# Example Problem with Complete Solution

2D-8 : Determining System Properties Using Thermodynamic Tables 10 pts
Complete the following table by determining the values of all the blank entries. The system contains only R-134a. 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: T (oC) P (kPa) v (m3/kg) u (kJ/kg) h (kJ/kg) x (kg vap/kg tot)
a.) -15     369.85
b.) 43 728
c.)   250 0.049
d.) 50 1547
e.)   976     318.7
Find: Fill in all the blank values in the table, above.
Assumptions: - No solid phase exists in any of these 5 systems
Equations / Data / Solve:
Part a.) Given : T -15 oC U 369.85 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)
163.94 -15 7.4469E-04 0.12067 180.02 369.85 180.14 389.63
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 kg vap/kg V 0.12067 m3/kg
P 163.94 kPa H 389.63 kJ/kg
Part B.) Given : T 43 oC P 728 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)
1016.6 40 8.7204E-04 0.019966 255.52 399.13 256.41 419.43
1159.9 45 8.8885E-04 0.017344 262.91 401.40 263.94 421.52
700 26.71 8.3320E-04 0.029365 236.41 392.64 236.99 413.20
750 29.08 8.3959E-04 0.027375 239.76 393.84 240.39 414.37
We could interpolate to determine the saturation properties at 728 kPa, but there isn't much point !  Since the system temperature is higher than the saturation temperature at EITHER 700 kPa or 750 kPa, the system temperature must also be higher than the interpolated value of Tsat(728 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 700 and 800 kPa, but not 728 kPa.  These two tables include rows for 40oC and 50oC, but not for 43oC.  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 Lesson 2C on page 18.
The data required for the double interpolation tables are :
P*(kPa) T (oC) V (m3/kg) U (kJ/kg) H (kJ/kg)
700 40 0.031696 404.53 426.72
700 50 0.033322 413.35 436.67
800 40 0.027036 402.97 424.59
800 50 0.028547 412.00 434.84
Here is the double interpolation table for V : 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 ) 700 728 800
40 0.031696 0.030391 0.0270357
43 0.032184 0.030869 0.027489
50 0.033322 0.031985 0.028547
V 0.030869 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 ) 700 728 800
40 404.53 404.09 402.97
43 407.18 406.76 405.68
50 413.35 412.97 412.00
U 406.76 kJ/kg
Here is the double interpolation table for H : 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 ) 700 728 800
40 426.72 426.13 424.59
43 429.71 429.14 427.67
50 436.67 436.16 434.84
H 429.14 kJ/kg
Part c.) Given : P 250 kPa V 0.049 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 > Vsat vap Then : The system contains a superheated vapor.
If : V < Vsat liq Then : The system contains a subcooled liquid.
If : Vsat vap > V > Vsat 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)
250 -4.28 7.6406E-04 0.080685 194.08 375.91 194.27 396.08
Because V lies between Vsat liq and Vsat vap, the system is the two-phase envelope and T = Tsat.
T -4.284 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. Eqn 1
We can determine x from the saturation data and the known value of u for the system using : Eqn 2
x 0.604 kg vap/kg
Now, we can plug x back into Eqn 1 and apply it to the unknown properties, U and H.
U 303.82 kJ/kg H 316.07 kJ/kg
Part d.) Given : T 50 oC P 1547 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 1547 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 50oC does appear in the Saturation Temperature Table and therefore does not require an interpolation.
P*(50oC) 1317.9 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 50oC at both 1400 kPa and 1600 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)
1400 50 9.0646E-04 270.32 271.59 V 9.0517E-04 m3/kg
1547 50 9.0517E-04 270.13 271.53 U 270.13 kJ/kg
1600 50 9.0470E-04 270.06 271.51 H 271.53 kJ/kg
Part e.) Given : P 976 kPa H 318.7 kJ/kg
This part of the problem is very similar to part a.
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)
950 37.50 8.6412E-04 0.0214415 251.86 397.95 252.69 418.32
1000 39.39 8.7007E-04 0.020316 254.63 398.85 255.50 419.16
Unfortunately, the system pressure of 976 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 976 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)
976 38.479 8.6721E-04 0.0208563 253.30 398.42 254.15 418.76
Because H lies between Hsat liq and Hsat vap, the system is the two-phase envelope and T = Tsat.
T 38.5 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. Eqn 3
We can determine x from the saturation data and the known value of U for the system using : Eqn 4
x 0.392 kg vap/kg
Now, we can plug x back into Eqn 1 and apply it to the unknown properties, V and U.
V 8.6721E-04 m3/kg U 253.30 kJ/kg
Verify: The assumption that there is no ice in the is confirmed by the fact that the the states in part (a) and part (b) are both located in the R-134a Tables. The R-134a Tables only consider states in which no solid R-134a can exist at equilibrium.
Answers : T (oC) P (kPa) V (m3/kg) U (kJ/kg) H (kJ/kg) x (kg vap/kg tot)
a.) -15 163.9 0.120671 369.85 389.63 1
b.) 43 728 0.030869 406.76 429.14 N/A - Superheated
c.) -4.284 250 0.049 303.82 316.07 0.604
d.) 50 1547 9.0517E-04 270.13 271.53 N/A - Subcooled
e.) 0 976 8.6721E-04 253.30 318.70 0.392 