Two-Stage R-134a Compression Refrigeration with Flash Chamber: 04-BS-10, Undated paper, Q3

Question 3: Two-Stage R-134a Compression Refrigeration with Flash Chamber (Part A – 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.

[Part (d) is printed "in kJ/K"; the quantity asked for is a rate of exergy destruction and is reported below in kJ/s.]

Given. $P_{low}=0.14$ MPa (evaporator), $P_{flash}=0.5$ MPa, $P_{high}=1.0$ MPa (condenser). $\eta_c=0.90$ for both compressors. Mass flow through the condenser (and high-pressure compressor) $\dot m_{cond}=0.25$ kg/s — part (a) then asks for the DIFFERENT, smaller flow rate through the evaporator/low-pressure compressor, found from the flash-chamber mass+energy balance. $T_0=300$ K. The printed apparatus diagram shows a single flash chamber with TWO inlets (the throttled condensate, state 6, and the low-pressure compressor discharge, state 2) and two outlets (saturated vapour, state 3, to compressor 2 and saturated liquid, state 7, to the second throttle), so the chamber is an equilibrium separator.

State	Description	P (MPa)	T (°C)	h (kJ/kg)	s (kJ/kg·K)
1	Evaporator exit, sat. vapor	0.14	−18.76	387.32	1.7402
2	Low-P compressor exit (actual)	0.5	24.97	416.37	1.7500
3	Flash chamber vapor exit	0.5	15.73	407.47	1.7197
4	High-P compressor exit (actual)	1.0	43.15	423.40	1.7247
5	Condenser exit, sat. liquid	1.0	39.39	255.50	—
6	Throttled into flash chamber	0.5	15.73	255.50	—
7	Flash chamber liquid exit	0.5	15.73	221.50	—
8	Throttled into evaporator	0.14	−18.76	221.50	—

Find. (a) $\dot m_{evap}$; (b) $\dot Q_L$; (c) COP; (d) $\dot X_{dest}$ in both compressors.

Fig. Q3 — P–h state points, two-stage flash-chamber cycle. Low-pressure loop $8\to1\to2$, high-pressure loop $3\to4\to5\to6$, flash chamber mixes $2$ and $6$ into vapor $3$ / liquid $7$.

Fig. Q3b — the same eight state points on the T–s diagram the question asks for, with the R-134a saturation dome. $8\to1$ evaporator, $1\to2$ low-pressure compressor, $3\to4$ high-pressure compressor, $4\to5$ condenser, $5\to6$ and $7\to8$ throttles; the flash chamber separates the mixed stream at 0.5 MPa into saturated vapour $3$ and saturated liquid $7$.

Approach

Fix the eight states (R-134a), then close the flash chamber with a combined mass-and-energy balance to solve for the unknown low-pressure (evaporator) flow rate, since only the high-pressure (condenser) flow rate is given.

Low-pressure compressor, actual exit (state 2). $s_{2s}=s_1=1.7402$ kJ/kg·K $\Rightarrow h_{2s}=CP(0.5\ \text{MPa},s_1)$. $$h_2=h_1+\frac{h_{2s}-h_1}{\eta_c}=387.32+\frac{25.31}{0.90}=416.37\ \text{kJ/kg}.$$
Flash-chamber balance — evaporator mass flow rate, part (a). Steady-flow mass and energy balance on the flash chamber ($\dot m_{evap}$ through state 2 in, $\dot m_{cond}$ through state 6 in; $\dot m_{evap}$ leaves as liquid state 7, $\dot m_{cond}$ leaves as vapor state 3): $$\dot m_{evap}\,h_2+\dot m_{cond}\,h_6=\dot m_{evap}\,h_7+\dot m_{cond}\,h_3$$ $$\dot m_{evap}=\dot m_{cond}\,\frac{h_3-h_6}{h_2-h_7}=0.25\times\frac{407.47-255.50}{416.37-221.50} =0.25\times0.7799=\boxed{0.1950\ \text{kg/s}}.$$
High-pressure compressor, actual exit (state 4). $s_{4s}=s_3=1.7197$ kJ/kg·K $\Rightarrow h_{4s}=CP(1.0\ \text{MPa},s_3)$. $$h_4=h_3+\frac{h_{4s}-h_3}{\eta_c}=407.47+\frac{14.35}{0.90}=423.40\ \text{kJ/kg}.$$
Refrigeration capacity, part (b). All the evaporator flow passes through $8\to1$, with $h_8=h_7$ (isenthalpic throttle): $$\dot Q_L=\dot m_{evap}(h_1-h_8)=0.1950\times(387.32-221.50)=\boxed{32.33\ \text{kJ/s}}.$$
Total compressor work and COP, part (c). $$\dot W_{c1}=\dot m_{evap}(h_2-h_1)=0.1950\times29.05=5.664\ \text{kW};\quad \dot W_{c2}=\dot m_{cond}(h_4-h_3)=0.25\times15.92=3.981\ \text{kW}.$$ $$\dot W_{total}=5.664+3.981=9.645\ \text{kW}\quad\Rightarrow\quad COP=\frac{\dot Q_L}{\dot W_{total}}=\frac{32.33}{9.645}=\boxed{3.352}.$$
Exergy destruction in both compressors, part (d). Each compressor is adiabatic, so $\dot X_{dest}=\dot m\,T_0\,\Delta s$ across it: $$\dot X_{c1}=\dot m_{evap}\,T_0(s_2-s_1)=0.1950\times300\times(1.7500-1.7402)=\boxed{0.573\ \text{kJ/s}}$$ $$\dot X_{c2}=\dot m_{cond}\,T_0(s_4-s_3)=0.25\times300\times(1.7247-1.7197)=\boxed{0.378\ \text{kJ/s}}$$

Quantity	Value
(a) $\dot m_{evap}$	0.1950 kg/s
(b) $\dot Q_L$	32.33 kJ/s
(c) COP	3.352
(d) $\dot X_{dest}$: low-P comp. / high-P comp.	0.573 / 0.378 kJ/s

Topic. Two-stage vapor-compression refrigeration with a flash chamber (R-134a).

Key relations. Flash-chamber mass+energy balance $\dot m_1h_{in,1}+\dot m_2h_{in,2}=\dot m_1h_{out,1}+\dot m_2h_{out,2}$; per-compressor $\eta_c=(h_{s}-h_{in})/(h_{out}-h_{in})$; $COP=\dot Q_L/\dot W_{total}$; $\dot X_{dest}=\dot m T_0\Delta s$.

Why this works. Flashing part of the condensed liquid off at the intermediate pressure and routing only saturated vapor into the high-pressure compressor means that compressor never has to handle the (larger) evaporator mass flow at all — it only compresses the flash vapor, which is why $\dot m_{evap}<\dot m_{cond}$ and the two compressor duties in part (c) use different mass flow rates.

Common pitfall. Assuming the given 0.25 kg/s is the evaporator flow rate (matching part (a)'s literal wording) rather than recognizing it as the condenser/high-pressure-loop rate that part (a) is actually asking you to convert — the two are equal only in a single-stage cycle.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., §11-5 (multistage compression refrigeration systems).

Question 3 of 9: Two-Stage R-134a Compression Refrigeration with Flash Chamber

Question 3: Two-Stage R-134a Compression Refrigeration with Flash Chamber (Part A – 20 marks)

Approach