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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

From Problem 1.71, consider the wire leads connecting the transistor to the circuit board. The leads are of thermal conductivity \(k\), thickness \(t\), width \(w\), and length \(L\). One end of a lead is maintained at a temperature \(T_{c}\) corresponding to the transistor case, while the other end assumes the temperature \(T_{b}\) of the circuit board. During steady-state operation, current flow through the leads provides for uniform volumetric heating in the amount \(\dot{q}\), while there is convection cooling to air that is at \(T_{\infty}\) and maintains a convection coefficient \(h\). (a) Derive an equation from which the temperature distribution in a wire lead may be determined. List all pertinent assumptions. (b) Determine the temperature distribution in a wire lead, expressing your results in terms of the prescribed variables.

Determine the parallel plate separation distance \(L\), above which the thermal resistance associated with molecule-surface collisions \(R_{t, m-s}\) is less than \(1 \%\) of the resistance associated with molecule-molecule collisions, \(R_{t, m-m}\) for (i) air between steel plates with \(\alpha_{t}=0.92\) and (ii) helium between clean aluminum plates with \(\alpha_{t}=0.02\). The gases are at atmospheric pressure, and the temperature is \(T=300 \mathrm{~K}\).

Consider the conditions of Problem \(3.149\) but now allow for a tube wall thickness of \(5 \mathrm{~mm}\) (inner and outer diameters of 50 and \(60 \mathrm{~mm}\) ), a fin-to-tube thermal contact resistance of \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the fact that the water temperature, \(T_{w}=350 \mathrm{~K}\), is known, not the tube surface temperature. The water-side convection coefficient is \(h_{w}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat transfer per unit tube length \((\mathrm{W} / \mathrm{m})\) to the water. What would be the separate effect of each of the following design changes on the heat rate: (i) elimination of the contact resistance; (ii) increasing the number of fins from four to eight; and (iii) changing the tube wall and fin material from copper to AISI 304 stainless steel \((k=20\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) ?

Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of \(\dot{q}=\) \(2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The rod is concentric with a hollow ceramic cylinder, creating an enclosure that is filled with air. The thermal resistance per unit length due to radiation between the enclosure surfaces is \(R_{\mathrm{rad}}^{\prime}=0.30 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\), and the coefficient associated with free convection in the enclosure is \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Construct a thermal circuit that can be used to calculate the surface temperature of the rod, \(T_{r}\). Label all temperatures, heat rates, and thermal resistances, and evaluate each thermal resistance. (b) Calculate the surface temperature of the rod for the prescribed conditions.

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 \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).
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