Question 1 of 9: Two-Stage Compression Refrigeration with Flash Chamber (R-134a)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-10, Thermodynamics — May 2013. 3 hours, Closed-Book Exam (approved
calculator and one double-sided 8.5x11-inch aid sheet permitted). Part A: answer 2 of Questions
1-3 (20 marks each). Part B: answer 4 of Questions 4-9 (15 marks each), for a 100-mark paper.
Only the first two Part-A and first four Part-B questions as they appear in the answer book are
marked. All nine questions (Part A complete, Part B complete) are solved below for
completeness.
Reference texts: Cengel & Boles, Thermodynamics: An Engineering
Approach, 8th ed. All state properties (R-134a, water/steam, air, N₂, CO₂) were
computed from high-accuracy equations of state in
place of printed property-table interpolation; every boxed numeric result.
Given. R-134a two-stage compression system with flash-tank (economizer) intercooling between the low-pressure (LP) and high-pressure (HP) compressors. Evaporator pressure $P_1=0.14$ MPa; flash-chamber pressure $P_{flash}=0.5$ MPa; condenser pressure $P_{cond}=1.0$ MPa. Condenser exit is saturated liquid; evaporator exit is saturated vapor. Both compressors have isentropic efficiency $\eta_c=90\%$. Mass flow through the condenser (and HP compressor) $\dot m_{high}=0.25$ kg/s. Dead-state temperature for exergy $T_0=310$ K.
State
Description
P (MPa)
1
Evaporator exit, sat. vapor
0.14
2
LP compressor exit (actual)
0.5
3
Flash-chamber vapor (sat.), HP compressor inlet
0.5
4
HP compressor exit (actual), condenser inlet
1.0
5
Condenser exit, sat. liquid
1.0
6
After throttle 5→flash chamber
0.5
7
Flash-chamber liquid (sat.)
0.5
8
After throttle 7→evaporator
0.14
Find. (a) $\dot m_{evap}$; (b) $\dot Q_L$; (c) COP; (d) exergy destruction in the two compressors.
Fig. Q1 — T–s state points for the two-stage R-134a cycle with flash-chamber intercooling (dashed segments 5→6 and 7→8 are irreversible isenthalpic throttling, not drawn as cycle lines).
Approach
Fix the eight state points from the given pressures and saturation/throttling conditions, then close the flash chamber with a combined mass-and-energy balance (its outlet vapor fraction is fixed by the requirement that the vapor leaving equals $\dot m_{high}$ and the liquid leaving equals $\dot m_{evap}$); heat removed and COP follow from the evaporator and total compressor work, and exergy destruction in each compressor follows from $T_0$ times its entropy generation.
Fix states 1 and 3 (saturation lines). State 1 is saturated vapor at 0.14 MPa: $T_1=-18.76\ ^\circ\text{C}$, $h_1=387.32$ kJ/kg, $s_1=1.7402$ kJ/kg·K. State 3 is saturated vapor at the flash pressure (0.5 MPa, $T_3=15.73\ ^\circ\text{C}$): $h_3=407.47$ kJ/kg, $s_3=1.7197$ kJ/kg·K.
LP compressor (1→2), actual, $\eta_c=90\%$. Isentropic exit enthalpy at 0.5 MPa with $s_{2s}=s_1$ gives $h_{2s}=413.47$ kJ/kg. Actual work is larger by the efficiency:
$$h_2=h_1+\frac{h_{2s}-h_1}{\eta_c}=387.32+\frac{413.47-387.32}{0.90}=\boxed{416.37\ \text{kJ/kg}}\quad(T_2=24.97\ ^\circ\text{C}),\ s_2=1.7500\ \text{kJ/kg}\cdot\text{K}.$$
HP compressor (3→4), actual, $\eta_c=90\%$. Same procedure from state 3 to 1.0 MPa: $h_{4s}=421.80$ kJ/kg, so
$$h_4=h_3+\frac{h_{4s}-h_3}{\eta_c}=\boxed{423.40\ \text{kJ/kg}}\quad(T_4=43.15\ ^\circ\text{C}),\ s_4=1.7247\ \text{kJ/kg}\cdot\text{K}.$$
Condenser exit and throttling into the flash chamber (5→6). State 5 is saturated liquid at 1.0 MPa, $h_5=255.50$ kJ/kg. Throttling is isenthalpic ($h_6=h_5$), so at 0.5 MPa the quality of stream 6 is
$$x_6=\frac{h_6-h_{f,\,0.5\,\text{MPa}}}{h_{fg,\,0.5\,\text{MPa}}}=\frac{255.50-221.50}{407.47-221.50}=0.1828.$$
Flash-chamber mass + energy balance ⇒ $\dot m_{evap}$. The tank separates the combined inflow (stream 6 at $\dot m_{high}$, quality $x_6$, and stream 2 at $\dot m_{evap}$) into saturated vapor (state 3, which must equal $\dot m_{high}$ since it feeds the HP compressor/condenser loop) and saturated liquid (state 7, which must equal $\dot m_{evap}$ since it returns through the evaporator loop). Substituting these mass constraints into the energy balance $\dot m_{high}h_6+\dot m_{evap}h_2=\dot m_{high}h_g+\dot m_{evap}h_f$ (both at 0.5 MPa) and solving for $\dot m_{evap}$:
$$\dot m_{evap}=\dot m_{high}\frac{h_g-h_6}{h_2-h_f}=0.25\times\frac{407.47-255.50}{416.37-221.50}=\boxed{0.1950\ \text{kg/s}}.$$
Liquid leaving the flash chamber and throttling into the evaporator (7→8). $h_7=h_{f,\,0.5\,\text{MPa}}=221.50$ kJ/kg; throttling to 0.14 MPa is isenthalpic, $h_8=h_7=221.50$ kJ/kg (quality $x_8=0.2181$).
Heat removed from the refrigerated space (evaporator, part b).
$$\dot Q_L=\dot m_{evap}(h_1-h_8)=0.1950\times(387.32-221.50)=\boxed{32.33\ \text{kJ/s}}.$$
Total compressor work and COP (part c).
$$\dot W_{LP}=\dot m_{evap}(h_2-h_1)=5.664\ \text{kW},\qquad \dot W_{HP}=\dot m_{high}(h_4-h_3)=3.981\ \text{kW}.$$
$$\text{COP}=\frac{\dot Q_L}{\dot W_{LP}+\dot W_{HP}}=\frac{32.33}{9.645}=\boxed{3.35}.$$
Exergy destruction in the compressors (part d). Both compressors are adiabatic, so exergy destruction is $T_0$ times the entropy generated in each:
$$\dot X_{dest,LP}=T_0\,\dot m_{evap}(s_2-s_1)=310\times0.1950\times(1.7500-1.7402)=0.592\ \text{kW}.$$
$$\dot X_{dest,HP}=T_0\,\dot m_{high}(s_4-s_3)=310\times0.25\times(1.7247-1.7197)=0.391\ \text{kW}.$$
$$\dot X_{dest,total}=\boxed{0.983\ \text{kW}}.$$
