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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

The air inside a chamber at \(T_{\infty, i}=50^{\circ} \mathrm{C}\) is heated convectively with \(h_{i}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) by a 200 -mm-thick wall having a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a uniform heat generation of \(1000 \mathrm{~W} / \mathrm{m}^{3}\). To prevent any heat generated within the wall from being lost to the outside of the chamber at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) with \(h_{o}=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a very thin electrical strip heater is placed on the outer wall to provide a uniform heat flux, \(q_{\sigma^{\prime}}\) (a) Sketch the temperature distribution in the wall on \(T-x\) coordinates for the condition where no heat generated within the wall is lost to the outside of the chamber. (b) What are the temperatures at the wall boundaries, \(T(0)\) and \(T(L)\), for the conditions of part (a)? (c) Determine the value of \(q_{o}^{\prime \prime}\) that must be supplied by the strip heater so that all heat generated within the wall is transferred to the inside of the chamber. (d) If the heat generation in the wall were switched off while the heat flux to the strip heater remained constant, what would be the steady-state temperature, \(T(0)\), of the outer wall surface?

A brass rod \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter extends horizontally from a casting at \(200^{\circ} \mathrm{C}\). The rod is in an air environment with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=30\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the temperature of the rod 25,50 , and \(100 \mathrm{~mm}\) from the casting?

A truncated solid cone is of circular cross section, and its diameter is related to the axial coordinate by an expression of the form \(D=a x^{3 / 2}\), where \(a=1.0 \mathrm{~m}^{-1 / 2}\). The sides are well insulated, while the top surface of the cone at \(x_{1}\) is maintained at \(T_{1}\) and the bottom surface at \(x_{2}\) is maintained at \(T_{2}\). (a) Obtain an expression for the temperature distribution \(T(x)\). (b) What is the rate of heat transfer across the cone if it is constructed of pure aluminum with \(x_{1}=0.075 \mathrm{~m}\), \(T_{1}=100^{\circ} \mathrm{C}, x_{2}=0.225 \mathrm{~m}\), and \(T_{2}=20^{\circ} \mathrm{C}\) ?

The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material, as shown in the schematic. For wires of diameter \(D=125 \mu \mathrm{m}\) and a convection coefficient of \(h=700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the minimum separation distance between the two legs of the sting, \(L=L_{1}+L_{2}\), to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and alumel wires. Evaluate the thermal conductivity of copper and constantan at \(T=300 \mathrm{~K}\). Use \(k_{\mathrm{Ch}}=19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{Al}}=29 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the thermal conductivities of the chromel and alumel wires, respectively.

In a test to determine the friction coefficient \(\mu\) associated with a disk brake, one disk and its shaft are rotated at a constant angular velocity \(\omega\), while an equivalent disk/shaft assembly is stationary. Each disk has an outer radius of \(r_{2}=180 \mathrm{~mm}\), a shaft radius of \(r_{1}=20 \mathrm{~mm}\), a thickness of \(t=12 \mathrm{~mm}\), and a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). A known force \(F\) is applied to the system, and the corresponding torque \(\tau\) required to maintain rotation is measured. The disk contact pressure may be assumed to be uniform (i.e., independent of location on the interface), and the disks may be assumed to be well insulated from the surroundings. (a) Obtain an expression that may be used to evaluate \(\mu\) from known quantities. (b) For the region \(r_{1} \leq r \leq r_{2}\), determine the radial temperature distribution \(T(r)\) in the disk, where \(T\left(r_{1}\right)=T_{1}\) is presumed to be known. (c) Consider test conditions for which \(F=200 \mathrm{~N}\), \(\omega=40 \mathrm{rad} / \mathrm{s}, \tau=8 \mathrm{~N} \cdot \mathrm{m}\), and \(T_{1}=80^{\circ} \mathrm{C}\). Evaluate the friction coefficient and the maximum disk temperature.
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