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Chapter 2: Problem 151

Heat is generated uniformly at a rate of \(4.2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\) in a spherical ball \((k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of diameter \(24 \mathrm{~cm}\). The ball is exposed to iced-water at \(0^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperatures at the center and the surface of the ball.

Short Answer

Expert verified
Question: Determine the temperatures at the center and the surface of a spherical ball with the given properties and conditions. Answer: The temperature at the surface of the ball is \(T_s = -1.05^{\circ}\mathrm{C}\). The temperature at the center of the ball is \(T_c = -1.05^{\circ}\mathrm{C}\).

Step by step solution

01

Calculate the temperature distribution

We can use the following equation for steady-state heat transfer in a spherical coordinate system: $$\frac{d}{dr}\left(r^2 \frac{d T(r)}{dr}\right) = -\frac{q_{\text{gen}}}{k} r^2$$ First, we calculate \(r^2 \frac{dT}{dr}\): $$\int \frac{d}{dr}\left(r^2 \frac{d T(r)}{dr}\right) = \int -\frac{q_{\text{gen}}}{k}r^2 dr$$ $$r^2 \frac{dT}{dr} = -\frac{q_{\text{gen}}}{3k}r^3 + C_1$$ Now, we can find \(\frac{dT}{dr}\): $$\frac{dT}{dr} = -\frac{q_{\text{gen}}}{3k}r + \frac{C_1}{r^2}$$ Integrating this expression will give us the temperature distribution: $$\int dT(r) = \int \left(-\frac{q_{\text{gen}}}{3k}r + \frac{C_1}{r^2}\right) dr$$ $$T(r) = -\frac{q_{\text{gen}}}{6k}r^2 + C_1 \ln{r} + C_2$$
02

Obtain the constants and solve for temperature

To find the constants \(C_1\) and \(C_2\), we need to apply the boundary conditions of the problem: 1. At the surface of the ball \((r=R)\), the temperature has to satisfy the convective boundary condition: $$h(T_s - T_{\infty}) = k \left. \frac{dT}{dr} \right|_{r=R}$$ 2. The temperature must be finite at the center \((r=0)\), which means that \(C_1\) must be zero to prevent a singularity: $$C_1 = 0$$ The temperature distribution becomes simplified: $$T(r) = -\frac{q_{\text{gen}}}{6k}r^2 + C_2$$ Now we apply the first boundary condition: $$1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}(T_s) = 45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot \frac{d T}{d r} \Big|_{r = R}$$ Calculate the temperature gradient at the surface: $$\frac{dT}{dr} = -\frac{q_{\text{gen}}}{3k}r = -\frac{4.2 \times 10^6 \mathrm{~W} / \mathrm{m}^{3}}{3 \times 45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} \cdot 0.12 \mathrm{~m} = -28 \mathrm{~K} / \mathrm{m}$$ Now we can find \(T_s\): $$T_s = T_{\infty} + \frac{k}{h} \left(\frac{dT}{dr}\right)_{r=R} = 0 + \frac{45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}{1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}} (-28 \mathrm{~K} / \mathrm{m}) = -1.05^{\circ}\mathrm{C}$$ Finally, we can find the temperature at the center of the ball: $$T(0) = -\frac{q_{\text{gen}}}{6k}(0)^2 + C_2 = -1.05^{\circ}\mathrm{C}$$ This leads us to conclude that: 1. The temperature at the surface of the ball is \(T_s = -1.05^{\circ}\mathrm{C}\). 2. The temperature at the center of the ball is \(T_c = -1.05^{\circ}\mathrm{C}\).

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Key Concepts

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

Spherical Heat Transfer
Understanding heat transfer in spherical objects is essential for a range of engineering applications, from nuclear fuel pellets to biomedical implants. Unlike planar or cylindrical geometries, spherical objects have a unique set of mathematical conditions due to their geometry. The heat transfer within a sphere can be described by three modes: conduction, convection, and radiation. In the provided exercise, the heat is generated uniformly within the sphere, and it's assumed that the transfer is steady-state and occurs primarily through conduction.

Spherical symmetry implies that the temperature within the sphere depends only on the radial position, and not on angular positions. The conservation of energy for heat conduction in spherical coordinates leads to the differential equation given in the solution steps. Solving this equation requires integrating twice and applying appropriate boundary conditions to find the constants. This step-by-step process simplifies the complex nature of heat transfer into manageable mathematical solutions.
Temperature Distribution
Temperature distribution refers to how temperature varies within an object. In a spherical object with steady-state heat generation, temperature distribution is vital for assessing the thermal performance and structural integrity of the object. It is influenced by several factors, such as the material's thermal conductivity, the internal heat generation rate, and the boundary conditions.

In our exercise, we derived an equation that predicts the change in temperature across the radius of the sphere. The integration process provided a general solution showing a quadratic dependence of temperature on the radius. The resulting equation can be used to calculate temperature at any point within the sphere, enabling us to conduct a thorough thermal analysis of the system. With the application of boundary conditions, this equation becomes particularly powerful in predicting performance and ensuring the safety and reliability of the spherical object subjected to a thermal load.
Convective Boundary Condition
The convective boundary condition is pivotal in determining how a physical system exchanges heat with its surroundings. When an object's surface is exposed to fluid flow, such as air or water, convective heat transfer plays a crucial role. The condition describes the relationship between the heat transfer coefficient, temperature difference between the surface and the fluid, and the thermal conductivity of the material.

In the exercise, we applied the convective boundary condition to the surface of the sphere to determine the surface temperature. The boundary condition allowed the integration of the heat transfer coefficient (\(h\)) and the ambient temperature (\(T_{\text{infinity}}\)) into our calculations. The heat transfer coefficient is a measure of the convective heat transfer capability of the surrounding fluid and is crucial for accurately predicting the surface temperature. Thus, understanding and applying the right convective boundary condition is a key step in solving real-world heat transfer problems in spherical geometry.

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Most popular questions from this chapter

Consider a large plane wall of thickness \(I_{\text {, }}\) thermal conductivity \(k\), and surface area \(A\). The left surface of the wall is exposed to the ambient air at \(T_{\infty}\) with a heat transfer coefficient of \(h\) while the right surface is insulated. The variation of temperature in the wall for steady one-dimensional heat conduction with no heat generation is (a) \(T(x)=\frac{h(L-x)}{k} T_{\infty}\) (b) \(T(x)=\frac{k}{h(x+0.5 L)} T_{\infty}\) (c) \(T(x)=\left(1-\frac{x h}{k}\right) T_{\infty}\) (d) \(T(x)=(L-x) T_{\infty}\) (e) \(T(x)=T_{\infty}\)

A pipe is used for transporting boiling water in which the inner surface is at \(100^{\circ} \mathrm{C}\). The pipe is situated in a surrounding where the ambient temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The pipe has a wall thickness of \(3 \mathrm{~mm}\) and an inner diameter of \(25 \mathrm{~mm}\), and it has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.003 \mathrm{~K}^{-1}\) and \(T\) is in \(\mathrm{K}\). Determine the outer surface temperature of the pipe.

Consider a spherical shell of inner radius \(r_{1}\) and outer radius \(r_{2}\) whose thermal conductivity varies linearly in a specified temperature range as \(k(T)=k_{0}(1+\beta T)\) where \(k_{0}\) and \(\beta\) are two specified constants. The inner surface of the shell is maintained at a constant temperature of \(T_{1}\) while the outer surface is maintained at \(T_{2}\). Assuming steady one- dimensional heat transfer, obtain a relation for \((a)\) the heat transfer rate through the shell and ( \(b\) ) the temperature distribution \(T(r)\) in the shell.

Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(12 \mathrm{~m}^{2}\). The left side of the wall at \(x=0\) is subjected to a net heat flux of \(\dot{q}_{0}=700 \mathrm{~W} / \mathrm{m}^{2}\) while the temperature at that surface is measured to be \(T_{1}=80^{\circ} \mathrm{C}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the temperature of the right surface of the wall at \(x=L\).

Consider a function \(f(x)\) and its derivative \(d f l d x\). Does this derivative have to be a function of \(x\) ?
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