Two-Stage Compression Refrigeration with Flash Chamber (R-134a) — Evaporator Flow from Given Condenser Flow: 04-BS-10, May 2014, Q3

Question 3: Two-Stage Compression Refrigeration with Flash Chamber (R-134a) — Evaporator Flow from Given Condenser Flow (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. R-134a two-stage compression, evaporator pressure $P_{evap}=0.14$ MPa, flash-chamber pressure $P_{mid}=0.5$ MPa, condenser pressure $P_{cond}=1.0$ MPa. Compressor I (LP) discharges directly into the flash chamber; the flash chamber's saturated-vapor outlet feeds Compressor II (HP); its saturated-liquid outlet is throttled to the evaporator. Both compressors have isentropic efficiency $\eta_c=0.90$. Condenser (= high-side) mass flow rate $\dot m_{high}=0.25$ kg/s. Evaporator exit is saturated vapor; condenser exit is saturated liquid.

State	Description	P	h (kJ/kg)
1	Evaporator exit, sat. vapor, $-18.76\ ^\circ$C	0.14 MPa	387.32
2	Compressor I exit (actual, $\eta_c=0.90$)	0.5 MPa	416.37
3 (g)	Flash-chamber vapor exit, sat. vapor	0.5 MPa	407.47
4	Compressor II exit (actual, $\eta_c=0.90$)	1.0 MPa	423.40
5	Condenser exit, sat. liquid	1.0 MPa	255.50
6	After throttle, into flash chamber	0.5 MPa	255.50
7 (f)	Flash-chamber liquid exit, sat. liquid	0.5 MPa	221.50
8	After throttle, into evaporator	0.14 MPa	221.50

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 flash chamber (2→3 mixing at the flash chamber and 6→7 liquid/vapor separation are drawn as straight connectors rather than single-stream process lines; 5→6 and 7→8 are irreversible isenthalpic throttling).

Approach

Fix states 1 and 3(g) from the given saturation pressures, compress each stage with the stated isentropic efficiency to get the actual states 2 and 4, then close the flash chamber with a combined mass-and-energy balance to solve for the low-side (evaporator) mass flow rate from the given high-side (condenser) flow rate — the reverse direction from a problem that gives evaporator capacity directly. Evaporator duty and COP follow.

Fix states 1 and 3(g) (saturation). State 1, saturated vapor at 0.14 MPa ($-18.76\ ^\circ$C): $h_1=387.32$ kJ/kg, $s_1=1.7402$ kJ/kg·K. State 3, saturated vapor at 0.5 MPa: $h_{3}=h_g=407.47$ kJ/kg, $s_g=1.7197$ kJ/kg·K.
Compressor I (1→2), actual. Isentropic exit at 0.5 MPa with $s=s_1$: $h_{2s}=413.47$ kJ/kg. Actual: $h_2=h_1+(h_{2s}-h_1)/\eta_c=387.32+(413.47-387.32)/0.90=416.37$ kJ/kg.
Compressor II (3→4), actual. Isentropic exit at 1.0 MPa with $s=s_g$: $h_{4s}=421.80$ kJ/kg. Actual: $h_4=h_g+(h_{4s}-h_g)/\eta_c=407.47+(421.80-407.47)/0.90=423.40$ kJ/kg.
Condenser and flash-chamber liquid states. Condenser exit, saturated liquid at 1.0 MPa: $h_5=255.50$ kJ/kg; throttling 5→6 into the flash chamber is isenthalpic, $h_6=h_5$. Flash-chamber liquid, saturated liquid at 0.5 MPa: $h_7=h_f=221.50$ kJ/kg.
Flash-chamber energy balance ⇒ evaporator mass flow rate (part a). The flash chamber receives Compressor-I's discharge ($\dot m_{evap}$ at $h_2$) and the throttled condenser liquid ($\dot m_{high}$ at $h_6$), and delivers saturated vapor to Compressor II ($\dot m_{high}$ at $h_g$, since that vapor is what ultimately supplies the condenser loop) and saturated liquid to the evaporator's expansion valve ($\dot m_{evap}$ at $h_f$): $$\dot m_{evap}\,h_2+\dot m_{high}\,h_6=\dot m_{high}\,h_g+\dot m_{evap}\,h_f$$ $$\dot m_{evap}=\dot m_{high}\,\frac{h_g-h_6}{h_2-h_f}=0.25\times\frac{407.47-255.50}{416.37-221.50}=\boxed{0.19497\ \text{kg/s}}.$$
Evaporator heat removal (part b). After the second throttle (7→8), $h_8=h_7=221.50$ kJ/kg: $$\dot Q_{evap}=\dot m_{evap}(h_1-h_8)=0.19497\times(387.32-221.50)=\boxed{32.33\ \text{kJ/s}}.$$
Coefficient of performance (part c). $$\dot W_{c,I}=\dot m_{evap}(h_2-h_1)=0.19497\times(416.37-387.32)=5.664\ \text{kW}.$$ $$\dot W_{c,II}=\dot m_{high}(h_4-h_g)=0.25\times(423.40-407.47)=3.981\ \text{kW}.$$ $$\text{COP}=\frac{\dot Q_{evap}}{\dot W_{c,I}+\dot W_{c,II}}=\frac{32.33}{5.664+3.981}=\boxed{3.352}.$$

Quantity	Result
(a) $\dot m_{evap}$	0.19497 kg/s
(b) $\dot Q_{evap}$	32.33 kJ/s
(c) COP	3.352

Concept & Theory

Topic. Two-stage vapour-compression refrigeration with a flash-chamber economizer, mass flow given on the high (condenser) side rather than the low (evaporator) side.

Key relations. Actual compressor exit from isentropic efficiency: $h_{out}=h_{in}+(h_{out,s}-h_{in})/\eta_c$. Flash-chamber combined mass+energy balance: $\dot m_{evap}h_2+\dot m_{high}h_6=\dot m_{high}h_g+\dot m_{evap}h_f$.

Why this works. The flash chamber's total mass balance ($\dot m_{evap}+\dot m_{high}$ in $=\dot m_{high}+\dot m_{evap}$ out) is automatically satisfied for any split, so only the energy balance is independent — whichever of $\dot m_{evap}$ or $\dot m_{high}$ is given, the same single equation solves for the other by simply moving the known mass flow to the other side algebraically.

Common pitfall. Reading the evaporator/condenser pressure limits (1 MPa and 0.14 MPa, stated as round numbers in the question) as usable directly, rather than realizing 0.14 MPa fixes the evaporator saturation temperature at $-18.76\ ^\circ$C (not a round $-20\ ^\circ$C) — always pull the saturation state from the given pressure, never from a mentally rounded temperature.

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

Question 3 of 9: Two-Stage Compression Refrigeration with Flash Chamber (R-134a) — Evaporator Flow from Given Condenser Flow

Question 3: Two-Stage Compression Refrigeration with Flash Chamber (R-134a) — Evaporator Flow from Given Condenser Flow (20 marks)

Approach