Question
Download Solution PDFDetermine the COP (coefficient of performance) of a heat pump if the rate of heat rejected is 360 kJ/min and power supplied is 2 kW.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
A heat pump is a device used to warm places by transferring thermal energy from a cooler space to a warmer space using the refrigeration cycle.
The coefficient of performance or COP of a heat pump is a ratio of useful heating provided to work (energy) required.
\({COP_{HP}={Heat\: Rejected \over Work\: Provided}} =\frac{Q_1}{W}\)
Calculation:
Given:
Rate of heat rejected, Q1 = 360 kJ/min = \( {360 \over 60}\) kJ/sec = 6 kW, power supplied, W = 2 kW
coefficient of performance of a heat pump
COP = \({Q_1 \over W}\)
= \( {6\over 2}\)
= 3
Additional Information
The coefficient of performance or COP of a refrigerator is a ratio of heat absorbed to work (energy) required.
\(\mathbf{COP_{RE}={Heat\: Absorbed \over Work\: Provided}}\)\(\mathbf{= {Q_2 \over W}}\)
∴ COPHP = COPRE + 1
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