Vapour Compression Refrigeration Questions and Answers

Answer: A

Heat is extracted at about 28.3 kJ/s.

In the ideal cycle the refrigerant enters the evaporator at the enthalpy of saturated liquid at 800 kPa (the throttle keeps enthalpy constant) and leaves as saturated vapour at 120 kPa:

\[ \begin{aligned} h_{1} &= h_{g,120} = 237\ \text{kJ/kg} \\ h_{4} &= h_{f,800} = 95.5\ \text{kJ/kg} \\ Q_{L} &= 0.2(237-95.5) = 28.3\ \text{kJ/s} \end{aligned} \]

Using h_f at 120 kPa (22.5) instead of at 800 kPa gives 42.9 kJ/s, which ignores the throttling. 34.4 and 14.6 do not follow from the table.

Answer: C

The compressor needs about 7.9 kW.

Compression is isentropic, so the exit entropy equals the inlet entropy s_g at 120 kPa, 0.95 kJ/kg.K. At 800 kPa this matches the superheated entry, where h = 276.45 kJ/kg:

\[ \begin{aligned} W &= \dot{m}(h_{2}-h_{1}) \\ &= 0.2(276.45-237) = 7.89\ \text{kW} \end{aligned} \]

Taking the exit as saturated vapour at 800 kPa (267.3) gives 6.1 kW, which is not offered and ignores the superheat. 39.5 forgets the mass flow rate of 0.2 kg/s (39.45 kJ/kg is the specific work), and 1.83 and 5.94 do not follow from the table.

Answer: C

The refrigerant mass flow is about 0.813 kg min⁻¹.

The 1 hp motor supplies 0.746 kW. With the compressor 80% efficient, the work delivered to the refrigerant is 0.597 kW, and the COP gives the refrigerating effect:

\[ \begin{aligned} W &= 0.746 \times 0.80 = 0.597\ \text{kW} \\ Q_{e} &= 4.2 \times 0.597 = 2.507\ \text{kW} \\ \dot{m} &= \frac{2.507}{185} = 0.01355\ \text{kg/s} \\ &= 0.813\ \text{kg/min} \end{aligned} \]

Ignoring the compressor efficiency gives about 1.02 kg/min, which is not offered; 1.634 and 1.090 overstate the flow, and 0.240 understates it.

Answer: B

The stroke is about 33.7 mm.

The suction volume flow is \(0.813 \times 0.15 = 0.122\) m³/min. At 1440 rpm, with two single acting cylinders each making one suction stroke per revolution:

\[ \begin{aligned} V_{cyl} &= \frac{0.122}{1440 \times 2} = 4.23 \times 10^{-5}\ \text{m}^{3} \\ A &= \tfrac{\pi}{4}(0.04)^{2} = 1.257 \times 10^{-3}\ \text{m}^{2} \\ L &= V_{cyl}/A = 0.0337\ \text{m} \end{aligned} \]

That is 33.7 mm, taking the volumetric efficiency as 100%. Putting the whole flow through one cylinder doubles the stroke to 67.4 mm, and halving it to 16.8 mm would treat the cylinders as double acting.

Answer: D

The cycle with wet vapour before compression is not a type of vapour compression cycle in the key. In the vapour compression cycle the vapour is compressed in the compressor, and it enters either dry saturated or superheated, and may leave dry saturated or superheated. Compression of wet vapour is avoided because liquid droplets damage the compressor, so it is not a standard cycle, and the other three options are standard types.

Slugging, the entry of liquid into the cylinder, is the damage that wet compression risks, so practical machines are designed to take only dry or superheated vapour.

Answer: D

The refrigerant leaves the compressor as superheated vapor.

In the standard vapor compression cycle, the refrigerant enters the compressor as dry saturated (or slightly superheated) vapor from the evaporator. Compression raises its pressure and, being close to isentropic, raises its temperature above the saturation temperature at the condenser pressure, so it leaves superheated.

Saturated liquid is the state at the condenser exit, before the expansion valve. Wet vapor is the state after the expansion valve, entering the evaporator. Dry saturated vapor is the usual state at the compressor inlet, not the outlet.

Answer: B

The dryness fraction at entry to the evaporator is 0.25.

Number the states: 1 evaporator exit (saturated vapour, \(h_{1} = 250\) kJ/kg), 2 compressor exit (\(h_{2} = 300\) kJ/kg), 3 condenser exit, 4 after throttling. Throttling keeps enthalpy constant, so \(h_{4} = h_{3}\).

\[ \begin{aligned} COP &= \frac{h_{1} - h_{4}}{h_{2} - h_{1}} = \frac{250 - h_{4}}{50} = 3 \\ h_{4} &= 250 - 150 = 100\ \text{kJ/kg} \end{aligned} \]

State 4 is a wet mixture at evaporator pressure, between saturated liquid (\(h_{f} = 50\)) and saturated vapour (\(h_{g} = 250\)):

\[ \begin{aligned} x_{4} &= \frac{h_{4} - h_{f}}{h_{g} - h_{f}} \\ &= \frac{100 - 50}{250 - 50} = 0.25 \end{aligned} \]

0.2 divides by 250 instead of by the latent enthalpy 200 (50/250). 0.3 and 0.35 do not follow from the given enthalpies; any error in the COP step changes \(h_{4}\) and every later figure.