College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Unit Five Homework Solutions, September 30, 2010
1. Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are
10 MPa, 450oC, and 80 m/s. The exit conditions are 10 kPa, 92% quality, and 50 m/s. The
mass flow rate of the steam is 12 kg/s. Determine (a) the change in kinetic energy, (b) the
power output, and (c) the turbine inlet area.
Here we have a steady problem with one inlet and one outlet. We are also told that the turbine is
adiabatic so that the heat transfer is zero. The general first law and mass balance equations are
shown below.
dE system
dt
2
2




V
V
 Q  W u 
m o  ho  o  gz o  
m i  hi  i  gz i 




2
2
outlet

 inlet 



dmsystem
dt
For the steady problem we know that

 m i   m o
inlet
dE system
dt

outlet
dmsystem
dt
 0 . We are given no elevation data to
compute the potential energy changes, but these are usually small so we will make the
assumption that they are negligible. Since we have only one inlet and one outlet we have only
 in  m out  m . Finally, the stipulation that the turbine is adiabatic
one unique mass flow rate: m
means that
follows:
Q = 0. With these assumptions, the first law energy balance can be written as

2


Vin2  
Vout

   hout 

Wu  m  hin 
2  
2 

The first question asks us to find the change in kinetic energy. If we interpret this as the change
in kinetic energy per unit mass, we can find this change as shown below.

2
2
2
Vin2 Vout
1  80 m   50 m   1 kJ  s 2
KE 

 
= 1.95 kJ/kg
 
 
2
2
2  s   s   1000 kg  m 2
We find the power output from the first law equation. Here we need the inlet and outlet enthalpy
values which are found from the property tables for water.
hin = h(10 MPa, 450oC) = 3242.4 kJ/kg from the superheat tables
hout = hf(10 kPa) + xout hfg(10 kPa) = 191.81 kJ/kg + (0.92)(2392.1 kJ/kg) =2393.5 kJ/kg (data from
the saturation table with pressure as the look-up variable).
We then find the power output as follows
Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Unit five homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 2
2 2

V
V  12 kg  3242.4 kJ 2393.5 kJ 1.95 kJ  1 MW  s


=
W u  m  hin  hout  in  out  




2
2
s
kg
kg
kg

 1000 kJ


10.2 MW
The relationship between mass flow rate and cross sectional area of the flow channel is given by

VA
 
the following relationship. m
. To apply this equation to the turbine inlet, we have to find the
v
specific volume at that point. From the superheat tables, we find that vin = v(10 MPa, 450oC) =
0.029782 m3/kg. Thus, the area of the turbine inlet is found as follows.
vm 0.029782 m 3 12 kg s
A  
= 0.00447 m2 .
kg
s 80 m
V
2. Argon gas enters an adiabatic turbine steadily at 900 kPa and 450oC with a velocity of 80
m/s and leaves at 150 kPa with a velocity of 150 m/s. The inlet area of the turbine is is 60
cm2. If the power output of the turbine is 250 kW, determine the exit temperature of the
argon.
The application of the first law and the mass-balance equation to this problem proceeds in exactly
the same way as in the previous problem. The resulting equation for the work will therefore be
the same.
2
2




V
V
W u  m  hin  in    hout  out 
2  
2 

This problem is different from the previous one in that we will use ideal gas properties for argon.
In addition, we are not given the mass flow rate; however we can find it from the data given on
the inlet velocity and area. We have to find the specific volume, using the ideal gas equation at
the inlet, to apply this formula. Using v = RT/P for the specific volume, with R = 0.2081 kJ/kg∙K
for argon gives the mass flow rate as follows.


VA VAP 80 m
m 


60 cm 2
v
RT
s


2
 10 2 m 
1 kJ
1
kg

 (900 kPa) kg  K
 2.874
3
 cm 
0.2081 kJ 723.15 K kPa  m
s


For noble gases like argon, the heat capacity is constant over a wide temperature range. (See
figure 4-24 on page 177.) Thus we can use the equation that hout – hin = cp(Tout – Tin). We can
find the value of cp for argon in Table A-1 to be 0.5203 kJ/kg∙K.
We can combine this ideal gas relationship for the enthalpy change with the first law equation to
get the following result.
2 2


V
V 
W u  m c p Tin  Tout    in  out 
2 
 2

 Tout

2

W u  Vin2 Vout

 

m  2
2 
 Tin 
cp
We can solve the final equation for Tout. Note that we could solve for the temperature in either
kelvins or degrees Celsius.
Unit five homework solutions
Tout
ME 370, L. S. Caretto, Fall 2010
Page 3

2

W u  Vin2 Vout

 

m  2
2 
 Tin 

cp
2
2
250 kW 1 kJ
1  80 m   150 m   kJ  s 2
 

 
 
2.874 kg kW  s 2  s   s   1000 kg  m 2


s
723.15 K 
0.5203 kJ
kg  K
Tout = 540 K
3. Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24oC and leaves at
0.8 MPa and 60oC. The mass flow rate of the refrigerant is 1.2 kg/s. Determine (a) the
power input to the compressor and (b) the volume flow rate of the refrigerant at the
compressor inlet.
The application of the first law and the mass-balance equation to this problem proceeds in almost
exactly the same way as in the first problem. The only difference is that we are given no data on
the inlet and outlet velocities with which to compute the kinetic energy terms. Based on our
experience, we expect that these terms will make only a small contribution to the overall energy
balance so we will assume that they are zero. With this assumption, plus all the assumptions that
were used in problem 1, the equation for the work can be written as follows.
W u  m hin  hout 
We can find the inlet and outlet enthalpies from the property data given, using R-134a tables.
hin = hg(-24oC) = 235.92 kJ/kg, from the saturation tables with temperature as the look-up variable
hout =h(0.8 MPa, 60oC) = 296.81 kJ/kg from the superheat tables.
We can now find the power input to the compressor.
1.2 kg  235.92 kJ 296.81 kJ  kW  s


W u  m hin  hout  

= –73.07 kW
s 
kg
kg
 1 kJ
The negative value for indicates that this is a power input to the compressor.
The volume flow rate is the product of the mass flow rate (kg/s) and the specific volume (m 3/kg).
The specific volume at the inlet is found from the saturation tables; vin = vg(-24oC) = 0.17395 m3/kg.
We can then find the volume flow rate at the inlet by multiplying this by the mass flow rate of
1.2 kg/s.
1.2 kg 0.117395 m 3 0.2087 m 3


V  mv 

s
kg
s
Unit five homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 4
4. Refrigerant-134a is throttled from the saturated liquid state at 700 kPa to a pressure of 160
kPa. Determine the temperature drop during this process and the final specific volume of
the refrigerant.
After this introduction to steady processes, you can simply approach this problem by starting with
the statement that the enthalpy is constant in a throttling process. To see how we reach this
conclusion, we start with our usual equations for the first law and the mass balance.
dE system
dt
 Q  W u 
2
2




V
V
o
m o  ho 
 gz o  
m i  hi  i  gz i 




2
2
outlet

 inlet 



dmsystem
dt

 m i   m o
inlet
outlet
Here our system is a valve. We assume that this is a steady flow, so
dE system
dt

dmsystem
dt
 0.
We are given no elevation data to compute the potential energy changes or velocity data to
compute kinetic energy changes, but these are usually small so we will make the assumption that
they are negligible. Since we have only one inlet and one outlet, we have only one unique mass
 in  m out  m . The valve has no work output device such as a shaft that rotates or
flow rate: m
electrical leads producing power, so we say that the useful work is zero. Finally, we assume that
the definition of a throttling process is one in which the valve is well-insulated so that the heat
transfer is zero. With these assumptions, the first law energy balance can be written as follows:
W u  Q  m hin  hout 
so that
0  0  m hin  hout 
or
hin  hout
We can find the inlet enthalpy from the R-134a tables: hin = hf(700 kPa) = 88.82 kJ/kg from the
saturation tables with temperature as the look-up variable. Since hout = hin and we are given Pout
= 160 kPa, we know that the final state is one where P = 160 kPa and h = 88.82 kJ/kg.
From the saturation tables at the outlet pressure of 160 kPa (0.16 MPa), we find that hf(160 kPa)
= 31.21 kJ/kg and hg(160 kPa) = 241.11 kJ/kg. Since hout = 88.82 kJ/kg < hg(160 kPa) = 241.11
kJ/kg and > hf(160 kPa) = 31.21 kJ/kg, the outlet enthalpy of 88.82 kJ/kg is in the mixed region at
this pressure.
Since the inlet is a saturated liquid, we know that T in = Tsat(Pin = 700 kPa) = 26.69oC. Similarly,
since the outlet is in the mixed region, we know that Tout = Tsat(Pout = 160 kPa) = -15.60oC. We
can then find the temperature drop.


Tin  Tout  26.69 o C  .  15.60 o C  42.29 o C
We can find the final specific volume from the final quality, where we use the final enthalpy to find
the quality. To compute the specific volume, we need the specific volumes of the saturated liquid
and saturated vapor at the final pressure of 160 kPa. From the saturation tables we find vf =
0.0007437 m3/kg and vg = 0.12348 kJ/kg. Thus the final specific volume can be found as follows.
Unit five homework solutions

ME 370, L. S. Caretto, Fall 2010

v  v f  x vg  v f  v f 
88.82 kJ 31.21 kJ

kg
kg
241.11 kJ 31.21 kJ

kg
kg
h  hf
hg  h f
Page 5
v g  v f   0.0007437 mkg 
3
v = 0.0344 m3/kg
 0.12348 m 3 0.0007437 m 3 





kg
kg


5. Refrigerant-134a at 1 MPa and 90oC is to be cooled to 1 MPa and 30oC in a condenser by
air. The air enters at 100 kPa and 27oC with a volume flow rate of 600 m3/min and leaves at
95 kPa and 60oC. Determine the mass flow rate of the refrigerant.
One approach that we will not take here is to
define two systems: the flowing air and the
flowing refrigerant. In this approach we would
compute the heat transfer from the air and set
that equal to the heat transfer to the
refrigerant. We will not take this approach.
Instead we will define one system as shown in
the diagram.
In this one system, we have two inlet flows and
two outlet flows. However, the air and R-134a
flows do not mix. Thus we conclude, for a
steady system, in which
dm system
dt
Air in
R-134a
in
R-134a
out
---system
boundary
= 0, that we
Air out
 air,in  m air,out m air , and m R ,in  m R ,out m R . (In this
have only two unique mass flow rates: m
problem we will use the subscript, “R” to indicate R-134a.
We can apply this result to our analysis of the first law equation.
2
2




V
V
 Q  W u 
m o  ho  o  gz o  
m i  hi  i  gz i 




2
2
outlet

 inlet 

dE system

dt
For this steady flow,
dE system
dt

 0 . We are given no elevation data to compute the potential
energy changes or velocity data to compute kinetic energy changes, but these are usually small
so we will make the assumption that they are negligible. The cooler has no work output device
such as a shaft that rotates or electrical leads producing power, so we say that the work is zero.
Finally, we assume that the cooler is designed so that the heat transfer from the cooler, defined
as a single system, is negligible, compared to the internal heat transfer between the air and the
refrigerant. So we will take the heat transfer to be zero. With these assumptions, the first law
energy balance can be written as follows:
0  m air,out hair,out  m R ,out hR ,out  m air,in hair,in  m R ,in hR ,in
 air,in  m air,out m air , and m R ,in  m R ,out m R from our analysis of the
Using the results that m
mass balance equation (and noting that the streams do not mix) our first law energy balance
becomes.
Unit five homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 6
m air hair,out  hair,in   m R hR ,in  hR ,out 
We can find the enthalpies of R-134a from the property tables: hR,in = hR(1 MPa, 90oC) = 324.64
kJ/kg from superheat tables on page 930. The value of hR,out = h(1 MPa, 30oC) is the enthalpy of
a compressed liquid. We can see this since the temperature of a saturated liquid at 1 MPa is
39.37oC, which is higher than the outlet temperature. We can approximate the outlet refrigerant
enthalpy as the enthalpy of a saturated liquid at the given temperature of 30 0C: hR,out = h(1 MPa,
30oC) ≈ hf(30oC) = 93.58 kJ/kg from Table A-11 on page 927.
We will assume that air behaves as an ideal gas at the temperatures and pressures of the air in
this problem. Since the temperature difference of the air is small and air is at low temperature,
we will use a constant heat capacity for air of 1.005 kJ/kg∙K, from Table A-2 on page 909. From
the same table we can find the gas constant for air: R = 0.2870 kJ/kg∙K.
We do not know the mass flow rate of the air, but we can compute it from the inlet volume flow
rate, Vin , and the inlet specific volume, vin = RTin/Pin.
m air 
Vair ,in
v air ,in

Vair ,in Pair ,in
RTair ,in
600 m 3 kg  K 100 kPa 1 kJ

 696.5 kg
3
min
min 0.287 kJ 300.15 K kPa  m
We can how substitute h = cpT for air into our first law and solve for the refrigerant flow rate.
m air hair,out  hair,in   m airc p ,air Tair,out  Tair,in   m R hR ,in  hR ,out 
In using this equation, we can compute the temperature difference in degrees Celsius or in
kelvins and get the same result.
696.5 kg 1.005 kJ
60  27 K
m air c p,air Tair ,out  Tair ,in 
min
kg  K
= 101.0 kg/min .
m R 

324.64 kJ 95.38 kJ
hR,in  hR,out 

kg
kg
6. Air enters the evaporator section of a window air conditioner at 14.7 psia and 90 oF with a
volume flow rate of 200 ft3/min. Refrigerant-134a at 20 psia with a quality of 30% enters the
evaporator at a rate of 4 lbm/min and leaves as a saturated vapor at the same pressure.
Determine (a) the exit temperature of the air and (b) the rate of heat transfer from the air.
This problem is similar to the previous one. For the first part of the problem we will analyze the
system as a combined system. We will make the same assumptions that we did in problem 5 –
steady state, air and refrigerant streams do not mix, negligible kinetic and potential energies, heat
and work both equal zero – so that we will arrive at the same equation that we had for the
previous problem.
m air hair,out  hair,in   m R hR ,in  hR ,out 
We can find the enthalpies of R-134a from the property tables: hR,out = hg(20 psia) = 102.73
Btu/lbm from the saturation table, A-12E. At the same pressure, the enthalpies of the saturated
liquid and vapor-liquid phase chanfe, hfg, are found from Table A-12E to compute the inlet
enthalpy from the given quality, xin = 0.3.
hin  h f (20 psia )  xin h fg (20 psia ) 
11.445 Btu
91.282 Btu 38.83 Btu
 (0.3)

lbm
lbm
lbm
Unit five homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 7
We will assume that air behaves as an ideal gas at the temperatures and pressures of the air in
this problem. Since the temperature difference of the air is small and air is at low temperature,
we will use a constant heat capacity of 0.240 Btu/lbm∙R, from Table A-2E. From the Table A-1E,
we can find the gas constant for air: R = 0.3704 psia∙ft3/lbm∙R.
We do not know the mass flow rate of the air, but we can compute it from the inlet volume flow
rate, Vin , and the inlet specific volume, vin = RTin/Pin. (Add 459.67 to the 90oF inlet temperature to
get the inlet temperature as 549.67 R.)
m air 
Vair,in
vair,in

Vair,in Pair,in
RTair,in

lbm  R
200 ft 3
14.7 psia
lb
 14.43 m
3
min
min 0.3704 psia  ft 549.67 R
We can how substitute h = cpT for air into our first law and solve for the outlet air temperature.
m air hair,out  hair,in   m airc p ,air Tair,out  Tair,in   m R hR ,in  hR ,out 
Tair ,out  Tair ,in 
m R hR,in  hR,out 
m air c p,air
 90 0 F 
4 lbm
min
 38.83 Btu 102.73 Btu 



lbm
 lbm
 = 16.2oF .
14.43 lbm 0.24 Btu
min
lbm  R
To get the heat transfer from the air to the refrigerant, we have to analyze either the air only or
the refrigerant only as the system. Here we choose to analyze the refrigerant only. With the
assumptions discussed previously – steady flow, negligible kinetic and potential energy changes,
no work – plus the fact that for the refrigerant only we have one inlet and one outlet, the first law
reduces to the following expression.
4 lbm
Q from air to refrigerant  Q  m R hR,out  hR,in  
min
 102.73 Btu 38.83 Btu 

 = 255.6 Btu/min .

lbm
lbm


Here the positive sign indicates that the direction of heat flow is from the air to the refrigerant.