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Chapter 3: Problem 38

A \(0.03\)-in-thick copper plate \(\left(k=223 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) is sandwiched between two \(0.15\)-in-thick epoxy boards \((k=\) \(0.15 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\) ) that are 7 in \(\times 9\) in in size. Determine the effective thermal conductivity of the board along its 9 -in-long side. What fraction of the heat conducted along that side is conducted through copper?

Short Answer

Expert verified
Question: Calculate the effective thermal conductivity of an assembly made up of a 0.03-inch-thick copper layer sandwiched between two 0.15-inch-thick epoxy layers. The thermal conductivity of copper is 223 Btu/(h·ft·°F) and the thermal conductivity of epoxy is 0.15 Btu/(h·ft·°F). Also, find the fraction of the heat conducted along one side that is conducted through the copper layer. Answer: To find the effective thermal conductivity of the assembly, follow these steps: 1. Calculate the thermal resistance of the copper layer and the epoxy layer. 2. Calculate the effective thermal resistance by considering the copper and epoxy layers as thermal resistances in parallel. 3. Calculate the effective thermal conductivity using the total thickness of the assembly and the effective thermal resistance. 4. Calculate the fraction of heat conducted through the copper layer by comparing its thermal resistance to the effective thermal resistance. By applying this method, you will determine the effective thermal conductivity and the fraction of heat conducted through the copper layer.

Step by step solution

01

Calculate the thermal resistance of each layer

The thermal resistance of each layer is given by the formula: $$R = \frac{L}{kA}$$ Where \(R\) is the thermal resistance, \(L\) is the thickness, \(k\) is the thermal conductivity, and \(A\) is the area of the layer. The area for each layer will be the same since the layers are placed one above the other. Here, we have \(A_{copper} = A_{epoxy}\). First, calculate the thermal resistance of the copper layer, \(R_{copper}\): $$R_{copper} = \frac{L_{copper}}{k_{copper} A} = \frac{0.03\,\text{in}}{223\,\frac{\text{Btu}}{\text{h} \cdot \text{ft} \cdot{ }^{\circ} \text{F}} \times A}$$ Next, calculate the thermal resistance of each epoxy layer, \(R_{epoxy}\): $$R_{epoxy} = \frac{L_{epoxy}}{k_{epoxy} A} = \frac{0.15\,\text{in}}{0.15\,\frac{\text{Btu}}{\text{h} \cdot \text{ft} \cdot{ }^{\circ} \text{F}} \times A}$$
02

Calculate the effective thermal resistance

Since the copper layer is sandwiched between the two epoxy layers, we can consider them as thermal resistances in parallel. The effective thermal resistance is given by the formula: $$\frac{1}{R_{effective}} = \frac{1}{R_{epoxy}} + \frac{1}{R_{copper}} + \frac{1}{R_{epoxy}} $$ Substitute the thermal resistance expressions from Step 1 and expand the equation to isolate the effective thermal resistance. $$R_{effective} = \frac{R_{copper}R_{epoxy}}{R_{copper}+R_{epoxy}}$$
03

Calculate the effective thermal conductivity

Now that we have the effective thermal resistance, we can calculate the effective thermal conductivity using the formula: $$k_{effective} = \frac{L_{total}}{R_{effective}}$$ Where the total thickness of the assembly is given by \(L_{total} = L_{epoxy} + L_{copper} + L_{epoxy} = 0.33\,\text{in}\). Calculate the effective thermal conductivity using the equation above.
04

Calculate the fraction of heat conducted through the copper layer

To find the fraction of heat conducted through the copper layer, we'll compare its thermal resistance to the effective thermal resistance. The fraction is given by the formula: $$\text{Fraction}_{copper} = \frac{R_{epoxy}}{R_{effective}}$$ Substitute the thermal resistance expressions from Steps 1 and 2 to calculate the fraction of heat conducted through the copper layer. After calculating the effective thermal conductivity in Step 3 and the fraction of heat conducted through the copper layer in Step 4, we will have our final answers.

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Key Concepts

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

Thermal Resistance
Thermal resistance is an important concept in heat transfer, which quantifies the temperature difference across a specific material when heat energy is transferred through it. Think of it like electrical resistance but for heat flow instead of electric current. The formula for thermal resistance \( R \) is \( R = \frac{L}{kA} \) where \( L \) is the thickness of the material (or layer), \( k \) stands for the thermal conductivity of the material, and \( A \) is the cross-sectional area over which the heat transfer occurs.

Understanding thermal resistance is essential since it helps us in designing and analyzing how insulating or conductive a material is within a system. It's particularly useful when comparing materials or systems, like in the exercise, where we see how a copper layer affects the overall resistance of a composite material.
Thermal Conductivity
Thermal conductivity \( k \) is a measure of a material's ability to conduct heat. It is defined as the quantity of heat, \( Q \) that passes in a unit of time, through a unit area, with a temperature gradient of one degree per unit of distance. The higher the thermal conductivity of a material, the greater its ability to transfer heat. This concept is pivotal in solving problems related to heat transfer in various materials.

Materials like copper are known for their high thermal conductivity, which makes them ideal for applications where efficient heat transfer is required, such as in heat exchangers, cooling systems, or electronic devices. On the other hand, materials with low thermal conductivity, like epoxy, are used as insulators to prevent heat flow.
Heat Transfer
Heat transfer is a phenomenon that can occur in three different ways: conduction, convection, and radiation. Our primary focus here is conduction, which is the transfer of heat through a material without any movement of the material itself—imagine how a metal spoon gets hot when left in a pot of boiling water.

Heat transfer through conduction occurs due to a temperature gradient, and the rate of this transfer is governed by the materials' thermal conductivity and the gradient's steepness. In the context of our problem, we're analyzing conductive heat transfer through layered materials to find out how effective they are when combined, hence calculating the effective thermal conductivity.
Composite Materials
Composite materials are composed of two or more distinct materials that when combined, exhibit properties unlike the individual constituents. This setup often aims to create a material with enhanced overall performance, such as higher strength, lighter weight, or in our case, a customized thermal conductivity profile.

When addressing a question about heat transfer in composite materials, as in the exercise provided, we consider the different layers of materials (copper plate and epoxy boards). By understanding how thermal resistance and thermal conductivity contribute to heat transfer in composites, we can effectively predict and manipulate the conductive behavior of such materials in practical applications, such as in thermal insulation or electronics.

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Most popular questions from this chapter

A spherical vessel, \(3.0 \mathrm{~m}\) in diameter (and negligible wall thickness), is used for storing a fluid at a temperature of \(0^{\circ} \mathrm{C}\). The vessel is covered with a \(5.0\)-cm-thick layer of an insulation \((k=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The surrounding air is at \(22^{\circ} \mathrm{C}\). The inside and outside heat transfer coefficients are 40 and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) all thermal resistances, in \(\mathrm{K} / \mathrm{W},(b)\) the steady rate of heat transfer, and \((c)\) the temperature difference across the insulation layer.

A hot surface at \(80^{\circ} \mathrm{C}\) in air at \(20^{\circ} \mathrm{C}\) is to be cooled by attaching 10 -cm-long and 1 -cm-diameter cylindrical fins. The combined heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer from the fin tip is negligible. If the fin efficiency is \(0.75\), the rate of heat loss from 100 fins is (a) \(325 \mathrm{~W}\) (b) \(707 \mathrm{~W}\) (c) \(566 \mathrm{~W}\) (d) \(424 \mathrm{~W}\) (e) \(754 \mathrm{~W}\)

Consider two finned surfaces that are identical except that the fins on the first surface are formed by casting or extrusion, whereas they are attached to the second surface afterwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat transfer? Explain.

Steam exiting the turbine of a steam power plant at \(100^{\circ} \mathrm{F}\) is to be condensed in a large condenser by cooling water flowing through copper pipes \(\left(k=223 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) of inner diameter \(0.4\) in and outer diameter \(0.6\) in at anerage temperature of \(70^{\circ} \mathrm{F}\). The heat of vaporization of water at \(100^{\circ} \mathrm{F}\) is \(1037 \mathrm{Btu} / \mathrm{lbm}\). The heat transfer coefficients are \(1500 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the steam side and \(35 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the water side. Determine the length of the tube required to condense steam at a rate of \(120 \mathrm{lbm} / \mathrm{h}\). Answer: \(1150 \mathrm{ft}\)

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?
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