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Chapter 3: Problem 206

A hot plane surface at \(100^{\circ} \mathrm{C}\) is exposed to air at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat loss from the surface is to be reduced by half by covering it with sufficient insulation with a thermal conductivity of \(0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming the heat transfer coefficient to remain constant, the required thickness of insulation is (a) \(0.1 \mathrm{~cm}\) (b) \(0.5 \mathrm{~cm}\) (c) \(1.0 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(5 \mathrm{~cm}\)

Short Answer

Expert verified
a) 0.1 cm b) 0.5 cm c) 1.0 cm d) 2.0 cm e) 5 cm Answer: The required thickness of insulation to reduce the heat loss by half is 1000 cm. None of the given options are correct. There seems to be a mistake in the question itself or a typographical error in the provided options.

Step by step solution

01

Review the heat transfer formula for plane layer

The heat transfer through a plane layer at steady state can be represented by the formula: \(q = k \frac{A(T_1 - T_2)}{d}\) where - \(q\) is the heat transfer rate (W), - \(k\) is the thermal conductivity (W/m K), - \(A\) is the surface area (m²), - \(T_1\) and \(T_2\) are the temperatures of the two parallel boundaries (°C), - and \(d\) is the thickness of the layer (m).
02

Calculate the initial heat transfer rate

First, we will find the initial heat transfer rate, \(q_{initial}\), using the given combined heat transfer coefficient, \(h\), and temperature difference. We have: \(q_{initial} = hA(T_1 - T_2)\) where - \(h = 20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), - \(T_1 = 100^{\circ}\mathrm{C}\), - \(T_2 = 25^{\circ}\mathrm{C}\). Note that the area, \(A\), will be the same for both situations (before and after insulation), so it will eventually cancel out in our calculations.
03

Introduce insulation and set up equation for required thickness

We are told that the heat loss from the surface should be reduced by half, so we will set up an equation for the final heat transfer rate, \(q_{final}\): \(q_{final} = \frac{1}{2}q_{initial}\) The insulation's thermal conductivity is given by \(k_{insulation} = 0.10 \mathrm{~W}/ \mathrm{m} \cdot \mathrm{K}\), so we can write the new heat transfer rate as: \(q_{final} = k_{insulation} \frac{A(T_1 - T_2)}{d_{required}}\)
04

Solving for the required thickness of insulation

Now, we can solve for \(d_{required}\): \(\frac{1}{2}q_{initial} = k_{insulation} \frac{A(T_1 - T_2)}{d_{required}}\) Divide both sides by \(A(T_1 - T_2)\), then multiply both sides by \(d_{required}\): \(\frac{1}{2} \frac{h}{k_{insulation}} = d_{required}\) Substitute the given values for \(h\) and \(k_{insulation}\): \(d_{required} = \frac{1}{2} \frac{20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}}{0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}\) Calculate \(d_{required}\): \(d_{required} = 10 \mathrm{~m}\) However, the given options are in centimeters, so we'll convert this value to centimeters: \(d_{required} = 1000 \mathrm{~cm}\)
05

Choose the correct answer

The required thickness of insulation to reduce the heat loss by half is \(d_{required} = 1000\mathrm{~cm}\). Therefore, none of the given options (a) \(0.1 \mathrm{~cm}\), (b) \(0.5 \mathrm{~cm}\), (c) \(1.0 \mathrm{~cm}\), (d) \(2.0 \mathrm{~cm}\), and (e) \(5 \mathrm{~cm}\) are correct. This seems to be a mistake in the question itself or a typographical error in the given options.

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Most popular questions from this chapter

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

The walls of a food storage facility are made of a 2 -cm-thick layer of wood \((k=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in contact with a 5 -cm- thick layer of polyurethane foam \((k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the temperature of the surface of the wood is \(-10^{\circ} \mathrm{C}\) and the temperature of the surface of the polyurethane foam is \(20^{\circ} \mathrm{C}\), the temperature of the surface where the two layers are in contact is (a) \(-7^{\circ} \mathrm{C}\) (b) \(-2^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(8^{\circ} \mathrm{C}\) (e) \(11^{\circ} \mathrm{C}\)

An 8-m-internal-diameter spherical tank made of \(1.5\)-cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located in a room whose temperature is \(25^{\circ} \mathrm{C}\). The walls of the room are also at \(25^{\circ} \mathrm{C}\). The outer surface of the tank is black (emissivity \(\varepsilon=1\) ), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\).

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.

Hot water \(\left(c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows through a 200-m-long PVC \((k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) pipe whose inner diameter is \(2 \mathrm{~cm}\) and outer diameter is \(2.5 \mathrm{~cm}\) at a rate of \(1 \mathrm{~kg} / \mathrm{s}\), entering at \(40^{\circ} \mathrm{C}\). If the entire interior surface of this pipe is maintained at \(35^{\circ} \mathrm{C}\) and the entire exterior surface at \(20^{\circ} \mathrm{C}\), the outlet temperature of water is (a) \(39^{\circ} \mathrm{C}\) (b) \(38^{\circ} \mathrm{C}\) (c) \(37^{\circ} \mathrm{C}\) (d) \(36^{\circ} \mathrm{C}\) (e) \(35^{\circ} \mathrm{C}\)
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