Refrigeration Fundamentals Questions and Answers

Answer: B

The COP is 1.5. The coefficient of performance of a refrigerator is the refrigeration effect divided by the work input: \[ COP = \frac{120}{80} = 1.5 \] The value 1.33 would result from dividing 80 by 60 and 2.0 and 3.0 do not follow from dividing 120 by 80, so only 1.5 is right.

Answer: B

Thermoelectric refrigeration produces low temperature by the reverse of the Seebeck effect, which is the Peltier effect: a current passed through a junction of two dissimilar conductors absorbs heat at one junction. Magnetic refrigeration works on the magnetocaloric effect, liquid gas refrigeration on the evaporation of a liquefied gas, and dry ice refrigeration on the sublimation of solid carbon dioxide, so only the thermoelectric method matches the effect named.

Answer: A

The heat rejected is about 84.40 kW.

A reversed Carnot cycle must use absolute temperatures: \(T_{L} = 243\) K and \(T_{H} = 293\) K.

\[ \begin{aligned} COP &= \frac{T_{L}}{T_{H} - T_{L}} = \frac{243}{50} = 4.86 \\ W &= \frac{70}{4.86} = 14.40\ \text{kW} \\ Q_{H} &= Q_{L} + W = 84.40\ \text{kW} \end{aligned} \]

14.41 kW is the compressor work, not the heat rejected. 116.67 kW is what the work comes to if the temperatures are left in degrees Celsius (COP = 30/50 = 0.6), and 186.67 kW is 70 plus that wrong work. Heat rejected must always be the refrigeration load plus the work input.

Answer: B

The COP is 10.

For a reversible heat pump, with temperatures in kelvin, \(T_{H} = 37 + 273 = 310\) K and \(T_{L} = 6 + 273 = 279\) K:

\[ \begin{aligned} COP_{HP} &= \frac{T_{H}}{T_{H} - T_{L}} \\ &= \frac{310}{31} = 10 \end{aligned} \]

Using the refrigerator formula \(T_{L}/(T_{H} - T_{L}) = 279/31 = 9\) gives one less, because \(COP_{HP} = COP_{R} + 1\). Using degrees Celsius instead of kelvin gives nonsense values. 8, 12 and 14 do not come from the correct temperatures.

Answer: C

The COP of the heat pump is 4.0.

A heat pump's COP is the heat delivered to the hot side divided by the work input. The work is the difference between heat rejected and heat added:

\[ \begin{aligned} W &= 1000 - 750 = 250\ \text{W} \\ COP_{HP} &= \frac{1000}{250} = 4.0 \end{aligned} \]

7.5 is the Carnot COP, \(300/40\), for these temperatures, an upper limit rather than the actual value. 3.0 is \(750/250\), the COP if the machine were judged as a refrigerator. 6.5 does not follow from the data.

Answer: A

The COP is 6.7.

For a reversed Carnot refrigerator, with \(T_{L} = -5 + 273 = 268\) K and \(T_{H} = 35 + 273 = 308\) K:

\[ \begin{aligned} COP &= \frac{T_{L}}{T_{H} - T_{L}} \\ &= \frac{268}{40} = 6.7 \end{aligned} \]

The 2.5 kJ/s heat load is not needed for the COP; it would be used to find the power input, \(2.5/6.7 = 0.37\) kW. The nearby values 6.6, 6.5 and 6.4 come from rounding or from using 273.15 inconsistently.

Answer: C

One ton of refrigeration is the heat needed to melt 1 ton of ice at 0°C in 24 hours.

The unit comes from the days when cooling was done with ice. Melting one short ton (2,000 lb) of ice absorbs 2,000 times its latent heat of 144 Btu/lb, spread over a day:

\[ \begin{aligned} \text{TR} &= \frac{2000 \times 144}{24} \\ &= 12{,}000\ \text{Btu/h} \approx 3.517\ \text{kW} \end{aligned} \]

Taking 1 hour instead would make the unit 24 times larger, 12 hours twice as large, and 1 minute 1,440 times as large. The day-long rate is what defines the ton.