Question 5.31 At steady state, a reversible refrigeration cycle discharges energy at the rate Ou to a hot reservoir at temperature Ty, while receiving energy at the rate Qc from a cold reservoir at temperature Tc. a. If Tk = 13°C and Tc=2°C, determine the coefficient of performance. b. If (x =10.5 kW, èc =8.75 kW, and Te = 0°C, determine Tv, in °C. c. If the coefficient of performance is 10 and TH = 27°C, determine T., in °C.

VHBKKA The Asker · Mechanical Engineering

Transcribed Image Text: 5.31 At steady state, a reversible refrigeration cycle discharges energy at the rate Ou to a hot reservoir at temperature Ty, while receiving energy at the rate Qc from a cold reservoir at temperature Tc. a. If Tk = 13°C and Tc=2°C, determine the coefficient of performance. b. If (x =10.5 kW, èc =8.75 kW, and Te = 0°C, determine Tv, in °C. c. If the coefficient of performance is 10 and TH = 27°C, determine T., in °C.
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Transcribed Image Text: 5.31 At steady state, a reversible refrigeration cycle discharges energy at the rate Ou to a hot reservoir at temperature Ty, while receiving energy at the rate Qc from a cold reservoir at temperature Tc. a. If Tk = 13°C and Tc=2°C, determine the coefficient of performance. b. If (x =10.5 kW, èc =8.75 kW, and Te = 0°C, determine Tv, in °C. c. If the coefficient of performance is 10 and TH = 27°C, determine T., in °C.
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WLE4DC

Hey, in case of any doubt please leave me acomment.Solution:Reversed refrigeration cycle:Part A) T_(H)=13C=13+273=286K,T_(C)=2C=2+273=275K. Find COP?{:[" COP "=T_(C)//(T_(H)-T_(C))=275//(286-275)=25." Answer. "],[" Maximum COP "=T_(H)//(T_(H)-T_(C))=286//(286-275)=26]:}Part B) Q_(H)=10.5kW,Q_(C)=8.75kW,T_(C)=0 ... See the full answer