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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 \(1.28 \mathrm{~kg} / \mathrm{m}^{3}\).

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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer

Heat transfer is the movement of thermal energy from one object to another. In refrigeration systems, this process is crucial as it helps remove unwanted heat. For the chilling room in the meat plant, heat is transferred away from the beef carcasses to cool them down from 35°C to 16°C.
In this case, the formula for calculating the heat removed is:

• \( Q_{carcasses} = m \cdot c_p \cdot (T_{initial} - T_{final}) \)

Here:

• \( m \) is total mass (77,000 kg for all carcasses)
• \( c_p \) is the specific heat of beef (3.14 kJ/kg°C)
• \( T_{initial} \) and \( T_{final} \) are initial and final temperatures

Understandably, total heat removal was calculated as 4,669,200 kJ. It's an essential piece in determining the refrigeration load of the system.

Thermodynamics

Thermodynamics is the science that deals with heat and temperature and their relation to energy and work. In cooling systems, thermodynamics principles help in understanding the energy exchanges that must occur to achieve desired temperatures.
This chilling room scenario uses these principles to calculate how much energy must be removed to reach a lower temperature efficiently over a set period.
By quantifying the heat exchange between the carcasses and their surroundings, we apply thermodynamic laws to solve practical problems in refrigeration.

Cooling Systems

Cooling systems like the chilling room work by removing heat to lower the temperature of an object or space. In our example, various elements contribute to the refrigeration load:

• Heat from carcasses
• Power consumed by fans and lights
• Heat gained through the room's envelope

The total heat load was calculated with the formula:

• \( Q_{total} = Q_{loss} + P_{fans} + P_{lights} + P_{envelope} \)

This showed the entire system's demands, resulting in a refrigeration load of 427,100 kJ/h. By understanding these components, one can design efficient systems that meet the cooling needs effectively.

Mass Flow Rate

Mass flow rate refers to the amount of mass moving through a given space per unit time, measured here for air within the chilling room. This allows for the determination of how much air is needed to remove the desired amount of heat.
Using the formula:

• \( \dot{m}_{air} = \frac{Q_{loss}}{c_{p_{air}} \cdot (T_{out} - T_{in})} \)

we calculated the mass flow rate of air to be 87,342 kg/h. Here,

• \( c_{p_{air} } \) is air's specific heat (1.0 kJ/kg°C)
• \( T_{out} \) and \( T_{in} \) are air's exit and entrance temperatures.

Converting to volume flow rate using the air's density (1.28 kg/m³), we find it's 68,235 m³/h. Proper airflow calculations ensure efficient cooling operations.