R-134a Ideal Vapor-Compression Refrigeration Cycle: 22-Agric-A3 Heat Engineering, May 2018, Q1

Problem 1: R-134a Ideal Vapor-Compression Refrigeration Cycle (25 points)

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. Ideal vapor-compression cycle (compressor, condenser, expansion valve, evaporator); refrigerant R-134a; saturated vapor enters the compressor at T₁ = 0°C; saturated liquid leaves the condenser at T₃ = 26°C; isentropic compression 1→2s; isenthalpic throttling 3→4; mass flow rate 0.08 kg/s.

State	Condition	T
1 (compressor inlet)	saturated vapor	0°C
2s (compressor exit)	superheated, isentropic from 1	—
3 (condenser exit)	saturated liquid	26°C
4 (evaporator inlet)	throttled from 3, h₄=h₃	0°C

Find. (a) The compressor power ḏ_c in kW; (b) the coefficient of performance, COP.

Approach. Fix states 1 and 3 from the saturation properties at the two reservoir temperatures, get state 2s by following the isentrope from state 1 up to the condenser pressure, get state 4 by throttling (constant h) from state 3 down to the evaporator pressure, then apply the steady-flow energy balance to the compressor and evaporator.

P–h diagram of the ideal cycle: 1→2s isentropic compression, 2s→3 condensation at constant P, 3→4 isenthalpic throttling, 4→1 evaporation at constant P.

Check: the source supplies only a saturation-property table for R-134a (Table A-10, reproduced on the exam's last page); state 2s is superheated, so no table entry in the source covers it directly. State 2s is fixed here from the R-134a Helmholtz-energy equation of state — the same real-gas surface an open-book candidate would read off a superheated R-134a table (e.g. Moran/Shapiro Table A-12 or Çengel Table A-13). This agrees with the source's own saturation values (Table A-10) to within about 0.7% at both 0°C and 26°C, so the same reference state is used consistently for all four cycle points and cancels out of every energy difference used below.

State 1 — saturated vapor entering the compressor. At T₁ = 0°C, the saturation pressure is p₁ = 2.93 bar, and $$h_1 = 398.60\ \text{kJ/kg}, \qquad s_1 = 1.7271\ \text{kJ/kg}\cdot\text{K}$$
State 3 — saturated liquid leaving the condenser. At T₃ = 26°C, the saturation pressure is p₃ = 6.85 bar, and $$h_3 = 235.97\ \text{kJ/kg}$$
State 2s — isentropic compression to the condenser pressure. Following the s₁ = 1.7271 kJ/kg·K isentrope up to p₃ = 6.85 bar lands in the superheated region, only $T_{2s}-T_{sat}(p_3) \approx 3.3^{\circ}\text{C}$ above the condensing temperature (R-134a's saturated-vapor line is nearly vertical over this modest pressure ratio, so little superheat is generated), giving $$h_{2s} = 416.21\ \text{kJ/kg}$$
State 4 — throttling to the evaporator pressure. The expansion valve is adiabatic with no work, so $h_4 = h_3 = 235.97$ kJ/kg (constant-enthalpy line down to p₁ = 2.93 bar, ending in the two-phase region).
Compressor power. A steady-flow energy balance on the adiabatic compressor gives $$\dot{W}_c = \dot{m}\left(h_{2s}-h_1\right) = 0.08\ \frac{\text{kg}}{\text{s}}\times(416.21-398.60)\ \frac{\text{kJ}}{\text{kg}} = \boxed{1.41\ \text{kW}}$$
Coefficient of performance. The refrigerating effect is $\dot{Q}_{in}=\dot{m}(h_1-h_4)=0.08\times(398.60-235.97)=13.01\ \text{kW}$, so $$\text{COP} = \frac{\dot{Q}_{in}}{\dot{W}_c} = \frac{13.01}{1.41} = \boxed{9.24}$$ For reference, the Carnot COP between the same two reservoirs is $T_C/(T_H-T_C)=273.15/26=10.51$, so the ideal cycle here reaches about 88% of the Carnot limit — plausible given the small superheat and the modest 26°C temperature lift.

Quantity	Result
Compressor power, ḏ_c	1.41 kW
Coefficient of performance, COP	9.24
(reference) Carnot COP, same reservoirs	10.51

Concept & Theory

Topic: The ideal (internally reversible) vapor-compression refrigeration cycle.

Key relations: Steady-flow energy balances on each component: $\dot{W}_c=\dot{m}(h_2-h_1)$ (compressor), $\dot{Q}_{in}=\dot{m}(h_1-h_4)$ (evaporator), $\dot{Q}_{out}=\dot{m}(h_2-h_3)$ (condenser); throttling is isenthalpic, $h_4=h_3$; $\text{COP}=\dot{Q}_{in}/\dot{W}_c$.

Why this works: "Ideal" here means every process is internally reversible except the throttling valve, which is inherently irreversible but has no shaft work — so compression follows a constant-entropy path and the condenser/evaporator exchange heat at constant pressure. Fixing all four states from just the two saturation temperatures is possible because states 1 and 3 sit exactly on the saturation dome (the problem statement gives saturated vapor/saturated liquid directly) and states 2s and 4 follow from the compressor's isentropic path and the valve's isenthalpic path respectively.

Common pitfall: Treating state 2 as isentropic from the condenser pressure's own saturation entropy instead of carrying s₁ forward, or forgetting that throttling conserves enthalpy (not entropy or temperature) between states 3 and 4.

Source: Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, Ch. 10 (vapor-compression refrigeration cycles).

Question 1 of 4: R-134a Ideal Vapor-Compression Refrigeration Cycle

Problem 1: R-134a Ideal Vapor-Compression Refrigeration Cycle (25 points)