Question
Download Solution PDFIn an ammonia vapour compression system, the pressure in the evaporator is 2 bar. Ammonia at exit is 0.85 dry and at entry its dryness fraction is 0.19. During compression, the work done per kg of ammonia is 150 kJ. The latent heat and specific volume at 2 bar are 1325 kJ/kg and 0.58 m3/kg, respectively. What will be its C.O.P?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
COP = \(\frac{Refrigeration~effect}{Compressor~work}~=~\frac{h_1~-~h_4}{h_2~-~h_1}\)
h4 = x4 × hfg
h1 = x1 × hfg
Calculation:
Given:
P1 = P4 = 2 bar, Dryness fractions: x1 = 0.85, x4 = 0.19, Compressor work, W = 150 kJ/kg, Latent heat, hfg = 1325 kJ/kg
∵ h4 = x4 × hfg
⇒ h4 = 0.19 × 1325 = 251.75 kJ/kg
and, h1 = x1 × hfg
⇒ h1 = 0.85 × 1325 = 1126.25 kJ/kg
∴ COP = \(\frac{Refrigeration~effect}{Compressor~work}~=~\frac{h_1~-~h_4}{h_2~-~h_1}\) = \(\frac{1126.25~-~251.75}{150}~=~5.83\)
Last updated on May 28, 2025
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