Question

A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the coefficient of performance is 10 and TC = -40°F, determine TH, in °F. (b) If TC = 0°C and TH = 30°C, determine the coefficient of performance. (c) If QC = 500 Btu, QH = 600 Btu, and TC = 20°F, determine TH, in °F. (d) If TC = 30°F and TH = 100°F, determine the coefficient of performance. (e) If the coefficient of performance is 8.9 and TC = -5°C, find TH, in °C.

Answer 1

Answer:

a) $T_{H} = 1.967\,^{\circ}F$, b) $COP_{R} = 9.105$, c) $T_{H} = 115.934\,^{\circ}F$, d) $COP_{R} = 6.995$, e) $T_{H} = 25.129\,^{\circ}C$

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

$COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}$

$10 = \frac{419.67\,R}{T_{H}-419.67\,R}$

The temperature of the hot reservoir is:

$10\cdot T_{H} - 4196.7 = 419.67$

$T_{H} = 461.637\,R$

$T_{H} = 1.967\,^{\circ}F$

b) The coefficient of performance is:

$COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}$

$COP_{R} = 9.105$

c) The temperature of the hot reservoir can be determined with the help of the following relation:

$\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}$

$\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5 = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5\cdot T_{H} - 2398.35 = 479.67$

$T_{H} = 575.604\,R$

$T_{H} = 115.934\,^{\circ}F$

d) The coefficient of performance is:

$COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}$

$COP_{R} = 6.995$

e) The temperature of the cold reservoir is:

$8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}$

$8.9\cdot T_{H} - 2386.535 = 268.15$

$T_{H} = 298.279\,K$

$T_{H} = 25.129\,^{\circ}C$