Two-Stage Compression Refrigeration with Flash Chamber (R-134a): 04-BS-10, December 2017, Q3

Question 3: Two-Stage Compression Refrigeration with Flash Chamber (R-134a) (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)	s (kJ/kg·K)
1	Evaporator exit, sat. vapor, $-18.76\ ^\circ$C	0.14 MPa	387.32	1.7402
2	Compressor I exit (actual, $\eta_c=0.90$)	0.5 MPa	416.37	1.7500
3 (g)	Flash-chamber vapor exit, sat. vapor	0.5 MPa	407.47	1.7197
4	Compressor II exit (actual, $\eta_c=0.90$)	1.0 MPa	423.40	1.7247
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; (d) $\dot S_{gen,I}$, $\dot S_{gen,II}$ [kW/K].

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. Evaporator duty and COP follow; part (d) reads the actual (irreversible) entropy rise directly off each compressor's inlet/outlet state.

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, with actual exit entropy $s_2=1.7500$ kJ/kg·K (read off the real-fluid EOS at $P_{mid}$, $h_2$).
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, with actual exit entropy $s_4=1.7247$ kJ/kg·K.
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 $h_g$) 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). $$\dot W_{c,I}=\dot m_{evap}(h_2-h_1)=0.19497\times(416.37-387.32)=5.664\ \text{kW}.$$ $$\dot W_{c,II}=\dot m_{high}(h_4-h_g)=0.25\times(423.40-407.47)=3.981\ \text{kW}.$$ $$\text{COP}=\frac{\dot Q_{evap}}{\dot W_{c,I}+\dot W_{c,II}}=\frac{32.33}{5.664+3.981}=\boxed{3.352}.$$
Entropy increase in both compressors (part d). Since each compressor is adiabatic (no heat exchange with the surroundings), the actual entropy rise across it equals its own entropy generation. Compressor I: $\Delta s_I=s_2-s_1=1.75002-1.74023=0.00980$ kJ/kg·K, at $\dot m_{evap}$. Compressor II: $\Delta s_{II}=s_4-s_g=1.72474-1.71969=0.00505$ kJ/kg·K, at $\dot m_{high}$. $$\dot S_{gen,I}=\dot m_{evap}\,\Delta s_I=0.19497\times0.00980=\boxed{0.00191\ \text{kW/K}}.$$ $$\dot S_{gen,II}=\dot m_{high}\,\Delta s_{II}=0.25\times0.00505=\boxed{0.00126\ \text{kW/K}}.$$ Both are strictly positive, consistent with the second law for an irreversible (90%-efficient) adiabatic compression.

Quantity	Result
(a) $\dot m_{evap}$	0.19497 kg/s
(b) $\dot Q_{evap}$	32.33 kJ/s
(c) COP	3.352
(d) $\dot S_{gen}$: Compressor I / Compressor II	0.00191 / 0.00126 kW/K

Concept & Theory

Topic. Two-stage vapour-compression refrigeration with a flash-chamber economizer, mass flow given on the high (condenser) side, extended with an entropy-generation check on each compressor.

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: $\dot m_{evap}h_2+\dot m_{high}h_6=\dot m_{high}h_g+\dot m_{evap}h_f$. For an adiabatic component, $\dot S_{gen}=\dot m(s_{out}-s_{in})\ge0$.

Why this works. The flash chamber's total mass balance 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. Because each compressor is adiabatic, its entropy generation is simply the mass-flow-weighted increase in specific entropy from inlet to actual (irreversible) outlet — no separate accounting for heat transfer to a reservoir is needed, unlike a non-adiabatic component.

Common pitfall. Reading the evaporator/condenser pressure limits (1 MPa and 0.14 MPa) 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). For part (d), forgetting to use each compressor's own actual mass flow rate ($\dot m_{evap}$ for Compressor I, $\dot m_{high}$ for Compressor II — they differ, since Compressor II only handles the flash-chamber vapor that ultimately supplies the condenser).

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 11 (Refrigeration Cycles) & Ch. 7 (Entropy).

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

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

Approach