Question 3 of 8: Ammonia vs. R134a — Comparative Vapour-Compression Refrigeration Cycle

Question 3: Ammonia vs. R134a — Comparative Vapour-Compression Refrigeration Cycle

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 ideal vapour-compression cycle skeleton (saturated liquid into the expansion valve, dry saturated vapour into the compressor) run at identical condenser/evaporator temperatures and identical cooling duty for two candidate refrigerants; the compressor floor efficiency of 90% applies equally to both, so it scales both fluids' actual work by the same factor and cannot change which fluid needs less power — it is evaluated anyway for a defensible absolute number.

Find. Which refrigerant, ammonia or R134a, gives the lower compressor power for the stated cycle and duty.

Approach. For each fluid: state 1 is saturated vapour at $T_{evap}$; state 3 is saturated liquid at $T_{cond}$, and $h_4=h_3$ across the (isenthalpic) expansion valve. Compress isentropically from state 1 to the condenser pressure to get $h_{2s}$, then apply the compressor efficiency to get the actual work; the mass flow rate follows from the refrigeration effect $\dot Q_L=\dot m(h_1-h_4)$, and the compressor power is $\dot m$ times the actual specific work.

1. Ammonia state points. $P_{evap}=226.25\text{ kPa}$, $P_{cond}=1311.66\text{ kPa}$ (matches the source's own saturation table to better than 0.1%). $h_1=1588.38\text{ kJ/kg}$, $s_1=6.3258\text{ kJ/kg}\cdot\text{K}$, $h_3=h_4=506.67\text{ kJ/kg}$. Isentropic compression to $P_{cond}$ gives $h_{2s}=1846.94\text{ kJ/kg}$, so $$w_{s,NH_3}=h_{2s}-h_1=258.56\text{ kJ/kg}, \qquad w_{a,NH_3}=\frac{w_{s,NH_3}}{\eta_c}=287.29\text{ kJ/kg}$$ The refrigeration effect per unit mass is $h_1-h_4=1081.71\text{ kJ/kg}$ (the printed appendix table gives $h_g(-16\,{}^{\circ}\text{C})-h_f(34\,{}^{\circ}\text{C})=1424.4-342.3=1082.1\text{ kJ/kg}$, within 0.04% — a good cross-check), so $$\dot m_{NH_3}=\frac{\dot Q_L}{h_1-h_4}=\frac{3.5}{1081.71}=0.003236\text{ kg/s}$$ $$\boxed{\dot W_{c,NH_3}=\dot m_{NH_3}\,w_{a,NH_3}=0.003236\times287.29=0.930\text{ kW}}$$
2. R134a state points. $P_{evap}=157.28\text{ kPa}$, $P_{cond}=862.63\text{ kPa}$. $h_1=389.02\text{ kJ/kg}$, $s_1=1.7379\text{ kJ/kg}\cdot\text{K}$, $h_3=h_4=247.54\text{ kJ/kg}$. Isentropic compression gives $h_{2s}=424.42\text{ kJ/kg}$, so $$w_{s,R134a}=35.40\text{ kJ/kg}, \qquad w_{a,R134a}=\frac{35.40}{0.90}=39.34\text{ kJ/kg}$$ Refrigeration effect per unit mass $h_1-h_4=141.48\text{ kJ/kg}$ (table cross-check: $237.74-97.31=140.43\text{ kJ/kg}$, within 0.7%), so $$\dot m_{R134a}=\frac{3.5}{141.48}=0.02474\text{ kg/s}$$ $$\boxed{\dot W_{c,R134a}=0.02474\times39.34=0.973\text{ kW}}$$
3. Comparison. Ammonia's much larger latent heat of vaporisation lets it move the same refrigeration duty with roughly $1/8$ the mass flow rate of R134a, more than offsetting its larger specific compression work: $$\boxed{\dot W_{c,NH_3}=0.930\text{ kW} \lt \dot W_{c,R134a}=0.973\text{ kW}}$$ Ammonia requires the lower compressor power (equivalently, $COP_{NH_3}=\dot Q_L/\dot W_{c,NH_3}=3.77$ versus $COP_{R134a}=3.60$).