Two-Evaporator R-134a Refrigeration System: 04-BS-10, December 2018, 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, saturated vapor exit, $\dot Q_{LT}=3$ tons. HT evaporator: saturated vapor exit at 320 kPa, $\dot Q_{HT}=2$ tons. $P_{cond}=1$ MPa, saturated liquid at condenser exit. $\eta_c=0.90$. No pressure drop across either evaporator or the condenser. Both evaporator feed lines are throttled independently from the same condenser-exit liquid; the higher-pressure evaporator's vapor is throttled down to the low-temperature evaporator's pressure before the two streams mix at the single compressor's suction.

Find. (a) $\dot m_{LT}$, $\dot m_{HT}$ [kg/min]; (b) $\dot W_c$ [kW]; (c) COP; (d) $\dot S_{gen,comp}$ [kW/K].

Fig. Q3 — T–s on the real R-134a dome: solid path is the LT-evaporator loop (3→5 throttle, 5→6 evaporation at $-18\ ^\circ$C, 6→1 mixing, 1→2 actual compression, 2→3 condensing); dashed path is the HT-evaporator branch (3→4 throttle, 4→7 evaporation at 320 kPa) whose vapor (state 7) is throttled down to the LT pressure before joining state 6 at the compressor inlet (state 1).

Approach

Fix the condenser-exit liquid state once, then run each evaporator's own throttle+evaporation leg independently (LT at $-18\ ^\circ$C, HT at 320 kPa) to get each branch's mass flow from its stated capacity. The HT-evaporator vapor is throttled down to the LT pressure and mixed with the LT vapor in an energy balance to fix the single compressor-inlet state, after which the compressor is a standard isentropic-efficiency calculation.

Condenser exit and evaporator capacities in kW. Saturated liquid at 1 MPa: $h_3=255.50$ kJ/kg. Converting tons to kW ($1\ \text{ton}=211/60=3.517$ kW): $$\dot Q_{LT}=3\times3.517=10.55\ \text{kW},\qquad \dot Q_{HT}=2\times3.517=7.033\ \text{kW}.$$
LT-evaporator branch (3→5→6), $P_{LT}=P_{sat}(-18\ ^\circ\text{C})=144.60$ kPa. Throttle is isenthalpic, $h_5=h_3=255.50$; saturated-vapor exit $h_6=387.79$ kJ/kg: $$\dot m_{LT}=\frac{\dot Q_{LT}}{h_6-h_5}=\frac{10.55}{387.79-255.50}\times60=\boxed{4.785\ \text{kg/min}}.$$
HT-evaporator branch (3→4→7), $P_{HT}=320$ kPa. $h_4=h_3=255.50$; saturated-vapor exit at 320 kPa, $h_7=400.04$ kJ/kg: $$\dot m_{HT}=\frac{\dot Q_{HT}}{h_7-h_4}=\frac{7.033}{400.04-255.50}\times60=\boxed{2.919\ \text{kg/min}}.$$
Mixing at the compressor inlet (state 1). State 7 is throttled down to $P_{LT}$ (isenthalpic, $h_8=h_7=400.04$) and combines with state 6: $$\dot m_1=\dot m_{LT}+\dot m_{HT}=4.785+2.919=7.704\ \text{kg/min}$$ $$h_1=\frac{\dot m_{LT}\,h_6+\dot m_{HT}\,h_8}{\dot m_1} =\frac{4.785(387.79)+2.919(400.04)}{7.704}=392.43\ \text{kJ/kg}.$$ At $P_{LT}=144.60$ kPa this fixes a superheated state, $T_1=-12.35\ ^\circ$C ($5.65\ ^\circ$C of superheat above $T_{sat}=-18\ ^\circ$C), $s_1=1.7576$ kJ/kg·K.
Compressor, actual (part b). Isentropic exit at 1 MPa: $h_{2s}=433.94$ kJ/kg; actual: $$w_{c,a}=\frac{h_{2s}-h_1}{\eta_c}=\frac{433.94-392.43}{0.90}=46.12\ \text{kJ/kg},\qquad h_2=h_1+w_{c,a}=438.55\ \text{kJ/kg}$$ $$\dot W_c=\dot m_1\,w_{c,a}=\frac{7.704}{60}\times46.12=\boxed{5.922\ \text{kW}}.$$
COP (part c) and compressor entropy increase (part d). $$\text{COP}=\frac{\dot Q_{LT}+\dot Q_{HT}}{\dot W_c}=\frac{10.55+7.033}{5.922}=\boxed{2.969}.$$ At $P=1$ MPa, $h_2=438.55$ kJ/kg fixes $s_2=1.7716$ kJ/kg·K; the compressor is adiabatic so its entropy generation equals its entropy rise directly: $$\dot S_{gen}=\dot m_1(s_2-s_1)=\frac{7.704}{60}(1.7716-1.7576)=\boxed{0.00180\ \text{kW/K}}.$$

Quantity	Result
(a) $\dot m_{LT}$, $\dot m_{HT}$	4.785, 2.919 kg/min
(b) $\dot W_c$	5.922 kW
(c) COP	2.969
(d) $\dot S_{gen}$	0.00180 kW/K

Concept & Theory

Topic. Two-evaporator vapor-compression refrigeration with a single compressor, each evaporator running at its own pressure and independently throttled from a common condenser-exit liquid line.

Key relations. Evaporator mass flow from capacity: $\dot m=\dot Q/(h_{out}-h_{in})$ with $h_{in}=h_{f,\,condenser\ exit}$ (isenthalpic throttle). Mixing-chamber energy balance for the combined compressor-inlet enthalpy: $\dot m_1 h_1=\sum \dot m_i h_i$. Actual compressor work from isentropic efficiency, entropy generation $=\dot m(s_2-s_1)$ since the compressor is adiabatic.

Why this works. The higher-pressure evaporator's vapor cannot join the lower-pressure evaporator's vapor directly — a real flow junction requires equal pressure at the mixing point, so the HT branch's vapor must first be throttled down to the LT evaporator's pressure. Only after both streams are at the same pressure does a simple energy-weighted mixing average for enthalpy apply.

Common pitfall. Mixing the two evaporator-exit vapors directly by mass-weighted enthalpy average without first equalizing their pressures — state 7 (320 kPa) is NOT at the same pressure as state 6 ($P_{LT}=144.6$ kPa), so it must be throttled first (state 8, $h_8=h_7$) before the mixing-chamber balance is valid.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 11 (Refrigeration Cycles).

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

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

Approach