College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Unit Eight Homework Solutions, November 2, 2010
1
The radiator of a steam heating system has a volume of 20 L and is filled with superheated
water at vapor at 200 kPa and 150oC. At this moment both the inlet and exit valves to the
radiator are closed. After a while the temperature of the steam drops to 40oC as a result of
heat transfer to the room air. Determine the entropy change of the steam during this
process in kJ/K.
If we assume that the radiator does not expand or contract significantly with the temperature
change, we have is a constant volume process. We can find the entropy change from the initial
state, where we are given the temperature and pressure and the final state where we know the
temperature and we also know that the specific volume will be the same as it is at the initial state.
We can also use the information on the radiator volume to find the system mass.
At the initial state of 200 kPa and 150oC, we find the following properties from Table A-6 on page
920: v1 = 0.95986 m3/kg and s1 = 7.2810 kJ/kg∙K. From Table A-4 on page 916, we see that, at
the final temperature T2 = 40oC, the specific volume, v2 = v1 is between vf(40oC) = 0.001008
m3/kg and vg(40oC) = 19.515 m3/kg. Thus the final state is in the mixed region. We find the final
quality from the following equation.
x2 
v2  v f ( 40o C )
v g (40 C )  v f ( 40 C )
o
o

0.95986 m
19.515 m
3
kg
3
kg
 0.001008 m
 0.001008 m
3
kg
3
 0.04913
kg
We can then use this quality to find the final entropy.
s 2  s f (40 o C )  x2 s fg (40 o C ) 
 7.6832 kJ  0.9499 kJ
0.5724 kJ
 
 (0.04913)
kg  K
kg  K
 kg  K 
To find the total entropy change we have to find the system mass. We can find this from the
radiator volume and the initial specific volume.
m3
V
1000 L
m 
 0.02084 kg
v1 0.95986 m 3
kg
20 L 
 0.9499 kJ 7.2810 kJ 
 = –0.132 kJ/K
S  ms 2  s1   0.02084 kg 

kg  K 
 kg  K
Note that we can have a negative entropy change and still satisfy the second law inequality that
dS ≥ dQ/T because the heat transfer is negative. (We did not actually compute Q, but we know
that heat has to leave the system to change the state from a vapor at 150oC to a mixture of liquid
and vapor at 40oC.)
Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Unit eight homework solutions
2
ME 370, L. S. Caretto, Fall 2010
Page 2
A well insulated rigid tank contains 2 kg of a saturated liquid-vapor mixture of water at 100
kPa. Initially, three quarters of the mass is in the liquid phase. An electrical resistance
heater placed in the tank is now turned on and kept on until all the liquid in the tank is
vaporized. Determine the entropy change of the steam during this process.
In this case the initial state is a pressure of 100 kPa and a quality of 0.25. Because it is a rigid
tank, with a fixed mass, the final state has the same specific volume as the initial state. From
Table A-5 on page 918, we find the properties of the saturated liquid and vapor states at 100 kPa
and use these to compute the initial specific volume and entropy. (Note that ¾ of the mass in the
liquid state is equivalent to a quality of 25%.)
s1  s f (100 kPa)  x1s fg (100 kPa) 
 6.0562 kJ  2.8168 kJ
1.3028 kJ
 (0.25)
 
kg  K
kg  K
 kg  K 
v1  v f (100 kPa)  x1 v g (100 kPa)  v f (100 kPa) 
0.001043 m 3

kg
 1.6941 m 3 0.001043 m 3  0.4243 m 3
 
(0.25)

kg
kg
kg


Since we are told that the final state is a saturated vapor and we reasoned that this is a constant
volume process, we have to find the final state as the saturated vapor state where the
temperature (or pressure) is such that vg = v1 = 0.4243 m3/kg. Looking at the vg data in Table A4, we see that this final state lies between a temperature of 145oC where vg = 0.44600 m3/kg, and
150oC, where vg = 0.39248 m3/kg. We can interpolate between these two specific volumes and
the corresponding values of sg to find the final entropy. This interpolation gives the following
result.
s2  s g (145 C ) 
o
s g (150o C )  s g (145o C )
v g (150 C )  v g
o
v
(145 C )
o
6.8371 kJ 6.8827 kJ

6.8827 kJ
kg  K
kg  K

3
0.39248 m
0.44600 m 3
kg  K

kg
kg
2

 v g (145o C ) 
 0.4243 m 3 0.44600 m 3  6.8649 kJ

 

kg
kg
kg  K


We can now compute the entropy change.
 6.8649 kJ 2.8168 kJ 
 = 8.10 kJ/K
S  ms 2  s1   2 kg 

kg  K 
 kg  K
3
A piston-cylinder device contains 2 lbm of refrigerant-134a at 120 psia and 100oF. The
refrigerant is now cooled at constant pressure until it exists as a liquid at 50oF. Determine
the entropy change of the refrigerant during this process.
At the initial state of 120 psia and 100oF, we find the entropy s1 = 0.22362 Btu/lbm∙ from Table A13E on page 981. Since the process occurs at constant pressure, the final pressure equals the
initial pressure of 120 psia. At this 120 psia and the final temperature of 50oF, we have a
compressed liquid because the temperature of 50oF is below the saturation temperature at 120
psia, which is 90.49oF. Here we make the approximation that the entropy of the liquid state
equals the entropy of the saturated liquid at the temperature of 50oF. From Table A-11E on page
Unit eight homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 3
978, we find this entropy to be 0.06039 Btu/lbm∙R. Hence we compute the entropy change as
follows.
 0.06039 Btu 0.22362 Btu 
 = –0.3264 Btu/R
S  ms 2  s1   2 lbm 

lb

R
lb

R
m
m


This is another example of a negative entropy change that presumably satisfies the second law
(dS ≥ dQ/T) because dQ < 0 for cooling.)
4
An insulated piston-cylinder device contains 0.05 m3 of saturated refrigerant 134a vapor at
0.8 MPa pressure. The refrigerant is now allowed to expand in a reversible manner until
the pressure drops to 0.4 MPa. Determine (a) the final temperature in the cylinder and (b)
the work done by the refrigerant.
Here we assume that the process in our “well-insulated cylinder” is adiabatic. Since we are given
that the expansion is reversible, we have reversible adiabatic process. We know that such a
process is an isentropic (constant entropy) process, so that the final entropy equals the initial
entropy. We can use this fact to find the answers to both parts of this problem.
First we find the data for the initial state. Since this is a saturated vapor at 0.8 MPa, we find the
following saturated vapor properties from Table A-12 on page 930: v1 = 0.025621 m3/kg, u1 =
246.79 kJ/kg, and s1 = 0.91835 kJ/kg∙K.
We can find the (constant) mass from the initial volume, V1 = 0.05 m3 and the initial specific
volume.
V1
0.05 m 3
m 
v1 0.025621 m 3
 1.952 kg
kg
For this closed system, the first law is Q = U + W. Since Q = 0, our first law reduces to W = -U
= m(u1 – u2). We can find u2, because we know the pressure P2 = 0.4 MPa and the entropy s2 =
s1 = 0.91835 kJ/kg∙K at the final state. From the data on saturation properties at 0.4 MPa (400
kPa) in Table A-12 on page 930, we see that the value of s2 lies between the values of sf =
0.24761 kJ/kg∙K and sg = 0.92691 kJ/kg∙K. Thus we know that we are in the mixed region. This
means that T2 = Tsat(P2 = 0.4 MPa) = 8.91oC .
We have to find the quality to determine the final internal energy.
x2 
s2  s f (0.4 MPa )
s fg (0.4 MPa )

 0.24761 kJ
kg  K
kg  K
 0.9874
0.67929 kJ
kg  K
0.91835 kJ
u2  u f (0.4 MPa )  x2 u fg (0.4 MPa ) 
63.62 kJ
171.45 kJ 232.91 kJ
 (0.9874)

kg
kg
kg
We now find the work from the first law.
 246.79 kJ 232.91 kJ 
W  m(u1  u2 )  (1.952 kg)

 = 27.08 kJ
kg
kg


Unit eight homework solutions
5
ME 370, L. S. Caretto, Fall 2010
Page 4
A piston-cylinder device contains 1.2 kg of saturated water vapor at 200oC. Heat is now
transferred to the steam and the steam expands reversibly and isothermally to a final
pressure of 800 kPa. Determine the heat transferred and the work done during this
process.
Here, since the process is reversible we know that dS = dQ/T. Since it is isothermal we can
integrate this equation to obtain S = (dQ/T) = (1/T)dQ = Q/T. Thus, the heat transferred in the
process can be found from the equation Q = TS = Tm(s2 – s1). We can find the two entropy
values since we have enough data to specify the initial and final states. Once we know the heat
transfer, we can find the work from the first law, Q = U + W. Solving this equation for the work
gives W = Q – m(u2 – u1).
At the initial state of saturated vapor at 200oC, we find the following properties from Table A-4 on
page 916: s1 = sg(200oC) = 6.4302 kJ/kg∙K and u1 = ug(200oC) = 2594.2 kJ/kg. At the final state
where T2 = T1 = 200oC and P2 = 800 kPa (0.8 MPa), we find the properties from Table A-6 on
page 894: s2 = s(200oC,800 kPa) = 6.8177 kJ/kg∙K and u2 = u(200oC,800 kPa) = 2631.1 kJ/kg.
We use the entropy values to compute the heat transfer at the constant temperature of 200oC =
473.15 K.)
 6.8177 kJ 6.4302 kJ 
Q  TS  Tms2  s1   (473.15 K )1.2 kg 

 = 220.0 kJ
kg  K 
 kg  K
We now use the first law to find the work.
 2631.1 kJ 2594.2kJ 
W  Q  mu2  u1   220.0 kJ  1.2 kg 

 = 175.7 kJ
kg
kg


6
A 50-kg copper block, initially at 80oC, is dropped into an insulated tank that contains
120 L of water at 25oC. Determine the final equilibrium temperature and the total entropy
change for this process.
Here we assume that the total system, consisting of the water and the copper block is isolated to
that there is no heat transfer or work between the tank (with the water and the block) and any
other system. In this case, the first law, Q = U + W becomes U = 0, or U = mCu(uCu,2 – uCu,1) +
mw(uw,2 + uw,1) = 0. We further assume that the change in internal energy can be modeled as the
product of a heat capacity and a temperature difference; i.e., du = cdT, where there is a negligible
difference between cp and cv for liquids and solids. From Table A-3 on page 915 we find the
following data on the heat capacity and density for copper and liquid water: at 25oC, w = 997
kg/m3 and cw = 4.18 kJ/kg∙oC; at 27oC, cCu = 0.386 kJ/kg∙oC. (Note that we could also have the
following units for the heat capacities: kJ/kg∙K.) Assuming that the heat capacities are constant
we have the following equation for our first law:
mCu(uCu,2 – uCu,1) + mw(uw,2 + uw,1) = mCucCu(TCu,2 – TCu,1) + mwcw(Tw,2 + Tw,1) = 0
The final temperature of the copper and the water will be the same: T Cu,2 = Tw,2 = T2. Substituting
T2 for TCu,2 and Tw,2, and solving for T2 gives the following result for the final temperature.
T2 
mw cwT1,w  mCu cCu T1,Cu
mw cw  mCu cCu
We can find the mass of the water from the density and the volume.
Unit eight homework solutions
ME 370, L. S. Caretto, Fall 2010
mw   wVw 
Page 5
997 kg 120 L
 120 kg
m 3 1000 L
m3
We now have all the values required to compute T 2.
T2 
mw cwT1,w  mCu cCu T1,Cu
mw cw  mCu cCu
4.18 kJ
0.386 kJ
( 25o C )  (50 kg)
(80o C )
o
o
kg C
kg C
= 27oC
4.18 kJ
0.386 kJ
(120 kg)
 (50 kg)
kgo C
kgo C
(120 kg)

We can now compute the entropy change. We assume that each substance undergoes an
internally reversible heat transfer such that dS = dQ/T = mcdT/T. The assumption that each
substance has a constant heat capacity allows us to integrate this equation giving S =
mcln(T2/T1). In this calculation the temperature must be in kelvins. We can apply this equation to
both substances, here using units of kelvins for the heat capacity.
T 
4.18 kJ  300.15  3.335 kJ

S w  mw cw ln  2   (120 kg)
ln 
kg  K  298.15 K 
kg  K
 T1,w 
 T
S Cu  mCu cCu ln  2
 T1,Cu

0.386 kJ  300.15 
3.140 kJ
  (50 kg)
  
ln 

kg  K
kg  K
 253.15 K 

The total entropy change is simply the sum of these two entropy changes.
Stotal  S w  SCu 
3.335 kJ 3.140 kJ  0.215 kJ

kg  K
kg  K
kg  K
We see that the total entropy change for this isolated system is positive satisfying the second law
of thermodynamics that requires S  0 for an isolated system. However, the entropy change for
the copper, which is cooled in this process, is negative.
7
An insulated piston-cylinder device contains 0.05 m3 of saturated refrigerant 134a vapor at
0.8 MPa pressure. Find the maximum work that can be done if the refrigerant expands to a
pressure of 0.4 MPa in an adiabatic process.
The maximum work occurs in a reversible process. If we assume that the “insulated” device has
no heat transfer, we have an adiabatic process so the maximum work will be in a reversible
adiabatic process. In such a process, the entropy is constant. This constant entropy expansion
has already been solved in problem 4 where we found the work. We know that this must be the
maximum work, Wmax = 27.08 kJ