Two-Stage Compression Refrigeration with a Flash Chamber: 04-BS-10, May 2016, Q3

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

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Given. Evaporator (low) pressure $P_{evap}=0.14$ MPa. Flash-chamber (intermediate) pressure $P_{flash}=0.5$ MPa. Condenser (high) pressure $P_{cond}=1.0$ MPa. Total mass flow through the condenser and high-pressure compressor $\dot m_{cond}=0.25$ kg/s. Evaporator exit and flash-chamber vapor outlet are both saturated vapor; condenser exit and flash-chamber liquid outlet are both saturated liquid. Both compressors isentropic.

State	Description	P	h (kJ/kg)
1	Evaporator exit, sat. vapor	0.14 MPa	387.32
2	LP compressor exit (isentropic)	0.5 MPa	413.47
3	Flash chamber, sat. vapor (bypasses to HP inlet)	0.5 MPa	407.47
4	HP compressor inlet (2 & 3 mixed)	0.5 MPa	412.37
5	HP compressor exit (isentropic)	1.0 MPa	427.14
6	Condenser exit, sat. liquid	1.0 MPa	255.50
8	Flash chamber, sat. liquid	0.5 MPa	221.50
9	Evaporator inlet (after 2nd throttle)	0.14 MPa	221.50

Find. (b) $\dot Q_L$ [kW]; (c) $\dot W_{in}$ [kW]; (d) COP.

Fig. Q3 — P–h state points for the two-stage cascade with a flash chamber (1→2 LP compressor, 2&3→4 adiabatic mixing at the flash pressure, 4→5 HP compressor, 5→6 condenser, 6→flash→(3,8) throttle+separation, 8→9 second throttle into the evaporator, 9→1 evaporator). State 3 (flash vapor) is not on the main loop shown — it bypasses directly into the mixing point at state 4.

Approach

Work outward from the two fixed saturation states (evaporator exit, condenser exit). The low-pressure compressor raises the evaporator vapor isentropically to the flash pressure (state 2). The condenser liquid throttles into the flash chamber, where it flashes into saturated vapor (state 3) and saturated liquid (state 8); the vapor fraction of that throttled stream is exactly the mass fraction of the total flow that bypasses the evaporator. Mixing states 2 and 3 in the proportions set by that flash quality gives the high-pressure compressor's inlet state (state 4), which is then compressed isentropically to the condenser pressure (state 5). Only the low-pressure branch (mass fraction $1-y$) flows through the evaporator and produces the refrigeration effect.

Evaporator exit and LP compressor (1→2). Saturated vapor at 0.14 MPa: $h_1=387.32$ kJ/kg, $s_1=1.7402$ kJ/kg·K ($T_1=-18.76\ ^\circ$C). Isentropic compression to 0.5 MPa: $h_2=413.47$ kJ/kg ($T_2=21.94\ ^\circ$C).
Condenser exit and flash chamber. Saturated liquid at 1.0 MPa: $h_6=255.50$ kJ/kg ($T_6=39.39\ ^\circ$C). Throttled into the flash chamber at 0.5 MPa ($h_7=h_6$), the flash quality (= vapor mass fraction, and also the fraction of the total flow that bypasses the evaporator) is $$y=\frac{h_6-h_{f,flash}}{h_{g,flash}-h_{f,flash}}=\frac{255.50-221.50}{407.47-221.50}=\boxed{0.1828}.$$ So $h_3=h_{g,flash}=407.47$ kJ/kg (vapor, $T_3=15.73\ ^\circ$C) and $h_8=h_{f,flash}=221.50$ kJ/kg (liquid), and $h_9=h_8=221.50$ kJ/kg after the second throttle into the evaporator.
Mass split and mixing (state 4). Low-pressure branch fraction of the total condenser flow: $\dot m_{low}=(1-y)\dot m_{cond}=(1-0.1828)\times0.25=0.20430$ kg/s. Energy balance on the adiabatic mixing chamber: $$h_4=(1-y)h_2+y\,h_3=0.8172\times413.47+0.1828\times407.47=\boxed{412.37\ \text{kJ/kg}}.$$
HP compressor (4→5), 0.5→1.0 MPa. Finding $s_4$ from $(P,h)=(0.5\ \text{MPa},\,412.37\ \text{kJ/kg})$ gives $s_4=1.7365$ kJ/kg·K. Isentropic compression to 1.0 MPa: $h_5=\boxed{427.14\ \text{kJ/kg}}$ ($T_5=46.55\ ^\circ$C).
Heat removed from the refrigerated space (part b). Only the low-pressure branch passes through the evaporator: $$\dot Q_L=\dot m_{low}(h_1-h_9)=0.20430\times(387.32-221.50)=0.20430\times165.82 =\boxed{33.877\ \text{kW}}.$$
Total compressor power input (part c). $$\dot W_{LP}=\dot m_{low}(h_2-h_1)=0.20430\times(413.47-387.32)=5.342\ \text{kW}.$$ $$\dot W_{HP}=\dot m_{cond}(h_5-h_4)=0.25\times(427.14-412.37)=3.692\ \text{kW}.$$ $$\dot W_{in}=\dot W_{LP}+\dot W_{HP}=5.342+3.692=\boxed{9.034\ \text{kW}}.$$
Coefficient of performance (part d). $$\text{COP}=\frac{\dot Q_L}{\dot W_{in}}=\frac{33.877}{9.034}=\boxed{3.750}.$$

Quantity	Result
(b) $\dot Q_L$	33.877 kW
(c) $\dot W_{in}$	9.034 kW
(d) COP	3.750

Concept & Theory

Topic. Two-stage vapor-compression refrigeration cycle with a flash chamber (economizer), R-134a, both compressors isentropic.

Key relations. Flash quality fixes the mass split: $y=(h_{f,cond,exit}-h_{f,flash})/(h_{g,flash}-h_{f,flash})$. Adiabatic mixing: $h_4=(1-y)h_2+y\,h_3$. Only the low-pressure branch produces refrigeration: $\dot Q_L=\dot m_{low}(h_1-h_9)$.

Why this works. The flash chamber lets the vapor that would otherwise re-form during the second throttle skip the evaporator entirely and rejoin the cycle directly at the intermediate pressure — this reduces both the mass flow that must be compressed the full range by the low-pressure compressor and the throttling irreversibility, which is why two-stage compression with a flash chamber gives a higher COP than a single throttle valve over the same overall pressure ratio.

Common pitfall. Applying the full condenser mass flow rate $\dot m_{cond}$ to the evaporator and low-pressure compressor — only the fraction $(1-y)$ that survives the flash separation as liquid actually flows through the evaporator; the vapor fraction $y$ bypasses it completely.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 11 (Refrigeration Cycles § Cascade and Multistage Compression Refrigeration Systems).

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

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

Approach