ENGR 2213 Thermodynamics
F. C. Lai
School of Aerospace and Mechanical
Engineering
University of Oklahoma
Reversible Steady-Flow Work
For a steady-flow device undergoing an internally
reversible process,
(δq)rev - (δw)rev = dh + d(ke) + d(pe)
Neglect the changes in kinetic and potential energies,
(δq)rev - (δw)rev = dh
(δq)rev = T ds = dh - v dp
(δw)rev = - v dp
wrev 
2
1 v
dp
Work
w
2
1 p
wrev 
dv
2
1 v
reversible work in closed systems
dp
reversible work associated with
an internally reversible process
an steady-flow device
► The larger the specific volume, the larger the
reversible work produced or consumed by the
steady-flow device.
Work
wrev 
2
1 v
dp
To minimize the work input during a compression
process
► Keep the specific volume of the working fluid
as small as possible.
To maximize the work output during an expansion
process
► Keep the specific volume of the working fluid
as large as possible.
Work
Why does a steam power plant usually have a
better efficiency than a gas power plant?
Steam Power Plant
► Pump, which handles liquid water that has a
small specific volume, requires less work.
Gas Power Plant
► Compressor, which handles air that has a
large specific volume, requires more work.
Work
Minimizing the Compressor Work
1. To approach an internally reversible process as
much as possible by minimizing the
irreversibilities such as friction, turbulence, and
non-quasi-equilibrium compression.
2. To keep the specific volume of the gas as small
as possible by maintaining the gas temperature
as low as possible during the compression
process.
This requires that the gas be cooled
as it is compressed.
Steady-Flow Work
wrev 
2
1 v
dp
Polytropic Processes
w rev 
1
2  C n
 1  
p
n

n 1
(pvn = constant)
dp 
1 n1
Cn p n 12
1
1
n
C 2
p

1
n
dp
n

(p1v1  p2 v 2 )
n 1
Steady-Flow Work
(pvn = constant)
n1 

nR(T1  T2 ) nRT1   p2  n 

1


  

n 1
p1 
n 1



Polytropic Processes
w rev
(pvk = constant)
k 1 

kR(T1  T2 ) kRT1   p2  k 

1


  

k 1
p1 
k 1



Isentropic Processes
w rev
Steady-Flow Work
Isothermal Processes
w rev 
2
 1
(pv = constant)
C
1 dp
dp  C 2
p
p
 p1 
 p1 
 C ln    RTln   p
 p2 
 p2 
1. n = k
2. 1 < n < k
3. n = 1
3
2 1
W
v
Example 1
Air entering a compressor at p1 = 100 kPa and
T1 = 20 ºC and exiting at p2 = 500 kPa. If the air
undergoes a polytropic process with n = 1.3,
determine the work and heat transfer per unit
mass of flow rate.
Example 1 (continued)
n1
n
T2  p2
 
T1  p1 
Polytropic processes
n1
n
 p2
T2  T1  
 p1 
1.31
 500  1.3
 293 
= 425 K

 100 
W nR

(T1  T2 )
m n 1
1.3

 0.287  293  425  = - 164.15 kJ/kg
1.3  1
Example 1 (continued)
Q W

 (h2  h1) 
m m
2
V2
Table A-17, T1 = 293 K,
T2 = 425 K,

2
2
V1
 g(z2  z1)
h1 = 293.17 kJ/kg
h2 = 426.35 kJ/kg
= - 164.15 + (426.35 – 293.17)
= - 30.97 kJ/kg
Isentropic Efficiency for Turbines
Isentropic efficiency for a turbine is defined as
the ratio of the actual performance of a turbine
to the performance that would be achieved by
undergoing an isentropic process for the same
inlet state and the same exit pressure.
Wactual
t 
Wisentropic
Isentropic Efficiency
Turbines
Q W

 (h2  h1) 
m m
W
 h1  h2
m
W
   h1  h2s
 m s
h1  h2
t 
h1  h2s
2
V2

2
2
V1
 g(z2  z1)
1
h
h1 – h2
h1 – h2s
2
2s
s
Example 2
Air enters a turbine at p1 = 300 kPa and
T1 = 390 K and exits at p2 = 100 kPa. Given
that the actual work output from the turbine is
74 kJ/kg and if the turbine operates adiabatically,
determine the isentropic efficiency for the turbine.
Wactual
Wactual
t 

Wisentropic h1  h2s
Example 2 (continued)
p2 pr 2

p1 pr1
Table A-17 T1 = 390 K
pr1 = 3.481, h1 = 390.88 kJ/kg
p2
pr2  pr1
= 3.481 (100/300) = 1.1603
p1
W
Table A-17 pr2 = 1.1603,
   h1  h2s
h2s = 285.27 kJ/kg
m
 s
= 390.88 – 285.27 = 105.6 kJ/kg
Wactual
74
t 

= 0.7
Wisentropic 105.6
Isentropic Efficiency for Compressors
Isentropic efficiency for a compressor is defined
as the ratio of the performance of a compressor
that would be achieved by undergoing an
isentropic process to the actual performance for
the same inlet state and the same exit pressure.
c 
Wisentropic
Wactual
Isentropic Efficiency
Compressors
Q W

 (h2  h1) 
m m
W
 h2  h1
m
W
   h2s  h1
 m s
h2s  h1
c 
h2  h1
2
V2

2
2
V1
 g(z2  z1)
2
2s
h
h1 – h2
h1 – h2s
1
s
Example 3
Air enters an insulated compressor at p1 = 95 kPa
and T1 = 22 ºC. Given that p2/p1 = 6 and ηc = 0.82,
determine the exit temperature for the air.
c 
Wisentropic
Wactual
h2s  h1

h2  h1
h2s  h1
h2 
 h1
c
Example 3 (continued)
p2 pr 2

p1 pr1
Table A-17 T1 = 295 K
pr1 = 1.3068, h1 = 295.17 kJ/kg
p2
pr2  pr1
= 1.3068 (6) = 7.841
p1
Table A-17
pr2 = 7.841,
h2s  h1
h2 
 h1 T = 490.29 K h = 493.0 kJ/kg
2s
2s
c
493.0  295.17

 295.17  536.4 kJ / kg
0.82
Table A-17
h2 = 536.4 kJ/kg,
T2 = 532 K
Example 4
0.5 kilogram of water executes a Carnot power
cycle. During the isothermal expansion, the water
is heated until it is a saturated vapor from an initial
state where the pressure is 1.5 MPa and the quality
is 25%. The vapor then expands adiabatically to
pressure of 100 kPa. Find
(a) the heat addition and rejection from this cycle.
(b) the cycle efficiency.
Example 4 (continued)
T
1
2
4
3
Given:
p1 = p2 = 1.5 MPa
p3 = p4 = 100 kPa
x1 = 0.25
W23 = 403.8 kJ/kg
S
Find:
Q12 = ?
Q34 = ?
η=?
Example 4 (continued)
Q12 = m(u2 – u1) + mp(v2 – v1)
= m(h2 – h1)
Table A-5
p1 = p2 = 1.5 MPa,
hf = 844.84 kJ/kg, hfg = 1947.3 kJ/kg, hg = 2792.2 kJ/kg
sf = 2.315 kJ/kg K, sfg = 4.1298 kJ/kg K, sg = 6.4448 kJ/kg K
h1 = hf + x1 hfg= 844.84 + 0.25(1947.3) = 1331.67 kJ/kg
s1 = sf + x1 sfg = 2.315 + 0.25(4.1298) = 3.3474 kJ/kg K
h2 = hg = 2792.2 kJ/kg
s2 = sg = 6.4448 kJ/kg K
Example 4 (continued)
Q12 = m(h2 – h1) = 0.5(2792.2 – 1331.67) = 730.27 kJ
Q34 = m(h4 – h3)
s3 = s2 = 6.4448 kJ/kg K
s4 = s1 = 3.3474 kJ/kg K
Table A-5
p3 = p4 = 100 kPa,
hf = 417.46 kJ/kg, hfg = 2258.0 kJ/kg, hg = 2675.5 kJ/kg
sf = 1.3026 kJ/kg K, sfg = 6.0568 kJ/kg K, sg = 7.3594 kJ/kg K
s3  sf 6.4448  1.3026
x3 

 0.849
sg  s f
6.0568
Example 4 (continued)
s4  sf 3.3474  1.3026
x4 

 0.338
sg  s f
6.0568
h3 = hf + x3 hfg= 417.46 + 0.849(2258) = 2334.5 kJ/kg
h4 = hf + x4 hfg = 417.46 + 0.338(2258) = 1180.66 kJ/kg
Q34 = m(h4 – h3) = 0.5(1180.66 – 2334.5) = -576.92 kJ
Q34
QL
576.92
  1
 1
 1
 0.21
QH
Q12
730.27