Ompressor fa refrigerator as superheated vapor at 0.14 MPa and -10°C at a rate of 0.05 kg/s and leaves at 0.8 MPa and 50°C. The refrigerant is cooled in the condenser to 26°C and 0.72 MPa and is throttled to 0.15 MPa. Disregarding any heat trans fer and pressure drops in the connecting lines between the components; determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, and (b) the coefficient of performance of the refrigerator. Solution: - P 0.14 MPa T = -10 C h=246.36 kJ/kg OF SA %3D SITY

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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SUBJECT: THERMODYNAMIC
COURSE: II
ASSI.LACTURE: NATIQ ABBAS
Example 2:-
Refrigerant-134a enters the compressor of a refrigerator as superheated
vapor at 0.14 MPa and -10°C at a rate of 0.05 kg/s and leaves at 0.8 MPa
and 50°C. The refrigerant is cooled in the condenser to 26°C and 0.72 MPa
and is throttled to 0.15 MPa. Disregarding any heat trans fer and pressure
drops in the connecting lines between the components; determine (a) the rate
of heat removal from the refrigerated space and the power input to the
compressor, and (b) the coefficient of performance of the refrigerator.
Solution: -
P 0.14 MPa
T=-10 C
212
h 246.36 kJ/kg
da
OF
SAMARRA
P 0.8 MPa
VER
h2 = 286.69 kJ/kg
O P 0.72 MPa
T= 26 C
hg = 87.83 kJ/kg
h3 =h = 87.83 KJ/kg
h4 = h3 (throttling) h4 87.83 kJ/kg
0.8 MPa
50°C
0.72 MPa/
26 C
0.15 MPa
0.14 MPa
-10°C
SUBIECT: THERMODYNAMIC
COURSE: II
ASSI.LACTURE: NATIQ ABBAS
RING
Transcribed Image Text:SUBJECT: THERMODYNAMIC COURSE: II ASSI.LACTURE: NATIQ ABBAS Example 2:- Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and -10°C at a rate of 0.05 kg/s and leaves at 0.8 MPa and 50°C. The refrigerant is cooled in the condenser to 26°C and 0.72 MPa and is throttled to 0.15 MPa. Disregarding any heat trans fer and pressure drops in the connecting lines between the components; determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, and (b) the coefficient of performance of the refrigerator. Solution: - P 0.14 MPa T=-10 C 212 h 246.36 kJ/kg da OF SAMARRA P 0.8 MPa VER h2 = 286.69 kJ/kg O P 0.72 MPa T= 26 C hg = 87.83 kJ/kg h3 =h = 87.83 KJ/kg h4 = h3 (throttling) h4 87.83 kJ/kg 0.8 MPa 50°C 0.72 MPa/ 26 C 0.15 MPa 0.14 MPa -10°C SUBIECT: THERMODYNAMIC COURSE: II ASSI.LACTURE: NATIQ ABBAS RING
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