College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Solutions to Unit Six Group Exercise – Unsteady Open Systems
1. A 0.3 m3 rigid tank is filled with saturated liquid water at
200oC. A valve at the bottom of the tank is opened, and
liquid is withdrawn from the tank. Heat is transferred to
the water such that the temperature in the tank remains
constant. Determine the amount of heat that must be
transferred by the time one-half of the total mass has been
withdrawn.
Rigid Tank
V = 0.3 m3
(a) See the diagram on the right to visualize the system.
(b) Why is this problem an open system? Why is this not
a steady flow system?
This is an open system because mass is withdrawn from the system. It is not a steady system,
because the amount of mass in the system (and probably the energy as well) is changing with
time.
(c) Start with the general mass balance and first law for unsteady flow systems and make
the necessary assumptions and simplifications to apply the first law and mass balance
to this problem.
The general first law equation for unsteady open system is shown below.


 



V2
V2
m2  u 
 gz   m1  u 
 gz  
 Q  Wu



2
2
 
2

1  system
2





Vo
Vi 2
  mo  ho 
 gzo    mi  hi 
 gzi 
2
2
outlet

 inlet 

We have no data for velocity or elevations to compute kinetic and potential energies. Based on
past experience, we will assume that the changes in these energy terms are negligible and will
ignore these terms. Also, the system under consideration here has no way to do useful work, so
the work term is zero. Finally, we note that there are no flow inlets and only one flow outlet. With
all these considerations we have the following first law equation for this problem.
m2u2  m1u1 system  Q  mouthout
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   mo  mout
inlet
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
outlet
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Solutions to exercise six
ME 370, L. S. Caretto, Fall 2010
Page 2
Combining the two equations to eliminate mout gives the following result for the heat transfer.
m2u2  m1u1 system  Q  m2  m1 system hout
(d) Find the properties that you will need to solve the problem.
We will need the energy properties, u1, u2, and hout. In addition, we will need the initial specific
volume, v1, to compute the initial mass, m 1 = V/v1.
Since the initial state is a saturated liquid at 200 oC, we find u1 = uf(200oC) = 850.46 kJ/kg and
v1 = vf(200oC) = 0.001157 m 3/kg from Table A-4, page 914.
Since the valve is draining liquid water from the tank and the temperature remains constant at the
temperature of the saturated liquid, we find that the outlet enthalpy is also a saturated liquid
property, hout = hf(200oC) = 852.26 kJ/kg, from the same table.
We do not know what the final state in the tank is. However we can find it from the problem
statement that half of the original mass leaves the tank. First, we have to find the original mass.
m1 
V
0.3 m 3

v1 0.001157 m 3
 259.3 kg
kg
The problem statement that mout = m1 / 2 allows us to compute mout and m2.
mout = m1/2 = (259.3 kg)/2 = 129.65 kg
m2 = m1 – mout = 259.3 kg - 129.65 kg = 129.65 kg
We know that the final state has a temperature of 200 oC, and we can now compute the final
specific volume from the final mass and the constant tank volume.
v2 
3
V
0.3 m 3

 0.002314 m
kg
m2 129.65 kg
We see that this specific volume is greater than the specific volume of the saturated liquid,
vf(200oC) = 0.001157 m 3/kg, but less than the volume of the saturated vapor, vg(200oC) =
0.12721 m3/kg. Thus we are in the mixed region and have to compute the quality, x2, to find the
final internal energy, u2.
x2 
v2  v f (T2  200o C )
v g (T2  200o C )  v f (T2  200o C )

0.002314 m
3
kg
3
0.12721 m
 0.001157 m
kg
3
kg
3
 0.001157 m
0.3 m 3
 0.0091787
kg
Now we can find the final internal energy, u2.
u2  u f (T2  200o C )  x2 u fg (T2  200o C ) 
 1743.7 kJ  866.46 kJ
850.46 kJ
 (0.0091787)
 
kg
kg
kg


Solutions to exercise six
ME 370, L. S. Caretto, Fall 2010
Page 3
(e) Obtain the answer for the heat transfer.
We use the results that we have found for the masses and the energy terms into our combined
first law and mass balance equation to compute the heat transfer.
Q  m2 u2  m1u1 system  mout hout  (129.65 kg) 866.46 kJ  
kg 

(259.3 kg) 850.46 kJ   (129.65 kg) 852.26 kJ 
kg 
kg 


Q = 2,308 kJ
2. A 4 ft3 rigid tank initially contains saturated water vapor at 250oF. The tank is connected
by a valve to a supply line that carries steam at 160 psia and 400 oF. Now the valve is
opened and steam is allowed to enter the tank. Heat transfer takes place with the
surrounding such that the temperature in the tank remains constant at 250 oF at all times.
The valve is closed when it is observed that one-half of the volume of the tank is occupied
by liquid water. Use the steps below to find the heat transfer.
(a) Draw a diagram for this system to help you visualize the overall process.
The diagram for this process is shown at the
right, where the dashed line is the system
boundary. Here we assume that the heat
transfer has the usual sign convention so that
heat input is positive and heat output is
negative. The initial condition inside the tank
of saturated vapor is indicated by the
specification that the initial quality, x1 = 1.
(b) Why is this an open system? Why is
this not a steady flow system?
This is an open system because mass is
added to the system. It is not a steady
system, because the amount of mass in the
system (and probably the energy as well) is
changing with time.
Pin = 160 psia
Tin = 400oF
Inflow
Q
Tank
V = 4 ft3
T1= 250oF
x1 = 1
(c) Use property data, the mass balance equation, and the statement that the final
condition is where half the volume of the tank is liquid water to find the final pressure
and the mass added.
We are told that the final state is a liquid-vapor mixture at 250oF. This means that the final
pressure must be the saturation pressure at 250oF. From the saturation tables we find this final
pressure, P2 = 29.844 psia .
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.
Solutions to exercise six
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` = vg(250oF) =
13.816 ft3/lbm, and the total tank volume, V = 4 ft3.
m1 
V

v1
4 ft 3
3
13.816 ft
 0.289 lbm
lbm
At the final state, half the volume of the tank is liquid and half the volume of the tank is vapor at
the constant temperature of 250oF. This means that the volume of liquid is 2 ft3 and the volume
of vapor is also 2 ft3. The mass of each phase is then found from the volume of the phase and
the specific volume of the phase. We already know vg at 250oF is 13.816 ft3/lbm.; vf at this
temperature is 0.01700 ft3/lbm. Thus the mass of each phase at the final state is given by the
following equations.
m f ,2 
mg , 2 
V f ,2
v f (250o F )
Vg , 2
v g (250o F )


2 ft 3
3
0.01700 ft
2 ft 3
3
13.816 ft
 117.64 lbm
lbm
 0.14 lbm
lbm
m2  m f , 2  mg , 2  117.64 lbm  0.14 lbm  117.78 lbm
From the mass-balance equation we then find the added mass as m in = m2 – m1 = 117.78 lbm –
0.289 lbm = 117.48 lbm added .
(d) Start with the general first law for unsteady flow systems and make the necessary
assumptions and simplifications to find the heat transfer for this problem.
The general first law equation for unsteady open system is shown below.


 



V2
V2
m2  u 
 gz   m1  u 
 gz  
 Q  Wu



2
2
 
2

1  system






Vo2
Vi 2
  mo  ho 
 gzo    mi  hi 
 gzi 
2
2
outlet

 inlet 

We have no data for velocity or elevations to compute kinetic and potential energies. Based on
past experience, we will assume that the changes in these energy terms are negligible and will
ignore these terms. Also, the system under consideration here has no way to do useful work, so
the work term is zero. Finally, we note that there are no flow outlets and only one flow inlet. With
all these considerations we have the following first law equation for this problem.
m2u2  m1u1 system  Q  minhin
Solutions to exercise six
ME 370, L. S. Caretto, Fall 2010
Page 5
In the general mass balance equation, shown below, we see that the left hand side is simply min,
because there are no outlets and only one inlet.
m2  m1 system   mi   mo
inlet

m2  m1  min
outlet
(e) Find the energy properties that you will need to find the heat transfer.
We will need the energy properties, u1, u2, and hin.
Since the initial state is a saturated vapor at 250oF, we find u1 = ug(250oF) = 1089.7 Btu/lbm. The
added steam has the constant properties of the line, so hin = h(160 psia, 400oF) = 1218.0 Btu/lbm.
The final state is in the mixed region. Since we have already found the masses of liquid and
vapor at this state, and since the final calculation requires the calculation of the product m 2u2, we
can obtain this final product as follows:
m2u2 = mf,2uf,2 + mg,2ug,2 = mf,2uf(250oF) + mg,2ug(250oF)
We already found ug(250oF) = 1089.7 Btu/lbm and we can find uf(250oF) = 218.54 Btu/lbm, so we
can compute m2u2 as follows.
m2u2 = mf,2uf(250oF) + mg,2ug(250oF) = (117.64 lbm)(218.54 Btu/lbm) + (0.14 lbm)(1087.7 Btu/lbm)
m2u2 = 25,861 Btu
(f) Obtain the answer for the heat transfer.
We use the results that we have found for the masses and the energy terms into our combined
first law and mass balance equation to compute the heat transfer.

Q  m2u2  m1u1 system  minhin  25,861 Btu  (0.289 lbm )1087.7 Btu

lb
m


 (117.49 lbm )1217.0 Btu
lbm 

Q = -1.175x105 Btu
The negative sign for the heat transfer shows that heat is actually removed from the system.
Does this seem correct to you? Let’s examine the process. We start with saturated vapor at
250oF and add steam at 400oF from the line. This means that between the initial tank contents
and the added steam we have a certain amount of steam that is at 250oF or higher. We end up
with a mixture that is mostly liquid (by mass) at 250oF. To get to this final state, we will have to
cool the initial and incoming steam. Thus the solution that the heat transfer is negative –
meaning that heat is removed to cool the system – seems correct.