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Transcript
```ENGR 224 - Thermodynamics
Problem :
Baratuci
HW #2
14-Apr-11
3.26 - Steam NIST / TFT Fundamentals - 2 pts
Complete the following table for water using either the NIST Webbook or the Thermal/Fluids Toolbox (TFT) Excel plug-in. Use the default
reference state for both the NIST and TFT.
T (oC)
P (kPa)
200
H (kJ/kg)
140
Phase Description
1800
950
500
800
80
x (kg vap/kg)
0.7
0
3162.2
This problem is designed to test how well you understand how to use the NIST Webbook tables of thermodynamic properties and/or
the Thermal/Fluid Properties (TFP) plug-in for Excel.
It also tests your understanding and ability to use quality, x.
Finally, it makes you think about the relationship between the data tables and the physical system (phases) that they represent.
Given :
Two pieces of data for each part, (a) through (e).
Find :
Complete the table.
Assumption:
The system is in an equilibrium state.
Solution :
Part a.)
Given :
P
200 kPa
x
0.7 kg vap/kg
NIST
Because we are given a quality we know that vapor and liquid both exist in the system in equilibrium. Therefore, we use the
Saturation properties — pressure increments option.
Results :
o
120.21 C
T
Hsat liq
504.70 kJ/kg
2706.2 kJ/kg
Hsat vap
ˆ
ˆ
ˆ
H
sat  x H sat   1  x  H sat
mix
Eqn 1
liq
vap
H
TFT
2045.8 kJ/kg
In order to determine Tsat at 200 kPa, use the following cell formula :
=TFProp("Water","SI_C","P",200000,"X",0.7,"T")
Eqn 2
T
120.23
o
C
In order to determine H at 200 kPa and x = 0.7 use the following cell formula :
=TFProp("Water","SI_C","P",200000,"X",0.7,"H")
Eqn 3
H
Part b.)
Given :
o
140 C
T
2045.7
H
kJ/kg
1800 kJ/kg
NIST
Because we are given a temperature and we need to determine which phases are present, the first thing we need to do is generate
a saturation temperature table using the NIST Webbook and the Saturation properties — temperature increments option.
The key results are :
Since :
ˆ  H
ˆ  H
ˆ
H
sat
sat
vap
liq
Temp.
(oC)
Pressure
(kPa)
140
361.54
H (kJ/kg)
Sat. Liq
589.16
Sat. Vap
2733.4
the system contains a saturated mixture and the pressure
must be equal to the saturation pressure.
P= Psat
361.54 kPa
Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is:
x
ˆ H
ˆ
H
sat liq
ˆ
ˆ
H
H
sat vap
Eqn 4
sat liq
x
Plugging values into Eqn 4 yields :
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, 3.26
0.5647 kg vap/kg
4/12/2011
TFT
Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present.
Hsat liq
Hsat vap
=TFProp("Water","SI_C","T",140,"X",0,"H")
=TFProp("Water","SI_C","T",140,"X",1,"H")
Hsat liq
Hsat vap
588.72 kJ/kg
2733.40 kJ/kg
ˆ  H
ˆ  H
ˆ
H
sat
sat
Since :
Eqn 5
Eqn 6
vap
liq
P
the system contains a saturated mixture and the pressure
must be equal to the saturation pressure.
=TFProp("Water","SI_C","T",140,"X",1,"P")
Eqn 7
361.29 kPa
P= Psat
Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is:
x
=TFProp("Water","SI_C","T",140,"H",1800,"X")
Eqn 7
x
Part c.)
Given :
P
950 kPa
0.5648 kg vap/kg
x
0 kg vap/kg
NIST
Because the quality is zero, we immediately know the system contains a saturated liquid.
This tells us that T = Tsat , H = Hsat liq and Psat = P = 950 kPa (given).
So, we can use 950 kPa in the Saturation properties — pressure increments option.
Pressure
(kPa)
950
Here is the relevant data :
Temp.
(oC)
H (kJ/kg)
Sat. Liq
Sat. Vap
752.74
2775.1
177.66
o
T = Tsat
177.66 C
H = Hsat liq
752.74 kJ/kg
TFT
Because the quality is zero, we immediately know the system contains a saturated liquid.
This tells us that T = Tsat , H = Hsat liq and Psat = P = 950 kPa (given).
Part d.)
x
=TFProp("Water","SI_C","P",950000,"X",0,"T")
Eqn 8
T
177.69 oC
H
=TFProp("Water","SI_C","P",950000,"X",0,"H")
Eqn 9
H
752.68 kJ/kg
Given :
T
o
80 C
P
500 kPa
NIST
Here we are given both the T and P. Let's use the given P, and the Saturation properties- pressure increments option, to
determine the Tsat associated with P. Then, we can determine the phases present by comparing the given T to Tsat(P).
Tsat(500 kPa) =
151.83 oC
Since T = 80oC is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is
undefined ! So, we need to look at either the Isothermal properties option or the Isobaric properties option in the NIST
Webbook. Either way, no interpolation is required because we know and enter botht he T and P values !
Here is the relevant data :
Temp.
(oC)
80
Pressure
(kPa)
500
H
(kJ/kg)
335.37
H
335.37 kJ/kg
Here we still need to determine Tsat associated with P.
TFT
Tsat(500 kPa) =TFProp("Water","SI_C","P",500000,"X",0,"T")
Eqn 10
Tsat(500 kPa) =
o
151.86 C
o
Since T = 80 C is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is
undefined ! But it is easy to use the TFT to determine H.
H
=TFProp("Water","SI_C","P",500000,"T",80,"H")
H
334.78 kJ/kg
Eqn 11
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, 3.26
4/12/2011
Part e.)
Given :
P
800 kPa
H
3162.2 kJ/kg
NIST
The first step here is to use the given P to obtain data using the Saturation properties — pressure increments option.
This will allow us to compare the given value of H to the values of Hsat liq and Hsat vap in order to determine the phase or phases
present in the system at equilibrium.
Since :
Temp.
(oC)
Pressure
(kPa)
800
Here is the relevant data :
170.41
H (kJ/kg)
Sat. Liq
Sat. Vap
720.86
2768.3
the system contains a superheated vapor, quality is
undefined and we must use the Isobaric properties
option to determine T.
ˆ  H
ˆ
H
sat
vap
This time we must enter a range of T values so that we bracket the given H value.
I chose to go from 100oC to 500oC by 10oC steps.
Here are the two rows in the resulting table that bracket the H value of 3161.7 kJ/kg.
Temp.
(oC)
340
???
350
Pressure
(kPa)
800
800
800
H
(kJ/kg)
3141.1
3162.2
3162.2
Now, we can determine H from this table by inspection !
T
350.00 oC
TFT
Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present.
Hsat liq
Hsat vap
=TFProp("Water","SI_C","P",800000,"X",0,"H")
=TFProp("Water","SI_C","P",800000,"X",1,"H")
Hsat liq
Hsat vap
Since :
720.68 kJ/kg
2768.67 kJ/kg
Eqn 13
the system contains a superheated vapor, quality is
undefined and we can directly calculate T using the
following formula.
ˆ  H
ˆ
H
sat
vap
T
Eqn 12
=TFProp("Water","SI_C","P",800000,"H",3161.7,"T")
Eqn 14
T
Verify :
None of the assumptions made in the solution of this problem can be verified based on the given information.
NIST Webbook
T (oC)
120.21
140
177.66
80
350.2
P (kPa)
200
361.54
950
500
800
H (kJ/kg)
2045.75
1800
752.74
335.37
3162.2
x (kg vap/kg)
0.7
0.565
0
N/A
N/A
Phase Description
Sat'd Mixture (VLE)
Sat'd Mixture (VLE)
Sat'd Liquid
Subcooled Liquid
Superheated Vapor
P (kPa)
200
361.29
950
500
800
H (kJ/kg)
2045.72
1800
752.68
334.78
3162.2
x (kg vap/kg)
0.7
0.565
0
N/A
N/A
Phase Description
Sat'd Mixture (VLE)
Sat'd Mixture (VLE)
Sat'd Liquid
Subcooled Liquid
Superheated Vapor
350.23 oC
TFT
T (oC)
120.23
140
177.69
80
350.2
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, 3.26
4/12/2011
ENGR 224 - Thermodynamics
Problem :
Baratuci
HW #2
14-Apr-11
3.29E - R-134a NIST/TFT Fundamentals - 2 pts
Complete the following table for R-134a using either the NIST Webbook or the Thermal/Fluids Toolbox (TFT) Excel plug-in. Use the default
reference state for both the NIST and TFT.
T
(oF)
P
(psia)
H
(Btu/lbm)
80
78
x
(lbm vap/lbm)
15
0.6
10
70
180
129.46
110
Phase Description
1.0
This problem is designed to test how well you understand how to use the NIST Webbook tables of thermodynamic properties
and/or the Thermal/Fluid Properties (TFP) plug-in for Excel.
It also tests your understanding and ability to use quality, x.
Finally, it makes you think about the relationship between the data tables and the physical system (phases) that they represent.
Given :
Two pieces of data for each part, (a) through (e).
Find :
Complete the table.
Assumptions:
The system is in an equilibrium state.
Solution :
Part a.)
P
80 psia
H
NIST :
The first step here is to use the given P to obtain data from the Saturation properties — pressure increments option.
This will allow us to compare the given value of H to the values of Hsat liq and Hsat vap in order to determine the phase or phases
present in the system at equilibrium.
Pressure
(psia)
80
Here is the relevant data :
Since :
78 Btu/lbm
Temp.
(oF)
H (Btu/lbm)
Sat. Liq
Sat. Vap
97.162
176.02
65.922
the system contains a subcooled liquid and temperature
must be less than the saturation temperature. We must
interpolate on Isobaric Properties .
ˆ  H
ˆ
H
sat
liq
This time we must enter a range of T values so that we bracket the given H value.
I chose to go from 0oF to 100oF by 10oF steps.
Here are the two rows in the resulting table that bracket the H value of 78 Btu/lbm .
Temp.
(oF)
0
???
10
Pressure
(psia)
80
80
80
H
(Btu/lbm)
75.994
78
79.107
Now, we can determine T from this table
by interpolation.
o
6.44 F
T
The quality of the water in the system is not defined because it is a subcooled liquid.
TFT :
Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present.
Hsat liq
Hsat vap
=TFProp("R-134a","EE_F","P",80,"X",0,"H")
=TFProp("R-134a","EE_F","P",80,"X",1,"H")
Hsat liq
Hsat vap
Since :
Eqn 2
Eqn 3
97.11 Btu/lbm
175.90 Btu/lbm
the system contains a subcooled liquid and temperature must be less
than the saturation temperature. We can now use the TFT to directly
evaluate T.
ˆ  H
ˆ
H
sat
liq
T
=TFProp("R-134a","EE_F","P",80,"H",78,"T")
o
15 F
Eqn 4
T
o
6.62 F
x
0.60 lbm vap/lbm
Part b.)
T
NIST :
Because the quality lies between 0 and 1, we immediately know the system contains a saturated mixture.
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, 3.29E
4/12/2011
This tells us that T = Tsat = 15oF (given) and P = Psat .
So, we can look up 15oF in the Saturation properties — temperature increments option.
Temp.
(oF)
Here is the relevant data :
Pressure
(psia)
29.739
15.00
H (Btu/lbm)
Sat. Liq
Sat. Vap
80.634
169.07
29.739 psia
P = Psat
Because both saturated liquid and saturated vapor are present in the system at equilibrium, we must use the quality to evaluate the
overall average specific internal energy, as follows.
The key equation that relates the specific internal energy of the system to the specific internal energy of the saturated liquid and of
the saturated vapor is:
ˆ
ˆ
ˆ
H
sat  x H sat   1  x  H sat
mix
Eqn 6
liq
vap
Plugging values into Eqn 2 yields :
TFT :
H
133.70 kJ/kg
Because the quality lies between 0 and 1, we immediately know the system contains a saturated mixture.
This tells us that T = Tsat = 15oF (given) and P = Psat .
In order to determine Tsat at 15oF, use the following cell formula :
=TFProp("R-134a","EE_F","T",15,"X",0.6,"P")
Eqn 7
P
29.739
psia
133.60
Btu/lbm
In order to determine H at 15oF and x = 0.6 use the following cell formula :
=TFProp("R-134a","EE_F","T",15,"X",0.6,"H")
Eqn 8
H
o
10 F
Part c.)
T
NIST :
Here we are given both the T and P. Let's use the given P, and the Saturation properties — pressure increments option, to
determine the Tsat associated with P. Then, we can determine the phases present by comparing the given T to Tsat(P).
P
70 psia
Tsat(70 psia) =
o
58.338 F
Since T = 10oF is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is
undefined ! So, we need to look at either the Isothermal properties option or the Isobaric properties option.
Temp.
(oF)
10
Here is the relevant data :
Pressure
(psia)
70
H
(Btu/lbm)
79.099
H
TFT :
79.099 Btu/lbm
Here we still need to determine Tsat associated with P.
Tsat(70 psia) =TFProp("R-134a","EE_F","P",70,"X",0,"T")
Eqn 9
Tsat(70 psia)
o
58.339 F
o
Since T = 10 F is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is
undefined ! But it is easy to use the TFT to determine H.
H
=TFProp("R-134a","EE_F","P",70,"T",10,"H")
Eqn 10
H
180 psia
H
79.045 Btu/lbm
Part d.)
P
129.46 Btu/lbm
NIST :
The first step here is to use the given P to generate data using the Saturation properties — pressure increments option.
This will allow us to compare the given value of U to the values of Usat liq and Usat vap in order to determine the phase or phases
present in the system at equilibrium.
Pressure
(psia)
180
Here is the relevant data :
Since :
ˆ H
ˆ  H
ˆ
H
sat
sat
liq
vap
Temp.
(oF)
117.73
H (Btu/lbm)
Sat. Liq
Sat. Vap
115.28
181.79
the system contains a saturated mixture and the temperature must be equal to the
saturation temperature.
T = Tsat
117.73 oF
Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is:
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, 3.29E
4/12/2011
x
ˆ H
ˆ
H
sat liq
ˆ
ˆ
H
H
sat vap
Eqn 11
sat liq
x
Plugging values into Eqn 11 yields :
TFT :
0.2132 kg vap/kg
Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present.
Hsat liq
Hsat vap
=TFProp("R-134a","EE_F","P",180,"X",0,"H")
=TFProp("R-134a","EE_F","P",180,"X",1,"H")
Hsat liq
Hsat vap
Eqn 12
Eqn 13
115.20 Btu/lbm
181.67 Btu/lbm
the system contains a saturated mixture and the temperature must be equal to the
saturation temperature.
Since :
ˆ H
ˆ  H
ˆ
H
sat
sat
T
=TFProp("R-134a","EE_F","P",180,"X",0,"T")
liq
vap
Eqn 13
o
117.74 F
T
Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is:
x
=TFProp("R-134a","EE_F","P",180,"H",129.46,"X")
Eqn 13
o
0.2145 F
x
o
110 F
Part e.)
T
1.0 lbm vap/lbm
NIST :
Because the quality is 1.0, we immediately know the system contains a saturated vapor.
This tells us that T = Tsat = 100oF (given), H = Hsat vap and P = Psat .
x
So, we can look up 110oF using the Saturation properties — temperature increments option.
Temp.
H (Btu/lbm)
Here is the relevant data :
Pressure
(oF)
(psia)
Sat. Liq
Sat. Vap
161.07
110
112.46
181.05
161.07 psia
P = Psat
TFT :
H = Hsat vap
181.05 Btu/lbm
In order to determine Psat at 110oF, use the following cell formula :
P
=TFProp("R-134a","EE_F","T",110,"X",1,"P")'
Eqn 14
P
161.05
psia
H
180.93
Btu/lbm
In order to determine H at 110oF and x = 1.0 use the following cell formula :
H
=TFProp("R-134a","EE_F","T",110,"X",1,"H")'
Eqn 15
Verify :
None of the assumptions made in the solution of this problem can be verified based on the given information.
NIST
T
(oF)
6.44
15
10
117.73
110
P
(psia)
80
29.74
70
180
161.07
H
(Btu/lbm)
78
133.70
79.10
129.46
181.05
x
(lbm vap/lbm)
N/A
0.6
N/A
0.2132
1.0
Subcooled Liquid
Sat'd Mixture (VLE)
Subcooled Liquid
Sat'd Mixture (VLE)
Saturated Vapor
P
(psia)
80
29.74
70
180
161.05
H
(Btu/lbm)
78
133.60
79.05
129.46
180.93
x
(lbm vap/lbm)
N/A
0.6
N/A
0.2145
1.0
Subcooled Liquid
Sat'd Mixture (VLE)
Subcooled Liquid
Sat'd Mixture (VLE)
Saturated Vapor
Phase Description
TFT
T
(oF)
6.62
15
10
117.74
110
Dr. Baratuci - ChemE 260
Phase Description
hw2-sp11.xlsm, 3.29E
4/12/2011
ENGR 224 - Thermodynamics
Baratuci
HW #2
14-Apr-11
Problem : 4.8 - Isobaric Expansion of R-134a - 6 pts
A piston-and-cylinder device with a set of stops contains 0.3 kg of steam at 1.0 MPa and 400oC. The location of the stops corresponds to
60% of the initial volume. Now, the steam is cooled. Determine the compression work if the final state is…
P2a = 1000 kPa
T2a = 250 oC
V2a > 0.6 V1
m = 0.3 kg
P1 = 1000 kPa
T1 = 400 oC
P2b = 500 kPa
V2b = 0.6 V1
a.)
b.)
c.)
1.0 MPa and 250oC
500 kPa
Determine the temperature at the final state in part (b)
The key to determining properties for state 1 is that we know both the pressure and temperature. So, we can determine every
other property, including the specific volume and internal energy.
The key for states 2a and 2b is whether the piston has fallen all the way down to rest on the stops or not.
Since P2a = P1, the piston cannot be resting on the pins. If the piston were resting on the pins, P2a < P1. Consequently, we
expect to find V2a > 0.6 * V1 and V2b = 0.6 * V1.
Boundary work can be determined from its definition if we assume the process is quasi-equilibrium.
Then, we can apply the 1st Law to the process to determine Q.
Given :
m
T1
P1
TFT Parameters :
Find :
a.)
b.)
c.)
Diagram :
See above.
T2a
P2a
P2b
V2b/V1
0.3 kg
o
400 C
1000 kPa
W ba
W bb
T2b
Dr. Baratuci - ChemE 260
Water
???
???
???
o
250 C
1000 kPa
500 kPa
0.6
SI_C
kJ
kJ
o
C
hw2-sp11.xlsm, 4.8
Qa
Qb
???
???
kJ
kJ
4/17/2011
Assumptions:
Solution :
- The steam in the cylinder is a closed system.
- Boundary work is the only form of work interaction.
- Changes in kinetic and potential energies are negligible.
- The expansion is a quasi-equilibrium process.
- The initial and final states are both equilibrium states.
As problems become more complex, I think it helps to organize the information you collect about each state in a table like the
one shown below. The bold green values represent the information obtained from the problem statement. Not every value
needs to be filled in, but the table gives you a concise way to keep track of what you know and what you do not know.
T
P
Tsat
Phase
X
v
V
U
H
1
400
1000
179.9
Super
2a
250
1000
179.9
Super
0.3066
0.091976
2956.8
3263.4
0.2327
0.069805
2709.4
2942.1
2b
151.9
500
151.9
Sat Mix
0.4892
0.1840
0.055186
1579.2
1671.2
o
C
kPa
o
C
kg vap/kg
m3/kg
m3
kJ/kg
kJ/kg
The overall objective of this problem is to determine Q and W. This almost always requires the application of the 1st Law.
Apply the integral form of the 1st Law to process 1-2a and 1-2b.
Q12  W12  U  Ekin  Epot
Eqn 1
If we assume that changes in kinetic and potential energies are negligible, then Eqn 1 simplifies to :

ˆ U
ˆ
Q12  W12  U  U 2  U1  m U
2
1

Eqn 2
Consider the steam inside the cylinder to be our system. If we assume that the process is a quasi-equilibrium process and
boundary work is the only type of work that crosses the system boundary, then we can use the definition of boundary work to
evaluate W12.
2
W12   P dV
Eqn 3
1
Part a.)
The process in part (a) is isobaric. This allows us to simplify Eqn 3, as follows.

ˆ V
ˆ
W12a   P dV  P  dV  P  V2a  V1   m P V
2a
1
2a
2a
1
1

Eqn 4
Combining Eqns 2 & 4 yields :







ˆ U
ˆ  mP V
ˆ V
ˆ m U
ˆ U
ˆ m H
ˆ H
ˆ
Q12a  W12a  m U
2a
1
2a
1
2a
1
2a
1

Eqn 5
Since we know both T1 and P1 and T2a and P2a, we can lookup any other intensive property for states 1 and 2a in the steam
tables. I used the TFT add-in to determine specific V, U and H for states 1 and 2a.
V
U
H
State 1
0.30659
2956.8
3263.4
State 2a
0.23268
2709.4
2942.1
m3/kg
kJ/kg
kJ/kg
We can now plug values into Eqns 4 & 5 to evaluate W12a and Q12a.
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, 4.8
W12a
-22.17
kJ
Q12a
-96.38
kJ
4/17/2011
Part b.)
This part is a bit trickier because the process is not isobaric and we need to determine what phase or phases are present in
state 2b. This is worrisome because we cannot apply Eqn 3 to evaluate boundary work unless we know the process path.
The process 1-2b consists of two parts. In the first part of the process, before the piston comes to rest on the pins, pressure is
constant. This part is an isobaric process and we can use Eqn 4 to evaluate the boundary work. In the second part of the
process in part (b), the volume does not change because the piston is resting on the pins.
This is an isochoric process and Wb = 0 kJ !
W12b
-36.79
kJ
Now, we can rearrange Eqn 2 algebraically to help us evaluate Q12b.

ˆ U
ˆ
Q12b  W12b  m U
2b
1

Eqn 6
So, all we need to do before we use Eqn 6 to finish this part of the problem is determine U2.
The problem is that we only know the value of one intensive variable for state 2b, P2b.
But we can determine V2b based on the given relationship: V2b = 0.6 * V1 . Since the mass inside the system is constant, the
specific volume in state 2b is also 0.6 times the sprecific volume in state 1.
0.18395 m3/kg
V2b
This gives us the 2nd intensive property value that we need to evalaute any other intensive property. In this case, we need U2b
so we can use Eqn 6 to evaluate Q12b.
1579.2 kJ/kg
U2b
Part c.)
Q12b
-450.1
kJ
We can determine the temperature in state 2b by using the TFT add-in to determine the specific volume of saturated liquid and
saturated vapor at P2b and comparing the actual specific volume, V2b.
At P2b =
500 kPa
Vsat liq =
Since :
ˆ
ˆ
ˆ
V
sat liq < V2b < Vsat vap
0.00109279 m3/kg
Verify :
None of the assumptions made in this problem solution can be verified.
a.)
Dr. Baratuci - ChemE 260
0.3748911 m3/kg
T2b = Tsat at P2b = 500 kPa
The TFT add-in yields the saturation temperature at 500 kPa :
c.)
Vsat vap =
W12a
-22.17
kJ
Q12a
-96.38
kJ
T2b
151.9
o
b.)
T2b
151.9
o
W12b
-36.79
kJ
Q12b
-450.1
kJ
C
C
hw2-sp11.xlsm, 4.8
4/17/2011
ENGR 224 - Thermodynamics
Problem :
Baratuci
HW #2
14-Apr-11
4.42 - Compression of Cooling Air by a Linear Spring - 5 pts
Steam at 75 kPa and 8% quality is contained in a spring-loaded piston-and-cylinder device as shown below, with an initial volume of 2 m3.
Steam is now heated until its volume is 5 m3 and its pressure is 225 kPa. Determine the heat transferred to and the work produced by the
steam during this process.
Steam
Linear Spring
The key to solving this problem is the linear nature of the relationship between P and V due to the linear nature of the spring.
This makes it relatively easy to evaluate the boundary work for the process. Because the process path is linear on a PV
Diagram, the boundary work (area under the path) is just the average pressure times the change in the volume. I demonstrate
why this is so in the solution, below. Since initial and final P and V values are given, Wb can be calculated without any other
Given the two intensive variables for the initial state, P1 and x1, we can evaluate all of the other properties, including V1 and U1.
This will allow us to determine the mass of water in the cylinder from the given initial volume and the specific volume we
determined from P1 and x1. This mass and the final volume lets us calculate the final specific volume. As a result, we know
the values of two intensive variables in the final state, P2 and specific volume, so we can evaluate U2.
Once we know U2, U1 and Wb, we can apply the 1st Law to evaluate Q for the process.
Given :
P1 =
V1 =
x1 =
75
2
0.08
kPa
m3
kg vap/kg
P2 =
V2 =
225
5
kPa
m3
Find :
Wb =
???
kJ
Q
???
kJ
Assumptions: - The steam in the cylinder is a closed system.
- The process occurs slowly enough that it is a quasi-equilibrium process.
- There is no friction between the piston and the cylinder wall.
- The spring force varies linearly with position.
- Changes in kinetic and potential energies are negligible.
- Boundary work is the only form of work that crosses the system boundary.
Solution :
Let's begin with the 1st Law. Ultimately, this is the equation that will give us the relationship between Q and W.
The 1st Law for a closed system with negligible change in kinetic and potential energies is :
Q - W = DU
Eqn 1
If we assume that boundary work is the only form of work that crosses the system boundary, then W becomes Wb in Eqn 1.
Q - Wb = DU
Eqn 2
2
We can evaluate Wb from the relationship :
Wb   P dV
Eqn 3
1
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, 4.42
4/12/2011
Because the spring is linear, P is a linear function of V. This means :
æ P - P1 ö
÷÷ ⋅ ( V - V )
P = P1 + ççç 2
1
èç V2 - V1 ÷ø÷
Eqn 4
2

  V2

 P  P1 
P2  P1  V22
 
 V1 V2    1  V1 V1  
Wb    P1   2
   V  V1   dV  P1  V2  V1  
V2  V1  2
1

 V2  V1 
  2
 

Eqn 5
Now, for some algebraic magic…
Wb  P1  V2  V1  
1 P2  P1

 V22  2 V1 V2  V12
2 V2  V1
Wb  P1  V2  V1  
1 P2  P1
2

  V2  V1 
2 V2  V1
Eqn 7
Wb  P1  V2  V1  
1
  P2  P1    V2  V1 
2
Eqn 8
Wb 


Eqn 6
P2  P1
  V2  V1 
2
Eqn 9
OK. That was a lot of work, but now we can plug in numbers and evaluate Wb.
Wb
450
kJ
So, all we need to do is evaluate U2 and U1 and we can use Eqn 10 to complete the solution of this problem.
We can evaluate U1 because we know both the pressure and the quality, P1 and x1.
Here is the relevant data :
TFT :
Chemical :
Units :
Temp.
Pressure
(kPa)
75
Water
SI_C
(oC)
91.78
U (kJ/kg)
Sat. Liq
Sat. Vap
384.92
2496.2
The key equation that relates the specific internal energy of the system to the specific internal energy of the saturated liquid
and of the saturated vapor is:
ˆ
ˆ
ˆ
U
sat  x U sat   1  x  U sat
mix
vap
Eqn 10
liq
U1
Plugging values into Eqn 10 yields :
553.82 kJ/kg
The key to evaluating U2 is to recognize that the mass of H2O inside the cylinder is the same throughout the process. This
allows us to modify Eqn 2, as follows.
(
ˆ -U
ˆ
Q = Wb + DU = Wb + m ⋅ U
2
1
)
Eqn 11
Since the initial state is completely determined because we know P1 and x1, we can evaluate V1 and use it to determine the
mass of H2O in the cylinder.
Here is the relevant data :
Temp.
Pressure
(kPa)
75
o
( C)
91.78
V (m3/kg)
Sat. Liq
Sat. Vap
0.001037
2.2171
ˆ
ˆ
ˆ
V
sat  x Vsat   1  x  Vsat
mix
vap
Eqn 12
liq
Plugging values into Eqn 12 yields :
V1
Now, we can determine the mass of H2O in the cylinder using :
m=
Plugging values into Eqn 13 yields :
m
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, 4.42
0.1783 m3/kg
V1
Eqn 13
V̂1
11.22 kg
4/12/2011
Now we know both the volume and the mass in the final state,
so we can evaluate the specific volume from :
V̂2 =
Plugging values into Eqn 15 yields :
V2
V2
m
Eqn 14
0.4458 m3/kg
Now we know the values of two intensive properties in the final state, P2 and specific volume, so we can evaluate any other
intensive property. In this case, we need to determine U2 so we can apply Eqn 12 to complete the solution of this problem.
Pressure
(kPa)
225
Here is the relevant data :
Since :
ˆ  V
ˆ  V
ˆ
V
sat
2
sat
liq
vap
Temp.
o
( C)
124.00
V (m3/kg)
Sat. Liq
Sat. Vap
0.001064
0.79325
U (kJ/kg)
Sat. Liq
Sat. Vap
520.34
2533.1
the system contains a saturated mixture and the temperature must be equal to the
saturation temperature.
In order to determine U2, we must first evaluate the quality, x2.
x2 
ˆ V
ˆ
V
2
sat liq
ˆ
ˆ
V
V
sat vap
Eqn 15
sat liq
0.5614 kg vap/kg
x2
Plugging values into Eqn 15 yields :
Next, we can use the following equation to evaluate U2.
ˆ x U
ˆ
ˆ
U
2
2
sat   1  x 2  U sat
vap
U2
Plugging values into Eqn 16 yields :
Eqn 16
liq
1650.3 kJ/kg
Finally, we can plug values into Eqn 11 to evaluate Q and complete this solution.
Verify :
None of the assumptions made in this problem solution can be verified.
Wb
Dr. Baratuci - ChemE 260
450
kJ
hw2-sp11.xlsm, 4.42
Q
12747.9 kJ
Q
12750 kJ
4/12/2011
ENGR 224 - Thermodynamics
Problem :
Baratuci
HW #2
14-Apr-11
4.59E - Determining H Using Heat Capacity Polynomials - 4 pts
Determine the enthalpy change of oxygen (O2), in Btu/lbm, as it is heated from 800 to 1500oR, using:
a.)
b.)
c.)
The empirical specific heat equation as a function of temperature (Table A-2Ec).
The CoP value at the average temperature (Table A-2Eb).
The CoP value at room temperature, 80oF, (TableA-2Ea).
The empirical specific heat equation will yield the most accurate estimate of the enthalpy change. Assuming a constant value of
Cp determined at the average temperature should yield a reasonable estimate of H as well. Using the Cp value at room
temperature should not be very accurate. We can compare this result to the value we get in part (a).
Given :
T1
Find :
H1-2 =
800 oR
Assumptions:
Solution :
Part a.)
???
1500 oR
T2
Btu/lbm
a.)
b.)
c.)
Oxygen behaves as an ideal gas
Oxygen behaves as an ideal gas with a constant heat capacity evaluated at Tavg.
Oxygen behaves as an ideal gas with a constant heat capacity evaluated at 80oF.
Let's begin by collecting the data we need from Table A-2Ec :
 oP  a  b T  c T 2  d T 3
C
Eqn 1
Where:
CP [=] Btu/lbmole-oR
32.00 lbm/lbmole
MW
T [=] oR
491-3240 oR
6.085
2.017E-03
-5.275E-07
5.372E-11
Temp (K)
a
b
c
d
The enthalpy change associated with a temperature change for an ideal gas can be determined from :

H
1 2 
T2
 C
o
P
dT
Eqn 2
T1
Combining Eqns 1 & 2 and integrating yields :
  a (T  T ) 
H
2
1
b 2
c
d
(T2  T12 )  (T23  T13 )  (T24  T14 )
2
3
4
Eqn 3
H
Plug in values for the temperatures and the constants to get :
Ĥ  Btu / lbm  
Btu/lbmole
  Btu / lbmole 
H
Eqn 4
MW  lb m / lbmole 
H1-2
Part b.)
5442.3
We can use Table A-2Eb to evaluate the heat capacity at Tavg :
Tavg
170.1
Btu/lbm
1150 oR
690.33 oF
CP
T (oF)
Btu/lbm-oR
600
690.33
700
0.239
Cp(1150oR)
0.242
Cp(1150oR)
0.2417 Btu/lbm-oR
Now we can use Eqn 2 to determine H1-2. It is easier to integrate this time because CP is assumed to be constant.
ˆo
Ĥ1 2  C
P,avg  T2  T1 
H1-2
Plugging values into Eqns 5 yields :
Part c.)
Eqn 5
We can use Table A-2Ea to evaluate the heat capacity at 80oF:
Cp(80oF)
Now, we can apply Eqn 5 again because CP is assumed to be a constant.
H1-2
169.2
Btu/lbm
0.2190 Btu/lbm-oR
153.3
Btu/lbm
This is more than a 10% error. This is unacceptable.
Verify :
None of the assumptions made in the solution of this problem can be verified based on the given information.
a.)
H1-2
170.1
Btu/lbm
b.)
H1-2
169.2
Btu/lbm
( Error ~ 0.5% )
c.)
H1-2
153.3
Btu/lbm
( Error > 10% )
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm , 4.59E
4/12/2011
ENGR 224 - Thermodynamics
Problem :
Baratuci
HW #2
14-Apr-11
4.60 - Determining U Using Heat Capacity Polynomials - 4 pts
Determine the internal energy change of hydrogen (H2), in kJ/kg, as it is heated from 200 to 800 K, using:
a.)
b.)
c.)
The empirical specific heat equation as a function of temperature (Table A-2c).
The CoV value at the average temperature (Table A-2b).
The CoV value at room temperature, 300 K, (TableA-2a).
The empirical specific heat equation will yield the most accurate estimate of the enthalpy change. Assuming a constant value of
CV determined at the average temperature should yield a reasonable estimate of U as well. Using the CV value at room
temperature should not be very accurate. We can compare this result to the value we get in part (a).
Given :
T1
Find :
U1-2 =
200 K
Assumption:
Solution :
Part a.)
???
T2
800 K
kJ/kg
a.)
b.)
c.)
Hydogen behaves as an ideal gas
Hydogen behaves as an ideal gas with a constant heat capacity evaluated at Tavg.
Hydogen behaves as an ideal gas with a constant heat capacity evaluated at 80oF.
Let's begin by collecting the data we need from Table A-2c :
  a  b T  c T2  d T3
C
o
P
Eqn 1
Where:
MW
Temp (K)
a
b
c
d
T [=] K
CP [=] J/mol-K
2.016 g/mol
273-1800 K
29.11
-1.916E-03
4.003E-06
-8.704E-10
The enthalpy change associated with a temperature change for an ideal gas can be determined from :

H
1 2 
T2
 C
o
P
dT
Eqn 2
T1
Combining Eqns 1 & 2 and integrating yields :
  a (T  T ) 
H
2
1
b 2
c
d
(T2  T12 )  (T23  T13 )  (T24  T14 )
2
3
4
Eqn 3
H
Plug in values for the temperatures and the constants to get :
17474.9
J/mol
Next, we have to determine U from H. Consider the differential form of the definition of enthalpy :

  U
   PV

H

But for ideal gases :
Eqn 4

Eqn 5
Combining Eqns 4 & 5 yields :
 IG  U
 IG  R TIG
H
Eqn 6
Solving Eqn 6 for U yields :
 IG  H
 IG  R TIG
U
Eqn 7
Plugging values into Eqn 7 yields :
Û  kJ / kg  
Part b.)

    R T
 PV
  J / mol 
U

1000  g / kg 
Eqn 8
MW  g / mol  1000  J / kJ 
R
U
8.314
12486.5
J/mol-K
J/mol
U1-2
6193.7
kJ/kg
We can use Table A-2b to evaluate the heat capacity at Tavg :
Tavg
Fortunately, there is an entry in Table A-2b at 500 K, so we can read CV :
CV
500 K
10.389
(kJ/kg-K)
The internal energy change associated with a temperature change for an ideal gas can be determined from :
Û1 2 
T2
 Cˆ
o
V
dT
Eqn 9
T1
Because CV is assumed to be constant, Eqn 9 is easy to integrate and obtain :
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm , 4.60
4/12/2011
ˆo
Û1 2  C
V ,avg  T2  T1 
Part c.)
Eqn 10
Plugging values into Eqn 10 yields :
U1-2
We can use Table A-2a to evaluate CV at 300 K:
CV(300 K)
Now, we can apply Eqn 6 again because CP is assumed to be a constant.
H1-2
6233.4
kJ/kg
10.1830 Btu/lbm-oR
6109.8
Btu/lbm
This is less than a 2% error. Not so bad. Why ? Because CV for H2 doesn't change so much with respect to T.
Verify :
None of the assumptions made in the solution of this problem can be verified based on the given information.
a.)
U1-2
6194
Btu/lbm
b.)
U1-2
6233
Btu/lbm
( Error ~ 0.64% )
c.)
U1-2
6110
Btu/lbm
( Error ~ 1.4% )
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm , 4.60
4/12/2011
ENGR 224 - Thermodynamics
Problem :
Baratuci
HW #2
14-Apr-11
WB-1 - Clapeyron & Clausius-Clapeyron Equations - 5 pts
Estimate the latent heat of vaporization, in Btu/lbm, of ammonia at -10oF using:
a.)
b.)
c.)
The Clapeyron equation
The Clausius-Clapeyron equation
The ammonia tables
The key to this problem is using the saturated NH3 tables to obtain P*(T) data for use in the Clapeyron and Clausius-Clapeyron
Equations. This data is used to estimate dP*/dT in the Clapeyron Equation and d[Ln(P*)] / d(1/T) in the Clausius-Clapeyron
Equation. The Clausius-Clapyron Equation is less accurate because it requires two more simplfying assumptions than the
Clapeyron Equation. But, the C-C Eqn does not require that you know the Vvap. So, it is a trade off.
Given :
T
o
-10 F
Assumptions:
P*(T) is approximately linear beteen 0 and -20 oF.
1-
Solution :
Part a.)

H
d P*
vap


dT
T V
vap
The Clapeyron Equation is :
Eqn 1
If P*(T) is approximately linear over a narrow temperature interval, then :
d P * P*

T
dT
Eqn 2
Now, we can substitute Eqn 2 into Eqn 1 and solve for Hvap, as follows.

H
P *
vap


T
T Vvap
Eqn 3


H
vap  T  Vvap
P *
T
Eqn 4
We can look up the vapor pressure of NH3 at 0oF and -20oF in the saturation temperature table and use this data to evaluate P* /
T.
T (oF)
T (oR)
P* (psia)
0
-20
459.67
439.67
30.397
18.279
P* / T
1 psia =
P* / T
o
0.6059 psia/ R
144 lbf / ft2
87.25 lbf / ft2-oR
Next, we must use the saturated NH3 Tables again to evaluate Vvap.
At -10oF :
Vsat liq
Vsat vap
0.023929 ft3 / lbm
11.502 ft3 / lbm
Now, we can plug values into Eqn 4 :
Vvap
11.47807 ft3 / lbm
T
Hvap
o
449.67 R
4.503E+05 ft-lbf / lbm
Hvap
578.70 Btu / lbm
Unit conversions:
1 Btu =
Dr. Baratuci - ChemE 260
778.17 ft-lbf
hw2-sp11.xlsm, WB-1
4/12/2011
Part b.)
The Clausius-Clapeyron Equation is :

  H
vap
Ln  P*   

R

 1
 C
 T

Eqn 5
This equation tells us that Ln[P*] vs. 1/T is linear and the slope is :
Slope 
Ln  P2*   Ln  P1* 
1 / T2  1 / T1


H
vap
Eqn 6
R
We can use the data in the table from part (a) to evaluate the slope.
Slope
o
-5139.39 R
Next we can solve Eqn 6 for Hvap and plug in values to evaluate it.

H
vap   R  Slope
Eqn 7
R
Hvap
Hvap
17.03 lbm/lbmole
MW
Part c.)
The saturated NH3 tables tell us that :
Verify :
None of the assumptions made in the solution of this problem can be verified based on the given
information.
a.)
Hvap
578.7 Btu / lbm
b.)
Hvap
599.6 Btu / lbm
c.)
Hvap
576.3 Btu / lbm
Hsat liq
Hsat vap
Hvap
o
1.987 Btu/lbmole- R
10211.97 Btu/lbmole
599.65 Btu / lbm
31.982 Btu / lbm
608.26 Btu / lbm
576.28 Btu / lbm
The Clausius is quite accurate, but the Clausius-Clapeyron Equation is off by about 4%.
The additional assumptions for the Clausius-Clapeyron Equation cause this increase in the error.
The molar volume of the saturated liquid is only about 2% of the molar volume of the saturated vapor, Vvap ~ Vsat vap is not a very
The Clausius-Clapeyron Equation also requires that the saturated vapor behave as an ideal gas.
V
The molar volume of the saturated vapor at -10oF is :
3
195.88 ft / lbmole
12.07 L/mole
Because the molar volume is much less than 20 L/mole, it is not accurate to treat the NH3 as an ideal gas. This is the principle
source of error in using the Clausius-Clapeyron Equation in this problem.
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, WB-1
4/12/2011
ENGR 224 - Thermodynamics
Problem :
WB-2 - Hypothetical Process Paths and the Latent Heat of Vaporization - 8 pts
the change in enthalpy in Joules for 20.0 g of heptane (C7H16) as it
changes from a saturated liquid at 300 K to a temperature of 370 K
and a pressure of 58.7 kPa. Calculate the H for each step in the
path. Do not use tables of thermodynamic properties, except to check
of vaporization of heptane at 300 K. Obtain the Antoine constants from
the NIST Webbook. Use the average heat capacity of heptane gas
(also from from the NIST Webbook) over the temperature range of
interest.
Assume heptane gas is an ideal gas at the relevant temperatures and
pressures.
Given:
Baratuci
HW #2
14-Apr-11
4
3
1
Saturated Liquid
P1, T1
Superheated Vapor
P2, T2
Superheated Vapor
P1, T2
2
Saturated Vapor
P1, T1
Step 1-2 is a bit tricky. We can use the Antoine Eqn with the Clausius-Clapeyron Eqn to estimate Hvap. Step 2-3 is
straightforward because the problem instructs us to use an average Cp value. The only difficulty will be that Cp values may not
be avail
m=
20 g
H1-2
???
J
T1 =
300 K
H2-3
???
J
x1 =
T2 =
P2 =
0.0 (sat'd liq)
370 K
58.7 kPa
H3-4
H1-4
???
???
J
J
Assumptions :
Find:
- Clausius-Clapeyron applies:
- The saturated vapor is an ideal gas
- The molar volume of the saturated vapor is much much greater than
the molar volume of the saturated liquid.
- The heat of vaporizatioon is constant over the temperature range of
interest.
- The superheated vapor also behaves as an ideal gas.
- The heat capacity of the vapor is nearly constant over the temperature range of
interest so that using the average value is a reasonable approximation.
Solution :
First we can observe that
H1-2 = Latent heat of vaporization at 300 K
We can estimate the heat of vaporization using the Clausius -Clapeyron Equation.
 vap
 H
Ln P*   

R

 1
  C
 T
Eqn 1
If we plot Ln P* vs. 1/T(K), the slope is - Hvap/R.
We can calculate the vapor pressures at two different temperatures using the Antoine equation. Use temperatures near the
temperature of interest, 300 K. Use the two points to estimate the slope over this small range of temperatures.
Ln Pb  Ln Pa
1 / Tb  1 / Ta
Eqn 2
Log10(P*) = A - (B / (T + C))
Eqn 3
Slope 
Antoine Equation:
P is in bar
T is in Kelvin
The Antoine constants from the NIST WebBook are:
A
4.02832
B
1268.636
C
-56.199
From the Antoine Eqn:
Ta =
Tb =
299.5 K
300.5 K
Dr. Baratuci - ChemE 260
Pa =
Pb =
6.52 kPa
6.85 kPa
hw2-sp11.xlsm , WB-2
Slope =
-4423.1 K
4/12/2011
Next we use this slope with Eqn 1 to determine the latent heat of vaporization at 300 K :
R=
n
8.314 J/mol K
m
MW
Eqn 4
Hvap =
36773 J/mol
MW =
100.20 g/mol
n
0.1996
H(1-2) =
mol
7,340 J
Next, let's consider the enthalpy change from states 2 to 3, saturated vapor to superheated vapor:
Because we assumed a constant heat capacity, we can evaluate H using:
 T
 C
H
p
Eqn 5
The heat capacities are tabulated in the NIST WebBook, under the Name Search option. Interpolate to estimate Cp at both T1
and T2. Then, average these two values of Cp to obtain the average heat capacity. This is equivalent to determining a linear
equation between T1 and T2 and integrating.
Gas phase heat capacity data from the NIST WebBook:
Solve:
Cp(T1) =
Cp(T2) =
Cp, avg =
Temperature (K)
300
400
500
Cp,gas (J/mol*K)
165.98
210.66
252.09
There are many different ways to estimate
Cp(T1) and Cp(T2).
166.0 J/mole-K
197.3 J/mole-K
181.6 J/mole-K
H(2-3) =
Now, just multiply by the number of moles, n, to get H2-3 :
12,713 J/mol
H(2-3) =
2,538 J
Last, we need to determine the enthalpy change from states 3 to 4, in which the pressure of the
superheated vapor is reduced.
Recall the assumption that the vapor behaves as an ideal gas. Because enthalpy is only a function
of T for ideal gases, and since T3 = T4 :
H(3-4) =
Finally:
H1-4 = H1-2 + H2-3 + H3-4 =
0 J
9,878 J
Verify :
None of the assumptions made in the solution of this problem can be verified based on the given information.
H1-2
7340
J
H3-4
0
J
H2-3
2538
J
H1-4
9,878
J
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm , WB-2
4/12/2011
ENGR 224 - Thermodynamics
Problem :
Baratuci
HW #2
14-Apr-11
WB-3 - Work and Heat Transfer for a Closed, 3-Step Cycle - 6 pts
A closed system undergoes a thermodynamic cycle consisting of the following processes:
Process 1-2:
Adiabatic compression from P1 = 50 psia, V1 = 3.0 ft3 to V2 = 1 ft3 along a path described by :
ˆ 1.4  constant
PV
Process 2-3:
Process 3-1:
Constant volume.
Constant pressure with U1 - U3 = 46.7 Btu.
There are no significant changes in kinetic or potential energies in any of these processes.
a.)
b.)
c.)
Sketch this cycle on a PV Diagram.
Calculate the net work for the cycle in Btu.
Calculate the heat transfer for process 2-3.
The sketch requires a lot of thought about the different types of processes involved.
Determine the Wb value for each step in the cycle and then add them up to get Wcycle.
Apply the 1st law to each step in the process and make use of the known value of U1 - U3 to evaluate Q23.
Given:
Step 1-2:
P1

50 psia
1.4
Step 2-3
V3 = V2 =
3
1.0 ft
Step 3-1:
P3 = P2
U1 - U3 =
46.7 Btu
Find:
Assumptions:
Solution :
Part a.)
a.)
b.)
c.)
3
3.0 ft
3
1.0 ft
V1
V2
Sketch this cycle on a PV Diagram.
W cycle
???
Btu
Q23
???
Btu
- The gas is a closed system
- Boundary work is the only form of work interaction
- Changes in kinetic and potential energies are negligible.
It might be easier to construct the PV Diagram after we complete the rest of the problem. But we might learn something by
constructing the diagram first. Let's try.
We know V1 > V2. We know step 1-2 is an adiabatic compression, so we know that P2 > P1 and W12 < 0. Because adiabatic
compression causes the fluid to heat up because of the addition of energy in the form of work, T2 > T1. That is all we need to
sketch path 1-2 on our PV Diagram. See below.
Because step 2-3 is isochoric, we know that the path for step 2-3 is a vertical line and W23 = 0. Because we also know that
step 3-1 is isobaric, we know that path 3-1 is a horizontal line at P3 = P1. This shows us that step 2-3 results in a decrease in
pressure at constant volume. The only way to accomplish this is to cool the system, so we expect T3 < T2 and Q23 < 0. Now,
we know everything we need to sketch path 2-3 on the PV Diagram.
Step 3-1 is an isobaric expansion in
which U increases by 46.7 Btu. Since
the volume increases, we know that W31
> 0. The addition of heat must be the
cause of the increase in U and the
expansion of the working fluid. So, we
expect Q 31 > 0.
P
T1
We are able to determine what the path
for this cycle looks like without calculating
anything !
T3
I 2
s
o
c
h
o
r
i
c 3
T2
PV1.4 = Constant
Isobaric
1
V
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, WB-3
4/12/2011
Part b.)
Since Wcycle = W12 + W23 + W31, we will work our way around the cycle and calculate each work term along the way.
In step 1-2: P V = C. We can use V or specific V here because the system is closed and the mass inside is constant. The
boundary work for a polytropic process is :
dV P2 V2  P1 V1


1 
1 V
2
2
W12   P dV  C 
1
Eqn 1
All we need to do is determine P2 and then we can use Eqn 1 to evaluate W12.
The way to go here is to evaluate the constant, C, from P1 and V1 :
P1 V1  C
Eqn 2
232.777 psia-ft
C
3
Now, apply the polytropic path equation to state 2, as follows.
P2 V2  C
P2 
Eqn 3
C
V2
P2
Eqn 4
232.8 psia
Now, we can plug values back into Eqn 1 to determine W12 :
1 ft2
1 Btu
2
144 in
778.17 ft-lbf
W 12
Step 2-3 is isochoric, so there is no baoundary work:
W 23
3
-206.94 psia-ft
-38.295 Btu
0 Btu
Step 3-1 is isobaric, so we can evaluate W23 from :
   P dV  P  dV  P  V  V   P V
W
31
1
3
Part c.)
1
1
3
3
Eqn 5
Here, P is either P1 or P3, because they are the same.
Plugging in values yields :
W 31
Now, we add up the work terms to get :
Wcycle =
3
100.00 psia-ft
18.505 Btu
-19.79
Btu
In order to determine Q23, we need to apply the 1st Law to the step.
Q23 - W23 = U3 - U2 = 0
Eqn 6
But, W23 = 0, so Eqn 6 becomes :
Q23 = U3 - U2
Eqn 7
Next, let's apply the 1st Law to step 1-2.
Q12 - W12 = U2 - U1
Eqn 8
Plugging in the value of W12 yields :
U2 - U1 =
38.295 Btu
We were given that :
U1 - U3 =
46.7 Btu
U3 - U2 = - ( U2 - U1 ) - ( U1 - U3 )
Therefore :
Eqn 9
U3 - U2 =
Q23
Therefore :
Verify :
None of the assumptions made in this problem solution can be verified.
a.)
See the PV Diagram, above.
b.)
Wcycle
-19.79
-84.995 Btu
-84.995 Btu
Btu
What kind of cycle is this ? This is a refrigeration or heat pump cycle because Wcycle < 0. For a cycle, Qcycle = Wcycle, so Qcycle
< 0 as well. The cycle rejects all the heat that is absorbs plus it converts a net amount of input work into heat and also rejects
that energy as heat. We cannot tell whether it is a refrigeration or heat pump cycle because the only difference is the goal or
desired quantity. If this cycle were a refrigerator, the goal would be to absorb Q23 from the refrigerated space. If the cycle
were a heat pump, the goal would be to reject Q31 to the heated space. Notice that more energy would be removed from the
refrigerated space than the net amount of work put into the cycle. If this cycle were a heat pump, it would deliver Q31 to the
heated space and Q31 is MUCH larger than Wcycle. This advantage is why heat pumps are ofter preferred to electrical
resistance heaters.
Q31
65.205 Btu
Applying the 1st Law to step 3-1 yields :
COPHP
Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, WB-3
3.29
4/12/2011
ENGR 224 - Thermodynamics
Baratuci
HW #2
14-Apr-11
Problem : WB-4 - Heat Conduction Through a Composite Wall - 3 pts
A composite plane wall consists of a 9 inch thick layer of brick and a 4 inch thick layer of insulation. The outer surface temperatures of the
brick and insulation are 1260oR and 560oR, respectively, and there is perfect contact at the interface between the brick and the insulation. At
steady-state, determine the heat conduction flux through the wall in Btu/h-ft2 and the temperature in oR at the interface between the brick
and the insulation.
Data :
Brick
1.4
k
Insulation
0.05
Btu/h-ft-oR
This problem is an application of Fourier's Law of Conduction.
The key is to assume that the process operates at steady-state. In this case, all of the heat that arrives at the interface
between the brick and the insulation by conduction through the brick must leave the interface as heat conduction through the
insulation. Otherwise, the temperature at the interface would rise or fall as energy accumulated or was depleted at the
interface.
Given :
Lbrick
9
0.75
1.4
1260
kbrick
Tbrick
Find :
q
???
Diagram :
in
ft
Lins
o
Btu/h-ft-oR
R
kins
Tins
Btu/h-ft2
Tint
Lbrick = 9 in
Tins = 1260 oR
4
0.3333
0.05
560
???
in
ft
Btu/h-ft-oR
R
o
o
R
Lins = 4 in
k = 0.05
Btu/h-ft2-oR
Tint = ? oR
k = 1.4
Btu/h-ft2-oR
Assumptions :
Solution :
Tins = 560 oR
1-
The heat transfer is a steady-state process. No variable changes with respect to time.
2-
The thermal conductivity of the brick is uniform and constant, not a function of temperature.
3-
The thermal conductivity of the insulation is uniform and constant, not a function of temperature.
The key to this problem is that, at steady-state, all of the heat that arrives at the interface between the brick and the insulation
by conduction through the brick must leave the interface as heat conduction through the insulation. Otherwise, the
temperature at the interface would rise or fall as energy accumulated or was depleted at the interface.
In terms or Fourier's Law of Conduction, this means :
 dT 
 dT 
q brick  k brick 
 q ins  k ins 


 dx  brick
 dx  ins
Eqn 1
If we assume that the thermal conductivity of brick is constant and the thermal conductivity of the insulation is also constant,
then the temperature profiles in the brick and in the insulation are linear. This allows us to replace the derivatives in Eqn 1
with the slope of the temperature profile in the brick and in the insulation.
 T 
 T 
k brick 
 k ins 


x


 brick
 x  ins
Eqn 2
Now, we can express the slopes of the temperature profiles in terms of the known temperatures on the outer faces of the wall
and the unknown temperature at the interface between the brick and insulation layers.
T
 Tint  Tins 
 Tint 
k brick  brick
  k ins 

 Lbrick

 L ins

Dr. Baratuci - ChemE 260
hw2-sp11.xlsm, WB-4
Eqn 3
4/12/2011
The only unknown in Eqn 3 is Tint, so we can solve for it.
k

k
k
k
Tint  ins  brick   Tins ins  Tbrick brick
L
L
L
L
brick 
ins
brick
 ins
Tins
Tint 
Eqn 4
k ins
k
 Tbrick brick
L ins
L brick
k ins k brick

L ins L brick
Plugging values into Eqn 5 yields :
Eqn 5
kbrick/Lbrick
kins/Lins
Tint
1.867 Btu/h-ft2-oR
0.150 Btu/h-ft2-oR
o
1207.9 R
Now that we know Tint, we can evaluate the temperature gradient or slope in each material and then use either part of Eqn 1 to
evaluate q.
(T/x)brick
qbrick
-69.42
o
R/ft
(T/x)brick
2
-97.19 Btu/h-ft
qins
Verify :
None of the assumptions made in this problem solution can be verified.
q
Dr. Baratuci - ChemE 260
2
-97.2 Btu/h-ft
Tint
hw2-sp11.xlsm, WB-4
o
-1943.80
R/ft
2
-97.19 Btu/h-ft
o
1208 R
4/12/2011
ENGR 224 - Thermodynamics
Baratuci
HW #2
14-Apr-11
Problem : WB-5 - Combined Convection and Radiation Heat Loss - 3 pts
A 3.0 m2 hot black surface at 80oC is losing heat to the surrounding air at 25oC by convection with a convection heat transfer coefficient of 12
W/m2-oC, and by radiation to the surrounding surfaces at 15oC. Determine the total rate of heat loss from the surface in W.
The total rate of heat loss from the black surface is the sum of the heat loss rates by convection and radiation.
The key assumption here is that the emissivity of the surface is 1.0 because it is described as "black".
Given :
AHT
h

Find :
Qtotal
3.0 m2
2
12 W/m -K
1
???
Diagram :
Ts
Tair
Tsurr
W
Tair = 25 oC
Ts = 80 oC
Assumptions :
1234-
Solution :
o
80 C
o
25 C
o
15 C
Tsurr = 15 oC
The heat transfer is a steady-state process. No variable changes with respect to time.
The surface only exchanges heat by radiation with the surroundings and only exchanges heat by
convection to the air.
The emissivity of the surface and of the surroundings are assumed to be 1.0.
The absorptivity, , of the surface and the surroundings are 1.0.
The key here is that the total rate of heat loss from the surface is the sum of the heat loss rate by



Q
Eqn 1
Heat transfer by convection can be determined using Newton's Law of Cooling.

Q
conv  h A HT  Ts  Tair 
Eqn 2
If we assume that the emissivity of the black surface and of the surroundings are both 1.0, then we can evaluate the radiation
heat transfer rate using :


4
Qrad    A HT Ts4  Tsurr

Eqn 3
Where  is the Stefan-Boltzmann Constant :

Plugging values into Eqns 3, 2 & 1 yields :
Qconv
1980 W
1473.0 W
Qtotal
3453 W
Ts
Tsurr
353.15 K
288.15 K
Verify :
None of the assumptions made in this problem solution can be verified.
Qtotal
Dr. Baratuci - ChemE 260
2 4
5.67E-08 W/m -K
3450 W
hw2-sp11.xlsm, WB-5
4/12/2011
ENGR 224 - Thermodynamics
Baratuci
HW #2
14-Apr-11
Problem : WB-6 - Minimum Insulation Thickness for a Hot Surface - 3 pts
A flat surface is covered with insulation with a thermal conductivity of 0.08 W/m-K. The temperature at the interface between the surface and
the insulation is 300oC. The outside of the insulation is exposed to air at 30oC and the convection heat transfer coefficient between the
insulation and the air is 10 W/m2-K. Ignoring radiation heat transfer, determine the minimum thickness of the insulation, in m, such that the
outside surface of the insulation is not hotter than 60oC at steady-state
This is a straight-forward application of Fourier's Law of Conduction and Newton's Law of Cooling.
The key is that, at steady-state, energy cannot accumulate at the surface where the air touches the insulation.
As a result, the rate at which heat is conducted out through the insulation must be equal to the rate at which heat is removed
by convection from the surface of the insulation.
Given :
kins
h
Find :
Lins
Tint
Tair
Ts
0.08 W/m-K
2
10 W/m -K
0 W
???
o
300 C
o
30 C
o
60 C
cm
Diagram :
Tair = 30oC
Ts = 60oC
Tint = 300oC
Lins
Assumptions :
Solution :
123-
The heat transfer is a steady-state process. No variable changes with respect to time.
The thermal conductivity of the brick is uniform and constant, not a function of temperature.
The thermal conductivity of the insulation is uniform and constant, not a function of temperature.
The key to this problem is that, at steady-state, all of the heat that arrives at the surface of the insulation by conduction through
the insulation must leave the surface as heat transfer by convection. Otherwise, the temperature at the surface would rise or
fall as energy accumulated or was depleted at the surface.
In terms or Fourier's Law of Conduction and Newton's Law of Cooling, this means :
 dT 
q ins  k ins 
  q conv  h  Ts  Tair 
 dx  ins
Eqn 1
If we assume that the thermal conductivity of the insulation is constant, then the temperature profile in the insulation is linear.
This allows us to replace the derivative in Eqn 1 with the slope of the temperature profile in the insulation.
 T 
k ins 
  h  Ts  Tair 
 x  ins
Eqn 2
Now, we can express the slopes of the temperature profiles in terms of the known temperature at the interface between the
surface and the insulation and the temperature at the outer surface of the insulation.
 T  Ts 
k ins  int
  h  Ts  Tair 
 L ins 
Eqn 3
The only unknown in Eqn 3 is Lins, so we can solve for it.
L ins 
k ins  Tint  Ts 


h  Ts  Tair 
Eqn 4
Plugging values into Eqn 4 yields :
Lins
Verify :
None of the assumptions made in this problem solution can be verified.
Lins
Dr. Baratuci - ChemE 260
0.064 m
0.064 m
hw2-sp11.xlsm, WB-6
4/12/2011
ENGR 224 - Thermodynamics
Baratuci
HW #2
14-Apr-11
Problem : WB-7 - 1st Law Analysis of Steam in a Closed System - 3 pts
As shown in the figure below, 5.0 kg of steam contained within a piston-and-cylinder device undergoes an expansion from state 1 where the
specific internal energy is 2709.9 kJ/kg to state 2 where the specific internal energy is 2659.6 kJ/kg. During the process, there is heat transfer
to the steam with a magnitude of 80 kJ. Also, a paddle wheel transfers energy to the steam by work in the amount of 18.5 kJ. There is no
significant change in the kinetic or gravitational potential energies of the steam. Determine the work done by the steam on the piston during
the process in kJ.
This problem is an application of the 1st Law.
Making and applying the proper assumptions are the keys to solving this problem correctly.
Given :
Q
W prop
m
Find :
Wb
Diagram :
See above.
Assumptions:
Solution :
U1
U2
80 kJ
-18.5 kJ
5 kg
???
2709.9 kJ/kg
2659.6 kJ/kg
kJ
- The steam is a closed system
- Boundary work and the work done by the paddle wheel are the only forms of work in this process.
- Changes in kinetic and potential energies are negligible.
- The process is a quasi-equilibrium process.
- The initial and final states are both equilibrium states.
Let's begin by writing the 1st Law, closed system, for this process.

ˆ U
ˆ
Q  Wtotal  m U
2
1

Eqn 1
Propeller work and boundary work are the only forms of work that cross the system boundary in this process.
Therefore :
Wtotal  Wprop  Wb
Combining Eqns 1 & 2 yields :

Eqn 2


ˆ U
ˆ
Q  Wprop  Wb  m U
2
1

Eqn 3
We can solve Eqn 3 for the unknown Wb, as follows.

ˆ U
ˆ
Wb  Q  Wprop  m U
2
1
Plugging values into Eqn 4 yields :
Eqn 4
Wb
Verify :
None of the assumptions made in this problem solution can be verified.
Wb
Dr. Baratuci - ChemE 260

350 kJ
350 kJ
hw2-sp11.xlsm, WB-7
4/12/2011
ENGR 224 - Thermodynamics
Baratuci
HW #2
14-Apr-11
Problem : 6.23 - Power Plant Efficiency - 2 pts
In 2001, the United States produced 51% of its electricity in the amount of 1.878 x 1012 kW-h from coal-fired plants. Taking the average
thermal efficiency to be 34%, determine the amount of thermal energy rejected by the coal-fired power plants in the United States.
Assume that the shaft work produced by turbines in electric power plants is 1.878 x 1012 kW-h.
Use the definition of the thermal efficiency of a heat engine to determine QH and then apply the 1st law to all of the power
generation facilities in the country, collectively, to determine QC.
Given :
W HE
Find :
QC

1.878E+12 kW-h
???
0.34
kW-h
Diagram :
Hot
QH = ??? kW-h
 = 0.34
HE
WHE = 1.878 x 1012 kW-h
QC = ??? kW‐h
Cold
3.646E+12 kW-h
- Assume that the shaft work produced by turbines in electric power plants is 1.878 x 1012 kW-h.
Assumptions:
Solution :
QC
Let's begin by writing the equation that defines the efficiency of a power cycle.

Desired Output WHE

Re quired Input
QH
Eqn 1
QH 
We can solve Eqn 1 for QH :
WHE

Plugging values into Eqn 2 yields :
Eqn 2
QH
5.524E+12 kW-h
We can determine the annual energy rejected, QC, by applying the 1st Law to all the heat engines in the US.
Solving Eqn 3 for QC yields :
QH = WHE + Q C
Eqn 3
Q C = QH - WHE
Eqn 4
Plugging values into Eqn 4 yields :
Verify :
QC
The big assumption on which this solution cannot be readily verified.
QC
3.65E+12
Dr. Baratuci - ChemE 260
3.646E+12
kW-h
This is an enormous amount of heat to
dump into the environment !
kW-h
hw2-sp11.xlsm, 6.23
4/12/2011
ENGR 224 - Thermodynamics
Baratuci
HW #2
14-Apr-11
Problem : 6.41 - Work, Heat and COP in a Refrigeration Cycle - 2 pts
A refrigerator used to cool a computer requires 3 kW of electrical
power and has a COP of 1.4. Calculate the cooling effect of this
refrigerator, in kW.
Hot
QH
Ref
Wcycle
QC
Cold
This is an application of the the definiton of COP for a refrigeration cycle.
Given :
COPR
1.4
Find :
QC
???
Assumptions :
1-
W
3
kW
kJ
When the refrigerator is operating, it operates at steady-state.
Equations / Data / Solve :
QC
QC

Wref
Q H  QC
The definition of COPR is :
COPR 
Solve Eqn 1 for QC :
QC  COPR  Wref
Eqn 1
Eqn 2
QC
Plugging values into Eqn 2 yields :
4.2
kW
We pay for just 3 kW of electrical power to operate our refrigerator and the device removes heat from the refrigerated space at
a rate of 4.2 kW. Pretty cool, eh ?
Verify :
None of the assumptions made in this problem solution can be verified.
QC
Dr. Baratuci - ChemE 260
4.2
kW
hw2-sp11.xlsm, 6.41
4/12/2011
ENGR 224 - Thermodynamics
Baratuci
HW #2
14-Apr-11
Problem : 6.55 - Heat Pump COP and Monthly Operating Cost - 2 pts
A house that was heated by electric resistance heaters consumed 1200 kW-h of electric energy in a winter month. If this house were heated
instead by a heat pump that has an average COP of 2.4, determine how much money the home owner would have saved that month.
Assume a price of \$0.085/kW-h for electricity.
This problem is an application of the definition of the coefficient of performance of a heat pump.
Notice that money "saved" is the difference between the monthly operating cost of the electrical resistance heater and the
monthly operating cost of the heat pump.
Given :
COPHP
Price
2.4
0.085
\$/kW-h
Savings
???
\$/month
Find :
W
1200
kW-h
Diagram :
Hot
QH
COPHP = 2.4
HP
WHP = 1200 kW-h
QC
Cold
Assumptions:
Solution :
1-
When the heat pump is operating, it operates at steady-state.
Let's begin by determining the cost to heat the house with the electrical resistance heaters.
In an electrical resistance heater, each kW-h of electrical energy is converted directly into a kW-h of heat that is transferred
into the house. The cost of this heat is just the product of the electrical energy used and the price of the electricity.
Cost  Wmonth kW  h / month  * Pr ice[\$ / kW  h]
Eqn 1
costresist
102
\$/month
Now, let's consider a heat pump that will also deliver the 1200 kW-h of heat into the house. In terms of our traditional tiefighter diagram, this means QH = 1200 kW-h for our heat pump. Now, we have to determine how much electrical work we
must supply to the heat pump in order to get QH = 1200 kW-h.
QH
1200 kW-h
Let's write the equation that defines the COP for a heat pump cycle.
COPHP 
WHP 
We can solve Eqn 2 for WHP, as follows :
QH
Desired Output

Re quired Input WHP
Eqn 2
QH
COPHP
Plugging values into Eqn 3 yields :
Eqn 3
W HP
500 kW
costHP
42.5 \$/month
We can determine the monthly energy consumption of the heat pump using Eqn 1.
We can now determine the savings that could be achieved with this heat pump from :
Savings = costresist - costHP
Verify :