/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Consider a plane composite wall ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 22

Consider a plane composite wall that is composed of two naterials of thermal conductivities \(k_{A}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{B}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thicknesses \(L_{A}=10 \mathrm{~mm}\) and \(L_{\mathrm{B}}=20 \mathrm{~mm}\). The contact resistance at the interface between the two materials is known to be \(0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Material A adjoins a fluid at \(200^{\circ} \mathrm{C}\) for which \(h=10\) W/m虏 \(\cdot \mathrm{K}\), and material B adjoins a fluid at \(40^{\circ} \mathrm{C}\) for which \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the rate of heat transfer through a wall that is \(2 \mathrm{~m}\) high by \(2.5 \mathrm{~m}\) wide? (b) Sketch the temperature distribution.

Short Answer

Expert verified
The rate of heat transfer is approximately 355.56 W. The temperature distribution is linear with a sharp drop at the interface.

Step by step solution

01

Formula for Overall Thermal Resistance

To find the rate of heat transfer, we first need to calculate the overall thermal resistance of the wall. The total thermal resistance \( R_{total} \) of the composite wall system includes the conductive resistance of both materials and the interface resistance. The formula for \( R_{total} \) is:\[ R_{total} = R_{A} + R_{B} + R_{interface} + R_{h,A} + R_{h,B} \]where\( R_{A} = \frac{L_{A}}{k_{A} \, A} \),\( R_{B} = \frac{L_{B}}{k_{B} \, A} \),\( R_{interface} = 0.30 \)\( R_{h,A} = \frac{1}{h_{A} \, A} \),and\( R_{h,B} = \frac{1}{h_{B} \, A} \).
02

Calculate Each Component's Resistance

The area \( A \) is given by the height and width of the wall, so \( A = 2 \times 2.5 = 5 \mathrm{~m}^2 \). Now, we calculate each resistance:1. \( R_{A} = \frac{0.01}{0.1 \times 5} = 0.02 \mathrm{~m}^2 \cdot \mathrm{K} / \mathrm{W} \)2. \( R_{B} = \frac{0.02}{0.04 \times 5} = 0.10 \mathrm{~m}^2 \cdot \mathrm{K} / \mathrm{W} \)3. \( R_{h,A} = \frac{1}{10 \times 5} = 0.02 \mathrm{~m}^2 \cdot \mathrm{K} / \mathrm{W} \)4. \( R_{h,B} = \frac{1}{20 \times 5} = 0.01 \mathrm{~m}^2 \cdot \mathrm{K} / \mathrm{W} \)
03

Total Thermal Resistance

Substitute the resistances into the total resistance equation:\[ R_{total} = 0.02 + 0.10 + 0.30 + 0.02 + 0.01 = 0.45 \mathrm{~m}^2 \cdot \mathrm{K} / \mathrm{W} \]
04

Calculate the Rate of Heat Transfer

Using Fourier鈥檚 law of heat conduction, the rate of heat transfer \( Q \) through the wall can be calculated as:\[ Q = \frac{T_{hot} - T_{cold}}{R_{total}} \]where \( T_{hot} = 200^{\circ} \mathrm{C} \) and \( T_{cold} = 40^{\circ} \mathrm{C} \). So:\[ Q = \frac{200 - 40}{0.45} = \frac{160}{0.45} \approx 355.56 \mathrm{~W} \]
05

Sketch the Temperature Distribution

The temperature distribution across the wall will be linear in each section and will show a sharp drop at the interface. Here's a rough sketch:- From \(200^{\circ} \mathrm{C} \) on the fluid side of material A, the temperature will decrease linearly across material A.- At the interface, a distinct temperature drop occurs due to the interface resistance.- Temperature continues to drop linearly across material B.- Finally, it reaches \(40^{\circ} \mathrm{C} \) on the fluid side of material B.

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Key Concepts

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

Composite Wall Analysis
When dealing with heat transfer problems that involve multiple layers or materials, a composite wall analysis becomes essential. Imagine a wall built of different materials layered together. Each material has distinct thermal properties which affect the overall behavior of the wall towards heat transfer.
  • Materials A and B in the exercise have thicknesses of 10 mm and 20 mm, respectively.
  • Their conductivities are specified as 0.1 W/m路K and 0.04 W/m路K.
The analysis helps you understand how heat is transferred through this layered structure by examining each material's effect individually. In the given problem, calculating the combined thermal resistance allows us to determine the overall rate of heat transfer through the composite wall. Understanding how each component contributes to the total thermal resistance is crucial. This analysis helps ensure efficient design and energy use in structures by predicting how they will respond to temperature differences across the layers.
Thermal Resistance
Thermal resistance is an essential concept to grasp when analyzing heat transfer through materials. Think of it as the opposition to heat flow. When heat travels through a material, some resistance is encountered based on the thermal properties of that material.
  • For Material A, the resistance is derived from its thickness and thermal conductivity, calculated as \( R_A = \frac{L_A}{k_A \times A} \).
  • Similarly, for Material B, this is \( R_B = \frac{L_B}{k_B \times A} \).
Along with materials' resistances, we consider contact resistance between them, which, in this case, is 0.30 \( \mathrm{m^2 \cdot K / W} \). Each layer of material and the interface between them add to the wall's total thermal resistance. Calculating the overall thermal resistance allows us to find out how the composite wall will insulate or conduct heat, ultimately determining the rate of heat transfer through the wall.
Conductive and Convective Heat Transfer
The two principal mechanisms of heat transfer in the problem are conduction and convection. Understanding each is vital for comprehensive heat transfer analysis.
**Conductive Heat Transfer:**Conduction is the process of heat being transferred through a material without any movement of the material itself. Within the composite wall, each material layer conducts heat at a rate determined by its thermal conductivity and thickness. The conductive resistance formula for each layer shows how effectively heat is passed through those solid materials.
**Convective Heat Transfer:**Convection refers to heat transfer between a solid surface and a fluid (like air or water) where the fluid is in motion. In the exercise, layers of materials A and B are in contact with fluids at specified temperatures. The convective heat transfer depends on the convective heat transfer coefficient (h values given in the problem) and the temperatures of adjoining fluids. Convective resistances \( R_{h,A} \) and \( R_{h,B} \) account for how quickly or slowly heat can be exchanged between a material and the fluids on either side.
By understanding these concepts, you can see how each part of the wall鈥攖he materials and interface鈥攊mpacts the overall thermal performance of the structure.

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Most popular questions from this chapter

A storage tank consists of a cylindrical secticn that has a length and inner diameter of \(L=2 \mathrm{~m}\) and \(D_{2}=1 \mathrm{~m}\), respectively, and two hemispherical end sections. The tank is constructed from 20-mm-thick glass (Pyrex) and is exposed to ambient air for which the temperature is \(300 \mathrm{~K}\) and the convection coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). The tank is used to store heated oil, which maintains the inser surface at a temperature of \(400 \mathrm{~K}\). Determine the electrical power that must be supplied to a heater submerged in the oil if the prescribed conditions are to be maintained. Radiation effects may be neglected, and the Pyrex may be assumed to have a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A spherical tank of \(3-\mathrm{m}\) diameter contains a liquifiedpetroleum gas at \(-60^{\circ} \mathrm{C}\). Insulation with a thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickncss \(250 \mathrm{~mm}\) is applied to the tank to roduce the heat gain. (a) Determine the radial position in the insulation layer at which the temperature is \(0^{2} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefficient on the outer surface is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) If the insulation is pervious to moisture from the atmospheric air, what conclusions can you reach about the formation of ice in the insulation? What effect will ice formation have on heat gain to the LP gas? How could this situation be avoided?

A hollow aluminem sphere, with an clectrical heater in the center, is used in tests to determine the thermal conductivity of insulating materials. The inner and outer radii of the sphere are \(0.15\) and \(0.18 \mathrm{~m}\), respectively. and testing is done under steady-state conditions with the inner surface of the aluminum maintained at \(250^{\circ} \mathrm{C}\). In a particular test, a spherical shell of insulation is cast on the outer surface of the sphere to a thickness of \(0.12 \mathrm{~m}\). The system is in a roorn for which the air temperature is \(20^{\circ} \mathrm{C}\) and the convection cocfficient at the outer surface of the insulation is \(30 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). If \(80 \mathrm{~W}\) are dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?

A commercial grade cubical freezer, \(3 \mathrm{~m}\) on a side, has a compesite wall consisting of an exterior sheet of 6.35-mm-thick plain carbon steel, an intermediate layer of 100 -rnm-thick cork insulation, and an inner sheet of 6.35-mm-thick aluminum alloy (2024). Adhesive interfaces berween the insulation and the metallic strips are each characterized by a thermul contact resistance of \(R_{L A}^{\prime \prime}=2.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the steady-state cooling load that must be maintained by the refrigerawor under conditions for which the outer and inner surface temperatures are \(2^{\circ} 2^{\circ} \mathrm{C}\) and \(-6^{\circ} \mathrm{C}\) respectively?

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2}+\mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the valie of the maximum temperature when the thickness is used?
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