Example Problem with Complete Solution

4A-3 : Quasi-Equilibrium Compression of Ammonia
4 pts
Ammonia vapor is compressed inside a cylinder by an external force acting on the piston. The ammonia is initially at 30°C and 500 kPa. The final pressure is 1400 kPa.
   
a.) Determine the work done by the ammonia in the process.
b.) What is the final temperature of the ammonia ?
                 
The following data have been measured for the process.    
                   
  P (kPa)
500
653
802
945
1100
1248
1400
  
  V (L)
1.25
1.08
0.96
0.84
0.72
0.60
0.50
  
Read : The key concept here is that boundary or PV work is represented by the area under the process path curve on a PV Diagram.  So, once we plot the given data on a PV Diagram, all we need to do is numerically integrate to determine the area under the curve and we will have know the work !
We can use the Ammonia Tables on the NIST WebBook to determine the final temperature because we know both Pfinal and Vfinal.  But we don't know the number of moles in the system.  Fortunately, we can use the initial state P1, V1, and T1 to determine the number of moles in this closed system.
  Figure 1.
Given:
P (kPa)
V (L)
500
1.25
653
1.08
802
0.96
945
0.84
1100
0.72
1248
0.60
1400
0.50
T1 =
30
 oC
Find: T2 =
???
 oC
Assumptions: - Each state in the data table is an equilibrium state.
- The process is a quasi-equilibrium process.
- The system is a closed system.
- The trapezoidal rule gives an acceptable estimate of the area under the process path in the PV Diagram
Part a.) The area of each trapezoid under the process path in the PV Diagram is the product of the average pressure for that trapezoid and the change in volume across the trapezoid.
Trapezoid
Pavg
(kPa)
ΔV (L)
W (J)
A
1324
0.10
132
B
1174
0.12
141
C
1023
0.12
123
D
874
0.12
105
E
728
0.12
87
F
577
0.17
98
 Wtotal =
686
 J
Part b.) If we knew the specific volume of the ammonia in the final state, we could use the ammonia tables to determine the temperature.  But, at this point all we know is the total volume in the final state.  Therefore, we need to determine the mass of ammonia in  the system is the same in the initial and final state.  So, we can use the information we have for the initial state to determine the mass of ammonia in the system.
First, look up the specific volume of ammonia in the initial state in the isothermal thermodynamic tables:
V1 =
0.28099
 m3/kg
 
Then, to calculate mNH3, use: Equation.
Eqn 1
 mNH3 = 0.00445 kg
Finally, calculate V2,hat using: Equation.
Eqn 2
 V2,hat = 0.112396 m3/kg
Now, use the isobaric thermodynamic tables at a pressure of 1400 kPa.  Specify a temperature range that you are sure brackets V2.  Here is a table that I cut-and-pasted from the NIST WebBook.
T
(C)
P
(MPa)
V
(m3/kg)
Phase
  
70
1.4
0.10881
vapor
  
72
1.4
0.1097
vapor
  
74
1.4
0.11059
vapor
  
76
1.4
0.11148
vapor
  
78
1.4
0.11236
vapor
   V2 falls between 78oC and 80oC
80
1.4
0.11323
vapor
   So, now, we must interpolate.

Answer:

 T2 = 78.1 oC
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