22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2016
Question 4 of 8: Freon-12 Refrigeration System from Test Data
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) — closed- and open-system energy balances, steam tables, vapour and gas power cycles, 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) — cylindrical composite-wall conduction, internal-flow and cross-flow convection correlations, natural convection with radiation, and the ε–NTU heat-exchanger method. Freon-12 property data are taken from the appendix supplied with the exam; steam, air and engine-oil data from standard tables.
Paper format: National Examination 07-Mec-A1, 3 hours, open book. Part A — Thermodynamics (Q1–4); Part B — Heat Transfer (Q5–8). Each answer carries equal value; a complete paper is any five (three from one part, two from the other). All eight questions are solved in full below.
Question 4: Freon-12 Refrigeration System from Test Data (equal value)
Figure 4 — T–s schematic of the measured Freon-12 cycle: 1→2 actual compression (heat is rejected, so 2 lies to the left of an adiabatic discharge), 2→3 condensation and subcooling to 15 °C, 3→4 throttling from the 20 °C valve inlet, 4→1 evaporation absorbing the refrigeration effect.
Given. Measured Freon-12 states — 150 kPa evaporator, 700 kPa condenser, compressor inlet/outlet 0/70 °C, condenser inlet/outlet 60/15 °C, valve inlet 20 °C, evaporator outlet −10 °C — and 3.35 kJ/kg rejected during compression. Find. The cycle COP and the isentropic (thermal) efficiency of the compression.
Approach. All enthalpies come from the Freon-12 appendix at the measured pressures and temperatures. The refrigeration effect is the evaporator enthalpy rise; the compressor work follows from an energy balance that credits the measured heat loss.
Refrigeration effect. The valve inlet is subcooled liquid at 20 °C, and throttling is isenthalpic, so $h_4=h_f(20\,{}^\circ\text{C})=54.83$ kJ/kg. With the evaporator outlet at −10 °C, $$q_L=h_\text{evap,out}-h_4=184.42-54.83=\boxed{129.6\ \text{kJ/kg}}.$$
Actual compressor work. An energy balance on the compressor with heat rejection $q_\text{out}=3.35$ kJ/kg gives $$w_\text{in}=(h_2-h_1)+q_\text{out}=(228.93-190.66)+3.35=41.62\ \text{kJ/kg}.$$
Coefficient of performance. $$\text{COP}=\frac{q_L}{w_\text{in}}=\frac{129.6}{41.62}=\boxed{3.11}.$$
Isentropic reference. From $s_1=0.7543$ kJ/kg·K, an isentropic compression to 700 kPa reaches $h_{2s}=220.31$ kJ/kg (interpolated in the superheat table), so the reversible work is $w_s=h_{2s}-h_1=29.65$ kJ/kg.
Thermal (isentropic) efficiency of the compression. $$\eta_\text{comp}=\frac{w_s}{w_\text{in}}=\frac{29.65}{41.62}=\boxed{71.2\%}.$$
Check
Energy closes on the compressor: the measured 38.3 kJ/kg enthalpy rise plus the 3.35 kJ/kg heat loss equal the 41.6 kJ/kg work input. "Thermal efficiency of the compression process" is read as the isentropic efficiency (reversible-adiabatic work ÷ actual work); if instead read as the fraction of work converted to enthalpy rise, $(h_2-h_1)/w_\text{in}=92\%$. All enthalpies use the exam-supplied Freon-12 tables; ±0.3 kJ/kg interpolation on $h_{2s}$ moves $\eta_\text{comp}$ by about ±0.5 point.
