Question 4 of 8: R-134a refrigeration plant — COP, mass flow and compressor size

Question 4: R-134a refrigeration plant — COP, mass flow and compressor size (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.

Find. The system diagram and the cycle on the attached p–h chart, the coefficient of performance, the refrigerant mass flow, and the bore and stroke.

Approach. Read the four state points off the attached R-134a chart, then form the refrigerating effect and compressor work per pound to get the COP; divide the capacity by the refrigerating effect for the mass flow; convert that to a suction volume with the specific volume at state 1, inflate it by the volumetric efficiency to a swept volume, and solve the cylinder geometry.

1. Fix the two pressures. Saturation pressures for R-134a are 33.1 psia at $20\,{}^{\circ}\text{F}$ and 119.0 psia at $90\,{}^{\circ}\text{F}$, a pressure ratio of 3.60. These locate the two horizontal lines of the cycle on the chart.
2. State 1 — compressor suction. Superheated $9\,{}^{\circ}\text{F}$ at the evaporating pressure, so $29\,{}^{\circ}\text{F}$ and 33.1 psia: $h_1 = 107.85$ Btu/lb, $s_1 = 0.2273$ Btu/lb °R, and the specific volume that will size the compressor, $v_1 = 1.445$ ft$^3$/lb.
3. State 2 — compressor discharge. Following the constant-entropy line from state 1 up to 119.0 psia gives $h_2 = 119.57$ Btu/lb at a discharge temperature of $107.7\,{}^{\circ}\text{F}$, comfortably below any R-134a discharge limit.
4. States 3 and 4 — condenser exit and valve exit. Subcooling $9\,{}^{\circ}\text{F}$ below $90\,{}^{\circ}\text{F}$ puts state 3 at $81\,{}^{\circ}\text{F}$ in the liquid region, where $h_3 = 38.51$ Btu/lb. The expansion valve is a throttle, so it is isenthalpic and $h_4 = h_3 = 38.51$ Btu/lb. At 33.1 psia the saturation enthalpies are $h_f = 18.47$ and $h_g = 105.99$ Btu/lb, so the valve delivers a wet mixture of quality $x_4 = (38.51-18.47)/(105.99-18.47) = 0.229$.
5. Part (a) — coefficient of performance. The refrigerating effect is the enthalpy the refrigerant picks up in the evaporator and the work is the rise across the compressor: $$q_{ref}=h_1-h_4=107.85-38.51=69.33\ \text{Btu/lb},\qquad w_c=h_2-h_1=119.57-107.85=11.72\ \text{Btu/lb}$$ $$\boxed{\;\text{COP}=\frac{q_{ref}}{w_c}=\frac{69.33}{11.72}=5.91\;}$$ The Carnot bound between the same two temperatures is $T_e/(T_c-T_e) = 479.67/(549.67-479.67) = 6.85$, so this cycle achieves 86% of Carnot — high, as it should be for a modest 3.6 pressure ratio with isentropic compression assumed.
6. Part (b) — refrigerant mass flow. Dividing the required capacity by the refrigerating effect per pound, $$\dot m=\frac{144\,000}{69.33}=\boxed{\;2077\ \text{lb}\,\text{h}^{-1}=34.6\ \text{lb}\,\text{min}^{-1}\;}$$ The compressor power follows as $\dot m w_c = 2077\times 11.72 = 24\,340$ Btu/hr, that is 9.57 hp, and the condenser must reject $\dot m (h_2-h_3) = 2077\times 81.06 = 168\,400$ Btu/hr — which is the capacity plus the work, as it must be.
7. Convert mass flow to a swept volume. The compressor ingests actual vapour at state 1, so the volume it must handle at suction is $$\dot V_{suc}=\dot m v_1 = 34.6\times 1.445 = 50.0\ \text{ft}^3\text{min}^{-1}$$ and the volumetric efficiency relates that to the geometric displacement: $$\dot V_{disp}=\frac{\dot V_{suc}}{\eta_v}=\frac{50.0}{0.70}=71.44\ \text{ft}^3\text{min}^{-1}$$
8. Part (c) — bore and stroke. Two cylinders, each of bore $D$ and stroke $L = 1.5D$, turning at $N = 900$ rev/min sweep $$\dot V_{disp}=n\,\frac{\pi D^2}{4}\,L\,N = 2\cdot\frac{\pi D^2}{4}\cdot 1.5D\cdot 900 = 2120.6\,D^3$$ with $D$ in feet. Solving for $D$, $$D^3=\frac{71.44}{2120.6}=0.033674\ \text{ft}^3 \quad\Rightarrow\quad D = 0.3232\ \text{ft}$$ $$\boxed{\;\text{bore }D = 3.88\ \text{in},\qquad \text{stroke }L = 1.5D = 5.81\ \text{in}\;}$$ Substituting back, $2(\pi/4)(0.3232)^2(0.4848)(900) = 71.44$ ft$^3$/min, so the geometry reproduces the required displacement exactly. A piston speed of $2LN/60 = 14.5$ ft/s is normal for a machine of this size and rating.