Example Problem with Complete Solution

2D-8 Determining System Properties Using Thermodynamic Tables 10 pts
Determine the missing property values in the table below for a system that contains pure R-134a.
                   
 
T (oC)
P (kPa)
v (m3/kg)
u (kJ/kg)
h (kJ/kg)
x (kg vap/kg tot)
    
a.)
30
836
     
b.)
-45
204.51
     
c.)
300
0.054
     
d.)
1124
244.0
      
e.)
65
1550
     
Read : The key to this problem is to recognize that all of the variables in the table are state variables, or properties, and that they are all intensive properties.  It is also important to assume that either one or two phases exist.  The triple point of R-134a is not common knowledge, but it is pretty safe to assume that it does not appear in this table.  We can verify this assumption later.  Also, since we have no data availabe about solid R-134a, we can assume that we have either a subcooled liquid, a superheated vapor or an equilibrium mixture of saturated vapor and saturated liquid in the system.  Gibbs Phase Rule tells us that for a pure substance in a single phase there are 2 degrees of freedom.  If two phases are present, then there is just 1 degree of freedom.  In either case, the two values of intensive properties given in each part of this problem will be sufficient to completely determine the values of all of the other intensive properties of the system.  So, we are in good shape to move forward on solving this problem.
Given : See the table in the problem statement.
Assumption : - No solid phase exists in any of these 5 systems
Lookup Data :
Saturation Data :
P*(kPa)
Tsat (oC)
vsat liq
(m3/kg)
vsat vap (m3/kg)
usat liq (kJ/kg)
usat vap (kJ/kg)
hsat liq (kJ/kg)
hsat vap (kJ/kg)
800
31.327
0.00084585
0.025625
94.825
246.82
95.501
267.32
900
35.526
0.00085811
0.022687
100.86
248.87
101.64
269.29
770.2
30
0.00084213
0.026642
92.93
246.16
93.578
266.67
39.117
-45
0.00069828
0.464730
-6.2836
204.51
-6.2563
222.69
300
0.672
0.00077366
0.067704
52.527
230.54
52.759
250.85
1100
42.969
0.00088185
0.018360
111.75
252.35
112.72
272.55
1200
46.315
0.00089351
0.016718
116.73
253.84
117.80
273.90
60
1.6818
0.00094979
0.011444
137.76
259.24
139.36
278.49
70
2.1168
0.00100380
0.008653
154.01
262.19
156.14
280.51
1500
55.233
0.00092836
0.013056
130.30
257.50
131.69
277.08
1600
57.906
0.00094010
0.012126
134.47
258.50
135.97
277.90
Subcooled Liquid Data :
P*
(kPa)
T
(oC)
v
(m3/kg)
u
(kJ/kg)
h
(kJ/kg)
800
30
0.00084199
92.904
93.577
900
30
0.00084151
92.817
93.574
Superheated Vapor Data :
P*
(kPa)
T
(oC)
v
(m3/kg)
u
(kJ/kg)
h
(kJ/kg)
1400
60
0.015005
264.47
285.47
1400
70
0.016060
274.62
297.11
1600
60
0.012373
260.90
280.69
1600
70
0.013430
271.76
293.25
Part a.) Given : T
30
 oC
P
836
 kPa
The first step in solving each part of this problem is to determine the state of the system.  Is it subcooled liquid, superheated vapor or a two-phase VLE mixture.
We could do this by determining the boiling point or saturation temperature at the system pressure.  But, since 836 kPa does not appear in the Saturation Pressure Table for R-134a, this would require an interpolation.  It is easier to determine the saturation pressure or vapor pressure based on the system temperature because 30°C does appear in the Saturation Temperature Table and therefore does not require an interpolation.
P*(30oC)
770.2
 kPa
Since the actual system pressure is ABOVE the vapor pressure, the system contains a subcooled liquid.
The quality of a subcooled liquid is undefined.  Therefore : x = N/A - Subcooled
The Subcooled Liquid Table for R-134 includes data fo 30°C at both 800 kPa and 900 kPa.
Therefore, a single-interpolation is required for each unknown property in the problem statement.
P*
(kPa)
T
(oC)
v
(m3/kg)
u
(kJ/kg)
h
(kJ/kg)
800
30
0.0008420
92.904
93.577
 v
0.0008418
 m3/kg
836
30
0.0008418
92.873
93.576
 u
92.873
 kJ/kg
900
30
0.0008415
92.817
93.574
 h
93.576
 kJ/kg
Part b.) Given : T
-45
 oC
u
204.51
 kJ/kg
We again begin by determining the state of the system.  In this case, it would be easiest to lookup the usat vap and usat liq at the given temperature.
If : u > usat vap Then : The system contains a superheated vapor.
If : u < usat liq Then : The system contains a subcooled liquid.
If : usat vap > u > usat liq Then : The system contains an equilibrium mixture of saturated liquid and saturated vapor.
Data :
P*(kPa)
Tsat (oC)
vsat liq
(m3/kg)
vsat vap (m3/kg)
usat liq (kJ/kg)
usat vap (kJ/kg)
hsat liq (kJ/kg)
hsat vap (kJ/kg)
39.117
-45
0.0006983
0.46473
-6.2836
204.51
-6.2563
222.69
Because u = usat vap, our system contains a saturated vapor.  The bonus here is that the quality is 1 and all the other answers to this part of the question come directly out of this table !
x 1  
P 39.117 kPa
v 0.46473 m3/kg
h 222.69 kJ/kg
Part c.) Given : P
300
 kPa
v
0.054
 m3/kg
We again begin by determining the state of the system.  In this case, it would be easiest to lookup the vsat vap and vsat liq at the given pressure.
If : v > v sat vap Then : The system contains a superheated vapor.
If : v < v sat liq Then : The system contains a subcooled liquid.
If : vsat vap > v > v sat liq Then : The system contains an equilibrium mixture of saturated liquid and saturated vapor.
Data :
P*(kPa)
Tsat (oC)
vsat liq
(m3/kg)
vsat vap (m3/kg)
usat liq (kJ/kg)
usat vap (kJ/kg)
hsat liq (kJ/kg)
hsat vap (kJ/kg)
300
0.672
0.0007737
0.067704
52.527
230.54
52.759
250.85
Because v lies between vsat liq and vsat vap, the system is the two-phase envelope and T = Tsat.
T 0.672 oC
In order to determine the values of the other properties of the system using the following equation, we will need to know the quality, x.
Equation. Eqn 1
We can determine x from the saturation data and the known value of u for the system using :
Equation. Eqn 2 x
0.795
 kg vap/kg total
Now, we can plug x back into Eqn 1 and apply it to the unknown properties, u and h.
u
194.09
 kJ/kg h 210.29 kJ/kg  

Part d.)

Given : P 1124 kPa This part of the problem is very similar to part b.
  h 244.0 kJ/kg
We again begin by determining the state of the system.  In this case, it would be easiest to lookup the hsat vap and hsat liq at the given temperature.
If : h > hsat vap Then : The system contains a superheated vapor.
If : h < hsat liq Then : The system contains a subcooled liquid.
If : hsat vap > h > hsat liq Then : The system contains an equilibrium mixture of saturated liquid and saturated vapor.
Data :
P*
(kPa)
Tsat
(oC)
vsat liq
(m3/kg)
vsat vap
 (m3/kg)
usat liq (kJ/kg)
usat vap (kJ/kg)
hsat liq (kJ/kg)
hsat vap (kJ/kg)
1100
42.969
0.0008819
0.018360
111.75
252.35
112.72
272.55
1200
46.315
0.0008935
0.016718
116.73
253.84
117.80
273.90
Unfortunately, the system pressure of 1124 kPa does not appear in the Saturation Pressure Table.
So, we will have to interpolate between the two rows in the table shown here to determine the saturation properties at 1124 kPa.
P*(kPa)
Tsat (oC)
vsat liq
(m3/kg)
vsat vap (m3/kg)
usat liq (kJ/kg)
usat vap (kJ/kg)
hsat liq (kJ/kg)
hsat vap (kJ/kg)
1124
43.772
0.0008846
0.0179659
112.95
252.71
113.94
272.87
Because h lies between hsat liq and hsat vap, the system is the two-phase envelope and P = P* = Psat.
P
43.8
 oC
In order to determine the values of the other properties of the system using the following equation, we will need to know the quality, x.
Equation. Eqn 3
We can determine x from the saturation data and the known value of u for the system using :
Equation. Eqn 4 x
0.818
 kg vap/kg total
Now, we can plug x back into Eqn 1 and apply it to the unknown properties, u and h.
v
0.014863
 m3/kg u
227.32
 kJ/kg
Part e.) Given : T
65
 oC
P
1550
 kPa
We again begin by determining the state of the system.  Unfortunately the system temperature is not listed in the Saturation Temperature Table and the system pressure is is not listed in the Saturation Pressure Table.  Either way we go, interpolation is required.
P*(kPa)
Tsat (oC)
vsat liq
(m3/kg)
vsat vap
 (m3/kg)
usat liq (kJ/kg)
usat vap (kJ/kg)
hsat liq (kJ/kg)
hsat vap (kJ/kg)
60
1.6818
0.0009498
0.0114440
137.76
259.24
139.36
278.49
70
2.1168
0.0010038
0.0086527
154.01
262.19
156.14
280.51
               
1500
55.233
0.00092836
0.0130560
130.3
257.5
131.69
277.08
1600
57.906
0.0009401
0.0121260
134.47
258.5
135.97
277.9
We could interpolate to determine the saturation properties at 1550 kPa, but there isn't much point !  Since the system temperature is higher than the saturation temperature at EITHER 1550 kPa or 1600 kPa, the system temperature must also be higher than the interpolated value of Tsat(1550 kPa).
Since the system temperature is greater than the saturation temperature at the system pressure, the system contains a superheated vapor.  Therefore, we must use data from the Superheated Vapor Table to determine the unknown properties of the system.
x N/A - Superheated
The Superheated Vapor Table includes tables for pressure of 1400 and 1600 kPa, but not 1550 kPa.  These two tables include rows for 60oC and 70oC, but not for 65oC.  Consequently a double interpolation is required for each unknown system propert, v, u and h.
The double interpolation can be done with the aid of tables like the ones developed in Thermo-CD.
The data required for the double interpolation tables are :
P*
(kPa)
T
(oC)
v
(m3/kg)
u
(kJ/kg)
h
(kJ/kg)
1400
60
0.015005
264.47
285.47
1400
70
0.016060
274.62
297.11
1600
60
0.012373
260.90
280.69
1600
70
0.013430
271.76
293.25
Here is the double interpolation table for v :
Pressure (kPa)
T( oC )
1400
1550
1600
I chose to interpolate on pressure first and then to interpolate on temperature.  If you do the interpolations in the opposite order, you will get a slightly different answer.  Either method is satisfactory.
60
0.015005
0.013031
0.012373
65
0.015533
0.013559
0.012902
70
0.016060
0.014088
0.013430
v
0.013559
 m3/kg
Here is the double interpolation table for u :
Pressure (kPa)
I chose to interpolate on pressure first and then to interpolate on temperature.  If you do the interpolations in the opposite order, you will get a slightly different answer.  Either method is satisfactory.
T( oC )
1400
1550
1600
60
264.47
261.79
260.90
65
269.55
267.13
266.33
70
274.62
272.48
271.76
u
267.13
 kJ/kg
Here is the double interpolation table for h :
Pressure (kPa)
T( oC )
1400
1550
1600
I chose to interpolate on pressure first and then to interpolate on temperature.  If you do the interpolations in the opposite order, you will get a slightly different answer.  Either method is satisfactory.
60
285.47
281.89
280.69
65
291.29
288.05
286.97
70
297.11
294.22
293.25
h
288.05
 kJ/kg
Answers :
T (oC)
P (kPa)
v (m3/kg)
u (kJ/kg)
h
(kJ/kg)
x (kg vap/kg tot)
a.)
30
836
0.0008418
92.873
93.576
N/A - Subcooled
b.)
-45
39.117
0.46473
204.51
222.69
1
c.)
0.672
300
0.054
194.09
210.29
0.795
d.)
0
43.772
0.014863
227.32
244
0.818
e.)
65
1550
0.013559
267.13
288.05
N/A - Superheated