 500 - 290
1.005 ln 500 - 0.287 ln 1000
290
=1
290 × 0.5474 – 0.6608
502.5 -
= 1 502.5 -
100
- 32.8712
= 535.3712 K
STEP:3
Availability of final state
Ψ2 = m Cp T2 - To Cp ln T2
- R ln P2
To
Po
=1
1.005 × 350 - 290
1.005 ln 350 - 0.287 ln 200
290
=1
=
100
351.75 - 290 0.1889 – 0.1989
351.75 - - 2.9
Ψ2 = 354.65 KJ
STEP :3
L.VIJAYAKUMAR /A.P-MECH
Page 20
W max = ψ1 – ψ2
= 535.3712 – 354.65
= 180.7212 KJ /Kg
11) A house hold refrigerator is maintained at a temperature of 275 K. Every time the
door is opened, warm material is placed inside ,introducing an average of 420 kJ, but
making only a small change in the temperature of the refrigerator. The door is
opened 20 times a day, and there refrigerator operates at 15% of the ideal COP. The
cost of work is Rs.2.50 per kW hr. What is the bill for the month of April for this
refrigerator? The atmosphere is at 303 K.
(AUC DEC ’07)
GIVEN:
Temp (Tlow) = 273 K
Temp (Thigh) = 303 K
COP of refrigerator = 15 % COP of ideal carnot ref
Heat supplied by the ref1 (Qr) = 420 KJ for one time
In one day = 20 times opened
Total heat rejected in sec = Q r – 20 times open
24 × 3600
= 420 × 20
24 × 3600
= 0.0972 KW
TO FIND:
Electricity charges of april month
SOLUTION:
STEP:1
COP ref =
TL
TH - TL
=
275
303 – 275
COP ref = 9.8214
L.VIJAYAKUMAR /A.P-MECH
Page 21
STEP:2
Actual COP of ref = 15 % COP of carnot ref
= 15 × 9.8214
100
= 1.4732
STEP:3
Work done (W) = heat rejected
C.O.Pref
= 0.0972
1.4732
= 0.06597
Work done (W) = 0.06597 × 3600 KW
(W) = 237.492 KW / hr
STEP:4
What is the bill for the month of april for this refrigerator?
Electricity charge of running of heat pump
= work × cost of work
= 237.492 × 2 .50
= 593.73 = 594
= RS 594
L.VIJAYAKUMAR /A.P-MECH
Page 22
12) An inventor claims to have developed an engine which receives 1000kJ
at a temperature of 160oC. It rejects heat at a temperature of 5oC and delivers
0.12kW h of mechanical work. Is this a valid claim? Justify your answer
Through Classius inequality.
(AUC MAY ’08)
GIVEN:
Heat receives (Q s ) = 1000 KJ
Work (w)
= 0.12 KW × 3600
= 432 KJ
Temphigh(T1) = 160oC +273 = 433K
Templow(T2) = 5oC+273 = 278K
TO FIND:
Is this a valid claim?
SOLUTION:
STEP:1
ŋ
carnot
= 1 - TL / TH
= 1 – 278 / 433
= 0.3579 × 100
ŋ
carnot
= 35 . 79%
STEP:2
ŋ
engine
= work = 432
Heat
100
= 0.432 × 100
L.VIJAYAKUMAR /A.P-MECH
Page 23
ŋ
engine
= 43.2%
STEP:3
ŋ
carnot
<
ŋ
engine
35 .79 < 43.2
RESULT:
Inventor claim is not valid
13) One kg of ice at -10oC is allowed to melt in atmosphere at 30oC. The ice melts
and the water so formed rises in temperature to that of atmosphere. Determine the
entropy change of ice, the entropy change of surrounding, and the entropy change of
universe and write your comment based on principle of increase in entropy. The
specific heat of ice is 2kJ/kg-K and its latent heat is 335kJ/kg.
(AUC MAY ’08)
GIVEN:
Mass = 1Kg
tempice = -10oC + 273 = 263 K
tempatm = 30oC.+ 273 = 303 K
specific heat of ice (Cpice) = 2 KJ / Kg.K
latent heat of ice (L) = 335 KJ / Kg
TO FIND:
a. Change in entropy of ice
b. Change in entropy surrounding (or) atm
c. Change in entropy universe
SOLUTION:
STEP:1
Heat absorbed by air from (SQ) = m Cpice (To – T1) + m L + m Cpw ( Ta - To )
= 1× 2× 0 – (-10)
+ 1 × 335
+ 1 × 4.187 30 – 0
= 20 + 335 + 125 . 61
(SQ) = 480.61 KJ
STEP:2
L.VIJAYAKUMAR /A.P-MECH
Page 24
Change in entropy of ice ( s)ice = m × Cpice × ln To
Tice
= 1 × 2 × ln 273
263
= 0.07463 KJ /K
STEP:3
Change in entropy of surrounding ( s) atm = - (SQ)
Tatm
= - (480.61)
303
= -1.5861 KJ/K
STEP:4
Change in entropy universe ( s)sy s = ( s)ice + ( s)f usion + ( s)liquid
( s)f usion = m × L = 1 × 335 = 1.2271 KJ/K
To
273
Where
( s)liquid = m × Cpw × ln Ta
To
= 1 × 4.187 × ln 303
273
= 0.4365 KJ/K
Change in entropy of universe ( s)univ = ( s)sy s + ( s)atm
= (0.07463 + 1.2271 + 0.4365) + (-1.5861)
( s)univ = 0.15213 KJ/K
L.VIJAYAKUMAR /A.P-MECH
Page 25
LATEST SOLVED ANNA UNIVERSITY
QUESTION
1.The interior lighting of refrigerator is provided by incandescent lamps whose switches are
actuated by the opening of the refrigerator door.consider a refrigerator whose 40 w light bulb
remains on continuously as a result of malfunction of the switch.If the refrigerator has a coefficient of performance of 1.3 and the cost of electricity of be RS 8 per kwh determine the
increase in energy consumption of the refrigerator and its cost per year if the switch is no fixed
(AUC MAY 2008)
Given:Work (W) = 40/1000w = 0.04
Cop = 1.3
Cost = RS 8 per kwh
To find:
1.Increase in energy consumption of refrigerator
2.Increase in cost of electricity
Solution:
Step:1
Cop = QR /W
1.3 = QR /40
1.3
× 40 = QR
52 watts/1000 = Q R
L.VIJAYAKUMAR /A.P-MECH
Page 26
0.052 kw = Q R
Step:2
Work(W) = QS - QR
QR – w = QS
0.052 – 0.04 = 0..012kw
Step:3
Increase in energy consumption = (Q S × 3600)
= (0.012 × 3600)
= 43.2 kwh
Step:4
Increase in cost of electricity = Increase in energy consumption
× cost per kwh
= 43.2 × 8
= Rs 345.6
L.VIJAYAKUMAR /A.P-MECH
Page 27
2.An inventor claims to have developed a heat engine which absorbs 1500KJ of heat from a
reservoir at 1000K and reject of 600KJ heat to a reservoir at 600K.Check whether the claim is
acceptance (or) not
(AUC MAY 2008)
Given:
Heat absorbs (or) heat supplied (Q S) = 1500KJ
Temp high (TH ) = 1000K
Heat rejected (Q R ) = 600KJ
Temp low (TL) = 600K
To find:
Whether the claim is acceptance (or) not
Solution:
Step:1
Maximum efficiency (ŋmax ) = TH – TL / TL
ŋmax = 0.4 × 100 = 40%
step:2
Efficiency of engine (ŋinv entor) = QS – QR / QS
ŋinv entor = 1500 – 1600 / 1500
= 0.6
× 100
ŋinv entor = 60%
Step:3
Maximum efficiency < inventor efficiency
40%
<
L.VIJAYAKUMAR /A.P-MECH
60%
Page 28
Therefore, his claim is not acceptance.
3. A heat engine receives reversibly 420KJ / cycle of heat from a source at 327˚C , and rejects
heat reversibly to a sink at 27˚C . There are no other heat transfers in (i) (ii) and (iii) below
compute ʃ dQ / T from these results show which case in irreversible, reversible and
impossible:
(i) 210KJ / cycle rejected
(ii) 105KJ / cycle rejected
(iii) 315KJ / cycle rejected
(AUC MAY 2009)
Given:
Heat receives (Q) = 420KJ
Temperature (T1) = 327˚C + 273 = 600K
Temperature (T2) = 27˚C + 273 = 300K
Condition(i) = 210KJ / cycle rejected
Condition(ii) = 105KJ / cycle rejected
Condition(iii) = 315KJ / cycle rejected
To find:
Results show which case in irreversible, reversible and impossible
Solution:
Step:1
Condition(i)
Q1 = 210KJ / cycle rejected
By clausius inequality ∫dQ/T = QS/T1 – Q1/T2
∫ dQ/T = 420/600 – 210/300
∫ dQ/T = 0
∫dQ/T =0 => so, the case is reversible
L.VIJAYAKUMAR /A.P-MECH
Page 29
Step:2
Condition(ii)
Q 2 = 105KJ / cycle rejected
By clausius inequality ∫ dQ/T = QS/T1 – Q2/T2
∫ dQ/T = 420/600
– 105/300
∫ dQ/T = 0.35
∫
dQ/T >0 => so, the case is impossible
Step:3
Condition(iii) Q3 = 315KJ / cycle rejected
By clausius inequality ∫ dQ/T = QS/T1 – Q3/T2
∫
∫
dQ/T = 420/600 – 315/300
∫
dQ/T = -0.35
dQ/T <0 => so, the case is irrevesible
L.VIJAYAKUMAR /A.P-MECH
Page 30
4. A carnot heat engine receives heat from a reservoir at 1173K at a rate of 800KJ/min and
rejects the waste heat to the ambient air at 300K. The entire work out of the heat engine is used
to drive a refrigerator that removes heat from the refrigerated space at 268K and transfer it to
the same ambient air at 300K. Determine the max rate of heat rejection to the ambient air (AUC MAY 2009)
Given: Carnot heat engine
Temperature(T1) = 1173K
Temperature(T2) = 800K
(QS1) = 800/60KJ / min
(QS1) = 13.3333KJ / sec
(T3) = 268K
To find:
Max.rate of heat rejection to the ambient air (Q R )
Solution:
Step:1 for heat engine
ŊH.E = 1 - (TL / TH)
ŊH.E = 1 - (300/1173)
ŊH.E = 0.7442 × 100
ŊH.E = 74.4245%
Step:2
ŊH.E = work done / heat supplied
Work done = ŊH.E
× QS1
= 0.7442
L.VIJAYAKUMAR /A.P-MECH
× 13.3333
Page 31
Work done(W) = 9.9226KW
Step:3
Work done(W) = Q S1 – QR1
(QR1) = QS1 – w
(QR1) = 13.3333 – 9.9226
QR1 = 3.4107KJ / Sec
Step:4 [for refrigerator]
=>C.O.PRef = T3 / T2 – T3
= 268 / 300 – 268
C.O.PRef = 8.375
=>C.O.PRef = QS2 / W
C.O.PRef × W = QS2
8.375
× 9.9226 = QS2
QS2 = 83.0967 KJ/S
Maximum rate of heat rejection to the ambient air (Q R2) = QS2 + W 2
note:W 2 = W 1
= 83.0967 + 9.9226
(QR2) = 93.0193KJ / Sec
L.VIJAYAKUMAR /A.P-MECH
Page 32
5.A heat pump working on the carnot cycle takes in heat from a reservoir at 5˚C and delivers
heat to a reservoir at 60˚C. the heat pump is driven by a reversible heat engine which takes
heat from reservoir at 840˚C and reject heat to a reservoir at 60˚C. the reversible heat engine
also drives a machine that absorbs 30KW. If heat pump extract 17KJ/Sec from the reservoir at
5˚C
(AUC DEC 2010)
(1) the rate of heat supply from 840˚C
(2) the rate of heat rejection to 60˚C sink
Given:
Temperature(T1) = 60˚C + 273 => 333K
Temperature(T2) = 840˚C + 273 => 1113K
Temperature(T3) = 5˚C + 273 => 278K
(QS1) = 17KJ/Sec × 10000 => 17000J/Sec
To find:
(1) the rate of heat supply from 840˚C
(2) the rate of heat rejection to 60˚C sink
Solution:
Step:1
Ŋmax = TH – TL / TH
ŋmax = 0.7008 × 100 = 70.08%
ŋmax
= W 1 / QS1
ŋmax × QS1 = W 1
L.VIJAYAKUMAR /A.P-MECH
Page 33
0.7008
× 17000 = 11913.6J
W 1 = 11913.6J
W = QS1 – QR1
QR1 = QS1 – W => 17000 – 11913.6
QR1 = 5086.4J
Step:2
ŊH.P = TH – TL / TH => 333 – 278 / 333
ŊH.P = 0.1651 × 100
ŊH.Pump
= 16.51%
ŊH.P = W 2 / QS2
QS2 = W 2 / Ŋ => 30
× 1000 / 0.1651
The rate of heat supply from 840˚C => Q S2 = 181708.0557J
=> W 2 = QS2 – QR2
QR2 = QS2 – W 2
=181708.0557 – 30000
QR2 = 151708.0557J
Net heat transfer (Q) = Q R1 + QR2
= 5086.4 + 151708.0557
The rate of heat rejection to the 60˚C sink => Q=156.7944 × 103 KJ
L.VIJAYAKUMAR /A.P-MECH
Page 34
L.VIJAYAKUMAR /A.P-MECH
Page 35