College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Unit Three Homework Solutions, September 16, 2010
1
A classroom that normally contains 40 people is to be air conditioned with window air
conditioning units of 6 kW cooling capacity. A person at rest may be assumed to
dissipate heat at a rate of about 360 kJ/h. There are 10 light bulbs in the room, each with a
rating of 100 W. The rate of heat transfer to the classroom through the walls is 15,000
kJ/h. If the room air is to be maintained at a constant temperature of 21oC, determine the
number of air-conditioning units required.
In this problem, we define the system to be the air in the room. We can assume that the air
behaves as an ideal gas. If the temperature remains constant, there is no change in the internal
energy of the air. Since the volume of the room remains constant, there is no work done. 1 Thus
the first law reduces to the equation that Q = 0. We account for the different heat sources in the
total heat rate, Q, as follows. The heat input comes (a) from the 40 students, Qs = (40
people)(360 kJ/h•person)(kW•s /kJ)(h/3600 s) = 4 kW, (b) heat transfer through the wall, Q w =
(15,000 kJ/h)(kW/kJ•s)(h/3600 s) =4.167 kW, and (c) heat added by the light bulbs, Q b = 10(100
W) = 1000 W = 1 kW. Thus the total heat input is 4 + 4.167 + 1 = 9.167 kW. The heat removal
comes from the air conditioners which remove 5 kW each. Thus for the net Q to be zero we must
have N air conditioners such that (5 kW)N = 9.167 kW. To satisfy this requirement, we must have
two air conditioning units .
2
A 0.5 m3 rigid tank contains regrigerant-134a initially at 160 kPa and 40% quality. Heat is
now transferred to the refrigerant until the pressure reaches 700 kPa. Determine (a) the
mass of refrigerant in the tank and (b) the amount of heat transferred. Also, show the
process on a P-v diagram with respect to the saturation lines.
The mass of refrigerant may be found from the initial state using the equation that m = V / v 1
where V = 0.5 m3 and v is found from the temperature and the quality. Since we are given an
initial quality, x1 = 40%, we know that we are in the mixed region. The specific volume is found
from the specific volumes of the saturated liquid and vapor, which are found in Table A-12 on
page 930: vf(160 kPa) = 0.0007437 m3/kg and vg(160 kPa) = 0.12348 m3/kg. We then find the
initial specific volume as follows.
3
  (0.4) 0.12348 m3   0.049838 m3
v  (1  x)v f  xvg  (1  0.4) 0.0007437 m
kg
kg 
kg



With this specific volume, we then find the refrigerant mass as follows:
m
V
0.5 m3

v 0.049838 m3
= 10.03 kg
kg
1
Note that is both the mass and volume remain constant, the specific volume remains constant. Since both
the temperature and specific volume are constant, the internal energy is a constant, regardless of whether or
not we assume that air is an ideal gas.
Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Unit three homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 2
To compute the heat transfer we apply the first law, Q = U + W. We assume that there is no
volume change in the “rigid” tank. If there is no volume change, no work is done. With W = 0, Q
= U. We find U = m(u2 – u1) where the specific internal energies are found from the property
tables. At the initial state we find u from the quality in the same way that we found the volume.
u  (1  x)u f  xug  (1  0.4) 31.09 kJ   (0.4) 221.35 kJ   107.19 kJ
kg 
kg 
kg


When we try to find the final state (P =
700 kPa, v = 0.049838 m3/kg) in the
superheat table, A-13, on page 930,
we see that the final specific volume in
the table for P = 700 kPa = 0.70 MPa
(at a temperature of 160oC) is only
0.048597 m3/kg. Furthermore, the
specific volume is increasing with
temperature so that the final state is at
a higher temperature than the final
temperature in the table.
Problem
4-282
Plot for
Problem
10000
Saturation
Path
1000
Pressure (kPa)
Because this is a constant volume
process, the final state has the same
specific volume as the initial state
(0.049838 m3/kg) and a given pressure
of 700 kPa. Knowing these data we
can plot the constant volume path for
this process as shown in the figure at
the left. We see that the initial state (1)
is in the mixed region and the increase
in pressure, at constant volume, brings
the final state into the gas region.
2
1
100
10
0.0001
0.001
0.01
0.1
The internal energy at the final state
3
can be found by extrapolation beyond
Specific volume (m /kg)
the last value in the table, using the
same formula as for interpolation, with the last two points in the table.
u  357.41kJ
kg

367.29 kJ
0.048597 m
kg
3
kg
 357.41kJ
 0.049838 m3  0.047306 m3   376.79 kJ

kg
kg 
kg

kg
 0.047306 m
1
3
kg
We can now find the heat transfer as follows.
Q  m(u 2  u1 )  (10.03 kg) 376.79 kJ  107.19 kJ  = 2,705 kJ
kg
kg 

3
A 20 ft3 rigid tank contains saturated regrigerant-134a vapor at 160 psia. As a result of
heat transfer from the refrigerant, the pressure drops to 50 psia. Show the process on a Pv diagram with respect to the saturation lines, and determine (a) the final temperature, (b)
the amount of refrigerant that has condensed, and (c) the heat transfer.
The solution of this problem is similar to the previous one. It is a constant volume process so that
the work is zero and the heat transfer equals the change in internal energy. The P-v diagram for
this problem is shown below. The process starts at the saturated vapor line at 120 psia and as
heat is removed, some refrigerant condenses, moving the process into the mixed region at the
final state where P2 = 50 psia..
Unit three homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 3
Since the final state is in the mixed region, we know that the final temperature is the saturation
temperature at the final pressure of 50 psia. From table A-12E on page 977, we find that the final
temperature, Tsat(50 psia) = 40.23oF .
V
20 ft 3
m 
v 0.29316 ft 3
 68.222 lbm
Problem 4-29E
3
1000
Saturation
Path
Pressure (psia)
The amount of refrigerant that has
condensed is the total mass minus the
mass that is vapor at the final state.
This is m – x2m. We can find the total
mass from the specific volume at the
initial state and the total volume of the
container: m = V/v1, where v1 is the
volume of the saturated vapor, vg, at the
initial pressure of 160 psia. From Table
A-12E on page 977 we find that vg(160
psia) = 0.29316 ft3/lbm, so that the total
mass is given by the following equation.
100
2
lbm
The specific volume at the final state is
the same as the initial specific volume,
so we can find the quality at this final
state from the saturated liquid and
vapor specific volumes at the final
pressure of 50 psia, taken from Table A12E.
x
1
v  vf
vg  v f

10
0.0001
0.01
0.1
1
10
3
Specific volume (ft /lbm )
3
0.28316 ft
0.94791 ft
0.001
lbm
3
lbm
3
 0.01252 ft
 0.01252 ft
lbm
3
 0.3000
lbm
The mass condensed can now be found.
mcondensed = m – mx2 = (1 – x2) m = ( 1 – 0.3000 ) ( 68.222 lbm ) = 47.75 lbm .
We have to find the internal energy at the initial and final states to compute the heat transfer. The
initial internal energy is simply ug(160 psia) = 108.50 Btu/lbm (Table A-12E). The internal energy
at the final state is found from the saturation properties and the quality, x2 = 0.3000.
u 2  (1  x2 )u f ( P2  50 psia )  x2u g ( P2  50 psia ) 
  (0.3000)100.69 Btu
  47.396 Btu
(1  0.3000) 24.832 Btu

lbm 
lbm 
lbm


The heat transfer can now be found from the first law.
Q = U + W = m(u2 – u1) + W = (68.222 lbm)(47.396 – 108.50) Btu/lbm + 0 = –4 ,169 Btu .
Unit three homework solutions
4
ME 370, L. S. Caretto, Fall 2010
Page 4
An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant
pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45
min through a resistor placed in the water. If one half of the liquid is evaporated during
this constant-pressure process and the paddle-wheel work amounts to 400 kJ, determine
the voltage of the source. Also, show the process on a P-v diagram with respect to the
saturation lines.
In this problem we assume that the insulation on the piston-cylinder device reduces the heat
transfer to a negligible amount so that we may assume that Q = 0. The evaporation of the liquid
comes from the addition of work from the resistor and the paddle wheel. At the same time, the
piston is expanding, at constant pressure, so that the water is doing work. Since the pressure is
constant, this work is simply equal to P(V2 – V1).
For this problem the first law becomes:
Q = 0 = U + W = m(u2 – u1) + P(V2 – V1) + W resistor + W paddle
We can write the volumes as the product of mass times specific volume and use the fact that P =
P1 = P2 to introduce the enthalpy, h = u + Pv.
–W resistor – W paddle = m(u2 – u1) + Pm(v2 – v1) =m[(u2 + P2v2) – (u1 + P1v1)] = m(h2 – h1)
We can find the mass from the initial state where we know V1 = 5 L and v1 = vf(P1 = 175 kPa).
Using Table A-5 on page 916 to find vf, we compute the mass as follows.
V
V1
5L
m 1 

v1 v f (175 kPa) 0.001057 m3
kg
0.001m3
 4.730 kg
L
Since the paddle wheel and resistance work are inputs, these are negative. Thus the values of
W resistor and W paddle are –EIt and –400 kJ, where E is the voltage that we want to find and I is the
given current of 8 A. The enthalpy at the initial state is simply hf at 175 kPa. The enthalpy at the
final state is the enthalpy of a mixture with a quality of 50% at the same 175 kPa pressure. We
thus have h1 = hf(175 kPa) = 487.01 kJ/kg, where we use Table A-5 for the saturation data.
Using the same table and the value of x2 = 50% we find h2 as follows.
h2  h f  xh fg   487.01 kJ   (0.5) 2213.1 kJ   1593.56 kJ
kg 
kg 
kg


Substituting the work terms and enthalpy and mass into our first law gives an equation for the
unknown voltage.
–W resistor – W paddle = EIt + 400 kJ = m(h2 – h1) = (4.730 kg)(1593.56 – 487.01) kJ/kg
Combining the numerical data gives an equation for the work done by the resistor.
EIt = 4834 kJ = 4834 kW•s
We can solve this to find the voltage applied to the resistor.
E
4834 kW  s 4834 kW  s 1 m 1000 V  A

= 224 V .
I t
(8 A)( 45 m) 60 s
kW
Unit three homework solutions
5
ME 370, L. S. Caretto, Fall 2010
Page 5
A piston-cylinder device initially contains steam at 200 kPa, 200oC, and 0.5 m3. At this
state a linear spring (F  x) is touching the piston but exerts no force on it. Heat is now
slowly transferred to the steam causing the pressure and the volume to rise to 500 kPa
and 0.6 m3, respectively. Show the process on a P-v diagram with respect to the
saturation lines and determine (a) the final temperature, (b) the work done by the steam,
and (c) the total heat transferred.
The path for this process is shown in the diagram below. The initial point of the path is (P1, V1) =
(200 kPa, 0.5 m3); the final point is (P2, V2) = (500 kPa, 0.6 m3).
P2
P
P1
V
V1
V2
The linear path results from the linear relationship between spring force and displacement and
the linear relationship between displacement and volume. The equilibrium point of the spring
occurs at the initial volume, since the spring is touching the piston but exerting no force at this
point. The pressure due to the spring is then given by the following equation.
V  V1
F k ( x  xe )
A  k V  V1
Pspring  

A
A
A
A2
k
The total pressure, P, acting on the water will be the sum of the constant pressure from the
weight of the piston (and atmospheric pressure), P1, and Pspring.
P  P1  Pspring  P1  k
V  V1
A2
This equation is seen to provide a linear relationship between pressure and volume as shown in
the diagram above. The work is the area under the path which is the area of a trapezoid: W = (P1
+ P2)(V2 – V1)/2. Using the values of pressure and volume given in the problem allows us to find
the work.
W 
P1  P2
( 200  500) kPa
kJ

(V2  V1 ) 
0.6 m 3  0.5 m 3 
 35 kJ
2
2
kPa  m 3
The heat transfer is given by the first law: Q = U + W = m(u2 – u1) + W. We have to find the
values of u1 and u2 from the property tables for water. In addition, we can find the mass from the
initial volume, V1 = 0.5 m3, and the initial specific volume, v1.
At the initial state of P1 = 200 kPa and T1 = 200oC, we find the following properties in the
superheat table, Table A-6 on page 918: v1 = 1.08049 m3/kg and u1 = 2654.6 kJ/kg. We can then
find the mass as follows.
V1
0.5 m3
m 
v1 1.08049 m3
 0.46275 kg
kg
Unit three homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 6
From this system mass, we can compute the specific volume at the final state where V2 = 0.6 m3.
3
V2
0.6 m3
v2 

 1.2966 m
kg
m 0.46275 kg
At the final pressure of 500 kPa, the specific volume, v2 = 1.2966 m3/kg occurs between
temperatures of 1100oC and 1200oC in Table A-6 on page 920. Interpolating between these two
points gives the temperature and internal energy at the final state.
u  4259.0 kJ
T  1100o C 
kg

4470.0 kJ
 4259.0 kJ
3
3
kg 
1.2966 m kg  1.26728 m kg   4325.8 kJ kg `
3


1.35972 m
 1.26728 m
kg
kg
kg
3
1200o C  1100o C
1.2966 m3  1.26728 m3   1131.7 o C `
3
3
kg
kg 

1.35972 m
 1.26728 m
kg
kg
So, the final temperature, T2 = 1132oC .
The heat transfer is found from the first law as usual.
Q = m(u2 – u1) + W = (0.46275 kg)(4325.8 – 2654.6)kJ/kg + 35 kJ = 808 kJ .
A piston-cylinder device initially
contains 0.8 m3 of saturated water
vapor at 250 kPa. At this state,
the piston is resting on a set of
stops and the mass of the piston
is such that a pressure of 300 kPa
is required to move it. Heat is
now slowly transferred to the
steam until the volume doubles.
Show the process on a P-V
diagram with respect to saturation
lines and determine (a) the final
temperature, (b) the work done
during this process, and (c) the
total heat transfer.
The P-v diagram for this process is
shown at the right. During the first
part of the process, from point 1 to
point 2, the pressure is too low to lift
the piston, so the process occurs at
constant volume. Once the pressure
reaches 300 kPa, the piston starts to
rise and the remainder of the
expansion (a doubling of volume)
occurs at constant pressure.
6
Problem 4-40
1000
900
Pressure (kPa)
6
800
Saturation
700
Path
600
500
400
2
3
300
200
1
100
0
0
0.5
1
1.5
Specific volume (m3/kg)
2
Unit three homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 7
The total heat transfer is found from the first law as Q = U + W = m(u3 – u1) + W, where W is
found as the area under the path: W = P2-3(V3 – V1) and P2-3 is the constant pressure of 300 kPa
= P2 = P3. We know that V1 = 0.8 m3 and V3 = 2 V1 = 1.6 m3. Thus, the work is found as follows.

W  P23 (V3  V1 )  (300 kPa) 1.6 m3  0.8 m3
 kPakJ m
3
W = 240 kJ
The mass is found from the initial volume of 0.8 m3 and the initial specific volume which is the
specific volume of a saturated vapor at 250 kPa. This is found from Table A-5 on page 916.
m
V1
V1
0.8 m3


v1 v g (250 kPa) 0.71873 m3
 1.113 kg
kg
Since the mass is constant and the volume doubles, the final specific volume is twice the initial
specific volume: v3 = 2v2 = 1.43746 m3/kg. The final state with this specific volume and a
pressure of 300 kPa (0.3 MPa) is found in the superheat tables, Table A-6 on page 918, to occur
at a temperature between 600oC and 700oC. The final temperature is found by interpolation.
T3  600o C 
700o C  600o C
1.49580 m
3
kg
 1.34139 m
1.437464 m3  1.34139 m3   662.2 o C `

kg
kg 

3
kg
So the final temperature, T3 = 662oC .
The initial internal energy for the saturated vapor state is found from Table A-5, u1 = ug(250 kPa)
= 2536.8 kJ/kg. The internal energy at the final state is found by an interpolation similar to the
one used for the temperature.
u3  3301.6 kJ
kg

3479.5 kJ
 3301.6 kJ
3
3
kg 

1.437464 m kg  1.34139 m kg   3412.3 kJ kg
3


1.49580 m
 1.34139 m
kg
kg
kg
3
The heat transfer is found from the usual equation for the first law.
Q = m(u3 – u1) + W = (1.113 kg)(3412.3 – 2536.8)kJ/kg + 240 kJ = 1214 kJ .