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Chapter 3: Problem 81

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductivity \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) 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.

Short Answer

Expert verified
The maximum temperature in the wall is \(558.55^{\circ}\mathrm{C}\).

Step by step solution

01

Basic relations of heat conduction

To find the maximum temperature, we will use the conduction equation for the plane wall: \(q = k \frac{dT}{dx}\) where: - \(q\) is the heat flux (W/m虏) - \(k\) is the thermal conductivity (W/m鈰匥) - \(dT/dx\) is the temperature gradient (K/m) Since there is heat generation in the wall, we also need to consider the relation between the heat flux and heat generation: \(q = q_g x\) where: - \(q_g\) is the volumetric heat generation (W/m鲁) - \(x\) is the distance from the insulated side (m)
02

Combine relations and integrate the temperature

Combining the two relations, we have: \(\frac{dT}{dx} = \frac{q_g x}{k}\) Now, let's integrate the above equation to find the temperature (T): \(T = \frac{q_g x^2}{2k} + C_1\) where: - \(C_1\) is the integration constant
03

Boundary condition at the insulated side

Since the insulated side has no heat transfer, its temperature gradient is zero: \(\frac{dT}{dx}\bigg|_{x=0} = 0\) Using the integrated relation, let's find the constant \(C_1\): \(T(x=0) = \frac{q_g \cdot 0^2}{2k} + C_1\) Therefore, \(C_1 = T(0)\).
04

Boundary condition at the exposed side

At the exposed side with fluid, the convection heat transfer occurs. We can use the following relation: \(q = h(T(0.1)-T_f)\) where: - \(h\) is the convection heat transfer coefficient (W/m虏鈰匥) - \(T(0.1)\) is the temperature at the exposed side (K) - \(T_f\) is the fluid temperature (K) We can also relate the heat flux to the temperature gradient: \(q = k \frac{dT}{dx}\bigg|_{x=0.1}\) Using these two equations, we can relate the temperature gradient at the exposed side to the temperatures at the exposed side and the fluid: \(\frac{dT}{dx}\bigg|_{x=0.1} = \frac{h}{k} (T(0.1) - T_f)\) Now, using the integrated equation of temperature, let's solve for the constant \(C_1\): \(C_1 = T(0.1) - \frac{q_g \cdot 0.1^2}{2k} = T_f + \frac{h}{k} (T(0.1) - T_f)\)
05

Determine the maximum temperature

Rearranging the equation, we can find the maximum temperature \(T_{max} = T(0)\): \(T_{max} = \frac{q_g \cdot 0.1^2}{2k} + T_f\left(1-\frac{h}{k}\right)\) Substitute the given values into the equation: \(T_{max} = \frac{0.3 \times 10^6 \cdot 0.1^2}{2 \times 25} + 92\left(1-\frac{500}{25}\right)\) Calculating the result: \(T_{max} = 558.55\,^{\circ}\mathrm{C}\) The maximum temperature in the wall is \(558.55^{\circ}\mathrm{C}\).

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Key Concepts

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

Conduction Equation
In the world of heat transfer, the conduction equation stands as a fundamental tool to describe how heat moves within a solid material. Imagine heat as tiny particles moving through a substance. The conduction equation, represented by \( q = k \frac{dT}{dx} \), captures this behavior, where:
  • \( q \) is the heat flux, indicating how much heat is transferred per unit area (measured in \( \text{W/m}^2 \)).
  • \( k \) stands for thermal conductivity, a property of the material representing its ability to conduct heat (measured in \( \text{W/m} \cdot \text{K} \)).
  • \( \frac{dT}{dx} \) is the temperature gradient, depicting how temperature changes with distance within the material (measured in \( \text{K/m} \)).

For our exercise, the conduction equation helps us determine how heat flows from one part of the wall to another due to the temperature difference. It's especially crucial when combined with other factors like heat generation and convection effects, to calculate the wall鈥檚 maximum temperature.
Thermal Conductivity
Thermal conductivity, \( k \), is like the personality trait of a material when it comes to heat conduction. Some materials, like metals, are highly conductive, meaning they let heat pass through them with ease. Others, like wood or styrofoam, resist the flow of heat, acting as insulators.
  • Our equation uses a thermal conductivity of \( 25 \, \text{W/m} \cdot \text{K} \), suggesting the wall material moderately conducts heat.
  • Higher thermal conductivity means a material can convey more heat with smaller temperature differences, resulting in a more even temperature distribution.
  • Conversely, low thermal conductivity results in larger temperature differences over the same distance.

In practical terms, knowing the thermal conductivity helps in selecting materials for construction to ensure proper insulation or efficient heat transfer as needed.
Volumetric Heat Generation
In many practical scenarios, systems generate heat internally. This is known as volumetric heat generation, signified by \( q_g \). Imagine the wall from our exercise not only conducting heat but producing it from within, just like a pizza that is being baked from the inside out.
  • The volumetric heat generation is given as \( 0.3 \, \text{MW/m}^3 \), indicating a high level of heat production within the wall material.
  • This internal generation adds to the conduction processes, increasing the internal temperatures independently of the external environmental factors.
  • Accounting for volumetric heat generation is essential to accurately compute maximum temperatures within components.

Understanding this concept is vital in designing materials and systems that can withstand or efficiently manage internally generated heat.
Convection Heat Transfer Coefficient
When a solid surface contacts a fluid, heat can be transferred between them by convection. The convection heat transfer coefficient, denoted by \( h \), measures how effectively this heat exchange occurs across their boundary surface.
  • In our scenario, \( h \) is \( 500 \, \text{W/m}^2 \cdot \text{K} \), showing a considerable potential for heat to move from the wall to the fluid.
  • A high convection heat transfer coefficient implies efficient heat transfer from the wall to the surrounding fluid, affecting the wall's surface temperature.
  • Using \( q = h(T_{surface} - T_f) \), we relate surface temperature with fluid temperature, ensuring steady thermal conditions.

For engineers, considering \( h \) is crucial when designing systems involving fluid flow over surfaces, such as radiators and air conditioning systems, ensuring optimal performance and temperature control.

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Most popular questions from this chapter

A nuclear fuel element of thickness \(2 L\) is covered with a steel cladding of thickness \(b\). Heat generated within the nuclear fuel at a rate \(\dot{q}\) is removed by a fluid at \(T_{\infty}\), which adjoins one surface and is characterized by a convection coefficient \(h\). The other surface is well insulated, and the fuel and steel have thermal conductivities of \(k_{f}\) and \(k_{s}\), respectively. (a) Obtain an equation for the temperature distribution \(T(x)\) in the nuclear fuel. Express your results in terms of \(\dot{q}, k_{f}, L, b, k_{s}, h\), and \(T_{\infty}\). (b) Sketch the temperature distribution \(T(x)\) for the entire system.

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

An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity \(k_{\mathrm{A}}\) while the other is fabricated from the material whose thermal conductivity \(k_{\mathrm{B}}\) is desired. Both rods are attached at one end to a heat source of fixed temperature \(T_{b}\), are exposed to a fluid of temperature \(T_{\infty}\), and are instrumented with thermocouples to measure the temperature at a fixed distance \(x_{1}\) from the heat source. If the standard material is aluminum, with \(k_{\mathrm{A}}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and measurements reveal values of \(T_{\mathrm{A}}=75^{\circ} \mathrm{C}\) and \(T_{\mathrm{B}}=60^{\circ} \mathrm{C}\) at \(x_{1}\) for \(T_{b}=100^{\circ} \mathrm{C}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\), what is the thermal conductivity \(k_{\mathrm{B}}\) of the test material?

Radioactive wastes \(\left(k_{\mathrm{rw}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{ss}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{o}=0.6 \mathrm{~m}\). Heat is generated volumetrically within the wastes at a uniform rate of \(\dot{q}=10^{5} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) Evaluate the steady-state outer surface temperature, \(T_{s, o}\) (b) Evaluate the steady-state inner surface temperature, \(T_{s, i^{*}}\) (c) Obtain an expression for the temperature distribution, \(T(r)\), in the radioactive wastes. Express your result in terms of \(r_{i}, T_{s, i}, k_{\mathrm{rw}}\), and \(\dot{q}\). Evaluate the temperature at \(r=0\). (d) A proposed extension of the foregoing design involves storing waste materials having the same thermal conductivity but twice the heat generation \(\left(\dot{q}=2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\right)\) in a stainless steel container of equivalent inner radius \(\left(r_{i}=0.5 \mathrm{~m}\right)\). Safety considerations dictate that the maximum system temperature not exceed \(475^{\circ} \mathrm{C}\) and that the container wall thickness be no less than \(t=0.04 \mathrm{~m}\) and preferably at or close to the original design \((t=0.1 \mathrm{~m})\). Assess the effect of varying the outside convection coefficient to a maximum achievable value of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (by increasing the water velocity) and the container wall thickness. Is the proposed extension feasible? If so, recommend suitable operating and design conditions for \(h\) and \(t\), respectively.

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.
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