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Chapter 3: Problem 112

A thin flat plate of length \(L\), thickness \(t\), and width \(W \geqslant L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{o}\). The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of \(q_{o}^{\prime \prime}\) (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.

Short Answer

Expert verified
The steady-state temperature distribution T(x) in the plate can be described by the equation \(T(x) = \frac{T(L)-T_{o}}{L} x + T_{o}\). The rate of heat transfer from the plate to the heat sinks is given by the expression \(Q = k \frac{T(L)-T_{o}}{L} \cdot tW\).

Step by step solution

01

Model the heat conduction in the plate

First, we need to analyze the heat conduction in the plate. Since heat conduction is steady-state, we can model this process using Fourier's law of heat conduction: \[\frac{d^2T}{dx^2} = 0\] where T(x) is the temperature distribution we want to find.
02

Solve the differential equation

We now need to solve the second-order differential equation: \[\frac{d^2T}{dx^2} = 0\] We can integrate the equation twice: \[ \frac{dT}{dx} = C_1\] Integrating again, \[T(x) = C_1 x + C_2\] Now we need to find the constants C_1 and C_2 using boundary conditions.
03

Apply boundary conditions

We are given that the plate is thermally joined to two large heat sinks that are maintained at a temperature T_{o}. This implies that the boundary conditions for the problem are: 1. At x=0, T(0)=T_{o}. 2. At x=L, T(L)=T_{o}. Now we can solve for C_1 and C_2 by applying these conditions. Applying the first boundary condition: \[T(0) = C_1 \cdot 0 + C_2\] \[C_2 = T_{o}\] Applying the second boundary condition: \[T(L) = C_1 L + T_{o}\] \[C_1 = \frac{T(L)-T_{o}}{L}\] Substituting the values of C_1 and C_2 back into the equation for T(x), we get: \[T(x) = \frac{T(L)-T_{o}}{L} x + T_{o}\]
04

Obtain heat transfer rate

Now we need to find the heat transfer rate from the plate to the heat sinks. We know that: Net heat flux, \(q_o^{\prime \prime} = -k \frac{dT}{dx}\) At x = 0: \[q_{o}^{\prime \prime} = -k \frac{dT}{dx} = k C_1\] Using the previously calculated value of C_1: \[q_{o}^{\prime \prime} = k \left(\frac{T(L)-T_{o}}{L}\right)\] Finally, to obtain the rate of heat transfer, we need to find the total heat transfer through the plate (area = tW): \[Q = (q_{o}^{\prime \prime}) tW = k \frac{T(L)-T_{o}}{L} \cdot tW\] Thus, the expression for the rate of heat transfer from the plate to the heat sinks is: \[Q = k \frac{T(L)-T_{o}}{L} \cdot tW\]

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Key Concepts

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

Steady-state temperature distribution
When discussing the steady-state temperature distribution in a material, such as a thin flat plate, it's essential to understand that this refers to a condition where the temperature within the material does not change over time.
In other words, the temperature at any given point remains constant even as heat flows through the material. This phenomenon occurs because the heat entering any section of the material is precisely equal to the heat leaving it.
We can describe this behavior using Fourier's Law of Heat Conduction. In the case of a steady-state temperature distribution, the differential equation is written as:\[\frac{d^2T}{dx^2} = 0\]Where \(T(x)\) signifies the temperature distribution across the plate along direction \(x\). In simple terms, this equation indicates that the temperature gradient is uniform, meaning the temperature change per unit distance is the same across the plate.
Solving this equation helps us predict the temperature at any location within the plate, hence drawing a complete map of how the temperature is distributed across the material.
Boundary conditions
Boundary conditions are essential as they provide the specific constraints needed to solve differential equations. They essentially help to determine the unknown constants after integrating these equations.
In our exercise, the boundary conditions are influenced by the connection of the plate to the heat sinks. These conditions are:
  • At the beginning of the plate (\(x=0\)), the temperature is maintained at \(T_{o}\) because it is in direct contact with the heat sink.
  • Similarly, at the end of the plate (\(x=L\)), the temperature is also \(T_{o}\) because it is connected to another heat sink.
By applying these boundary conditions to the temperature distribution function:\[T(x) = C_1 x + C_2\]We solve for the constants \(C_1\) and \(C_2\). This process ensures that the temperature distribution aligns with the physical constraints of having both ends of the plate maintained at a constant temperature.
Heat transfer rate
Once we have determined the temperature distribution along the plate, we can proceed to calculate the heat transfer rate. This refers to the quantity of thermal energy transferred from the plate to the heat sinks over time.
The heat transfer rate is computed using the net heat flux, which is given by the expression:\[q_{o}^{\prime \prime} = -k \frac{dT}{dx}\]Here, \(k\) is the thermal conductivity of the plate's material, and \(\frac{dT}{dx}\) is the gradient of the temperature distribution. At \(x=0\), using the expression for \(\frac{dT}{dx}\) achieved from the integration process, we find:\[q_{o}^{\prime \prime} = k \left(\frac{T(L)-T_{o}}{L}\right)\]The total heat transfer rate \(Q\) through the plate's surface is then calculated as:\[Q = q_{o}^{\prime \prime} \, tW\]Where \(t\) is the thickness and \(W\) is the width of the plate.
This formula indicates the efficiency of heat transfer based on factors such as material properties and geometric dimensions.

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Most popular questions from this chapter

Copper tubing is joined to a solar collector plate of thickness \(t\), and the working fluid maintains the temperature of the plate above the tubes at \(T_{o}\). There is a uniform net radiation heat flux \(q_{\text {rad }}^{\prime \prime}\) to the top surface of the plate, while the bottom surface is well insulated. The top surface is also exposed to a fluid at \(T_{\infty}\) that provides for a uniform convection coefficient \(h\). (a) Derive the differential equation that governs the temperature distribution \(T(x)\) in the plate. (b) Obtain a solution to the differential equation for appropriate boundary conditions.

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?

Annular aluminum fins of rectangular profile are attached to a circular tube having an outside diameter of \(50 \mathrm{~mm}\) and an outer surface temperature of \(200^{\circ} \mathrm{C}\). The fins are \(4 \mathrm{~mm}\) thick and \(15 \mathrm{~mm}\) long. The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) If there are 125 such fins per meter of tube length, what is the rate of heat transfer per unit length of tube?

The outer surface of a hollow sphere of radius \(r_{2}\) is subjected to a uniform heat flux \(q_{2}^{\prime \prime}\). The inner surface at \(r_{1}\) is held at a constant temperature \(T_{s, 1}\). (a) Develop an expression for the temperature distribution \(T(r)\) in the sphere wall in terms of \(q_{2}^{\prime \prime}, T_{s, 1}, r_{1}, r_{2}\), and the thermal conductivity of the wall material \(k\). (b) If the inner and outer tube radii are \(r_{1}=50 \mathrm{~mm}\) and \(r_{2}=100 \mathrm{~mm}\), what heat flux \(q_{2}^{\prime \prime}\) is required to maintain the outer surface at \(T_{s, 2}=50^{\circ} \mathrm{C}\), while the inner surface is at \(T_{s, 1}=20^{\circ} \mathrm{C}\) ? The thermal conductivity of the wall material is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?
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