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A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat transfer from well water at 9°C and discharges energy by heat transfer to the building at a rate of 120,000 kM/h. Over a period of 14 days, an electric meter records that 2000 kW · h of electricity is provided to the heat pump. Determine

a. the amount of energy that the heat pump recieves over the 14 dayperiod from the well water by ehat transer, in kJ.
b. the heat pump's coeffficient of performance
c. the coeffficient of performance of a reversible heat pump cycleoperating between hot and col reservoirs at 20°C and 10°C.

Respuesta :

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

[tex]Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ[/tex]

b) The coefficient of performance is:

[tex]COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6[/tex]

c) The coefficient of performance of a reversible heat pump is:

[tex]COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l} }[/tex]

Th=20+273=293 K

Tl=10+273=283K

Replacing:

[tex]COP_{reversible}=\frac{293}{293-283}=29.3[/tex]

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