Refrigerant selection: ammonia vs. R-134a: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2017, Q3

Question 3 — Refrigerant selection: ammonia vs. R-134a (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. Standard vapour-compression cycle: $T_\text{evap}=-16$ °C, $T_\text{cond}=34$ °C, saturated liquid into the valve, dry saturated vapour into the compressor, $\eta_c=0.90$, refrigeration duty $\dot Q_L = 3.5$ kW. Find. which refrigerant needs less compressor power. Properties are read from the appended ammonia and R-134a tables.

State property (from exam tables)	Ammonia	R-134a
Evaporator (−16 °C): $h_1=h_g$	1424.4 kJ/kg	237.74 kJ/kg
Evaporator (−16 °C): $s_1=s_g$	5.5600	0.9298
Condenser (34 °C): $P_\text{cond}$	1311.6 kPa	8.625 bar
Condenser (34 °C): $h_3=h_4=h_f$	342.3 kJ/kg	97.31 kJ/kg

Figure 3 — Standard vapour-compression loop, circulating compressor → condenser → expansion valve → evaporator → compressor. States 1 (evaporator exit, saturated vapour), 2 (compressor discharge), 3 (condenser exit, saturated liquid) and 4 (valve exit, wet) are the same for both fluids; only the property values differ.

Approach. For each fluid, evaluate the four cycle states, size the mass flow from the required refrigeration effect $h_1-h_4$, find the isentropic discharge enthalpy by interpolating the superheat table at $P_\text{cond}$ and $s_1$, apply the 90% compressor efficiency, and compare $\dot W = \dot m\,w_\text{act}$.

Ammonia — isentropic discharge. Interpolating the superheated-ammonia table at $P_\text{cond}=1311.6$ kPa (between 12 and 14 bar) for $s = s_1 = 5.5600$ gives $h_{2s}=1683.8$ kJ/kg. With $\eta_c=0.90$: $$w_\text{act} = \frac{h_{2s}-h_1}{\eta_c} = \frac{1683.8-1424.4}{0.90} = 288.3\ \text{kJ/kg}$$
Ammonia — mass flow and power. The refrigeration effect is $q_L = h_1-h_4 = 1424.4-342.3 = 1082.1$ kJ/kg, so $$\dot m = \frac{\dot Q_L}{q_L} = \frac{3.5}{1082.1} = 3.23\times10^{-3}\ \text{kg/s},\qquad \dot W = \dot m\,w_\text{act} = 932\ \text{W}$$ Ammonia: $\dot W \approx 932$ W, COP $= q_L/w_\text{act} = 3.75$
R-134a — isentropic discharge. Interpolating the superheated-R-134a table at $P_\text{cond}=8.625$ bar (between 8 and 9 bar) for $s=s_1=0.9298$ gives $h_{2s}=272.9$ kJ/kg, so $$w_\text{act} = \frac{272.9-237.74}{0.90} = 39.05\ \text{kJ/kg}$$
R-134a — mass flow and power. Here $q_L = 237.74-97.31 = 140.4$ kJ/kg: $$\dot m = \frac{3.5}{140.4} = 24.9\times10^{-3}\ \text{kg/s},\qquad \dot W = 973\ \text{W}$$ R-134a: $\dot W \approx 973$ W, COP $= 3.60$
Compare. Ammonia (932 W, COP 3.75) requires less compressor power than R-134a (973 W, COP 3.60) for the same 3.5 kW duty: Ammonia gives the lower compressor power (≈ 41 W less).

Check — superheat interpolation.

Both discharge enthalpies come from a double interpolation (pressure between adjacent table columns, then temperature for the target entropy). A neighbouring-row read shifts $h_{2s}$ by ≈ ±1 kJ/kg (ammonia) and ±0.3 kJ/kg (R-134a), i.e. a few watts on each power — well inside the 41 W margin, so the ranking (ammonia lower) is robust.

Quantity	Ammonia	R-134a
Refrigeration effect $q_L$	1082.1 kJ/kg	140.4 kJ/kg
Actual compression work	288.3 kJ/kg	39.05 kJ/kg
Mass flow $\dot m$	3.23 g/s	24.9 g/s
Compressor power $\dot W$	≈ 932 W	≈ 973 W
COP	3.75	3.60
Lower power: ammonia

Concept & Theory

Topic. Ideal vapour-compression refrigeration cycle with a non-ideal (90%) compressor; refrigerant comparison at fixed evaporator/condenser temperatures.

Key relations. Evaporator $q_L=h_1-h_4$; throttle $h_4=h_3=h_f(T_\text{cond})$; isentropic discharge $s_{2s}=s_1$; compressor efficiency $w_\text{act}=(h_{2s}-h_1)/\eta_c$; $\dot m=\dot Q_L/q_L$; $\text{COP}=q_L/w_\text{act}$.

Why this works. Fixing the two saturation temperatures fixes states 1 and 3 for either fluid; the only fluid-specific work is the isentropic enthalpy rise to the condenser pressure, derated by the compressor efficiency. Ammonia’s very large latent heat gives a huge refrigeration effect per kilogram, so despite its large per-kilogram work its tiny mass flow yields the smaller power — equivalently, its higher COP wins.

Common pitfall. Comparing per-kilogram works directly (ammonia’s is far larger) instead of the power for equal duty; or reading the condenser pressure at the wrong saturation temperature.

Source. Çengel & Boles, Thermodynamics §11 (refrigeration cycles; the ideal and actual vapour-compression cycle, COP, refrigerant selection).

Question 3 of 8: Refrigerant selection: ammonia vs. R-134a

Question 3 — Refrigerant selection: ammonia vs. R-134a (Part A, equal value)