College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Unit Six Homework Solutions, October 7, 2010
1
Consider an 8 L evacuated rigid bottle that is surrounded by the atmosphere at 100 kPa
and 17oC. A valve at the neck of the bottle is now opened and the atmospheric air is
allowed to flow into the bottle. The air trapped in the bottle eventually reaches thermal
equilibrium with the atmosphere as a result of heat transfer through the wall of the bottle.
The valve remains open during the process so that the trapped air also reaches
mechanical equilibrium with the atmosphere. Determine the net heat transfer through the
wall of the bottle during this filling process.
If we define the bottle as our system, we see that we have an unsteady problem because mass
enters through the one inlet and there are no outlets for the mass to exit. The general first law
equation for unsteady open systems is shown below.


 



V2
V2
 gz   m1  u 
 gz  
 Q  Wu
 m2  u 



2
2
 
2

1  system






Vi 2
Vi 2
  mi  hi 
 gzi    mi  hi 
 gzi 
2
2
outlet

 inlet 

We see that there is no mechanism for useful work in this system so we set W u = 0 and make the
usual assumption that kinetic and potential energy terms are zero. This gives the following
expression for the first law.
m2u2  m1u1 system  m2u2  m1u1  Q  minhin
Since the cylinder is initially evacuated, we have m 1 = 0, so that m2 = min = m. This gives the
following result for the first law.
Q  m2 u2  m1u1  min hin  m2 u2  0  min hin  mu2  hin 
We can compute the mass from the given data for the final state. The temperature and pressure
in the bottle at the final state are the same as those of the atmosphere due to the thermal and
mechanical equilibrium. Thus, P2 = 100 kPa and T2 = 17oC = 290.15 K. We find the gas constant
for air from Table A-1: R = 0.2870 kJ/kg∙K = 0.2870 kPa∙m3/kg∙K. We then find the mass as
follows:
(100 kPa)(8 L)
m  min  m2 
m3
1000 L
P2Vbottle

 0.0096 kg
0.2870 kPa  m 3
RT2
(290.15 K )
kg  K
Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Unit six homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 2
We can use the ideal gas properties for air in Table A-17. For a temperature of 290.15, which is
both the temperature of the inlet air and the final temperature in the cylinder, we find u2 =
206.91 kJ/kg and hin = 290.16 kJ/kg. We use these values and the mass to find the heat transfer.
Q  mu2  hin   (0.0096 kg) 206.91 kJ  290.16 kJ  = –0.8 kJ
kg
kg 

The minus sign shows that heat is transferred out of the bottle.
2
An insulated rigid tank is initially evacuated. A valve is opened and atmospheric air at
95 kPa and 17oC enters the tank until the pressure in the tank reaches 95 kPa, at which
point the valve is closed. Determine the final temperature of the air in the tank. Assume
constant specific heats.
If we define the tank as our system, we see that we have an unsteady problem because mass
enters through the one inlet and there are no outlets for the mass to exit. The general first law
equation for unsteady open systems is shown below.
2
2
 


 
V
V


m2  u 
 gz  m1 u 
 gz  
 Q  Wu


 
 
2
2
2

1  system
 
2
2




V
V

mo  ho  o  gz o  
mi  hi  i  gz i 




2
2
outlet

 inlet 



We see that there is no mechanism for useful work in this system so we set W u = 0 and make the
usual assumption that kinetic and potential energy terms are zero. This gives the following
expression for the first law.
m2u2  m1u1 system  m2u2  m1u1  Q  minhin
Since the cylinder is initially evacuated, we have m 1 = 0, so that m2 = min = m. This gives the
following result for the first law.
Q  m2 u2  m1u1  min hin  m2 u2  0  min hin  mu2  hin 
If we further assume that the “insulated” tank is so well insulated that the heat transfer is zero, the
first law reduced to the following final form.
Q  mu2  hin   0

u2  hin
We are told to use constant heat capacities; this allows us to compute changes in u or h from
changes in temperature. But, here we have a difference between an internal energy and an
enthalpy. We can use the relationship that the enthalpy definition, h = u + Pv, becomes h = u +
RT for an ideal gas. Thus, we can rewrite the u2 = hin equation as follows.
u2  hin  uin  RTin

u2  uin  RTin
We can use the result that u = cvT for an ideal gas with constant heat capacity to obtain an
equation to solve for the final temperature.
Unit six homework solutions
u2  uin  cv T2  Tin   RTin
ME 370, L. S. Caretto, Fall 2010

T2  Tin 
Page 3
cp
c R
R
Tin  v
Tin  Tin  kTin
cv
cv
cv
Here, we have used the definition of k as the ratio of heat capacities, k = c p/cv. From Table A-2(a)
we find the value of k for air at 300 K is 1.4, so T2 = kTin = 1.4(290.15 K) = 406 K = 133oC. We
see that the average temperature is (290 K+ 406 K)/2 = 348 K, We can check the value of k as a
function of temperature from Table A2(b) and we see that it is still essentially 1.4 at this average
temperature.
3
A 0.2 m3 rigid tank initially
Inflow
Pin = 1 MPa
contains refrigerant-134a at
oC
T
=
100
in
8oC. At this state 70% of the
mass is in the vapor phase
and the rest is in the liquid
Q
phase. The tank is
connected by a valve to a
Tank
supply line the refrigerant at
V = 0.2 m3
1 MPa and 100oC flows
T1= 8oC
steadily. Now the valve is
x1 = 0.7
opened slightly and the
refrigerant is allowed to
enter the tank. When the
pressure in the tank reaches
800 Pa, the entire refrigerant in the tank exists as a saturated vapor. At this point the valve
is closed. Determine (a) the final temperature in the tank, (b) the mass of refrigerant that
has entered the tank and (c) the heat transfer between the system and the surroundings.
Since the final state is a saturated vapor, we know that the temperature must be the saturation
temperature at the final pressure of 800 kPa. Thus, T2 = Tsat(P2 = 800 kPa) = 31.31oC , from
Table A-12, page 928.
If we define the tank as our system, we see that we have an unsteady problem because mass
enters through the one inlet and there are no outlets for the mass to exit. The general first law
equation for unsteady open systems is shown below.
2
2
 


 
V
V


m2  u 
 gz  m1 u 
 gz  
 Q  Wu


 
 
2
2
2

1  system
 
2
2




V
V
o

mo  ho 
 gz o  
mi  hi  i  gz i 




2
2
outlet

 inlet 



We see that there is no mechanism for useful work in this system so we set W u = 0 and make the
usual assumption that kinetic and potential energy terms are zero. This gives the following
expression for the first law.
m2u2  m1u1 system  m2u2  m1u1  Q  minhin
To find the mass added we simplify the general the mass balance equation for this problem
where there is only one inlet. This gives the following result.
Unit six homework solutions
ME 370, L. S. Caretto, Fall 2010
m2  m1 system

 mi   mo
inlet

Page 4
m2  m1  min
outlet
The initial mass, m1, is found from knowing the initial specific volume, v1, which is found from the
initial temperature and quality as follows

v1  v f (T1  8 o C )  x1 v g (v f (T1  8 o C )  v f (T1  8 o C )
 0.0007887 m
3
kg

3
3
  0.037170 m 3
 (0.7) 0.052762 m
 0.0007887 m
kg
kg
kg


With this initial specific volume, we find the initial mass as follows.
m1 
0.2 m 3
V

v1 0.037170 m 3
 5.3807 kg
kg
At the final state, the specific volume is that of the saturated vapor; i.e., v2 = vg(P2 = 800 kPa) =
0.025621 m3/kg. We can use this specific volume to find the final mass in the tank.
m2 
0.2 m 3
V

v 2 0.025621 m 3
 7.8061 kg
kg
We can now find the added mass from our mass balance equation.
min  m2  m1  7.8061 kg  5.38068 kg = 2.43 kg .
In order to compute the heat transfer we have to find the values of the energy properties. These
are all found from the tables for R-134a.
u1  u f (T1  8 o C )  x1u fg (T1  8 o C )  62.39 kJ
u2 = ug(P2 = 800 kPa) = 246.79 kJ/kg
kg
 (0.7)172.19 kJ   182.92 kJ
kg 
kg

hin = h(1 MPa, 100oC) = 335.06 kJ/kg
Substituting these property values and the values for the initial, final and added masses into the
first law gives the heat transfer.
Q  m2 u 2  m1u1 system  min hint  (7.8061 kg) 246.79 kJ  
kg 

(5.38068 kg)182.92 kJ   (2.4254 kg) 335.06 kJ 
kg 
kg 


Q = 130 kJ
The positive sign for Q indicates that heat is added to the system.
Unit six homework solutions
4
ME 370, L. S. Caretto, Fall 2010
Page 5
An insulated 60 ft3 rigid tank contains air at 75 psia and 120oF. A valve connected to the
tank is now opened and air is allowed to escape until the pressure inside drops to 30 psia.
The air temperature during this process is maintained constant by an electrical resistance
heater placed in the tank. Determine the electrical work done during this process.
If we define the tank as our system, we see that we have an unsteady problem because mass
leaves through the one outlet and there are no inlets for the mass to enter. The general first law
equation for unsteady open systems is shown below.
2
2
 


 
V
V
m2  u 
 gz   m1  u 
 gz  
 Q  Wu


 
 
2
2
2

1  system
 
2
2




V
V
o

mo  ho 
 gz o  
mi  hi  i  gz i 




2
2
outlet

 inlet 



We see that there is an electrical work input for the resistance “heater”. We will assume that the
heat transfer is negligible for this insulated tank. We also make the usual assumption that kinetic
and potential energy terms are zero. This gives the following expression for the first law.
m2u2  m1u1 system  m2u2  m1u1  Wu  mouthout
Our sign convention for work always assumes that W u is a work output. For this problem we
expect W u to be negative since the problem statement that there is an input of electric power.
In the general mass balance equation, shown below, we see that the left hand side is simply
-mout, because there are no inlets and only one outlet.
m2  m1 system   mi   mi  mout
inlet

mout  m1  m2
outlet
We can find the initial and final mass in the tank from the ideal gas law. Table A-1E gives the gas
constant for air as R = 0.3704 psia∙ft3/lbm∙R.
m1 
m2 
P1Vtan k
(75 psia )( 60 ft 3 )

 20.95 lbm
0.3704 psia  ft 3
RT1
(579.67 R)
lbm  R
P2Vtan k
(30 psia )( 60 ft 3 )

 8.38 lbm
0.3704 psia  ft 3
RT2
(579.67 R)
lbm  R
The mass that left the tank is simply the difference between the initial and final mass: m out = m1 –
m2 = 20.95 lbm – 8.38 lbm = 12.57 lbm. We can use the ideal gas properties for air from Table A17E. Since T1 = T2 = Tout= 120oF = 579.69 R, we have u1 = u2 =98.90 Btu/lbm and hout =
138.66 Btu/lbm. Plugging these properties and the masses found above into our first law equation
gives the work.
Unit six homework solutions
ME 370, L. S. Caretto, Fall 2010

Wu  m1u1  m2 u2  mout hout  ( 20.95 lbm ) 98.90 Btu
lbm 

  (12.57 lb )138.66 Btu

 (8.38 lbm ) 98.90 Btu


m 
lb
lb
m
m


Wu = –500 Btu
As expected, the work is negative indicating that there is a work input of 500 Btu from the
resistance heater.
Page 6