P-h Diagram for R170
Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter47: High-pressure, Low-pressure, And Absorption Chilled-water Systems
Section: Chapter Questions
Problem 13RQ: The subcooling temperature in a condenser can be measured by taking the difference between the A....
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Question
I couldn't upload P-h Diagram for R170 (Ethane) due to upload image limit.
Please find it on google. I really need help on this
![The cold storage room is maintained at T; = -70 °C by two stage vapor-compression refrigeration unit. In the
low temperature cycle, R170 (ethane) is used as the working fluid. In the high temperature cycle, R134a is used
as the working fluid. After selecting reasonable pressures for all heat exchangers, calculate evaporator
capacity, COP of the system, and compressor powers. Show refrigeration cycles on P-h diagrams i
Draw the T-s diagrams of R134a and R170 cycles as well. Draw schematic
of refrigeration system with its components.
ULI
Notes
1) Isentropic efficiency of compressors are 0.75.
2) Refrigerants leave the condensers as saturated liquid and exit from the evaporators as saturated vapor.
3) When determining the cooling requirement, only the wall conduction load will be taken into account.
(consider 5 walls, neglect heat loss through the floor)
4) The inner and outer walls are normally covered with steel sheets, but their conduction resistance can be
neglected due to high heat conductivity constant k of steel.
See the Lecture Note titled “Refrigeration" regarding two stage vapor-compression refrigeration cycles.
Cold Storage
5 cm thick polyurethane foam
insulation material
is used for all storage walls.
Kurethane = 0.035 W/m.°C
2m
T = -70 °C
h; = 10 W/m?.°C
T, = 25 °C
ho = 10 W/m?.°C
%3D
1 m
1 m
Heat loss through
the floor is neglected.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feabcba1c-0269-4c8f-b29a-c34fcc486852%2F1345689d-6555-4655-8d01-2c3df54f6446%2Fldj4xpe_processed.png&w=3840&q=75)
Transcribed Image Text:The cold storage room is maintained at T; = -70 °C by two stage vapor-compression refrigeration unit. In the
low temperature cycle, R170 (ethane) is used as the working fluid. In the high temperature cycle, R134a is used
as the working fluid. After selecting reasonable pressures for all heat exchangers, calculate evaporator
capacity, COP of the system, and compressor powers. Show refrigeration cycles on P-h diagrams i
Draw the T-s diagrams of R134a and R170 cycles as well. Draw schematic
of refrigeration system with its components.
ULI
Notes
1) Isentropic efficiency of compressors are 0.75.
2) Refrigerants leave the condensers as saturated liquid and exit from the evaporators as saturated vapor.
3) When determining the cooling requirement, only the wall conduction load will be taken into account.
(consider 5 walls, neglect heat loss through the floor)
4) The inner and outer walls are normally covered with steel sheets, but their conduction resistance can be
neglected due to high heat conductivity constant k of steel.
See the Lecture Note titled “Refrigeration" regarding two stage vapor-compression refrigeration cycles.
Cold Storage
5 cm thick polyurethane foam
insulation material
is used for all storage walls.
Kurethane = 0.035 W/m.°C
2m
T = -70 °C
h; = 10 W/m?.°C
T, = 25 °C
ho = 10 W/m?.°C
%3D
1 m
1 m
Heat loss through
the floor is neglected.
![Licensed for single user. © 2009 ASHRAE, Inc.
100
20
150
200
250
300
350
400
450
500
550
R-134a
600
20
1050 --
1000
(1,1,1,2-tetrafluoroethane)
10
Reference state:
900
600
p= 500 kg/ma
-800
700 -
h = 200.0 kJ/kg, s= 1.00 kJ/(kg K)
for saturated liquid at 0°C
8
400
10
6
300
8
200
6
c.p.
150
4
100
2
80
60
60
2
1
40
0.8
40
30
0.6
30 4
1
0.8
20
20
0.4
0.6
15
10
0.4
10
0.2
8.0
+T=-10°C
6.0
0.2
-20 4
0.1
4.0
0.08
-30
3.0
0.1
0.06
2.0
0.08
-40
0.04
1.5
0.06
50
1.0
0.04
0.02
20
0.80
0.60
0.02
0.01
100
0.40
150
200
250
300
350
400
0.01
600
Properties computed with: NIST REFPROP
450
500
550
ENTHALPY, KJ/kg
version 7.0
Based on formulation of Tillner-Roth and Baehr (1994)
Fig. 8
Pressure-Enthalpy Diagram for Refrigerant 134a
INT |
1100
1200
1250
.091 = 1 IN T E.
001
F110
08
F0L T I
09-
-50 1450
E 0,00- =1
Oc tei
4.
-30 == 1400 =
20
.30 kJ/(kg K)
- 00](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feabcba1c-0269-4c8f-b29a-c34fcc486852%2F1345689d-6555-4655-8d01-2c3df54f6446%2Fypaqxdh_processed.png&w=3840&q=75)
Transcribed Image Text:Licensed for single user. © 2009 ASHRAE, Inc.
100
20
150
200
250
300
350
400
450
500
550
R-134a
600
20
1050 --
1000
(1,1,1,2-tetrafluoroethane)
10
Reference state:
900
600
p= 500 kg/ma
-800
700 -
h = 200.0 kJ/kg, s= 1.00 kJ/(kg K)
for saturated liquid at 0°C
8
400
10
6
300
8
200
6
c.p.
150
4
100
2
80
60
60
2
1
40
0.8
40
30
0.6
30 4
1
0.8
20
20
0.4
0.6
15
10
0.4
10
0.2
8.0
+T=-10°C
6.0
0.2
-20 4
0.1
4.0
0.08
-30
3.0
0.1
0.06
2.0
0.08
-40
0.04
1.5
0.06
50
1.0
0.04
0.02
20
0.80
0.60
0.02
0.01
100
0.40
150
200
250
300
350
400
0.01
600
Properties computed with: NIST REFPROP
450
500
550
ENTHALPY, KJ/kg
version 7.0
Based on formulation of Tillner-Roth and Baehr (1994)
Fig. 8
Pressure-Enthalpy Diagram for Refrigerant 134a
INT |
1100
1200
1250
.091 = 1 IN T E.
001
F110
08
F0L T I
09-
-50 1450
E 0,00- =1
Oc tei
4.
-30 == 1400 =
20
.30 kJ/(kg K)
- 00
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