Question

The cargo space of a refrigerated truck whose inner dimensions are \(12 \mathrm{m} \times 2.3 \mathrm{m} \times 3.5 \mathrm{m}\) is to be precooled from \(25^{\circ} \mathrm{C}\) to an average temperature of \(5^{\circ} \mathrm{C}\). The construction of the truck is such that a transmission heat gain occurs at a rate of \(120 \mathrm{W} /^{\circ} \mathrm{C}\). If the ambient temperature is \(25^{\circ} \mathrm{C}\) determine how long it will take for a system with a refrigeration capacity of \(11 \mathrm{kW}\) to precool this truck.

Short answer

Answer: It will take approximately 27.07 hours.

Step-by-step solution

Calculate the temperature difference required to precool the truck

First, we need to determine how much the temperature needs to be decreased. This is simply the difference between the initial temperature (\(25^{\circ} \mathrm{C}\)) and the final desired temperature (\(5^{\circ} \mathrm{C}\)): \(\Delta T = T_{initial} - T_{final} = 25^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C} = 20^{\circ} \mathrm{C}\)

Calculate the volume of the inner space

The volume of the cargo space can be calculated using the dimensions provided: \(Volume = Length \times Width \times Height = 12 \mathrm{m} \times 2.3 \mathrm{m} \times 3.5 \mathrm{m} = 96.6 \mathrm{m^3}\)

Calculate the heat transfer required to decrease the temperature

The total heat transfer (Q) required to decrease the temperature is given by the temperature difference and the transmission heat gain: \(Q = \Delta T \times transmission\ heat\ gain = 20^{\circ} \mathrm{C} \times 120 \frac{\mathrm{W}} {^{\circ} \mathrm{C}} = 2400 \mathrm{W}\)

Calculate the net cooling capacity

Since the system has a refrigeration capacity of \(11 \mathrm{kW}\), and there is a heat gain of \(2400 \mathrm{W}\) due to the truck's construction, we can calculate the net cooling capacity as follows: \(Net\ cooling\ capacity = Refrigeration\ capacity - heat\ gain = 11000 \mathrm{W} - 2400 \mathrm{W} = 8600 \mathrm{W}\)

Calculate the time required to precool the truck

To calculate the time (t) required to precool the truck, we can use the formula: \(t = \frac{Q \times Volume}{Net\ cooling\ capacity} = \frac{2400 \mathrm{W} \times 96.6 \mathrm{m^3}}{8600 \mathrm{W}} \approx 27.067 \mathrm{h}\) It will take approximately 27.07 hours to precool the truck from \(25^{\circ} \mathrm{C}\) to \(5^{\circ} \mathrm{C}\) using the given refrigeration system.