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Rasek [7]
3 years ago
11

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using 0.96 kg of

refrigerant-134a as the working fluid. Is it 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 I s 22 kJ. If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine he minimum pressure in the cycle.
Engineering
1 answer:
Eduardwww [97]3 years ago
8 0

Answer:

P_m_i_n= 356.9 KPa

Explanation:

Coefficient of performance (COP) of refrigeration cycle is given by:

COP=\frac{1}{\frac{T_H}{T_L}-1 }

We are given:

T_H=1.2T_L

COP=\frac{1}{1.2-1 }

COP= 5

We can also write Coefficient of performance (COP) of refrigeration cycle  as:

COP_R=\frac{Q_L}{W_i_n}

Amount of heat absorbed by low temperature reservoir can be found as:

Q_L=COP_R * W_i_n

Q_L=5 * 22 KJ

Q_L=110 KJ

According to first law of thermodynamics amount of heat rejected by hot reservoir is given by:

Q_H=Q_L + W_i_n

Q_H=110 KJ + 22 KJ

Q_H=132 KJ

We are given the mass of 0.96 kg. So,

q_H=\frac{Q_H}{m}

q_H=\frac{132 KJ}{0.96Kg}

q_H=137.5 KJ/Kg

Since it is a saturated liquid-vapour mixture q_H=h_f_g.

q_H=h_f_g=137.5 KJ/Kg

From Refrigerant 134-a tables T_H at h_f_g=137.5 KJ/Kg is 61.3 C. (We calculated this by interpolation)

Converting T_H from Celsius to Kelvin:

61.3^{o} C+273 = 334.3^{o} K

T_H= 334.3^{o} K

We are given:

T_H=1.2T_L

T_L=\frac{T_H}{1.2}

T_L=\frac{334.3}{1.2}

T_L=278.58^{o} K

Converting T_L from Kelvin to Celsius:

278.58^{o} K-273 = 5.58^{o} C

T_L= 5.58^{o} C

From Refrigerant 134-a tables P_m_i_n at T_L=5.58^{o} C is 356.9 KPa. (We calculated this by interpolation).

P_m_i_n= 356.9 KPa

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