College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Unit Ten Homework Solutions, November 18, 2010
1
Steam enters an adiabatic turbine at 800 psia and 900oF and leaves at a pressure of 40
psia. Determine the maximum amount of work that can be delivered by this turbine.
Assume that the turbine is a steady flow device with negligible changes in kinetic and potential
energies. For this adiabatic turbine, with one inlet and one outlet, the first law gives the work as
w = hin – hout. The maximum work in an adiabatic process occurs when the outlet entropy is the
same as the inlet entropy. Use the steam tables to find the properties.
In this case, hin = h(800 psia 900oF) = 1456.0 Btu/lbm and sin = s(800 psia 900oF) = 1.6413
Btu/lbm∙R. At the outlet pressure of 40 psia, the entropy value of 1.6413 Btu/lbm∙R is seen to be
in the mixed region, so we have to compute the quality to find the outlet enthalpy.
xout 
sout  s f ( 40 psia )
s fg ( 40 psia )

1.6413 Btu
hout  h f (40 psia )  xout h fg (40 psia ) 
lbm  R
 0.39213 kJ
1.28448 kJ
kg  K
 0.9725
kg  K
 933.69 Btu  1144.2 Btu
236.14 Btu
 
 (0.9725)
lbm
lbm
lbm


We can now compute the maximum work.
w  hin  hout 
2
1456.0 Btu 1144.2 Btu

lbm
lbm
w
311.9 Btu
lbm
Air is compressed steadily by a 5-kW compressor from 100 kPa and 17oC to 600 kPa and
167oC at a rate of 1.6 kg/min. During this process some heat transfer takes place between
the compressor and the surrounding medium at 17oC. Determine the rate of entropy
change of air during this process.
It appears that we have a large amount of extraneous information in this problem. Since we are
only asked to find the entropy change of the air and the process is steady, the desired entropy
change is given by the following equation if we use the air tables to find the entropy change,
accounting for the variable heat capacity.
 o
 P 
o
Sair  m air sout  sin   m air  sair
(T2 )  sair
(T1 )  R ln  2 
 P1 

Using the values of so from the air tables and R = 0.287 kJ/kg∙K for air gives the answer as
follows.
Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Error! Reference source not found. homework solutionsME 370, L. S. Caretto, Fall 2010 Page 2
 o
 P  1.6 kg min  2.0877 kJ
o
Sair  m air  sair
( 440.15 K )  sair
( 290.15 K )  R ln  2  
min 60 s  kg  K
 P1 


3
1.66802 kJ 0.287 kJ  600 kPa  kW  s


ln 
kg  K
kg  K
 100 kPa  1 kJ
0.00250 kW
Sair  
K
Air enters a nozzle steadily at 280 kPa and 77oC with a velocity of 50 m/s and exits at 85
kPa and 320m/s. The heat losses from the nozzle to the surrounding medium at 20oC are
estimated to be 3.2 kJ/kg. Determine (a) the exit temperature and (b) the total entropy
change for this process.
Here we have a steady flow system with one inlet and one outlet. There is no useful work in the
nozzle. We neglect changes in potential energy and write the first law as follows


V2 
V2 
   h 

q   h 
2  out 
2 in

We use the air tables to account for the temperature variation of the heat capacity and we note
that q = -3.2 kJ/kg since this is a heat loss. With the inlet enthalpy, hin = h(350 K) = 350.49 kJ/kg
from the air tables, we can solve our first law for the outlet enthalpy as follows.
2
hout  q  hin 
2
Vin2  Vout
2
2
 50 m   320 m 

 

3.2 kJ 350.49 kJ  s   s 
kJ  s 2



kg
kg
2
1000 kg  m 2
This gives hout = 297.34 kJ/kg. Interpolating in the air tables to find this enthalpy, we find that it
occurs at Tout = 297.2 K .
The total entropy change, stotal is the sum of the entropy change of the air, sair, plus that of the
surroundings, ssurround. We find the entropy change of the air from the usual equation for the air
tables.
P 
 85 kPa 
o
o
o
o

sair  sair
(T2 )  sair
(T1 )  R ln  2   sair
( 297.2 K )  sair
(350 K )  R ln 
 280 kPa 
 P1 
1.6924 kJ 0.287 kJ  85 kPa  0.1775 kJ
 


ln 
kg  K
kg  K
kg  K
 280 kPa 
The entropy change of the surroundings (per kilogram of air) is the heat transfer divided by the
constant temperature of 293.15 K for the surroundings. Note that the heat transfer to the
surroundings is the negative of the heat transfer to the nozzle: qsurround = 3.2 kJ/kg.
s surr 
qsurr
Tsurr
3.2 kJ
0.0109 kJ
kg


293.15 K
kg  K
Adding the entropy change of the air and the surroundings gives the total entropy change.
Error! Reference source not found. homework solutionsME 370, L. S. Caretto, Fall 2010 Page 3
stotal  sair  ssurr 
4
0.1775 kJ 0.0109 kJ

kg  K
kg  K
stotal 
0.1884 kJ
kg  K
Steam enters an adiabatic turbine at 8 MPa and 500oC with a mass flow rate of 3 kg/s and
leaves at 30 kPa. The isentropic efficiency of the turbine is 90%. Neglecting the kinetic
energy change of the steam, determine (a) the temperature at the turbine exit and (b) the
power output of the turbine.
We assume that we have a steady flow process in which the kinetic and potential energies are
negligible. We are given that the turbine is adiabatic and we note that is has only one inlet and
one outlet. We thus have the simple form of the first law, w = hin – hout. This equation applies to
both the actual and the ideal process.
We first compute the ideal work, ws = hin – hout,s. From the steam tables we find that hin = h(8
MPa, 500oC) = 3399.5 kJ/kg and sin = s(8 MPa, 500oC) = 6.7266 kJ/kg∙K. The ideal outlet state
of 30 kPa and sout,s = sin is seen to be in the mixed region, so we have to compute the quality to
find the outlet enthalpy.
xout 
sout  s f (30 kPa)
s fg (30 kPa)

 0.9441 kJ
kg  K
kg  K
 0.84745
6.8234 kJ
kg  K
6.7266 kJ
hout  h f (30 kPa)  xout h fg (30 kPa) 
 2335.3 kJ  2268.3 kJ
289.27 kJ
 (0.84745)
 
kg
kg
kg


We can now compute the maximum work.
ws  hin  hout,s 
3399.5 kJ 2268.3 kJ 1131.2 kJ


kg
kg
kg
Multiplying this by the isentropic efficiency gives the actual work and multiplying that result by the
mass flow rate gives the power output of the turbine.
3 kg
90%  1131.2 kJ MW  s
W  m w  m  s ws 
s
kg
1000 kJ
W  3.054 MW
Applying the first law to the actual process gives hout,a as follows.
hout,a  hin  w  hin 
W 3399.5 kJ
3.054 MW
2381.4 kJ



3 kg MW  s
m
kg
kg
s 1000 kJ
At the outlet pressure of 30 kPa, the value of hout = 2380.9 kJ/kg is seen to be in the mixed region
so the final temperature, Tout = Tsat(30 kPa) = 69.09oC .
5
Refrigerant 134a enters an adiabatic compressor as a saturated vapor at 120 kPa at a rate
of 0.3 m3/min and exits at 1 MPa pressure. If the isentropic efficiency of the compressor is
80%, determine (a) the temperature of the refrigerant at the exit of the compressor and (b)
the power input in kW. Also, show the process on a T-s diagram with respect to the
saturation lines.
Error! Reference source not found. homework solutionsME 370, L. S. Caretto, Fall 2010 Page 4
We assume a steady process with negligible changes in kinetic and potential energies. We are
given that the compressor is adiabatic and we see that it has one inlet and one outlet. Thus the
 ( hin  hout ) . We use the property tables for
first law for this compressor reduces to W  m
refrigerant 134a. On a per unit mass basis we can compute the ideal work, ws = hin – hout,s and
the actual work wa = hin – hout.
At the inlet state, a saturated vapor at 120 kPa, we find the following properties: hin = hg(120 kPa)
= 236.97 kJ/kg; sin = sg(120 kPa) = 0.94779 kJ/kg∙K; vin = vg(120 kPa) = 0.16212 m3/kg. We can
use the inlet volume flow rate and the inlet specific volume to compute the mass flow rate.
0.3 m 3 min
V
min 60 s 0.03084 kg
m  in 

vin 0.16212 m 3
s
kg
At the ideal outlet state of Pout = 1 MPa and sout,s = sin = 0.94779 kJ/kg∙K, we find the ideal outlet
state to be between 40oC and 50oC. We would get the ideal outlet enthalpy by interpolation.
hout
282.74 kJ 271.71 kJ

271.71 kJ
kg
kg


0.9525 kJ 0.9179 kJ
kg

kg  K
kg  K
 0.94779 kJ 0.9179 kJ  281.25 kJ



kg  K 
kg
 kg  K
We can now compute the ideal work per unit mass.
ws  hin  hout,s 
236.97 kJ 281.25 kJ
44.27 kJ


kg
kg
kg
The negative work confirms out understanding that this is a work input device so that we have to
divide the ideal work by the isentropic efficiency to find the actual work. Multiplying the actual
work by the mass flow rate gives the power.
0.03084 kg  44.27 kJ  kW  s
 

s
kg  1 kJ
m ws


W  m w 

s
80%
W  1.707 kW
Applying the first law to the actual process gives hout,a as follows.
hout,a  hin  w  hin 
 1.707 kW   292.31 kJ
W 236.97 kJ


0.03084 kg kW  s
m
kg
kg
s
1 kJ
At the outlet pressure of 1 MPa, the value of hout = 292.31 is between temperatures of 50oC and
60oC. Interpolation gives the corresponding outlet temperature as follows:
Tout
60o C  50o C
 50 C 
293.38 kJ 282.74 kJ

kg
kg
o
 292.31 kJ 282.74 kJ 



kg
kg


Tout = 59.0oC .
Error! Reference source not found. homework solutionsME 370, L. S. Caretto, Fall 2010 Page 5
6
Air is compressed steadily by a 5-kW compressor from 100 kPa and 290 K to an outlet
pressure at a rate of 1.6 kg/min. During this process the heat transfer between the
compressor and the surrounding medium is negligible. Determine the maximum pressure
that can be delivered by the compressor. Do the calculation for (a) constant heat
capacities and (b) using air tables.
The maximum pressure would occur in an ideal process. Such a process would be a reversible
process. Since there heat transfer is negligible, we can assume that the process is adiabatic so
an ideal process will be an adiabatic reversible or isentropic process. This means that sout = sin.
The first law analysis of this problem uses the equation for a steady open system with one inlet
and one outlet. If we ignore changes in kinetic and potential energies and use the assumption
that the negligible heat transfer can be assumed to be zero we have the following result. (Note
that we set the power to -5 kW since it is a power input.)
 5 kW 60 s 1 kJ
W
kJ
W  m hin  hout   hout  hin 
 hin 
 hin  187.5
1.6 kg min kW  s
m
kg
min
From the air tables on page 934 we find the following data at Tin = 290 K: hin= h(290 K) = 290.16
kJ/kg and Pr(Tin = 290 K) = 1.2311. From the first law above we find outlet enthalpy as h out = hin +
187.5 kJ/kg = 290.16 kJ/kg + 187.5 kJ/kg = 477.66 kJ/kg We can find the ideal (isentropic) outlet
pressure from the equation Pout = PinPr(Tout)/Pr(Tin). We already know Pin = 100 kPa and hout =
477.66 kJ/kg. We have to find the value of Pr that corresponds to this value of hout from the air
tables. We can interpolate in the air tables to find the following value of P r(Tout)
Pr Tout   6.742 
7.268  6.742
482.49 kJ 472.24 kJ

kg
kg
 477.66 kJ 472.24 kJ 

  6.9899

kg
kg


This gives the ideal outlet pressure as Pout = PinPr(Tout)/Pr(Tin) = (100 kPa)(6.9899)/(1.2311) so the
ideal outlet pressure is Pout,ideal = 567.7 kPa .
Using constant heat capacities we have the first law result as follows.
W
W  m hin  hout   m c p Tin  Tout   Tout  Tin 
m c p
Using the low temperature value of cp = 1.005 kJ/kg gives
Tout  Tin 
 5 kW
60 s 1 kJ
W
 290 K 
 476.6 K
1
.
6
kg
1
.
005
kJ

mc p
min kW  s
min kg  K
The isentropic pressure for this outlet temperature and the given inlet conditions is found from the
following equation using k = 1.4 for air.
Tout
P
 Tin  out
 Pin



k 1
k

Pout
T
 Pin  out
 Tin
k
1.4
 k 1
 476.6 K  1.4 1
  100 kPa

 290 K 

Pout,ideal = 568.9 kPa
Error! Reference source not found. homework solutionsME 370, L. S. Caretto, Fall 2010 Page 6
Repeat both calculations to determine the actual outlet pressure if the compressor has the
same actual work input (5 kW) and an isentropic efficiency of 85%.
When we have an isentropic efficiency, we have to use the basic idea that the outlet pressure is
the same for both the actual process and the hypothetical isentropic process. We can thus
repeat the computations above; using the hypothetical isentropic process. In this process the first
compute isentropic work from the actual work that is given. For a work input device, s = ws/w so
ws = s w.
ws   s w   s
 5 kW 60 s 1 kJ
W
kJ
 85% 
 159.45
1.6 kg min kW  s
m
kg
min
Using the air tables we have the same values for hin= h(290 K) = 290.16 kJ/kg and Pr(Tin = 290 K)
= 1.2311 found above. We then use the first law to find hout,s = hin – (-159.4 kJ/kg) = 290.16 kJ/kg
+ 159.4 kJ/kg = 449.54 kJ/kg. The value of Pr(Tout,s) that corresponds to this value of hout is found
by interpolation as follows.
Pr Tout, s   5.332 
5.775  5.332
451.80 kJ 441.61 kJ

kg
kg
 449.54 kJ 441.61 kJ 

  5.6765

kg
kg


This gives the outlet pressure for the hypothetical isentropic process as Pout = PinPr(Tout,s)/Pr(Tin) =
(100 kPa)6765)/(1.2311) so the outlet pressure is Pout = 461.1 kPa .
Using constant heat capacities we have the first law result for the hypothetical isentropic outlet
state for this work input device as follows.
W
 W
W s  m hin  hout, s   m c p Tin  Tout, s   Tout, s  Tin  s  Tin  s
m c p
m c p
Using the low temperature value of cp = 1.005 kJ/kg gives
Tout,s  Tin 
 sW
m c p
 290 K 
85% 5 kW  60 s
1kJ
 448.6 K
1.6 kg 1.005 kJ min kW  s
min kg  K
The isentropic pressure for this outlet temperature and the given inlet conditions is found from the
following equation using k = 1.4 for air.
Tout,s
P 
 Tin  out,s 
 Pin 
k 1
k
k
 Pout,s
1.4
 T  k 1
 448.6 K 1.41

 Pin  out   100 kPa
 290 K 
 Tin 
Pout,ideal = 460.3 kPa