Question 4 of 8: Freon-12 Vapour-Compression Refrigeration Cycle

Question 4: Freon-12 Vapour-Compression Refrigeration Cycle

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Mass flow rate 0.04 kg/s; compressor inlet 150 kPa/$-10\,{}^{\circ}\text{C}$; compressor exit 1.2 MPa/75°C; expansion-valve inlet 1.15 MPa/40°C (subcooled liquid, since $T_{\text{sat}}$ at 1.15 MPa is about $47\,{}^{\circ}\text{C}$); evaporator exit 175 kPa/$-15\,{}^{\circ}\text{C}$ (a distinct, slightly higher-pressure state than the compressor inlet — a realistic suction-line pressure drop); actual compressor power 1.9 kW.

Find. (a) $\Delta s$ across the compressor; (b) refrigeration capacity $\dot Q_{evap}$; (c) COP; (d) heat lost by the refrigerant per unit mass while passing through the compressor.

Approach. Read every state from the paper's own R-12 appendix (Appendix 2, superheated Freon-12; Appendix 3, saturated Freon-12), interpolating where a state falls between tabulated rows. The valve is isenthalpic, so the evaporator-inlet enthalpy equals the valve-inlet enthalpy. Compressor entropy change, the evaporator's own energy balance, COP, and the compressor's energy balance (with heat loss) follow directly. State numbers follow the figure: 1 compressor inlet, 2 compressor exit, 3 valve inlet, 4 evaporator inlet, 5 evaporator exit.

1. State properties from the appendix.
   • State 1 (150 kPa, $-10\,{}^{\circ}\text{C}$; superheated, $T_{sat}\approx-20\,{}^{\circ}\text{C}$): tabulated directly in the 0.15 MPa column, $h_1=184.619$ kJ/kg, $s_1=0.7313$ kJ/kg·K.
   • State 2 (1.2 MPa, $75\,{}^{\circ}\text{C}$): midway between the 70 and $80\,{}^{\circ}\text{C}$ rows of the 1.20 MPa column, $h_2=(222.687+230.398)/2=226.543$ kJ/kg, $s_2=(0.7293+0.7514)/2=0.7404$ kJ/kg·K.
   • State 3 (1.15 MPa, $40\,{}^{\circ}\text{C}$): compressed liquid ($T_{sat}$ at 1.15 MPa is about $47\,{}^{\circ}\text{C}$), taken as saturated liquid at the same temperature, $h_3\approx h_f(40\,{}^{\circ}\text{C})=74.527$ kJ/kg.
   • State 5 (175 kPa, $-15\,{}^{\circ}\text{C}$): interpolating the saturation table between $-20\,{}^{\circ}\text{C}$ (0.1509 MPa) and $-15\,{}^{\circ}\text{C}$ (0.1825 MPa) gives $T_{sat}(175\text{ kPa})=-16.19\,{}^{\circ}\text{C}$ and $h_g=180.315$ kJ/kg, so the vapour is superheated by 1.19 K; with the vapour specific heat implied by the 0.10 MPa column, $(185.707-179.361)/10=0.635$ kJ/kg·K, $$h_5=180.315+0.635\times1.19=181.068\text{ kJ/kg}$$
2. Part (a) — entropy change across the compressor. $$\boxed{\Delta s = s_2-s_1 = 0.7404-0.7313 = 0.0090\text{ kJ/kg}\cdot\text{K}}$$ A small positive change, consistent with an irreversible (non-isentropic) real compression; it is small because the heat the compressor loses (part (d)) carries entropy out.
3. Part (b) — refrigeration capacity. The valve is throttled at constant enthalpy, so $h_4=h_3=74.527$ kJ/kg. The refrigeration effect is the enthalpy rise across the evaporator itself, from its inlet (state 4) to its own exit (state 5) — not to the compressor inlet, which lies downstream of a suction-line pressure drop: $$\dot Q_{evap} = \dot m\,(h_5-h_4) = 0.04\times(181.068-74.527) = 0.04\times106.541$$ $$\boxed{\dot Q_{evap} = 4.26\text{ kW}}$$
4. Part (c) — coefficient of performance. COP uses the ACTUAL compressor power (given directly), not an ideal isentropic estimate: $$\boxed{\text{COP} = \frac{\dot Q_{evap}}{\dot W_c} = \frac{4.262}{1.9} = 2.24}$$
5. Part (d) — heat lost through the compressor casing. The compressor's own enthalpy rise, $$\Delta h_{comp} = h_2-h_1 = 226.543-184.619 = 41.924\text{ kJ/kg}$$ is LESS than the actual specific work supplied, $$w_{in} = \frac{\dot W_c}{\dot m} = \frac{1.9}{0.04} = 47.500\text{ kJ/kg}$$ The steady-flow energy equation $q-w=\Delta h$ (with $w=-w_{in}$, work done on the fluid) gives $$q = \Delta h_{comp} - w_{in} = 41.924-47.500$$ $$\boxed{q = -5.58\text{ kJ/kg (a heat LOSS of about 5.6 kJ/kg to the surroundings)}}$$ The extra work supplied beyond the enthalpy rise leaves the refrigerant as heat through the compressor casing, rather than showing up as additional gas heating.