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Chapter 3: Problem 101

A thin flat plate of length \(L\) thickness \(t\), and width \(W=L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{e}\). 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 \(q^{*}\) - (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 temperature distribution is \(T(x) = -\frac{q^*}{2k}x^2 + \frac{q^*}{2k}Lx + T_e\), and the heat transfer rate is \(q^* L^2\).

Step by step solution

01

Understand the Problem

We are given a thin flat plate attached to two heat sinks which receive heat flux from the top surface. The problem asks for a differential equation describing steady-state temperature distribution and then solving for the temperature distribution and heat transfer rate.
02

Set Up the Differential Equation

Considering the one-dimensional heat conduction along the length of the plate and no heat generation within the plate, the relevant form of Fourier鈥檚 law is used. Assume heat balance over an infinitesimal element: \[ q_x(x) - q_x(x + dx) + q^* dx = 0 \] Using Fourier鈥檚 law, \[ q_x = -k \frac{dT}{dx} \].Thus, \[ -\frac{d}{dx}(k \frac{dT}{dx}) + q^* = 0 \].If the thermal conductivity \(k\) is constant, this simplifies to,\[ \frac{d^2T}{dx^2} = -\frac{q^*}{k} \].This is the differential equation governing temperature distribution.
03

Solve the Differential Equation for Temperature Distribution

The differential equation, \[ \frac{d^2T}{dx^2} = -\frac{q^*}{k} \] can be integrated to find the steady-state temperature distribution. The general solution is: \[ \frac{dT}{dx} = -\frac{q^*}{k}x + C_1 \]. Integrating again, \[ T(x) = -\frac{q^*}{2k}x^2 + C_1x + C_2 \].
04

Apply Boundary Conditions for Constants

Use boundary conditions to determine constants \(C_1\) and \(C_2\). At \(x = 0\), the boundary condition is \(T(0) = T_e\), leading to \(C_2 = T_e\).At \(x = L\), the boundary condition is also \(T(L) = T_e\), solving which we can find \(C_1\):\[ -\frac{q^*}{2k}L^2 + C_1L + T_e = T_e \], simplifying to \[ C_1 = \frac{q^*}{2k}L \].Thus, the temperature distribution is: \[ T(x) = -\frac{q^*}{2k}x^2 + \frac{q^*}{2k}Lx + T_e \].
05

Calculate Rate of Heat Transfer

The rate of heat transfer \(Q\) from the plate can be calculated using the heat flux and the area,\[ Q = q^* \times W \times L \] where \(W = L\) for the plate, thus,\[ Q = q^* L^2 \].

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Key Concepts

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

Steady-State Temperature Distribution
In heat conduction, the steady-state temperature distribution is a key concept when analyzing how heat flows through materials like a flat plate. When we say 'steady-state,' it means the temperature of the system no longer changes with time. Instead, any heat entering a particular section of the plate is precisely balanced by heat leaving it.

This problem involves a plate attached to two heat sinks with a uniform heat flux being applied and a corresponding temperature distribution described by a differential equation. The uniform heat flux is represented by the term \(q^*\), and the resulting temperature profile does not change with time, meaning that the distribution \(T(x)\) along the plate remains constant.

When solving for the temperature distribution, the goal is to use boundary conditions鈥攕uch as the known temperatures at the ends of the plate鈥� to find specific solutions. This results in a clear mathematical depiction of how temperature varies across the plate.
Fourier's Law
Fourier's Law is central to understanding heat conduction. It describes how heat flows through a material and is mathematically expressed as:
  • \( q_x = -k \frac{dT}{dx} \)
This formula states that the heat flux \( q_x \), which is the rate at which heat energy moves through a surface per unit area, is proportional to the negative gradient of the temperature, and \( k \) is the thermal conductivity.

This means that heat flows from areas of high temperature to areas of lower temperature. In our plate problem, since the plate is homogenous and \( k \) is constant, the differential form of Fourier鈥檚 Law simplifies the temperature change across the material. By applying it to derive the differential equation, we effectively describe how temperature is distributed throughout the plate under the given conditions.

Understanding this concept helps us predict how different materials will react to heat inputs and is foundational in solving many heat transfer problems.
Heat Transfer Rate
The heat transfer rate is an essential measure in heat conduction, indicating how much heat is conveyed through a material per unit time.

For the given plate problem, we determined the rate of heat transfer \(Q\) using the heat flux \(q^*\) and the surface area over which it transfers. This can be described using the equation:
  • \( Q = q^* \times A \)
where \( A \) is the area of the top surface of the plate. Given that the plate's width \(W = L\), the area is \( L \times L \), simplifying the formula to:
  • \( Q = q^* L^2 \)
This expression conveys how larger areas or higher heat fluxes increase the total heat transferred. Calculating \(Q\) helps in understanding not just the speed but also the capacity of the heat transfer through the plate, making it a critical aspect of thermal management in engineering designs.

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Most popular questions from this chapter

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between shect metal pancls. Consider a wall made from fiberglass insulation of thermal conductivity \(k_{1}=0.046 \mathrm{~W} / \mathrm{m}-\mathrm{K}\) and thickness \(I_{-}=50 \mathrm{~mm}\) and steel parels, each of thermal conductivity \(k_{p}=60\) W/m \(: \mathrm{K}\) and thickness \(L_{7}=3 \mathrm{~mm}\). If the wall separates refrigerated air at \(T_{x, j}=4^{\circ} \mathrm{C}\) from ambient air at \(T_{2,0}=25^{\circ} \mathrm{C}\), what is the heat gain per unit surface area? Cocfficicnts associated with natural convection at the inner and outer surfaces may be approximated as \(h_{y}=h_{g}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductiv ity \(25 \mathrm{~W} / \mathrm{m}\), K having uniform volumetric heat generation of \(0.3 \mathrm{MW} / \mathrm{m}^{3}\) is insulated on one side, while the other side is exposed to a fluid at \(92^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the wall and the fluid is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the maximum temperature in the wall.

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?

The rear window of an automobile is defogged by attaching a thin, transparent, film-type heating clement to its inner surface. By electrically heating this element, a uniform heat flux may be established at the inner surface. (a) For 4-mm-thick window glass, determine the electrical power required per unit window area to maintain an inner surface temperature of \(15^{\circ} \mathrm{C}\) when the interior air temperature and convection cocfficient are \(T_{=i}=25^{\circ} \mathrm{C}\) and \(h_{i}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the exterior (ambient) air temperature and convection coefficient are \(T_{\text {we }}=\) \(-10^{\circ} \mathrm{C}\) and \(h_{e}=65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) In practice \(T_{w e}\) and \(h_{y}\) vary according to weather conditions and car speed. For values of \(h_{\mathrm{m}}=2,20,65\), and \(100 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\), determine and plot the electrical power requirement as a function of \(T_{z,}\) for \(-30 \leq T_{x p} \leq\) \(0^{\circ} \mathrm{C}\). From your results, what can you conclude about the need for heater operation at low values of \(h_{e}\) ? How is this conclusion affected by the value of \(T_{\text {w, }, 0}\) ? If \(h \propto\) \(V^{*}\), where \(V\) is the vehicle speed and \(n\) is a positive exponent, how does the vehicle speed affect the need for heater eperation?

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?
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