Two-Evaporator R-134a Refrigeration System: 04-BS-10, December 2019, Q3

Question 3: Two-Evaporator R-134a Refrigeration System (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. LT evaporator: $-18^\circ$C sat. vapor exit, 3 tons. HT evaporator: $10^\circ$C sat. vapor exit, 2 tons. Condenser: 1 MPa, sat. liquid exit. Compressor isentropic; both evaporators feed the SAME compressor but at different pressures, so the HT branch's vapor must be throttled down to the LT branch's (lower) pressure before the two streams mix at the compressor inlet.

State	Description	P	h (kJ/kg)
2/3/4	Condenser exit, sat. liquid (feeds both throttles)	1000 kPa	255.50
6	LT evaporator exit, sat. vapor	144.60 kPa	387.79
7	HT evaporator exit, sat. vapor	414.61 kPa	404.32
8	HT branch, throttled down to $P_{LT}$	144.60 kPa	404.32
1	Compressor inlet, mixed 6+8	144.60 kPa	393.94
2$'$	Compressor exit, isentropic	1000 kPa	435.82

Find. (a) $\dot m_{LT},\dot m_{HT}$ [kg/min]; (b) $\dot W_c$ [kW]; (c) $\dot Q_{out}$ [kW].

Fig. Q3 — T–s on the real R-134a dome. Both evaporator legs throttle from the same condenser-exit liquid; the HT branch's higher-pressure vapor (7) is throttled again down to the LT branch's pressure before the two streams mix (state 1) at the single compressor's inlet.

Approach

Get each evaporator's saturation pressure from its exit temperature, size each branch's mass flow from its stated tonnage, throttle the HT branch's vapor down to the LT branch's pressure (isenthalpic), mix the two streams by an energy balance to fix the compressor-inlet state, then compress isentropically to the condenser pressure.

Evaporator saturation pressures. $$P_{LT}=P_{sat}(-18^\circ\text{C})=144.60\ \text{kPa},\qquad P_{HT}=P_{sat}(10^\circ\text{C})=414.61\ \text{kPa}.$$
Mass flow through each evaporator (part a). Both throttle from the same condenser-exit liquid ($h=255.50$ kJ/kg), so each branch's capacity fixes its own flow: $$\dot m_{LT}=\frac{3\times211/60\ \text{kW}}{h_6-h_{3}}=\frac{10.55}{387.79-255.50}=0.07975\ \text{kg/s}=\boxed{4.785\ \text{kg/min}}$$ $$\dot m_{HT}=\frac{2\times211/60\ \text{kW}}{h_7-h_{4}}=\frac{7.033}{404.32-255.50}=0.04726\ \text{kg/s}=\boxed{2.836\ \text{kg/min}}.$$
Mixing at the compressor inlet. The HT branch is throttled to $P_{LT}$ ($h_8=h_7=404.32$, isenthalpic) and mixes with the LT branch: $$h_1=\frac{\dot m_{LT}h_6+\dot m_{HT}h_8}{\dot m_{LT}+\dot m_{HT}}=\frac{0.07975(387.79)+0.04726(404.32)}{0.12701}=393.94\ \text{kJ/kg}.$$
Compressor power (part b). Isentropic from state 1 ($P_{LT}$) to the condenser pressure: $$\dot W_c=(\dot m_{LT}+\dot m_{HT})(h_{2'}-h_1)=0.12701\times(435.82-393.94)=\boxed{5.319\ \text{kW}}.$$
Condenser heat rejection (part c). Energy balance on the whole cycle: everything absorbed by both evaporators plus the compressor work must leave through the condenser: $$\dot Q_{out}=\dot Q_{LT}+\dot Q_{HT}+\dot W_c=(3+2)\times\frac{211}{60}+5.319=17.583+5.319=\boxed{22.90\ \text{kW}}.$$

Quantity	Result
(a) $\dot m_{LT}$, $\dot m_{HT}$	4.785 kg/min, 2.836 kg/min
(b) $\dot W_c$	5.319 kW
(c) $\dot Q_{out}$	22.90 kW

Concept & Theory

Topic. Two-evaporator, single-compressor vapor-compression refrigeration serving two different load temperatures.

Key relations. Each evaporator branch throttles independently from the shared condenser-exit liquid; a branch operating at a HIGHER saturation pressure than another cannot simply mix with it at the compressor inlet — it must first be throttled down to the lower pressure, an extra irreversible step invisible in the problem's plain state-numbering.

Why this works. The compressor has a single, well-defined suction pressure; both evaporator streams must arrive at that SAME pressure before mixing is physically meaningful (mixing at unequal pressures is undefined). The HT evaporator's higher operating pressure is only useful for raising ITS OWN evaporating temperature, not for feeding the compressor at that higher pressure directly.

Common pitfall. Mixing states 6 and 7 directly (ignoring the pressure mismatch) — this understates the compressor's actual suction pressure and produces a compressor work number that doesn't correspond to any real single-suction-pressure machine.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 11 (Refrigeration Cycles, Multistage/Multi-evaporator Systems).

Question 3 of 9: Two-Evaporator R-134a Refrigeration System

Question 3: Two-Evaporator R-134a Refrigeration System (20 marks)

Approach