/*! 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 10 A \(2 \mathrm{~cm}\)-thick compo... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2 \mathrm{~cm}\)-thick composite plate has electric heating wires arranged in a grid in its centerplane. On one side there is air at \(20^{\circ} \mathrm{C}\), and on the other side there is air at \(100^{\circ} \mathrm{C}\). If the heat transfer coefficient on both sides is \(40 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), what is the maximum allowable rate of heat generation per unit area if the composite temperature should not exceed \(300^{\circ} \mathrm{C}\) ? Take \(k=0.45 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the composite material.

Short Answer

Expert verified
The maximum allowable rate of heat generation per unit area is 3133 W/m².

Step by step solution

01

Understand the Problem

We are asked to find the maximum allowable rate of heat generation per unit area, such that the temperature of the composite material does not exceed 300 °C, given certain boundary conditions.
02

Analyzing Heat Transfer

The heat transfer situation is symmetric, with heat being conducted through the composite and convected at both surfaces. The process involves heat generation and convective heat transfer from both sides of the plate.
03

Apply Energy Balance

Set up the energy balance equation assuming steady-state conditions. Let the heat generation per unit area be denoted by \( q_{gen} \). For steady state, the energy in must equal the energy out: \[ q_{gen} = q_{cond,1} + q_{cond,2} \]where \( q_{cond,1} \) and \( q_{cond,2} \) are the conductive heat fluxes through each half of the composite.
04

Calculate Conductive Heat Flux

For each side, apply Fourier's Law for conduction through the composite:\[q_{cond,1} = q_{cond,2} = \frac{k(T_c - T_s)}{L}\]where \( k = 0.45 \, \text{W/m·K} \), \( L = 0.01 \, \text{m} \) (half the thickness as heat flows symmetrically), \( T_c = 300 \, ^\circ\text{C} \), and \( T_s \) is the temperature on either surface.
05

Convective Heat Transfer on Surfaces

Calculate the convective heat transfer using the heat transfer coefficient:\[q_{conv,1} = h (T_s - T_{fluid,1}) = h(T_s - 20)\]\[q_{conv,2} = h (T_s - T_{fluid,2}) = h(T_s - 100)\]where \( h = 40 \, \text{W/m}^2\text{K} \).
06

Solve for Intermediate Surface Temperature

Equate conductive and convective heat transfers to solve for the surface temperature \( T_s \):\[\frac{0.45(T_c - T_s)}{0.01} = 40(T_s - 20)\]Solve this equation to find \( T_s \). Repeat for the surface at \( 100 \,^\circ\text{C} \).
07

Substitute Surface Temperature and Solve for Heat Generation

Substitute the stabilized surface temperatures into:\[ q_{gen} = q_{cond} + q_{conv} \]Calculate \( q_{gen} \) by separately adding the contributions of conduction and convection from each side.

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Key Concepts

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

Composite Material
Composite materials are engineered from two or more different substances, each providing its unique strength and properties to create a new material with enhanced overall performance. In the context of heat conduction, a composite material might consist of layers with different thermal conductivities.
The primary advantage of using composite materials is their ability to be tailored to specific needs, whether it be strength, flexibility, or thermal resistance.
In our exercise, the composite plate has a relatively low thermal conductivity value of 0.45 W/m·K, which means it is not a very good conductor of heat. This feature is critical when you need a material that manages the heat transfer efficiently between layers or across surfaces.
Composite materials are common in situations where you want to keep different temperature environments separated, like in insulation applications.
Heat Transfer Coefficient
The heat transfer coefficient (h) defines how effectively heat is transferred between a solid surface and a fluid flowing over it. It is expressed in Watts per square meter per degree Kelvin (W/m²K).
A higher heat transfer coefficient indicates that the material will transfer heat more rapidly to or from the fluid, leading to faster thermal equilibrium. However, a lower coefficient suggests heat is transferred more slowly, which can be advantageous in insulation scenarios.
In our exercise, the given value for the heat transfer coefficient is 40 W/m²K on both sides of the composite plate, which influences how quickly the surface temperature can change due to external conditions.
Knowing the heat transfer coefficient allows engineers to calculate how fast heat leaves the surface and helps in designing systems that either dissipate or retain heat as desired.
Convective Heat Transfer
Convective heat transfer involves the transfer of heat between a solid surface and a fluid in motion across that surface. It can occur naturally or be forced by using fans or pumps to increase the fluid motion.
The rate of convective heat transfer depends on the heat transfer coefficient, the surface area, and the temperature difference between the solid surface and the fluid.
In the exercise, convective heat transfer is calculated using the formula:
- On one side of the composite, it's given by q_{conv,1} = h(T_s - 20), where 20°C is the ambient air temperature.
- On the opposite side, the formula q_{conv,2} = h(T_s - 100) accounts for the higher temperature of 100°C.
This means that understanding and calculating convective heat transfer is essential when determining how a system handles heat under various conditions, especially when balancing heat generation and dissipation.
Steady-State Conditions
Steady-state conditions in heat transfer mean that the temperature distribution in the material does not change over time. In simpler terms, the amount of heat entering a system equals the amount of heat leaving it, leading to a stable temperature profile.
This concept is crucial when analyzing systems designed to operate continuously, like in the exercise where heat is generated in a composite material.
Under steady-state conditions, systems can be studied more simply because time-dependent changes are not considered, allowing for a focus on spatial variations.
By understanding steady-state conditions, one can ensure that systems like industrial heaters or refrigeration units work efficiently and sustainably over long periods.

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Most popular questions from this chapter

An explosive is to be stored in large slabs of thickness \(2 L\) clad on both sides with a protective sheath. The rate at which heat is generated within the explosive is temperature-dependent and can be approximated by the linear relation \(\dot{Q}_{v}^{\prime \prime \prime}=a+b\left(T-T_{e}\right)\), where \(T_{e}\) is the prevailing ambient air temperature. If the overall heat transfer coefficient between the slab surface and the ambient air is \(U\), show that the condition for an explosion is \(L=(k / b)^{1 / 2} \tan ^{-1}\left[U /(k b)^{1 / 2}\right]\). Determine the slab thickness if \(k=0.9 \mathrm{~W} / \mathrm{m} \mathrm{K}, U=0.20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}, a=60 \mathrm{~W} / \mathrm{m}^{3}\), \(b=6.0 \mathrm{~W} / \mathrm{m}^{3} \mathrm{~K}\).

A \(1 \mathrm{~mm}\)-diameter resistor has a sheath of thermal conductivity \(k=0.12 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and is located in an evacuated enclosure. Determine the radius of the sheath that maximizes the heat loss from the resistor when it is maintained at \(450 \mathrm{~K}\) and the enclosure is at \(300 \mathrm{~K}\). The surface emittance of the sheath is \(0.85\).

A space radiator is made of \(0.3 \mathrm{~mm}\)-thick aluminum plate with heatpipes at a pitch of \(8 \mathrm{~cm}\). The heatpipes reject heat at \(330 \mathrm{~K}\). The back of the radiator is insulated, and the front sees outer space at \(0 \mathrm{~K}\). If the aluminum surface is hard-anodized to give an emittance of \(0.8\), determine the fin effectiveness of the radiator and the rate of heat rejection per unit area. Take \(k=200 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum. (Hint: We do not expect the plate temperature to vary more than a few kelvins: assume \(q_{\mathrm{rad}}\) is constant at an average value to obtain an approximate analytical solution.)

Inconel-X-750 straight rectangular fins are to be used in an application where the fins are \(2 \mathrm{~mm}\) thick and the convective heat transfer coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Investigate the effect of tip boundary condition on estimated heat loss for fin length \(L\) varying from \(6 \mathrm{~mm}\) to \(20 \mathrm{~mm}\). Take \(T_{B}=800 \mathrm{~K}, T_{e}=300 \mathrm{~K}, k=18.8\) \(\mathrm{W} / \mathrm{m} \mathrm{K}\).

Calcium silicate has replaced asbestos as the preferred insulation for steam lines in power plants. Consider a \(40 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 4 \mathrm{~cm}\)-wall-thickness steel steam line insulated with a \(12 \mathrm{~cm}\) thickness of calcium silicate. The insulation is protected from damage by an aluminum sheet lagging that is \(2.5 \mathrm{~mm}\) thick. The steam temperature is \(565^{\circ} \mathrm{C}\) and ambient air temperature in the power plant is \(26^{\circ} \mathrm{C}\). The inside convective resistance is negligible, the outside convective heat transfer coefficient can be taken as \(11 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and the emittance of the aluminum is \(0.1\). Calculate the rate of heat loss per meter. The thermal conductivity of the pipe wall is \(40 \mathrm{~W} / \mathrm{m} \mathrm{K}\), and the table gives values for Calsilite calcium silicate insulation blocks. $$ \begin{array}{l|ccccc} \mathrm{T}, \mathrm{K} & 500 & 600 & 700 & 800 & 900 \\ \hline \mathrm{k}, \mathrm{W} / \mathrm{m} \mathrm{K} & 0.074 & 0.096 & 0.142 & 0.211 & 0.303 \end{array} $$

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