Complete
the following table by
determining the values
of all the blank
entries. The system
contains only R134a. 





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 R134a 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 R134a, 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 (^{o}C) 
P (kPa) 
v (m^{3}/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 
^{o}C 


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 U_{sat vap} and U_{sat liq} at the given temperature. 











If : 
U > U_{sat vap} 

Then : 
The system contains a
superheated vapor. 


If : 
U < U_{sat liq} 

Then : 
The system contains a
subcooled liquid. 


If : 
U_{sat
vap} > U > U_{sat
liq} 
Then : 
The system
contains an equilibrium mixture of saturated liquid and saturated vapor. 






Data : 



















P*(kPa) 
T_{sat} (^{o}C) 
V_{sat
liq} (m^{3}/kg) 
V_{sat
vap} (m^{3}/kg) 
U_{sat liq} (kJ/kg) 
U_{sat
vap} (kJ/kg) 
H_{sat
liq} (kJ/kg) 
H_{sat
vap} (kJ/kg) 


163.94 
15 
7.4469E04 
0.12067 
180.02 
369.85 
180.14 
389.63 












Because U = U_{sat 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 
m^{3}/kg 

P 
163.94 
kPa 



H 
389.63 
kJ/kg 










Part B.) 
Given : 
T 
43 
^{o}C 


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) 
T_{sat} (^{o}C) 
V_{sat
liq} (m^{3}/kg) 
V_{sat
vap} (m^{3}/kg) 
U_{sat liq} (kJ/kg) 
U_{sat
vap} (kJ/kg) 
H_{sat
liq} (kJ/kg) 
H_{sat
vap} (kJ/kg) 


1016.6 
40 
8.7204E04 
0.019966 
255.52 
399.13 
256.41 
419.43 


1159.9 
45 
8.8885E04 
0.017344 
262.91 
401.40 
263.94 
421.52 


700 
26.71 
8.3320E04 
0.029365 
236.41 
392.64 
236.99 
413.20 


750 
29.08 
8.3959E04 
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 T_{sat}(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 40^{o}C and 50^{o}C, but not for 43^{o}C. 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 (^{o}C) 
V (m^{3}/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( ^{o}C ) 
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 
m^{3}/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( ^{o}C ) 
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( ^{o}C ) 
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 
m^{3}/kg 











We again begin by
determining the state of the system.
In this case, it would be easiest to lookup the V_{sat vap} and V_{sat 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 : 
V_{sat
vap} > V > V_{sat
liq} 
Then : 
The system
contains an equilibrium mixture of saturated liquid and saturated vapor. 






Data : 



















P*(kPa) 
T_{sat} (^{o}C) 
V_{sat
liq} (m^{3}/kg) 
V_{sat
vap} (m^{3}/kg) 
U_{sat liq} (kJ/kg) 
U_{sat
vap} (kJ/kg) 
H_{sat
liq} (kJ/kg) 
H_{sat
vap} (kJ/kg) 


250 
4.28 
7.6406E04 
0.080685 
194.08 
375.91 
194.27 
396.08 












Because V lies between V_{sat liq} and V_{sat
vap}, the system is the twophase envelope
and T = T_{sat}. 


















T 
4.284 
^{o}C 











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 
^{o}C 

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 twophase 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 R134a, this would require an
interpolation. It is easier to
determine the saturation pressure or vapor pressure based on the system
temperature because 50^{o}C does appear in the Saturation Temperature Table and
therefore does not require an interpolation. 

















P*(50^{o}C) 
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 R134 includes data fo 50^{o}C at both 1400
kPa and 1600 kPa. 

Therefore, a
singleinterpolation is required for each unknown property in the problem
statement. 












P*(kPa) 
T (^{o}C) 
V (m^{3}/kg) 
U (kJ/kg) 
H (kJ/kg) 





1400 
50 
9.0646E04 
270.32 
271.59 

V 
9.0517E04 
m^{3}/kg 

1547 
50 
9.0517E04 
270.13 
271.53 

U 
270.13 
kJ/kg 

1600 
50 
9.0470E04 
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
H_{sat vap} and H_{sat liq} at the given temperature. 











If : 
H > H_{sat vap} 

Then : 
The system contains a
superheated vapor. 


If : 
H < H_{sat liq} 

Then : 
The system contains a
subcooled liquid. 


If : 
H_{sat
vap} > H > H_{sat
liq} 
Then : 
The system
contains an equilibrium mixture of saturated liquid and saturated vapor. 






Data : 



















P*(kPa) 
T_{sat} (^{o}C) 
V_{sat
liq} (m^{3}/kg) 
V_{sat
vap} (m^{3}/kg) 
U_{sat liq} (kJ/kg) 
U_{sat
vap} (kJ/kg) 
H_{sat
liq} (kJ/kg) 
H_{sat
vap} (kJ/kg) 


950 
37.50 
8.6412E04 
0.0214415 
251.86 
397.95 
252.69 
418.32 


1000 
39.39 
8.7007E04 
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) 
T_{sat} (^{o}C) 
V_{sat
liq} (m^{3}/kg) 
V_{sat
vap} (m^{3}/kg) 
U_{sat liq} (kJ/kg) 
U_{sat
vap} (kJ/kg) 
H_{sat
liq} (kJ/kg) 
H_{sat
vap} (kJ/kg) 


976 
38.479 
8.6721E04 
0.0208563 
253.30 
398.42 
254.15 
418.76 












Because H lies between H_{sat liq} and H_{sat
vap}, the system is the twophase envelope
and T = T_{sat}. 

















T 
38.5 
^{o}C 











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 
8.6721E04 
m^{3}/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 R134a Tables. The R134a Tables only consider states in which no solid R134a can exist at equilibrium. 










Answers : 

T (^{o}C) 
P (kPa) 
V (m^{3}/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 
9.0517E04 
270.13 
271.53 
N/A  Subcooled 


e.) 
0 
976 
8.6721E04 
253.30 
318.70 
0.392 










