Question 6.6: A Carnot Refrigeration Cycle Operating in the Saturation Dom...

A Carnot refrigeration cycle is executed in a closed system. The mass fraction of the refrigerant that vaporizes during the heat addition process and the pressure at the end of the heat rejection process are to be determined.

Assumptions     The refrigerator operates on the ideal Carnot cycle.

Analysis     Knowing the high and low temperatures, the coefficient of performance of the cycle is

\mathrm{COP}_{\mathrm{R}}=\frac{1}{T_{H} / T_{L}-1}=\frac{1}{(20+273  \mathrm{~K}) /(-8+273  \mathrm{~K})-1}=9.464

The amount of cooling is determined from the definition of the coefficient of performance to be

Q_{L}=\mathrm{COP}_{\mathrm{R}} \times W_{\mathrm{in}}=(9.464)(15  \mathrm{~kJ})=142  \mathrm{~kJ}

The enthalpy of vaporization of R-134a at -8^{\circ} \mathrm{C} is h_{f g}=204.59  \mathrm{~kJ} / \mathrm{kg} (Table A-11). Then the amount of refrigerant that vaporizes during heat absorption becomes

Q_{L}=m_{\text {evap }} h_{f g   @  -8^{\circ} \mathrm{C}} \rightarrow m_{\text {evap }}=-\frac{142  \mathrm{~kJ}}{204.59  \mathrm{~kJ} / \mathrm{kg}}=0.694  \mathrm{~kg}

Therefore, the fraction of mass that vaporized during the heat addition process to the refrigerant is

\text { Mass fraction } =\frac{m_{\text {evap }}}{m_{\text {total }}}=\frac{0.694 \mathrm{~kg}}{0.8  \mathrm{~kg}}=0.868 \text { (or } 86.8 \text { percent })

The pressure at the end of the heat rejection process is simply the saturation pressure at heat rejection temperature,

P_{4}=P_{\text {sat } @ 20^{\circ} \mathrm{C}}=572.1  \mathrm{kPa}

Discussion     The reversed Carnot cycle is an idealized refrigeration cycle, thus it cannot be achieved in practice. Practical refrigeration cycles are analyzed in Chap. 11 .