Answer:
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
Explanation:
From figure
[tex] Q_H [/tex] is heat rejection process
[tex] Q_L [/tex] is heat transferred from the refrigerated space
[tex] T_H [/tex] is high temperature = 50 °C + 273 = 323 K
[tex] T_L [/tex] is low temperature = -20 °C + 273 = 253 K
[tex] W_{net} [/tex] is net work of the cycle (the difference between compressor's work and turbine's work)
Coefficient of performance of a Carnot refrigerator [tex] (COP_{ref}) [/tex] is calculated as
[tex] COP_{ref} = \frac{T_L}{T_H - T_L} [/tex]
[tex] COP_{ref} = \frac{253 K}{323 K - 253 K} [/tex]
[tex] COP_{ref} = 3.61 [/tex]
From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table
[tex] Q_H = 122.5 kJ/kg [/tex]
From coefficient of performance definition
[tex] COP_{ref} = \frac{Q_L}{Q_H - Q_L} [/tex]
[tex] Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L[/tex]
[tex] Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)} [/tex]
[tex] Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)} [/tex]
[tex] Q_L = 95.93 kJ/kg [/tex]
Energy balance gives
[tex] W_{net} = Q_H - Q_L [/tex]
[tex] W_{net} = 122.5 kJ/kg - 95.93 kJ/kg [/tex]
[tex] W_{net} = 26.57 kJ/kg [/tex]
Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)
[tex] x = \frac{s_1 - s_f}{s_g - s_f} [/tex]
From figure
[tex] s_1 = s_4 = 1.165 kJ/(K kg) [/tex]
Replacing with table values
[tex] x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)} [/tex]
[tex] x = 0.37 [/tex]
Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get
[tex] h_1 = h_f + x \times (h_g - h_f) [/tex]
[tex] h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg [/tex]
[tex] h_1 = 241.58 kJ/kg [/tex]
By energy balance, [tex] W_{t} [/tex] turbine's work is
[tex] W_{t} = |h_1 - h_4| [/tex]
[tex] W_{t} = |241.58 kJ/kg - 249.7 kJ/kg| [/tex]
[tex] W_{t} = 8.12 kJ/kg [/tex]
Finally, [tex] W_{c} [/tex] compressor's work is
[tex] W_{c} = W_{net} + W_{t}[/tex]
[tex] W_{c} = 26.57 kJ/kg + 8.12 kJ/kg[/tex]
[tex] W_{c} = 34.69 kJ/kg [/tex]
