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Chapter 4: Problem 133

The chilling room of a meat plant is \(15 \mathrm{~m} \times 18 \mathrm{~m} \times\) \(5.5 \mathrm{~m}\) in size and has a capacity of 350 beef carcasses. The power consumed by the fans and the lights in the chilling room are 22 and \(2 \mathrm{~kW}\), respectively, and the room gains heat through its envelope at a rate of \(14 \mathrm{~kW}\). The average mass of beef carcasses is \(220 \mathrm{~kg}\). The carcasses enter the chilling room at \(35^{\circ} \mathrm{C}\), after they are washed to facilitate evaporative cooling, and are cooled to \(16^{\circ} \mathrm{C}\) in \(12 \mathrm{~h}\). The air enters the chilling room at \(-2.2^{\circ} \mathrm{C}\) and leaves at \(0.5^{\circ} \mathrm{C}\). Determine \((a)\) the refrigeration load of the chilling room and \((b)\) the volume flow rate of air. The average specific heats of beef carcasses and air are \(3.14\) and \(1.0 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the density of air can be taken to be \(1.28 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
Based on the given information, the refrigeration load (a) of the chilling room at a meat plant is 427,100 kJ/h, and the volume flow rate of air (b) is 68,235 m³/h.

Step by step solution

01

Calculate the heat removed from carcasses

First, we need to calculate the total heat removed from the 350 carcasses that enter the room. We can use the formula: \(Q_{carcasses} = m \cdot c_p \cdot (T_{initial} - T_{final})\) where \(Q_{carcasses}\) is the heat removed from carcasses, \(m\) is the total mass of carcasses, \(c_p\) is the specific heat of carcasses, and \(T_{initial}\) and \(T_{final}\) are the initial and final temperatures of carcasses, respectively. We know the mass of each carcass is 220 kg and there are 350 carcasses, so \(m = 220 \mathrm{~kg} \times 350 = 77,000 \mathrm{~kg}\). The specific heat, \(c_p\), is 3.14 kJ/(kg°C), while the initial and final temperatures are 35°C and 16°C, respectively. Therefore, the heat removed from the carcasses can be calculated as follows: \(Q_{carcasses} = 77,000 \mathrm{~kg} \times 3.14 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot{(^{\circ}C)}} \times (35^{\circ} C - 16^{\circ} C) = 4,669,200 \mathrm{~kJ}\)
02

Calculate the heat loss over time

The heat removed from the carcasses occurs over 12 hours, so we need to determine the heat loss per unit of time: \(Q_{loss} = \frac{Q_{carcasses}}{t}\) where \(Q_{loss}\) is the heat loss per unit of time and \(t\) is the cooling time in hours. We know the cooling time is 12 hours, so we can find the heat loss over time: \(Q_{loss} = \frac{4,669,200 \mathrm{~kJ}}{12 \mathrm{~h}} = 389,100 \frac{\mathrm{kJ}}{\mathrm{h}}\)
03

Determine the total heat load and refrigeration load

Now we can determine the total heat load, which includes the heat loss from carcasses, heat added by fans and lights, and heat gained through the envelope. The total heat load is given by: \(Q_{total} = Q_{loss} + P_{fans} + P_{lights} + P_{envelope}\) where \(P_{fans}\), \(P_{lights}\), and \(P_{envelope}\) are the power consumed by fans, lights, and heat gained through the envelope, respectively. Plugging in the values, we get: \(Q_{total} = 389,100 \frac{kJ}{h} + 22,000 \frac{kJ}{h} + 2,000 \frac{kJ}{h} + 14,000 \frac{kJ}{h} = 427,100 \frac{kJ}{h}\) The refrigeration load \((a)\) is thus 427,100 kJ/h.
04

Calculate the mass flow rate of air

Next, we need to find the mass flow rate of air, which can be obtained using the equation: \(\dot{m}_{air} = \frac{Q_{loss}}{c_{p_{air}} \cdot (T_{out} - T_{in})}\) where \(\dot{m}_{air}\) is the mass flow rate of air, \(c_{p_{air}}\) is the specific heat of air, and \(T_{out}\) and \(T_{in}\) are the temperatures of air leaving and entering the room, respectively. We know the specific heat of air is 1.0 kJ/(kg°C), and the temperatures of the air entering and leaving the chilling room are -2.2°C and 0.5°C, respectively. Using these values, we can find the mass flow rate of air: \(\dot{m}_{air} = \frac{389,100 \frac{kJ}{h}}{1.0 \frac{kJ}{kg \cdot{(^{\circ} C)}} \times (0.5^{\circ}C - (-2.2^{\circ} C))} = 87,342 \frac{kg}{h}\)
05

Calculate the volume flow rate of air

Finally, we can find the volume flow rate of air \((b)\) using the formula: \(\dot{V}_{air} = \frac{\dot{m}_{air}}{\rho_{air}}\) where \(\dot{V}_{air}\) is the volume flow rate of air and \(\rho_{air}\) is the density of air. We are given the density of air as 1.28 kg/m³. Plugging in the values, we obtain: \(\dot{V}_{air} = \frac{87,342 \frac{kg}{h}}{1.28 \frac{kg}{m^3}} = 68,235 \frac{m^3}{h}\) The volume flow rate of air \((b)\) is 68,235 m³/h.

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