--- title: "Actual VCR cycle with compressor efficiency" subject: "Applied Thermodynamics" chapter: 3 key: "semester_3/applied_thermodynamics:vcr-actual-compressor-efficiency-numerical" source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/vcr-actual-compressor-efficiency-numerical site: BCT Engineering --- # Actual VCR cycle with compressor efficiency Applied Thermodynamics · Chapter 3: Refrigeration - Asked in: year not given, 8 marks - Video: [Ideal and Actual Vapor-compression Refrigeration Cycles](https://www.youtube.com/watch?v=6F100qUXHlw) — Dr Kidd Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/vcr-actual-compressor-efficiency-numerical - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-3 ## Question A refrigeration plant of 10 kW capacity uses R-134a between 0.14 MPa and 0.8 MPa. Vapour leaves the evaporator as dry saturated ($h = 239.16$ kJ/kg) and gains heat in the suction line so that it enters the compressor with $h_1 = 242.2$ kJ/kg. The isentropic compression ends at $h_{2s} = 278.6$ kJ/kg. The isentropic efficiency of the compressor is 75% and the motor-drive efficiency is 90%. Saturated liquid at 0.8 MPa has $h_3 = 95.47$ kJ/kg. Find (a) the actual compressor work per kg, (b) the mass flow rate, (c) the electrical power input, (d) the actual COP, and (e) the heat rejected in the condenser. ## Short answer Actual compressor work per kg = 48.5 kJ/kg; mass flow rate = 0.0696 kg/s; shaft power input = 3.38 kW, electrical input = 3.75 kW; actual COP = 2.96 (shaft) / 2.66 (electrical); heat rejected = 13.6 kW. ## Answer #### Refrigerating effect Only the heat absorbed in the evaporator is useful cooling. The suction-line heat gain comes from outside the cold space, so: $$ q_L = h_{evap\ exit} - h_4 = 239.16 - 95.47 = 143.69\ \text{kJ/kg} $$ #### (a) Compressor work per kg $$ \begin{aligned} w_s &= h_{2s} - h_1 = 278.6 - 242.2 = 36.4\ \text{kJ/kg}\\ w_a &= \frac{w_s}{\eta_s} = \frac{36.4}{0.75} = 48.53\ \text{kJ/kg} \end{aligned} $$ Actual discharge enthalpy: $h_2 = 242.2 + 48.53 = 290.7$ kJ/kg. #### (b) Mass flow rate $$ \dot m = \frac{10}{143.69} = 0.0696\ \text{kg/s} $$ #### (c) Power $$ \begin{aligned} P_{comp} &= \dot m\,w_a = 0.0696 \times 48.53 = 3.38\ \text{kW}\\ P_{elec} &= \frac{P_{comp}}{\eta_{motor}} = \frac{3.38}{0.9} = 3.75\ \text{kW} \end{aligned} $$ #### (d) Actual COP Based on shaft power: $\text{COP} = 10/3.38 = 2.96$. Based on electrical input: $\text{COP} = 10/3.75 = 2.66$. (For comparison, the ideal COP with no losses is $143.69/36.23 = 3.97$.) #### (e) Condenser heat rejection $$ Q_H = \dot m (h_2 - h_3) = 0.0696 \times (290.7 - 95.47) = 13.59\ \text{kW} $$ (Energy check: $10 + 0.21$ (suction-line gain) $+ 3.38 = 13.59$ kW.) **Answer:** (a) 48.5 kJ/kg; (b) 0.0696 kg/s; (c) 3.38 kW at the shaft, 3.75 kW electrical; (d) 2.96 (shaft) or 2.66 (electrical); (e) 13.6 kW. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this plant, which enthalpy difference gives the refrigerating effect q_L? A. h1 - h4 (compressor inlet minus evaporator inlet) B. h(evaporator exit, dry saturated) - h4 = 239.16 - 95.47 C. h2 - h3 D. h2s - h1 Answer: B. Cooling happens in the evaporator, so only heat gained up to the evaporator exit counts. Suction-line heat gain is from outside the cold space and is not useful cooling. (Key point: Refrigerating effect) 2. How is the actual compressor work per kg obtained from the isentropic work? A. w_a = w_s x eta_s = 36.4 x 0.75 B. w_a = w_s / eta_s = 36.4/0.75 = 48.53 kJ/kg C. w_a = w_s / eta_motor D. w_a = h2s - h4 Answer: B. Isentropic efficiency is ideal work divided by actual work, so actual work is larger: 36.4/0.75 = 48.53 kJ/kg. (Key point: Actual compressor work) 3. After finding m = 10/143.69 = 0.0696 kg/s and shaft power 3.38 kW, how is the electrical input found? A. 3.38 x 0.9 = 3.04 kW B. 3.38 / 0.9 = 3.75 kW C. 3.38 / 0.75 = 4.51 kW D. 10 / 2.96 = 3.38 kW Answer: B. The motor delivers shaft power with efficiency 0.9, so electrical input = shaft power / eta_motor = 3.75 kW. (Key point: Mass flow and electrical power) 4. Which statement about the results is correct? A. COP based on electrical input (2.66) is higher than on shaft power (2.96) B. Condenser heat Q_H = m(h2 - h3) uses the actual discharge enthalpy h2 = 290.7 kJ/kg, giving about 13.6 kW C. Condenser heat uses h2s = 278.6 because the isentropic state is the real exit D. The ideal COP of 3.97 is lower than the actual COP Answer: B. The real compressor discharges at h2 = h1 + w_a = 290.7 kJ/kg, so Q_H = 0.0696 x 195.23 = 13.6 kW. Actual COPs are below the ideal 3.97. 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