A heat pump cycle is used to maintain the interior of a building at 21°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 kJ/h. Over a period of 14 days, an electric meter records that 1490 kW ⋅ h of electricity is provided to the heat pump. Determine: 1. the amount of energy that the heat pump receives over the 14-day period from the well water by heat transfer, in kJ. 2. the heat pump’s coefficient of performance. 3. the coefficient of performance of a reversible heat pump cycle operating between hot and cold reservoirs at 21°C and 9°C.

Respuesta :

Answer:

1)Qa=34956 MJ

2)COP=7.51

COP= 24.5

Explanation:

Given that

Qr= 120,000 KJ/h

We know that

1 day =  24 hr

14 day =24 x  14 hr

14 day = 336 hr

Qr= 120000 x 336 KJ

Qr= 40320 MJ

W= 1490 KW.h    ( 1 h = 3600 s)

W= 5364 MJ

We know that

From first law of thermodynamics

Qr= Qa+W

Qa= 40320 - 5364 MJ

1)Qa=34956 MJ

COP = Qr/W

COP= 40320 / 5364

2)COP=7.51

3)COP of ideal heat pump

[tex]COP=\dfrac{T_H}{T_H-T_L}[/tex]

[tex]COP=\dfrac{273+21}{21-9}[/tex]

COP= 24.5