Question

Refrigerant R-12 is used in a Carnot refrigerator

operatingbetween saturated liquid and vapor during the heat
rejectionprocess. If the cycle has a high temperature of 50 deg C
and a lowtemperature of -20 deg C, find the heat transferred from
therefrigerated space, the work required, the coefficient
ofperformance and the quality at the beginning of the heat
additioncycle.

Answer 1

Heat transferred from the refrigerated space = 95.93 kJ/kg

Work required = 18.45 kJ/kg

Coefficient of performance = 3.61

Quality at the beginning of the heat addition cycle = 0.37

Step-by-step explanation:

From figure

`Q_H` is heat rejection process

`Q_L` is heat transferred from the refrigerated space

`T_H` is high temperature = 50 °C + 273 = 323 K

`T_L` is low temperature = -20 °C + 273 = 253 K

`W_(net)` is net work of the cycle (the difference between compressor's work and turbine's work)

Coefficient of performance of a Carnot refrigerator
`(COP_(ref))` is calculated as

`COP_(ref) = (T_L)/(T_H - T_L)`

`COP_(ref) = (253 K)/(323 K - 253 K)`

`COP_(ref) = 3.61`

From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table

`Q_H = 122.5 kJ/kg`

From coefficient of performance definition

`COP_(ref) = (Q_L)/(Q_H - Q_L)`

`Q_H * COP_(ref) = (COP_(ref) + 1) * Q_L`

`Q_L = (Q_H * COP_(ref))/((COP_(ref) + 1))`

`Q_L = (122.5 kJ/kg * 3.61)/((3.61 + 1))`

`Q_L = 95.93 kJ/kg`

Energy balance gives

`W_(net) = Q_H - Q_L`

`W_(net) = 122.5 kJ/kg - 95.93 kJ/kg`

`W_(net) = 26.57 kJ/kg`

Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)

`x = (s_1 - s_f)/(s_g - s_f)`

From figure

`s_1 = s_4 = 1.165 kJ/(K kg)`

Replacing with table values

`x = (1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg))/(1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg))`

`x = 0.37`

Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get

`h_1 = h_f + x * (h_g - h_f)`

`h_1 = 181.6 kJ/kg + 0.37 * 162.1 kJ/kg`

`h_1 = 241.58 kJ/kg`

By energy balance,
`W_(t)` turbine's work is

`W_(t) = |h_1 - h_4|`

`W_(t) = |241.58 kJ/kg - 249.7 kJ/kg|`

`W_(t) = 8.12 kJ/kg`

Finally,
`W_(c)` compressor's work is

`W_(c) = W_(net) + W_(t)`

`W_(c) = 26.57 kJ/kg + 8.12 kJ/kg`

`W_(c) = 34.69 kJ/kg`