| 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 |
|
|
|
|
|
|
|
|
|
|
|
