Question

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.

Answer 1

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).

$P_m_i_n= 356.9 KPa$