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geration cycle gerant-134a as

geration cycle gerant-134a as

Elements Of Electromagnetics
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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Subject Thermodynamics. Instructions: Don't round off in the process. Just round off in the final answer with 2 decimals only. Use 273.15 K to convert Celsius to Kelvin

An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800kPa and the evaporator at -12°C. Determine the amount of power (kW) required to service a 5792.6kW cooling load. T 2 3 /0.8 МPа -12°C 4s 4 Qi T =-12°C h =hg @ -1rc = 243.34 kJ/kg S1 = Sg @ -12°c =0.93925 kJ/kg - K sat. vapor P = 0.8 MPa s2 = S1 h2 = 273.71kJ/kg P3 = 0.8 MPa sat. liquid h3 = h5 @1MP2 =95.48 kJ/kg h4 = h3 = 95.48 kJ/kg (throttling)
Transcribed Image Text:An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800kPa and the evaporator at -12°C. Determine the amount of power (kW) required to service a 5792.6kW cooling load. T 2 3 /0.8 МPа -12°C 4s 4 Qi T =-12°C h =hg @ -1rc = 243.34 kJ/kg S1 = Sg @ -12°c =0.93925 kJ/kg - K sat. vapor P = 0.8 MPa s2 = S1 h2 = 273.71kJ/kg P3 = 0.8 MPa sat. liquid h3 = h5 @1MP2 =95.48 kJ/kg h4 = h3 = 95.48 kJ/kg (throttling)
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