Question Solved1 Answer Quiz 1: A heat pump unit is used to heat water for a hot-water supply. Assume that the unit uses R-22 and operates on the ideal vapor compression cycle. The evaporator temperature is 10 °C and the condenser temperature is 60 °C. If the amount of hot water needed is 4.8 kg/min, the water temperature should be increased from 20 to 45 °C, and the value of Cp for water is 4.18 kJ/kg.K, determine the COP of the heat pump and amount of energy saved by using the heat pump instead of directly heating water using an electric heater.

1MNT90 The Asker · Mechanical Engineering

Transcribed Image Text: Quiz 1: A heat pump unit is used to heat water for a hot-water supply. Assume that the unit uses R-22 and operates on the ideal vapor compression cycle. The evaporator temperature is 10 °C and the condenser temperature is 60 °C. If the amount of hot water needed is 4.8 kg/min, the water temperature should be increased from 20 to 45 °C, and the value of Cp for water is 4.18 kJ/kg.K, determine the COP of the heat pump and amount of energy saved by using the heat pump instead of directly heating water using an electric heater.
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Transcribed Image Text: Quiz 1: A heat pump unit is used to heat water for a hot-water supply. Assume that the unit uses R-22 and operates on the ideal vapor compression cycle. The evaporator temperature is 10 °C and the condenser temperature is 60 °C. If the amount of hot water needed is 4.8 kg/min, the water temperature should be increased from 20 to 45 °C, and the value of Cp for water is 4.18 kJ/kg.K, determine the COP of the heat pump and amount of energy saved by using the heat pump instead of directly heating water using an electric heater.
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Ideal heat Thomp with (R-22) usas refivigetant,. We have at (i) Comptessor inlet (10^(@)C) sat vaipol From tables of R-22,quad h=255.81(kJ)/(kg) & quadS_(1)=0.9216(kJ)/(kg-c). Now 1-2 (Isentropic conplossion) where S_(1)=S_(2)T_("sat ")=60^(@)C=>P_("sat ")=2.43MPa=>P_(2)=P_(3)Now from saper heated table, P_(2)=243MP9,S_(2)=0.9216(kJ)/(kg-c) h_(2)=288kJ//kgquadT_(2)=83^(@)CNow 2.3 (con. Pressfic heat rejection) where h_(3)=h_(p) at 60^(@)Cquadh_(3)=123kJ//kg=h_(4) (Iseethats.)Now water get heated from 20^(@)C to 45^(@)C With m=4.8(kg)/(min) & c_(p)= ... See the full answer