Question

At steady state, a reversible heat pump cycle discharges energy at the rate to a hot reservoir at temperature TH, while receiving energy at the rate from a cold reservoir at temperature TC. (a) If TH = 16°C and TC = 2°C, determine the coefficient of performance. (b) If 10.5 kW, 8.75 kW, and TC = 0°C, determine TH, in °C. (c) If the coefficient of performance is 12 and TH = 27°C, determine TC, in °C.

Answer 1

A) β_max = 20.64

B) TH = 68.25°C

C) TC = 54.27°C

Step-by-step explanation:

A) We are given;

TH = 16°C = 16 + 273K = 289K

TC = 2°C = 2 + 273K = 275K

Formula for maximum cycle coefficient of performance is given as;

β_max = TH/(TH - TC)

β_max = 289/(288 - 275)

β_max = 20.64

B) We are given;

Heat rejected to system at hot reservoir; Q_H = 10.5 KW

Heat provided to system at cold reservoir; Q_C = 8.75 KW

Cold reservoir temperature; TC = 0°C = 0 + 273K = 273K

Formula for actual cycle COP is given as;

β_actual = Q_C/W_cycle

Where W_cycle is the work done and is given by;

W_cycle = Q_H - Q_C

W_cycle = 10.5 - 8.75 = 1.75 KW

Thus,

β_actual = 8.75/1.75

β_actual = 5

Actual cycle COP is defined as;

β_actual = TH/(TH - TC)

And we are looking for TH.

Thus,

TH = TC/(1 - (1/β_actual))

TH = 273/(1 - 1/5)

TH = 273/(4/5)

TH = 341.25K = 341.25 - 273°C = 68.25°C

C) We are given;

TH = 27°C = 27 + 273 = 300°C

β_max = 12

Thus, from,

β_max = TH/(TH - TC)

TC = TH(1 - (1/β_max))

TC = 300/(1 - 1/12)

TC = 327.27K = 327.27 - 273 °C = 54.27°C