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Chapter 3: Problem 205

Consider two walls, \(A\) and \(B\), with the same surface areas and the same temperature drops across their thicknesses. The ratio of thermal conductivities is \(k_{A} / k_{B}=4\) and the ratio of the wall thicknesses is \(L_{A} / L_{B}=2\). The ratio of heat transfer rates through the walls \(\dot{Q}_{A} / \dot{Q}_{B}\) is (a) \(0.5\) (b) 1 (c) \(2 \quad(d) 4\) (e) 8

Short Answer

Expert verified
Answer: The ratio of heat transfer rates through wall A to wall B is 2.

Step by step solution

01

Recall the formula for the heat transfer rate through a wall

The formula for the rate of heat transfer through a wall is given by: \(\dot{Q} = \frac{kA\Delta T}{L}\), where \(\dot{Q}\) is the heat transfer rate, \(k\) is the thermal conductivity, \(A\) is the surface area, \(\Delta T\) is the temperature drop across the wall, and \(L\) is the thickness of the wall.
02

Write the formula for heat transfer rates of walls A and B

Use the formula from Step 1 for wall A and wall B respectively: \(\dot{Q}_{A} = \frac{k_{A}A\Delta T}{L_{A}}\) \(\dot{Q}_{B} = \frac{k_{B}A\Delta T}{L_{B}}\)
03

Write the formula for the ratio of heat transfer rates

Divide the formula for wall A by the formula for wall B: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = \frac{\frac{k_{A}A\Delta T}{L_{A}}}{\frac{k_{B} A\Delta T}{L_{B}}}\)
04

Simplify the ratio

Cancel out the same terms and simplify: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = \frac{k_{A}L_{B}}{k_{B}L_{A}}\)
05

Substitute the given ratios

Substitute the provided ratios into the simplified equation: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = \frac{4k_{B}L_{B}}{k_{B}(2L_{B})}\)
06

Simplify the equation further

Simplify the equation by canceling out the terms: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = \frac{4}{2}\)
07

Calculate the ratio of heat transfer rates

Solve for the ratio: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = 2\) So, the ratio of heat transfer rates through the walls, \(\dot{Q}_{A} / \dot{Q}_{B}\) is (c) \(2\).

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Most popular questions from this chapter

The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?

Consider a pipe at a constant temperature whose radius is greater than the critical radius of insulation. Someone claims that the rate of heat loss from the pipe has increased when some insulation is added to the pipe. Is this claim valid?

A 3-m-diameter spherical tank containing some radioactive material is buried in the ground \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The distance between the top surface of the tank and the ground surface is \(4 \mathrm{~m}\). If the surface temperatures of the tank and the ground are \(140^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer from the tank.

Consider a 25-m-long thick-walled concrete duct \((k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section. The outer dimensions of the duct are \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\), and the thickness of the duct wall is \(2 \mathrm{~cm}\). If the inner and outer surfaces of the duct are at \(100^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer through the walls of the duct. Answer: \(47.1 \mathrm{~kW}\)

A row of 3 -ft-long and 1-in-diameter used uranium fuel rods that are still radioactive are buried in the ground parallel to each other with a center-to- center distance of 8 in at a depth of \(15 \mathrm{ft}\) from the ground surface at a location where the thermal conductivity of the soil is \(0.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). If the surface temperature of the rods and the ground are \(350^{\circ} \mathrm{F}\) and \(60^{\circ} \mathrm{F}\), respectively, determine the rate of heat transfer from the fuel rods to the atmosphere through the soil.
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