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

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`

Step-by-step explanation:

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

`COP_(R) = (T_(C))/(T_(H)-T_(C))`

`10 = (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) = (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:

`(Q_(C))/(Q_(H)-Q_(C)) = (T_(C))/(T_(H)-T_(C))`

`(500\,BTU)/(600\,BTU-500\,BTU) = (479.67\,R)/(T_(H)-479.67\,R)`

`5 = (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) = (489.67\,R)/(559.67\,R-489.67\,R)`

`COP_(R) = 6.995`

e) The temperature of the cold reservoir is:

`8.9 = (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`