Ammonia ice-making refrigeration plant: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2018, Q4

Question 4 — Ammonia ice-making refrigeration plant (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. A vapour-compression cycle with ammonia. Evaporation at −15 °C (saturated vapour out), condensation at 31 °C (saturated liquid out), isentropic compressor efficiency $\eta_c=0.95$, refrigerating capacity $\dot Q_L=70$ kW. Cooling water 20 → 27 °C, $c_{p,w}=4.186\ \text{kJ/kg}\cdot\text{K}$. Properties from the saturated- and superheated-ammonia tables appended to the paper.

State	Description	$h$ (kJ/kg)	$s$ (kJ/kg·K)
1	sat. vapour, −15 °C	1425.7	5.545
2s	isentropic exit, 31 °C ($\approx$12 bar)	1663.0	5.545
2	actual exit ($\eta_c=0.95$)	1675.7	—
3 = 4	sat. liquid 31 °C / after throttle	327.8	—

Find. the compressor power per kW of refrigeration, the COP, and the cooling-water flowrate.

Figure 4 — Vapour-compression cycle on a $\ln P$–$h$ plane: 1→2 compression (with $\eta_c$), 2→3 condensation to saturated liquid at 31 °C, 3→4 throttle (constant $h$), 4→1 evaporation at −15 °C.

Approach. Read $h_1$ (sat. vapour, −15 °C) and $s_1$; find $h_{2s}$ at the condenser pressure for $s=s_1$; apply $\eta_c$ for the real work; the throttle gives $h_4=h_3$ (sat. liquid, 31 °C). Then form $q_L$, the specific work, COP, and close the condenser energy balance for the water flow.

Refrigerating effect. With $h_1=1425.7$ and $h_4=h_3=327.8\ \text{kJ/kg}$: $$q_L=h_1-h_4=1425.7-327.8=1097.9\ \text{kJ/kg}$$
Compressor work. Isentropic exit at $P_\text{cond}\approx12$ bar ($T_\text{sat}=31$ °C) with $s_{2s}=s_1$ gives $h_{2s}=1663.0\ \text{kJ/kg}$; the actual work is larger by $1/\eta_c$: $$w=\frac{h_{2s}-h_1}{\eta_c}=\frac{1663.0-1425.7}{0.95}=249.8\ \text{kJ/kg},\qquad h_2=h_1+w=1675.7\ \text{kJ/kg}$$
Power per kW of refrigeration and COP. $$\frac{\dot W}{\dot Q_L}=\frac{w}{q_L}=\frac{249.8}{1097.9}=0.228\ \frac{\text{kW}}{\text{kW}},\qquad \text{COP}=\frac{q_L}{w}=\frac{1097.9}{249.8}=4.40$$ $\dot W/\dot Q_L=0.228$, COP $=4.40$
Mass flow and compressor power. $$\dot m=\frac{\dot Q_L}{q_L}=\frac{70}{1097.9}=0.0638\ \text{kg/s},\qquad \dot W=\dot m\,w=15.9\ \text{kW}$$
Cooling-water flowrate. The condenser rejects $\dot Q_H=\dot Q_L+\dot W=70+15.9=85.9$ kW (equivalently $\dot m(h_2-h_3)$). Setting this equal to the water sensible gain: $$\dot m_w=\frac{\dot Q_H}{c_{p,w}\,\Delta T_w}=\frac{85.9}{4.186(27-20)}=2.93\ \text{kg/s}$$ $\dot m_w\approx 2.93\ \text{kg/s}$

Quantity	Result
Refrigerating effect $q_L$	1098 kJ/kg
Specific compressor work $w$	250 kJ/kg
Power per kW refrigeration	0.228 kW/kW
Coefficient of performance	4.40
Ammonia mass flow / compressor power	0.0638 kg/s / 15.9 kW
Cooling-water flowrate	2.93 kg/s

Concept & Theory

Topic: the vapour-compression refrigeration cycle with a non-ideal compressor.

Key relations: $q_L=h_1-h_4$; isentropic $s_{2s}=s_1$; $w=(h_{2s}-h_1)/\eta_c$; $\text{COP}=q_L/w$; throttle $h_4=h_3$; condenser balance $\dot Q_H=\dot Q_L+\dot W=\dot m_w c_{p,w}\Delta T_w$.

Why this works: the four processes are fixed by the two saturation temperatures plus the isentropic-then-corrected compression. The condenser must reject both the absorbed refrigeration load and the compressor work (first law on the whole cycle), so the cooling-water duty is $\dot Q_L+\dot W$, not $\dot Q_L$ alone — a common slip.

Common pitfall: forgetting the compressor efficiency (using $h_{2s}$ for the actual state) or sizing the cooling water from $\dot Q_L$ only. Also, "power per kW of refrigeration" is simply the reciprocal of the COP.

Source: Çengel & Boles, Thermodynamics 9th ed., §11-2 to §11-4; ammonia property tables appended to the examination.

Question 4 of 8: Ammonia ice-making refrigeration plant

Question 4 — Ammonia ice-making refrigeration plant (Part A, equal value)