Two-Stage Compression Refrigeration with Flash Chamber: 04-BS-10, May 2018, Q3

Question 3: Two-Stage Compression Refrigeration with Flash Chamber (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. $P_{evap}=0.14$ MPa, $P_{mid}$ (flash chamber) $=0.5$ MPa, $P_{cond}=1.0$ MPa. Both compressors $\eta_c=0.90$. Mass flow through the condenser (= through the high-pressure compressor) $\dot m_{high}=0.25$ kg/s.

State	Description	P	h (kJ/kg)	s (kJ/kg·K)
1	Evaporator exit, sat. vapor	0.14 MPa	387.32	1.7402
2	LP compressor exit (actual)	0.5 MPa	416.37	1.7500
3	Flash chamber vapor, sat. vapor	0.5 MPa	407.47	1.7197
4	HP compressor exit (actual)	1.0 MPa	423.40	1.7248
5	Condenser exit, sat. liquid	1.0 MPa	255.50	1.1876
6	Flash chamber liquid (throttled from 5)	0.5 MPa	255.50	1.1936
7	Evaporator inlet (throttled from flash liquid)	0.14 MPa	221.50	1.0884

Find. (a) $\dot m_{evap}$ [kg/s]; (b) $\dot Q_{evap}$ [kJ/s]; (c) COP.

Fig. Q3 — T–s state points for the two-stage R-134a cycle with a flash chamber (1→2 LP compressor, 2&6→flash chamber, 3→4 HP compressor, 4→5 condenser, 5→6 throttle, 6→7 throttle, 7→1 evaporator). The flash chamber is an equilibrium separator: its vapor outlet (state 3) is pure saturated vapor at $P_{mid}$, not a flow-weighted mixture of streams 2 and 6.

Approach

Fix the evaporator-exit and condenser-exit states from the given saturation pressures, run each compressor isentropically then divide by $\eta_c$ for the actual work, and use an energy balance across the flash chamber (which behaves as an equilibrium two-phase separator) to size $\dot m_{evap}$ relative to the given high-pressure-side flow; COP then follows from the standard work-weighted definition.

Evaporator exit and LP compressor (1→2), 0.14→0.5 MPa. Saturated vapor at 0.14 MPa ($T_{sat}=-18.76\ ^\circ$C): $h_1=387.32$ kJ/kg, $s_1=1.7402$ kJ/kg·K. Isentropic exit at 0.5 MPa gives $h_{2s}=413.47$ kJ/kg; actual: $$h_2=h_1+\frac{h_{2s}-h_1}{\eta_c}=387.32+\frac{413.47-387.32}{0.90}=\boxed{416.37\ \text{kJ/kg}}.$$
Flash chamber energy balance → $\dot m_{evap}$ (part a). The flash chamber's vapor outlet is saturated vapor at 0.5 MPa ($h_g=407.47$ kJ/kg); its liquid outlet is saturated liquid at 0.5 MPa ($h_f=221.50$ kJ/kg), which is also the state entering the evaporator after the second throttle ($h_7=h_f$). The condenser liquid throttled into the flash chamber carries $h_6=h_5=255.50$ kJ/kg (sat. liquid at 1.0 MPa, unchanged across the throttle). A steady mass/energy balance on the chamber ($\dot m_{evap}\,h_2+\dot m_{high}\,h_6=\dot m_{high}\,h_2+ \dot m_{evap}\,h_f$, rearranged for the vapor/liquid split) gives: $$\dot m_{evap}=\dot m_{high}\,\frac{h_g-h_6}{h_2-h_f} =0.25\times\frac{407.47-255.50}{416.37-221.50}=0.25\times\frac{151.97}{194.87} =\boxed{0.1950\ \text{kg/s}}.$$
Evaporator heat removal rate (part b). Across the evaporator, $7\to1$: $$q_{evap}=h_1-h_7=387.32-221.50=165.82\ \text{kJ/kg}.$$ $$\dot Q_{evap}=\dot m_{evap}\,q_{evap}=0.1950\times165.82=\boxed{32.33\ \text{kJ/s}}.$$
Flash-chamber vapor and HP compressor (3→4), 0.5→1.0 MPa. Saturated vapor at 0.5 MPa ($T_{sat}=15.73\ ^\circ$C): $h_3=407.47$ kJ/kg, $s_3=1.7197$ kJ/kg·K. Isentropic exit at 1.0 MPa gives $h_{4s}=421.80$ kJ/kg; actual: $$h_4=h_3+\frac{h_{4s}-h_3}{\eta_c}=407.47+\frac{421.80-407.47}{0.90}=\boxed{423.40\ \text{kJ/kg}}.$$
Compressor work and COP (part c). LP compressor handles $\dot m_{evap}$; HP compressor handles $\dot m_{high}$: $$\dot W_{LP}=\dot m_{evap}(h_2-h_1)=0.1950\times(416.37-387.32)=5.66\ \text{kW}.$$ $$\dot W_{HP}=\dot m_{high}(h_4-h_3)=0.25\times(423.40-407.47)=3.98\ \text{kW}.$$ $$\text{COP}=\frac{\dot Q_{evap}}{\dot W_{LP}+\dot W_{HP}}=\frac{32.33}{5.66+3.98}=\frac{32.33}{9.64} =\boxed{3.352}.$$

Quantity	Result
(a) $\dot m_{evap}$	0.1950 kg/s
(b) $\dot Q_{evap}$	32.33 kJ/s
(c) COP	3.352

Concept & Theory

Topic. Two-stage vapor-compression refrigeration with a flash-chamber (flash intercooler) economizer, mass flow given on the high-pressure side.

Key relations. Actual compressor work from isentropic efficiency: $h_{2}=h_1+(h_{2s}-h_1)/\eta_c$. Flash-chamber mass/energy balance sizes the low-pressure mass flow relative to the high-pressure mass flow: $\dot m_{evap}=\dot m_{high}(h_g-h_6)/(h_2-h_f)$. COP $=\dot Q_{evap}/(\dot W_{LP}+\dot W_{HP})$.

Why this works. The flash chamber is an equilibrium two-phase separator: any vapor entering it flashes off as pure saturated vapor at the intermediate pressure, and the liquid leaves as pure saturated liquid at that same pressure — so the HP compressor always draws saturated vapor at $P_{mid}$ (state 3), never a flow-weighted mixture of the LP discharge and the throttled condensate. That mixing only matters for the mass-flow-rate balance that sizes $\dot m_{evap}$, not for fixing state 3's properties.

Common pitfall. Modeling the HP compressor's inlet as a flow-weighted enthalpy mix of the LP discharge and the flash-chamber's saturated vapor — this double-counts the separator's role and gives the wrong $h_3$; the correct state 3 is always the flash chamber's own saturated-vapor property at $P_{mid}$.

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

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

Question 3: Two-Stage Compression Refrigeration with Flash Chamber (20 marks)

Approach