22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2013
Question 4 of 8: Freon-12 Refrigeration Cycle
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — ideal-gas processes, gas power cycles, reheat Rankine and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — natural-convection and internal-flow correlations, lumped-capacitance transients, radiation exchange, and the LMTD/ε–NTU heat-exchanger method. Freon-12 property data are read from the appendix supplied with the exam; steam properties from standard tables. This is an open-book, equal-value paper; candidates answer any five of the eight questions (three from one Part and two from the other) — all eight are worked here as a study resource.
Given. $\dot m=0.04$ kg/s Freon-12; compressor power $\dot W_c=1.9$ kW ($w_\text{in}=47.5$ kJ/kg). States read from the exam Freon-12 appendix:
State
Condition
h (kJ/kg)
s (kJ/kg·K)
1 — compressor in
150 kPa, −10 °C (superheated)
190.56
0.7543
2 — compressor out
1.2 MPa, 75 °C (superheated)
226.54
0.7404
3 — valve in
1.15 MPa, 40 °C (subcooled liquid)
74.53
—
4 = evap in
after throttle (h = h₃)
74.53
—
evap out
175 kPa, −15 °C (superheated)
187.59
0.7428
Find. (a) $\Delta s$ across the compressor; (b) refrigeration capacity $\dot Q_L$; (c) COP; (d) heat lost per unit mass in the compressor.
Figure 4 — T–s sketch of the vapour-compression cycle: 1→2 non-adiabatic compression (the line leans slightly left — $s$ falls by 0.014 because heat is rejected during compression), 2→3 de-superheating, condensation and subcooling to 40 °C on the liquid side, 3→4 throttling (isenthalpic, so $s$ rises into the two-phase region), 4→1 evaporation and a small superheat out to state 1.
Approach. Read the four state enthalpies and entropies from the Freon-12 appendix, then apply an entropy difference (a), an evaporator energy balance (b), the COP definition (c), and a compressor energy balance (d).
(a) Entropy change during compression. Directly from the tabulated states,$$\Delta s_{1\to2}=s_2-s_1=0.7404-0.7543=\boxed{-0.0140\ \text{kJ/kg}\cdot\text{K}}.$$The slight decrease signals that the compressor rejects heat (it is not adiabatic).
(b) Refrigeration capacity. The evaporator absorbs heat between the throttle exit ($h_4=h_3=74.53$) and the evaporator outlet ($187.59$):$$\dot Q_L=\dot m\,(h_\text{evap,out}-h_4)=0.04(187.59-74.53)=\boxed{4.52\ \text{kW}}\;(\approx1.29\ \text{tons}).$$
(c) Coefficient of performance. With the measured compressor power,$$\text{COP}=\frac{\dot Q_L}{\dot W_c}=\frac{4.52}{1.9}=\boxed{2.38}.$$
(d) Heat lost in the compressor. A steady-flow balance on the compressor, $q=(h_2-h_1)-w_\text{in}$ with $w_\text{in}=\dot W_c/\dot m=47.5$ kJ/kg,$$q=(226.54-190.56)-47.5=-11.5\ \text{kJ/kg}\;\Rightarrow\;\boxed{q_\text{lost}\approx11.5\ \text{kJ/kg}}.$$
Check
The negative entropy change (a) and the 11.5 kJ/kg heat rejection (d) are the same physical fact seen two ways: a real compressor that loses heat to the surroundings can leave the gas at slightly lower entropy despite the work input. The evaporator outlet enthalpy is taken from the 150 kPa superheated column at −15 °C (the appendix has no 175 kPa entry there); the <25 kPa pressure gap shifts $h$ by well under 1 kJ/kg.
