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Chapter 4: Problem 12

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

Expert verified
Answer: The characteristic lengths (δ) for the given geometries are as follows: 1. Large plane wall (thickness 2L): δ = L 2. Very long cylinder (radius r_o): δ = r_o 3. Sphere (radius r_o): δ = r_o

Step by step solution

01

Step 1. Large Plane Wall: Determine Possible Length Scale and Find Relation

First, we determine relevant lengths for a large plane wall's heat transfer direction. Since only the thickness is given, the characteristic length will be proportional to the thickness of the wall. The characteristic length of a plane wall is defined as half of the wall's thickness, so for a wall of thickness \(2L\), the characteristic length \(\delta\) can be found with the relation: $$\delta = L$$
02

Step 2. Very Long Cylinder: Determine Possible Length Scale and Find Relation

For a very long cylinder, we will relate its characteristic length to its radius, \(r_o\). Since heat transfer primarily happens radially, the characteristic length will be proportional to the radius. The characteristic length of a very long cylinder is defined as its radius, so for a cylinder of radius \(r_o\), the characteristic length \(\delta\) can be found with the relation: $$\delta = r_o$$
03

Step 3. Sphere: Determine Possible Length Scale and Find Relation

For a sphere, we relate its characteristic length to its radius, \(r_o\). Given that the heat transfer occurs from the center outwards, the characteristic length will be proportional to the radius. The characteristic length of a sphere is also defined as its radius, so for a sphere with radius \(r_o\), the characteristic length \(\delta\) is found with the relation: $$\delta = r_o$$ To summarize, we found the characteristic lengths for each given geometry: 1. Large plane wall with thickness \(2L\): \(\delta = L\) 2. Very long cylinder with radius \(r_o\): \(\delta = r_o\) 3. Sphere with radius \(r_o\): \(\delta = r_o\)

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Most popular questions from this chapter

Copper balls $\left(\rho=8933 \mathrm{~kg} / \mathrm{m}^{3}, \quad k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\(, \)c_{p}=385 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, \alpha=1.166 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\( ) initially at \)180^{\circ} \mathrm{C}$ are allowed to cool in air at \(30^{\circ} \mathrm{C}\) for a period of $2 \mathrm{~min}\(. If the balls have a diameter of \)2 \mathrm{~cm}$ and the heat transfer coefficient is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the center temperature of the balls at the end of cooling is (a) \(78^{\circ} \mathrm{C}\) (b) \(95^{\circ} \mathrm{C}\) (c) \(118^{\circ} \mathrm{C}\) (d) \(134^{\circ} \mathrm{C}\) (e) \(151^{\circ} \mathrm{C}\)

A small chicken $(k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad \alpha=0.15 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) can be approximated as an \(11.25-\mathrm{cm}\)-diameter solid sphere. The chicken is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is to be cooked in an oven maintained at \(220^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). With this idealization, the temperature at the center of the chicken after a 90 -min period is (a) \(25^{\circ} \mathrm{C}\) (b) \(61^{\circ} \mathrm{C}\) (c) \(89^{\circ} \mathrm{C}\) (d) \(122^{\circ} \mathrm{C}\) (e) \(168^{\circ} \mathrm{C}\)

Lumped system analysis of transient heat conduction situations is valid when the Biot number is (a) very small (b) approximately one (c) very large (d) any real number (e) cannot say unless the Fourier number is also known

To warm up some milk for a baby, a mother pours milk into a thin-walled cylindrical container whose diameter is \(6 \mathrm{~cm}\). The height of the milk in the container is \(7 \mathrm{~cm}\). She then places the container into a large pan filled with hot water at \(70^{\circ} \mathrm{C}\). The milk is stirred constantly so that its temperature is uniform at all times. If the heat transfer coefficient between the water and the container is $120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine how long it will take for the milk to warm up from \(3^{\circ} \mathrm{C}\) to $38^{\circ} \mathrm{C}$. Assume the entire surface area of the cylindrical container (including the top and bottom) is in thermal contact with the hot water. Take the properties of the milk to be the same as those of water. Can the milk in this case be treated as a lumped system? Why? Answer: \(4.50\) min

Conduct the following experiment at home to determine the combined convection and radiation heat transfer coefficient at the surface of an apple exposed to the room air. You will need two thermometers and a clock. First, weigh the apple and measure its diameter. You can measure its volume by placing it in a large measuring cup halfway filled with water, and measuring the change in volume when it is completely immersed in the water. Refrigerate the apple overnight so that it is at a uniform temperature in the morning, and measure the air temperature in the kitchen. Then take the apple out and stick one of the thermometers to its middle and the other just under the skin. Record both temperatures every \(5 \mathrm{~min}\) for an hour. Using these two temperatures, calculate the heat transfer coefficient for each interval and take their average. The result is the combined convection and radiation heat transfer coefficient for this heat transfer process. Using your experimental data, also calculate the thermal conductivity and thermal diffusivity of the apple and compare them to the values given above.
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