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Consider the freezing of packaged meat in boxes with refrigerated air. How do \((a)\) the temperature of air, \((b)\) the velocity of air, \((c)\) the capacity of the refrigeration system, and \((d)\) the size of the meat boxes affect the freezing time?

Short Answer

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Answer: The freezing time of packaged meat is affected by the following factors: 1. Air temperature: Lower air temperatures result in faster heat transfer and faster freezing times, while higher air temperatures result in slower heat transfer and increased freezing times. 2. Air velocity: Higher air velocities increase the rate of heat transfer, leading to faster freezing times. Lower air velocities result in slower heat transfer and increased freezing times. 3. Refrigeration system capacity: A system with a higher capacity can remove heat more efficiently, reducing freezing time. A system with a lower capacity takes longer to reach and maintain the desired air temperature, increasing freezing time. 4. Meat box size: Larger meat boxes take longer to freeze due to their greater mass and larger surface area, while smaller meat boxes freeze more quickly due to their lower mass and smaller surface area.

Step by step solution

01

Effect of Air Temperature

The temperature of the refrigerated air is a critical factor in determining freezing time. Lower air temperatures will result in faster heat transfer from the meat, thus reducing the time it takes for the meat to reach the freezing point. Conversely, higher air temperatures will result in slower heat transfer and increase the freezing time.
02

Effect of Air Velocity

The velocity of the air flowing around the meat boxes also plays a significant role in the freezing process. Higher air velocities will result in an increased rate of heat transfer due to the reduction in the thickness of the thermal boundary layer around the meat. This will lead to faster freezing times. On the other hand, lower air velocities will result in slower heat transfer and increased freezing times.
03

Effect of Refrigeration System Capacity

The capacity of the refrigeration system will also impact the freezing time. A refrigeration system with a greater capacity can remove heat from the air more efficiently, which will reduce the time taken in reaching the desired air temperature and maintaining it, resulting in faster freezing times. Conversely, a refrigeration system with a lower capacity will take longer to reach and maintain the desired air temperature, increasing the freezing time.
04

Effect of Meat Box Size

The size of the meat boxes affects the freezing time because larger meat boxes have a greater mass and take longer to freeze. The larger the meat box, the more heat that needs to be extracted from the meat before it reaches the freezing point. It also has a larger surface area, which may affect the rate of heat transfer. Smaller meat boxes will freeze more quickly due to their lower mass and smaller surface area.

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Most popular questions from this chapter

Layers of \(23-\mathrm{cm}\)-thick meat slabs $(k=0.47 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and \)\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\( ) initially at a uniform temperature of \)7^{\circ} \mathrm{C}$ are to be frozen by refrigerated air at \(-30^{\circ} \mathrm{C}\) flowing at a velocity of \(1.4 \mathrm{~m} / \mathrm{s}\). The average heat transfer coefficient between the meat and the air is $20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming the size of the meat slabs to be large relative to their thickness, determine how long it will take for the center temperature of the slabs to drop to \(-18^{\circ} \mathrm{C}\). Also, determine the surface temperature of the meat slab at that time.

A long 35-cm-diameter cylindrical shaft made of stainless steel $304\left(k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=477 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\(, and \)\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\( ) comes out of an oven at a uniform temperature of \)500^{\circ} \mathrm{C}$. The shaft is then allowed to cool slowly in a chamber at \(150^{\circ} \mathrm{C}\) with an average convection heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the shaft 20 min after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. Solve this problem using the analytical one-term approximation method. Answers: \(486^{\circ} \mathrm{C}, 22,270 \mathrm{~kJ}\)

Stainless steel ball bearings $\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, \)k=15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, \quad c_{p}=0.480 \mathrm{KJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, \quad\( and \)\quad \alpha=3.91 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}\( ) having a diameter of \)1.2 \mathrm{~cm}$ are to be quenched in water. The balls leave the oven at a uniform temperature of $900^{\circ} \mathrm{C}\( and are exposed to air at \)30^{\circ} \mathrm{C}$ for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is $125 \mathrm{~W} / \mathrm{m}^{2},{ }^{\circ} \mathrm{C}$, determine how long they can stand in the air before being dropped into the water.

A large chunk of tissue at \(35^{\circ} \mathrm{C}\) with a thermal diffusivity of \(1 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) is dropped into iced water. The water is well-stirred so that the surface temperature of the tissue drops to \(0^{\circ} \mathrm{C}\) at time zero and remains at \(0^{\circ} \mathrm{C}\) at all times. The temperature of the tissue after 4 min at a depth of $1 \mathrm{~cm}$ is (a) \(5^{\circ} \mathrm{C}\) (b) \(30^{\circ} \mathrm{C}\) (c) \(25^{\circ} \mathrm{C}\) (d) \(20^{\circ} \mathrm{C}\) (e) \(10^{\circ} \mathrm{C}\)

When water, as in a pond or lake, is heated by warm air above it, it remains stable, does not move, and forms a warm layer of water on top of a cold layer. Consider a deep lake $\left(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)$ that is initially at a uniform temperature of \(2^{\circ} \mathrm{C}\) and has its surface temperature suddenly increased to \(20^{\circ} \mathrm{C}\) by a spring weather front. The temperature of the water \(1 \mathrm{~m}\) below the surface \(400 \mathrm{~h}\) after this change is (a) \(2.1^{\circ} \mathrm{C}\) (b) \(4.2^{\circ} \mathrm{C}\) (c) \(6.3^{\circ} \mathrm{C}\) (d) \(8.4^{\circ} \mathrm{C}\) (e) \(10.2^{\circ} \mathrm{C}\)

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