College of Engineering
Spring Session- 2016
THERMODYNAMICS II - ME 272
Dr. Saeed J. Almalowi, [email protected]
PROBLEM 2.1 STATEMENT
A piston-cylinder device contains 0.05 kg of steam at 1 MPa and 300oC. Steam now expands to a
final state of 200 kPa and 150oC, doing work. Heat losses from the system to the surrounding are
estimated to be 2 kJ during this process. Assuming the surrounding to be To = 25oC and Po =100
kPa,
Determine
a.
b.
c.
d.
e.
The exergy of the steam at the initial and the final state,
The exergy change of the steam,
The useful work,
The exergy destroyed, and
The second law efficiency?
Fig.2.1 A Piston-Cylinder
Ans. [X1=35 kJ, X2=25.4 kJ, Wu=5.3 kJ, Xdestroyed=4.3 kJ, and ηII = 55.2 %]
Solution:
1
X1 = m(u1 − uo ) + mPo (ν1 − νo ) − mTo (s1 − so ) + mV12 + mgz1
2
1
X2 = m(u2 − uo ) + mPo (νout − νo ) − mTo (s2 − so ) + mV22 + mgz2
2
State
Description of state
State (1) P1= 1000 kPa
T1=300oC
State(2) P2 =200 kPa,
T2=150oC
Surrounding State:
Dead state* To=25oC,
P0=100 kPa
T1>Tsat. @ P1=1000 kPa
Superheated steam
T2>Tsat. @ P2=200 kPa
Superheated steam
Compressed Liquid
ME 272-College of Engineering-Taibah University
Internal energy
kJ/kg
u1=2794
Entropy
kJ/kg.K
s1=7.122
Specific Volume
m3/kg
v1=0.2579
u2=2577
s2=7.279
v2=0.9597
uo=104.7
so= 0.3669
vo=0.001003
College of Engineering
Spring Session- 2016
THERMODYNAMICS II - ME 272
Dr. Saeed J. Almalowi, [email protected]
Dead state is the state has an equilibrium with its surrounding (environment).
Xheat − Xwork − Xdestroyed = ∆Xsys = X2 − X1
a. X1 = 0.05 × [(2794 − 104.7) + 100 × (0.2579 − 0.001003) − 298 × (7.122 − 0.3669)]
= 35.0kJ
b. X2 = 0.05 × [(2577 − 104.7) + 100 × (0.9597 − 0.001003) − 298 × (7.279 − 0.3669)]
= 25.4kJ
𝑐. ∆𝑋 = 𝑋2 − 𝑋1 = 25.4 − 35.0 = −9.58 𝑘𝐽
Wu = Wsys − Wsurr , Wsurr = mPo (v2 − v1 ) = 0.05 × 100 × (0.9597 − 0.2579) = 3.509 kJ
Wsys = Q + m(u1 − u2 ) = −2 + 0.05 × (2794 − 2577) = 8.85 kJ
Wu = 8.85 − 3.509 = 5.341 kJ
d. Xdestroyed = X2 − X1 − Wu = 9.58 − 5.341 = 4.239 kJ
Xdestroyed = To [m(s2 − s1 ) +
Q
2
] = 298 × [0.05 × (7.279 − 7.122) +
] = 4.33 kJ
To
298
e. Wrev = Xdestroyed + Wu = X1 − X2 = 9.58 kJ, Wrev > Wu , [Producing devic], ηII =
=
5.341
= 0.5575
9.58
Wu
Wrev
PROBLEM 2.2 STATEMENT
Steam enters a turbine steadily at 3MPa and 450oC at rate of 8 kg/s and exits at 0.2 MPa and
150C. The steam is losing heat to the surrounding air at 100 kPa and 25oC at a rate of 300 kW,
and the kinetic and potential energy changes are negligible.
Determine
a.
b.
c.
d.
e.
The actual power output,
The maximum possible power,
The second –law efficiency,
The exergy destroyed, and
The exergy of the steam at the inlet condition?
Solution:
1
2
𝑋̇in = ṁ(hin − ho ) − ṁTo (sin − so ) + ṁVin
+ ṁgzin , where Vo = 0, zo = 0
2
ME 272-College of Engineering-Taibah University
College of Engineering
Spring Session- 2016
THERMODYNAMICS II - ME 272
Dr. Saeed J. Almalowi, [email protected]
1
2
𝑋̇out = ṁ(hout − ho ) − ṁTo (sout − so ) + ṁVout
+ ṁgzout , where Vo = 0, zo = 0
2
Exergy for Fixed Volume
1 2
2
) + g(Zout − Zin )
Δψ = (hout − hin ) − To (sout − sin ) + (Vout
− Vin
2
∆Ẋ = 𝑚̇∆ψ
State
Description of state
State (1) P1= 3000 kPa
T1=450oC
State(2) P2 =200 kPa,
T2=150oC
Surrounding State:
Dead state* To=25oC,
P0=100 kPa
̇Ein = Ėout
T1>Tsat. @ P1=1000 kPa
Superheated steam
T2>Tsat. @ P2=200 kPa
Superheated steam
Compressed Liquid
Internal energy
kJ/kg
h1=3344
Entropy
kJ/kg.K
s1= 7.083
Specific Volume
m3/kg
v1=0.1079
h2=2769
s2=7.279
v2=0.9597
ho=104.8
so= 0.3669
vo=0.001003
ṁ 1 h1 = ṁ2 h2 + Ẇ + Q̇ → Ẇ = ṁ(h1 − h2 ) − Q̇ = 8 × (3344 − 2769) − 300 = 4300 kW
= Wsys
̇ u = ẆActaul = 4300 kW
a. W
̇ rev = ṁ[(h1 − h2 ) − To (s1 − s2 )] = 8 × [(3344 − 2769) − 298 × (7.083 − 7.279)]
b. W
= 5067.3 kW
It means the turbine is producing device.
c. ηII =
Ẇu
4300
=
= 0.848
Ẇrev 5067.3
d. Ẋdestroyed = I = Ẇrev − Ẇu = 5067.3 − 4300 = 767.3 kJ
𝑐. 𝑋̇in = 8 × [(3344 − 104.8) − 298 × (7.083 − 0.3669)] = 9902.42 𝑘𝑊
ψ̇in =
Xiṅ
9902.42
kJ
=
= 1238
[ Specific Exergy: Exergy per unit mass]
ṁ
8
kg
ME 272-College of Engineering-Taibah University