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17-Phys-B6 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. 17-Phys-B6 Applied Thermodynamics and Heat Transfer, National Examinations, May 2019 — a three-hour open-book examination; candidates are expected to bring both a thermodynamics text and a heat-transfer text to make use of the property tables and graphs the exam supplies. A complete examination is five questions — either three from Part A (Thermodynamics, Q1–Q4) and two from Part B (Heat Transfer, Q5–Q8), or two from Part A and three from Part B — every question carrying equal value; all eight are solved below as a complete study set.

Reference texts. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed. (polytropic closed-system processes, throttling, Rankine-cycle reheat/extraction turbines, air-standard Brayton-cycle energy balances, vapour-compression refrigeration); F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed. (composite plane-wall conduction with convection and radiation at both faces, combined entry-length internal convection, natural convection with radiation from a vertical plate, sh​ell-and-tube heat exchanger sizing via the LMTD correction-factor method). Ammonia, steam and R-134a property values were computed (Bell et al., IAPWS-95 / REFPROP-quality equations of state) and cross-checked against the printed saturated-ammonia appendix table on page 6 of the source exam, which it matched to 3–4 significant figures.

Question 4: R-134a Vapour-Compression Refrigeration Cycle (20 marks)

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. An actual (irreversible-compression, subcooled-liquid) R-134a cycle with every state pinned by two independent properties, so no assumed process (e.g. isentropic compression) is needed for the ACTUAL cycle — only for its ideal comparison.

Given data
StateDescriptionConditions
1compressor inlet0.14 MPa, $-10\,{}^{\circ}\text{C}$ (superheated)
2compressor outlet0.8 MPa, $50\,{}^{\circ}\text{C}$
3condenser outlet0.72 MPa, $26\,{}^{\circ}\text{C}$ (subcooled liquid)
4after throttle valve0.15 MPa, $h_4=h_3$
$\dot m=0.05$ kg/s

Find. (a) $\dot Q_L$, $\dot W_{in}$, compressor isentropic efficiency, $COP_{actual}$; (b) the ideal-cycle $\dot Q_L$ and $\dot W_{in}$ at the same evaporator/condenser pressures, and the differences.

Approach. For the ideal comparison, replace states 1 and 3 with the SATURATED states at the same two pressures (saturated vapour at $P_1$, saturated liquid at $P_2$) and compress isentropically from the new state 1 to $P_2$.

  1. Actual-cycle enthalpies. $$h_1=394.50\text{ kJ/kg},\quad h_2=434.84\text{ kJ/kg},\quad h_3=h_4=235.97\text{ kJ/kg}$$ Compressor-inlet entropy $s_1=1.7680$ kJ/kg·K gives the isentropic-exit enthalpy at 0.8 MPa, $h_{2s}=432.31$ kJ/kg.
  2. Part (a) — evaporator and compressor duties. $$\dot Q_L=\dot m(h_1-h_4)=0.05\times(394.50-235.97)$$ $$\boxed{\dot Q_L=7.93\text{ kW}}$$ $$\dot W_{in}=\dot m(h_2-h_1)=0.05\times(434.84-394.50)$$ $$\boxed{\dot W_{in}=2.02\text{ kW}}$$
  3. Part (a) — compressor isentropic efficiency and COP. $$\eta_C=\frac{h_{2s}-h_1}{h_2-h_1}=\frac{432.31-394.50}{434.84-394.50}$$ $$\boxed{\eta_C=0.937\ (93.7\%)}$$ $$COP_{actual}=\frac{\dot Q_L}{\dot W_{in}}=\frac{7.93}{2.02}$$ $$\boxed{COP_{actual}=3.93}$$
  4. Part (b) — ideal-cycle states at the same two pressures. Saturated vapour at 0.14 MPa: $h_{1,ideal}=387.32$ kJ/kg, $s_{1,ideal}=1.7909$ kJ/kg·K. Isentropic compression to 0.8 MPa: $h_{2,ideal}=423.53$ kJ/kg. Saturated liquid at 0.8 MPa: $h_{3,ideal}=h_{4,ideal}=243.65$ kJ/kg.
  5. Part (b) — ideal-cycle duties and the differences. $$\dot Q_{L,ideal}=0.05\times(387.32-243.65)$$ $$\boxed{\dot Q_{L,ideal}=7.18\text{ kW}},\qquad \dot W_{in,ideal}=0.05\times(423.53-387.32)$$ $$\boxed{\dot W_{in,ideal}=1.81\text{ kW}}$$ $$\Delta\dot Q_L=7.93-7.18=\boxed{0.74\text{ kW greater (actual)}}$$ $$\Delta\dot W_{in}=2.02-1.81=\boxed{0.21\text{ kW greater (actual)}}$$
s (kJ/kg·K)T (°C)R-134a — actual vapour-compression cycle 1→2→3→4→11234
Actual cycle: superheated compressor inlet (1) → irreversible compression to superheated (2) → subcooled-liquid condenser exit (3) → isenthalpic throttle to the wet region (4) → back to 1 through the evaporator.
Question 4 — results
QuantityActualIdeal (same pressures)
Heat removal $\dot Q_L$7.93 kW7.18 kW
Compressor power $\dot W_{in}$2.02 kW1.81 kW
COP3.933.97
Compressor isentropic efficiency: 93.7%  |  Actual exceeds ideal by 0.74 kW ($\dot Q_L$) and 0.21 kW ($\dot W_{in}$)