The property data for a steady state vapor-compression refrigeration cycle with refrigerant 134a as the working fluid are given in the table below. The refrigeration capacity of the cycle is 9.4 tons and the refrigerant mass flow rate is 12.4 kg/min. Determine the net changes in flow exergy rate of the refrigerant passing through the evaporator A (Ed) evap and condenser A(Ed) cond, both in kW. Let po = 1 bar, To = 24°C. (a) A (Ed) evap = Ex: 7.77 kW (b) A (Ed) cond = Ex: 0.999 kW T3, P3 = P2 == Qout 3 2 State p (bar) T (°C) h (kJ/kg) s (kJ/kg. K) www P2 1 1.4 -10.0 243.4 0.9606 2 7.0 64.0 300.8 1.0303 3 7.0 24.0 82.90 0.3113 4 1.4 -18.8 82.90 0.3301 4 == Expansion valve P4 Pl Evaporator ww Condenser Compressor W T₁, Pi m
The property data for a steady state vapor-compression refrigeration cycle with refrigerant 134a as the working fluid are given in the table below. The refrigeration capacity of the cycle is 9.4 tons and the refrigerant mass flow rate is 12.4 kg/min. Determine the net changes in flow exergy rate of the refrigerant passing through the evaporator A (Ed) evap and condenser A(Ed) cond, both in kW. Let po = 1 bar, To = 24°C. (a) A (Ed) evap = Ex: 7.77 kW (b) A (Ed) cond = Ex: 0.999 kW T3, P3 = P2 == Qout 3 2 State p (bar) T (°C) h (kJ/kg) s (kJ/kg. K) www P2 1 1.4 -10.0 243.4 0.9606 2 7.0 64.0 300.8 1.0303 3 7.0 24.0 82.90 0.3113 4 1.4 -18.8 82.90 0.3301 4 == Expansion valve P4 Pl Evaporator ww Condenser Compressor W T₁, Pi m
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:The property data for a steady state vapor-compression refrigeration cycle with refrigerant 134a as the working fluid are
given in the table below. The refrigeration capacity of the cycle is 9.4 tons and the refrigerant mass flow rate is 12.4 kg/min.
Determine the net changes in flow exergy rate of the refrigerant passing through the evaporator A (Ed) evap and condenser
A(Ed) cond, both in kW. Let po = 1 bar, To = 24°C.
(a) A (Ed) evap
= Ex: 7.77
kW
(b) A (Ed) cond
= Ex: 0.999
kW
T3, P3 = P2
==
Qout
3
2
State p (bar) T (°C) h (kJ/kg) s (kJ/kg. K)
www
P2
1 1.4
-10.0 243.4
0.9606
2
7.0
64.0 300.8
1.0303
3
7.0
24.0 82.90
0.3113
4
1.4
-18.8 82.90
0.3301
4
==
Expansion
valve
P4 Pl
Evaporator
ww
Condenser
Compressor
W
T₁, Pi
m
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