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Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of refrigerant-134a as the working fluid. It is known that the maximum absolute temperature in the cycle is 1.2 times the minimum absolute temperature, and the net work input to the cycle is 22 kJ. If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine the minimum pressure in the cycle.

Respuesta :

Chrisnando

Answer:

[tex] P_m_i_n = 442KPA [/tex]

Explanation:

We are given:

m = 1.06Kg

[tex] T_H = 1.2T_L[/tex]

T = 22kj

Therefore we need to find coefficient performance or the cycle

[tex] COP_R = \frac {1}{(T_R/T_l) -1} [/tex]

[tex] = \frac {1 }{1.2-1} [/tex]

= 5

For the amount of heat absorbed:

[tex] Q_l = COP_R Wm [/tex]

= 5 × 22 = 110KJ

For the amount of heat rejected:

[tex] Q_H = Q_L + W_m [/tex]

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= [tex] = \frac{132}{1.06} [/tex]

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

[tex] T_L = \frac{T_H}{1.2} [/tex];

[tex] T_L = \frac{342.5}{1.2} [/tex]

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at [tex]T_L[/tex] = 12.4°c we have 442KPa

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