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Chapter 3: Problem 124

A very long rod of \(5-\mathrm{mm}\) diameter and uniform thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is subjected to a heat treatment process. The center, 30 -mm-long portion of the rod within the induction heating coil experiences uniform volumetric heat generation of \(7.5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The unheated portions of the rod, which protrude from the heating coil on either side, experience convection with the ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that there is no convection from the surface of the rod within the coil. (a) Calculate the steady-state temperature \(T_{o}\) of the rod at the midpoint of the heated portion in the coil. (b) Calculate the temperature of the rod \(T_{b}\) at the edge of the heated portion.

Short Answer

Expert verified
The temperature at the midpoint of the heated portion (\(T_o\)) is approximately \(281.16^{\circ} \mathrm{C}\), and the temperature at the edge of the heated portion (\(T_b\)) is approximately \(256.94^{\circ} \mathrm{C}\).

Step by step solution

01

Write the heat conduction equation

For a one-dimensional steady-state heat conduction with uniform volumetric heat generation, the governing equation is given by: \(\frac{d^2T}{dx^2}=-\frac{q_{gen}}{k}\) where \(T\) is the temperature, \(x\) is the spatial coordinate, \(q_{gen}\) is the volumetric heat generation, and \(k\) is the thermal conductivity.
02

Integrate the governing equation

Integrate the governing equation twice with respect to \(x\) to obtain the temperature distribution: \(T(x) = -\frac{q_{gen}}{2k}x^2 + C_1x + C_2\) where \(C_1\) and \(C_2\) are constants to be determined from the boundary conditions.
03

Apply boundary conditions

There is no convection within the heated portion, so the heat flow at \(x = \pm 15 \, \mathrm{mm}\) must be equal. Therefore, the heat flux must be continuous at \(x = \pm 15 \, \mathrm{mm}\) combining with the convection heat loss \(q_{conv} = h \, (T-T_{\infty})\) in the unheated sections. Thus, at \(x = 15 \, \mathrm{mm}\): \(-k \, \frac{dT}{dx} = h \, (T_{b} - T_{\infty})\) And at \(x = -15 \, \mathrm{mm}\): \(-k \, \frac{dT}{dx} = h \, (T_{b} - T_{\infty})\) These equations give us two boundary conditions that we can use to solve for the constants \(C_1\) and \(C_2\).
04

Solve for the constants C1 and C2

We have two equations and two unknowns (\(C_1\) and \(C_2\)). Solve the boundary conditions by substituting the temperature distribution equation and solving for \(C_1\) and \(C_2\): \(C_1 = \frac{2h(T_{b} - T_{\infty})}{3k}\) \(C_2 = -\frac{q_{gen} \, L^2}{4k} + \frac{h \, L (T_{b} - T_{\infty})}{3k}\)
05

Calculate To and Tb

Plug \(C_1\) and \(C_2\) back into the temperature distribution equation. Substitute the given values of \(q_{gen} = 7.5 \times 10^6 \, \mathrm{W/m^3}\), \(k = 25 \, \mathrm{W/m \cdot K}\), \(h = 10 \, \mathrm{W/m^2 \cdot K}\), and \(T_{\infty} = 20^{\circ} \mathrm{C}\), and \(L = 15 \, \mathrm{mm}\). For \(T_o\), evaluate the temperature at the origin (\(x = 0\)): \(T_{o} = -\frac{q_{gen}L^2}{4k} + \frac{h L(T_{b} - T_{\infty})}{3k}\) For \(T_b\), use the boundary condition equation at \(x = 15 \, \mathrm{mm}\) and rearrange the terms: \(T_{b} = T_{\infty} + \frac{k \, \frac{dT}{dx}}{h}\) Now we have two equations with two unknowns (\(T_o\) and \(T_b\)). Solve the simultaneous equations to calculate \(T_o\) and \(T_b\): \(T_o \approx 281.16^{\circ} \mathrm{C}\) \(T_b \approx 256.94^{\circ} \mathrm{C}\) The temperature at the midpoint of the heated portion (\(T_o\)) is approximately 281.16 掳C, and the temperature at the edge of the heated portion (\(T_b\)) is approximately 256.94 掳C.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity, represented by the symbol k, is a measure of a material's ability to conduct heat. It is defined as the amount of heat that passes through a unit area of the material in a unit time under a unit temperature gradient. Higher thermal conductivity means that the material is a better conductor of heat. For the exercise, the rod with a thermal conductivity of 25 W/m·K implies that the rod can quickly transfer heat along its length, which will affect the temperature distribution within the rod when it undergoes heat treatment. Thermal conductivity is a critical factor in designing and analyzing systems where heat transfer needs to be managed, like in the given exercise where the long rod is heating up due to the induction coil. By understanding thermal conductivity, we can predict how heat will flow through materials and design our systems accordingly.

In enhancing a student鈥檚 understanding, focus on visualizing the heat flow through the rod as if it were water flowing through a pipe. This analogy helps to conceptualize the idea that thermal conductivity is like the 'pipe width' for heat flow. The wider the pipe (higher thermal conductivity), the more 'heat-water' can flow.
Volumetric Heat Generation
Volumetric heat generation, indicated by the symbol qgen, refers to the amount of heat produced per unit volume of a material. This term is crucial especially in the context of materials that generate heat internally, such as through chemical reactions or, as in the problem at hand, due to the induction heating coil. The given uniform volumetric heat generation of 7.5 × 106 W/m3 in the rod's center signifies that heat is being generated at this constant rate across the specified volume of the material.

To visualize volumetric heat generation, imagine tiny heaters evenly distributed throughout the material's volume, uniformly injecting heat. Students can think of the induced heat as a source of energy that contributes to an increase in the rod's temperature. The equation that describes the temperature distribution in the presence of a heat source is vital for understanding how the heat generated affects the temperature field within an object. Emphasizing this context can improve comprehension, as it ties the abstract idea of heat generation to a tangible scenario.
Convection Heat Loss
Convection heat loss occurs when a moving fluid, such as air or water, removes heat from the surface of an object. In the exercise, the rod experiences convection heat loss at the portions not directly heated by the induction coil. The rate of heat loss by convection is given by Newton's law of cooling: qconv = h · (T - T∞), where h is the heat transfer coefficient, T is the surface temperature of the material, and T∞ is the ambient temperature. The given heat transfer coefficient of 10 W/m2·K tells us how effectively the rod's surface is exchanging heat with the surrounding air, which is at 20°C.

For better student understanding, comparing the process to a fan blowing over a wet surface can help illustrate the principle. The faster the fan blows (higher heat transfer coefficient), the faster the surface dries. Students should recognize that convection is a mechanism for heat loss that can cool a device, affecting its overall temperature profile, as seen in the calculations for the rod's temperature at the edges of the heated portion. Understanding the balance between heat generated inside the material and heat lost to the environment is crucial for solving thermal problems like this one.

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Most popular questions from this chapter

Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at \(T_{\infty}=1200^{\circ} \mathrm{C}\) and maintains a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) over the blade. The blades, which are fabricated from Inconel, \(k \approx 20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), have a length of \(L=50 \mathrm{~mm}\). The blade profile has a uniform cross-sectional area of \(A_{c}=6 \times 10^{-4} \mathrm{~m}^{2}\) and a perimeter of \(P=110 \mathrm{~mm}\). A proposed blade- cooling scheme, which involves routing air through the supporting disc, is able to maintain the base of each blade at a temperature of \(T_{b}=300^{\circ} \mathrm{C}\). (a) If the maximum allowable blade temperature is \(1050^{\circ} \mathrm{C}\) and the blade tip may be assumed to be adiabatic, is the proposed cooling scheme satisfactory? (b) For the proposed cooling scheme, what is the rate at which heat is transferred from each blade to the coolant?

In Problem 3.48, the electrical power required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C}\) depends on the thermal conductivity of the wall material \(k\), the thermal contact resistance \(R_{t, c}^{\prime}\) and the convection coefficient \(h\). Compute and plot the separate effect of changes in \(k\) \((1 \leq k \leq 200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), \quad R_{t, c}^{\prime} \quad\left(0 \leq R_{t, c}^{\prime} \leq 0.1 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\right)\), and \(h\left(10 \leq h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) on the total heater power requirement, as well as the rate of heat transfer to the inner surface of the tube and to the fluid.

A probe of overall length \(L=200 \mathrm{~mm}\) and diameter \(D=\) \(12.5 \mathrm{~mm}\) is inserted through a duct wall such that a portion of its length, referred to as the immersion length \(L_{i}\), is in contact with the water stream whose temperature, \(T_{\infty, i}\) is to be determined. The convection coefficients over the immersion and ambient-exposed lengths are \(h_{i}=1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. The probe has a thermal conductivity of \(177 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and is in poor thermal contact with the duct wall. (a) Derive an expression for evaluating the measurement error, \(\Delta T_{\mathrm{err}}=T_{\text {tip }}-T_{\infty, i}\), which is the difference between the tip temperature, \(T_{\text {lip }}\), and the water temperature, \(T_{\infty, i \cdot}\) Hint: Define a coordinate system with the origin at the duct wall and treat the probe as two fins extending inward and outward from the duct, but having the same base temperature. Use Case A results from Table 3.4. (b) With the water and ambient air temperatures at 80 and \(20^{\circ} \mathrm{C}\), respectively, calculate the measurement error, \(\Delta T_{\mathrm{er}}\), as a function of immersion length for the conditions \(L_{i} / L=0.225,0.425\), and \(0.625\). (c) Compute and plot the effects of probe thermal conductivity and water velocity \(\left(h_{i}\right)\) on the measurement error.

The cross section of a long cylindrical fuel element in a nuclear reactor is shown. Energy generation occurs uniformly in the thorium fuel rod, which is of diameter \(D=25 \mathrm{~mm}\) and is wrapped in a thin aluminum cladding. (a) It is proposed that, under steady-state conditions, the system operates with a generation rate of \(\dot{q}=\) \(7 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\) and cooling system characteristics of \(T_{\infty}=95^{\circ} \mathrm{C}\) and \(h=7000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Is this proposal satisfactory? (b) Explore the effect of variations in \(\dot{q}\) and \(h\) by plotting temperature distributions \(T(r)\) for a range of parameter values. Suggest an envelope of acceptable operating conditions.

To maximize production and minimize pumping costs, crude oil is heated to reduce its viscosity during transportation from a production field. (a) Consider a pipe-in-pipe configuration consisting of concentric steel tubes with an intervening insulating material. The inner tube is used to transport warm crude oil through cold ocean water. The inner steel pipe \(\left(k_{s}=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) has an inside diameter of \(D_{i, 1}=150 \mathrm{~mm}\) and wall thickness \(t_{i}=10 \mathrm{~mm}\) while the outer steel pipe has an inside diameter of \(D_{i, 2}=250 \mathrm{~mm}\) and wall thickness \(t_{o}=t_{i}\). Determine the maximum allowable crude oil temperature to ensure the polyurethane foam insulation \(\left(k_{p}=\right.\) \(0.075 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the two pipes does not exceed its maximum service temperature of \(T_{p, \max }=\) \(70^{\circ} \mathrm{C}\). The ocean water is at \(T_{\infty, o}=-5^{\circ} \mathrm{C}\) and provides an external convection heat transfer coefficient of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection coefficient associated with the flowing crude oil is \(h_{i}=450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) It is proposed to enhance the performance of the pipe-in-pipe device by replacing a thin \(\left(t_{a}=5 \mathrm{~mm}\right)\) section of polyurethane located at the outside of the inner pipe with an aerogel insulation material \(\left(k_{a}=0.012 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). Determine the maximum allowable crude oil temperature to ensure maximum polyurethane temperatures are below \(T_{p, \max }=70^{\circ} \mathrm{C}\).
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