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

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

A thermocouple, with a spherical junction diameter of \(0.5 \mathrm{~mm}\), is used for measuring the temperature of hot air flow in a circular duct. The convection heat transfer coefficient of the air flow can be related with the diameter \((D)\) of the duct and the average air flow velocity \((V)\) as \(h=2.2(V / D)^{0.5}\), where \(D, h\), and \(V\) are in \(\mathrm{m}, \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{m} / \mathrm{s}\), respectively. The properties of the thermocouple junction are \(k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8500 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=320 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Determine the minimum air flow velocity that the thermocouple can be used, if the maximum response time of the thermocouple to register 99 percent of the initial temperature difference is \(5 \mathrm{~s}\).

The Biot number can be thought of as the ratio of (a) The conduction thermal resistance to the convective thermal resistance. (b) The convective thermal resistance to the conduction thermal resistance. (c) The thermal energy storage capacity to the conduction thermal resistance. (d) The thermal energy storage capacity to the convection thermal resistance. (e) None of the above.

The walls of a furnace are made of \(1.2\)-ft-thick concrete \(\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)\). Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at \(1800^{\circ} \mathrm{F}\) with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(116 \mathrm{~min}\)

Thick slabs of stainless steel \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and copper \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=117 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) are subjected to uniform heat flux of \(8 \mathrm{~kW} / \mathrm{m}^{2}\) at the surface. The two slabs have a uniform initial temperature of \(20^{\circ} \mathrm{C}\). Determine the temperatures of both slabs, at \(1 \mathrm{~cm}\) from the surface, after \(60 \mathrm{~s}\) of exposure to the heat flux.

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