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Obtain relations for the characteristic lengths of a large plane wall of thickness \(2 L\), a very long cylinder of radius \(r_{o}\), and a sphere of radius \(r_{o}\).

Short Answer

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Question: Determine the characteristic lengths for a large plane wall, a very long cylinder, and a sphere in terms of their geometrical parameters. Answer: The characteristic lengths for the given shapes are: - Plane wall: \(2L\) - Long cylinder: \(\frac{r_{o}}{2}\) - Sphere: \(\frac{1}{3} r_{o}\)

Step by step solution

01

Plane Wall

First, we will find the characteristic length for a large plane wall of thickness \(2L\). 1. The volume, \(V\), of the plane wall can be defined by the product of its length (\(2L\)), its width, and its height. 2. The surface area, \(A\), of the plane wall is the product of its width and its height. 3. The characteristic length, \(L_{ch}\), is the volume-to-surface area ratio: \(L_{ch} = \frac{V}{A}\). The thickness of the plane wall is the relevant length to consider. Therefore, the characteristic length of the plane wall is \(2L\).
02

Long Cylinder

Next, we'll find the characteristic length for a very long cylinder of radius \(r_{o}\). 1. The volume, \(V\), of the cylinder is given by \(V = \pi r_{o}^2 L\), where \(L\) is the length of the cylinder. 2. The surface area, \(A\), of the cylinder is given by \(A = 2\pi r_{o} L\). 3. The characteristic length, \(L_{ch}\), is the volume-to-surface area ratio: \(L_{ch} = \frac{V}{A}\). Plugging in the volume and surface area expressions, we get $$ L_{ch} = \frac{\pi r_{o}^2 L}{2\pi r_{o} L} = \frac{r_{o}}{2} $$ The characteristic length of the long cylinder is \(\frac{r_{o}}{2}\).
03

Sphere

Finally, let's find the characteristic length for a sphere of radius \(r_{o}\). 1. The volume, \(V\), of the sphere is given by \(V = \frac{4}{3} \pi r_{o}^3\). 2. The surface area, \(A\), of the sphere is given by \(A = 4\pi r_{o}^2\). 3. The characteristic length, \(L_{ch}\), is the volume-to-surface area ratio: \(L_{ch} = \frac{V}{A}\). Plugging in the volume and surface area expressions, we get $$ L_{ch} = \frac{\frac{4}{3} \pi r_{o}^3}{4\pi r_{o}^2} = \frac{1}{3} r_{o} $$ The characteristic length of the sphere is \(\frac{1}{3} r_{o}\). In summary, the characteristic lengths for the given shapes are as follows: - Plane wall: \(2L\) - Long cylinder: \(\frac{r_{o}}{2}\) - Sphere: \(\frac{1}{3} r_{o}\)

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Key Concepts

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

Plane Wall Heat Conduction
Understanding heat conduction through a plane wall is essential for various engineering applications. Heat conduction can be envisioned as the microscopic transfer of kinetic energy between particles due to a temperature difference.

For a large plane wall with a thickness of \(2L\), the characteristic length is simply the thickness itself. This is because the one-dimensional heat transfer mainly occurs through the wall's thickness. It's important to note that the characteristic length is a measure used to compare the heat transfer abilities of different materials or shapes.

In practical terms, a wall with a larger characteristic length, or thickness, will generally have a lower heat transfer rate given the same thermal conductivity. This is why insulation materials have considerable thickness to reduce heat loss or gain. An improved understanding can be reached by studying the relationship between the temperature gradient, the wall's thermal conductivity, and the resulting heat flux.
Cylinder Thermal Analysis
The analysis of heat transfer in cylindrical shapes is crucial in industries dealing with pipes and tubes, such as in HVAC systems or chemical processing.

In a very long cylinder, the characteristic length, based on the volume-to-surface area ratio, gives us the \(\frac{r_{o}}{2}\) value. Because heat transfer in a cylinder occurs radially, the radius affects how efficiently the heat will be conducted outwards or inwards. This characteristic length is crucial for determining how the cylindrical shape will influence the thermal resistance and heat flow rate.

For cylinders, especially those in a steady-state heat transfer scenario, the radial temperature distribution is also a factor to consider. The longer the cylinder, the closer the heat transfer process gets to a steady state, making the characteristic length a worthwhile measure for predicting thermal behavior over length.
Sphere Heat Transfer Properties
Spherical objects have distinct heat transfer characteristics due to their geometry. The process of heat transfer in a sphere involves a three-dimensional temperature field.

The characteristic length derived for a sphere with radius \(r_{o}\) is \(\frac{1}{3} r_{o}\). This indicates how the radius of the sphere plays a critical role in its heat transfer properties. Unlike plane walls or cylinders, the spherical shape has no edges or corners, which allows for a uniform heat transfer in all directions from the center.

While studying the heat transfer properties of a sphere, it's likewise important to understand how the changing surface area to volume ratio, as the sphere heats up or cools down, affects the rate of heat transfer. In summary, a sphere's small characteristic length implies a relatively fast response to a change in temperature compared to larger characteristic lengths of other shapes.

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Most popular questions from this chapter

A 5-cm-high rectangular ice block \((k=2.22 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=0.124 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at \(-20^{\circ} \mathrm{C}\) is placed on a table on its square base \(4 \mathrm{~cm} \times 4 \mathrm{~cm}\) in size in a room at \(18^{\circ} \mathrm{C}\). The heat transfer coefficient on the exposed surfaces of the ice block is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Disregarding any heat transfer from the base to the table, determine how long it will be before the ice block starts melting. Where on the ice block will the first liquid droplets appear?

Plasma spraying is a process used for coating a material surface with a protective layer to prevent the material from degradation. In a plasma spraying process, the protective layer in powder form is injected into a plasma jet. The powder is then heated to molten droplets and propelled onto the material surface. Once deposited on the material surface, the molten droplets solidify and form a layer of protective coating. Consider a plasma spraying process using alumina \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) powder that is injected into a plasma jet at \(T_{\infty}=15,000^{\circ} \mathrm{C}\) and \(h=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The alumina powder is made of particles that are spherical in shape with an average diameter of \(60 \mu \mathrm{m}\) and a melting point at \(2300^{\circ} \mathrm{C}\). Determine the amount of time it would take for the particles, with an initial temperature of \(20^{\circ} \mathrm{C}\), to reach their melting point from the moment they are injected into the plasma jet.

Chickens with an average mass of \(2.2 \mathrm{~kg}\) and average specific heat of \(3.54 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\) are to be cooled by chilled water that enters a continuous-flow-type immersion chiller at \(0.5^{\circ} \mathrm{C}\). Chickens are dropped into the chiller at a uniform temperature of \(15^{\circ} \mathrm{C}\) at a rate of 500 chickens per hour and are cooled to an average temperature of \(3^{\circ} \mathrm{C}\) before they are taken out. The chiller gains heat from the surroundings at a rate of \(210 \mathrm{~kJ} / \mathrm{min}\). Determine \((a)\) the rate of heat removal from the chicken, in \(\mathrm{kW}\), and \((b)\) the mass flow rate of water, in \(\mathrm{kg} / \mathrm{s}\), if the temperature rise of water is not to exceed \(2^{\circ} \mathrm{C}\).

A body at an initial temperature of \(T_{i}\) is brought into a medium at a constant temperature of \(T_{\infty}\). How can you determine the maximum possible amount of heat transfer between the body and the surrounding medium?

Refractory bricks are used as linings for furnaces, and they generally have low thermal conductivity to minimize heat loss through the furnace walls. Consider a thick furnace wall lining with refractory bricks \(\left(k=1.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=5.08 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\), where initially the wall has a uniform temperature of \(15^{\circ} \mathrm{C}\). If the wall surface is subjected to uniform heat flux of \(20 \mathrm{~kW} / \mathrm{m}^{2}\), determine the temperature at the depth of \(10 \mathrm{~cm}\) from the surface after an hour of heating time.

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