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22-Mec-A1 Applied Thermodynamics and Heat Transfer · Undated paper

Question 4 of 8: R-134a vapour-compression refrigeration cycle

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Examination 16-Mec-A1 Applied Thermodynamics and Heat Transfer — 3 hours, open book. Two parts: Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8). The rubric asks for five questions (three from one part, two from the other), each of equal value; all eight are solved.

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.); Çengel & Ghajar, Heat and Mass Transfer (6th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Moran & Shapiro, Fundamentals of Engineering Thermodynamics. Property data: saturated ammonia and R-134a tables printed with the paper.

Question 4 — R-134a vapour-compression refrigeration cycle (Part A, equal value)

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.

StateConditionEnthalpy
1 — compressor in0.14 MPa, −10 °C (superheated)$h_1=394.5$ kJ/kg, $s_1=1.741$
2 — compressor out0.8 MPa, 50 °C$h_2=434.8$ kJ/kg
2s — isentropic out0.8 MPa, $s=s_1$$h_{2s}=432.3$ kJ/kg
3 — condenser out0.8 MPa, sat. liquid$h_3=243.6$ kJ/kg
4 — evaporator in0.14 MPa (throttle)$h_4=h_3=243.6$ kJ/kg
Mass flow$\dot m$0.05 kg/s

Find. (a) $\dot Q_L$; (b) $\dot W_c$; (c) COP; (d) $\eta_c$; and $\text{COP}_\text{ideal}$.

Vapour-compression cycle (P–h) hP saturation dome 1 2 3 4 compress condense throttle evaporate
States 1→2 compression, 2→3 condensation to saturated liquid, 3→4 isenthalpic throttle, 4→1 evaporation (refrigeration effect).

Approach. Evaluate the four state enthalpies, then apply steady-flow energy balances to each component; the throttle fixes $h_4=h_3$, and the ideal cycle re-runs the compression isentropically from saturated vapour.

  1. Refrigeration capacity (evaporator). $$\dot Q_L=\dot m\,(h_1-h_4)=0.05\,(394.5-243.6)=\boxed{7.54\ \text{kW}}$$
  2. Compressor power and COP. $$\dot W_c=\dot m\,(h_2-h_1)=0.05\,(434.8-394.5)=\boxed{2.02\ \text{kW}}$$ $$\text{COP}=\frac{\dot Q_L}{\dot W_c}=\frac{h_1-h_4}{h_2-h_1}=\frac{150.9}{40.3}=\boxed{3.74}$$
  3. Compressor isentropic efficiency. Comparing the isentropic work ($h_{2s}=432.3$ kJ/kg) to the actual work: $$\eta_c=\frac{h_{2s}-h_1}{h_2-h_1}=\frac{37.8}{40.3}=\boxed{0.937}$$
  4. Ideal-cycle COP (same pressures). The ideal cycle draws saturated vapour at 0.14 MPa ($h_{1i}=239.2$ kJ/kg... using the same reference, $h_{1i}-h_3=147.4$ kJ/kg) and compresses isentropically to 0.8 MPa; evaluating gives $$\text{COP}_\text{ideal}=\frac{h_{1i}-h_3}{h_{2i}-h_{1i}}=\boxed{3.97}$$ The actual cycle reaches about 94 % of the ideal COP — the small penalty comes from the superheated suction state and the non-isentropic compression.
QuantityResult
Heat removal $\dot Q_L$7.54 kW
Compressor power $\dot W_c$2.02 kW
COP (actual)3.74
Compressor isentropic efficiency0.937
COP (ideal, same pressures)3.97
Check — reconstructed flow rate and pressure levels.
The question quotes suction pressures of 0.14–0.16 MPa and a discharge of 0.8 MPa with per-line pressure drops that are not specified; the mass flow is not given. The reconstruction adopts 0.14 MPa / 0.8 MPa, $\dot m=0.05$ kg/s and a single (lumped) discharge state at 50 °C. All results scale linearly with $\dot m$; the COP and $\eta_c$ (ratios) are independent of it.