A spherical container of inner radius \(r_{1}=2 \mathrm{~m}\), outer radius \(r_{2}=2.1 \mathrm{~m}\), and thermal conductivity \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). The container is gaining heat by convection from the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the inner surface temperature of the container to be \(0^{\circ} \mathrm{C},(a)\) express the differential equation and the boundary conditions for steady one- dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container by solving the differential equation, and \((c)\) evaluate the rate of heat gain to the iced water.

Step by step solution

01

Express the differential equation for steady one-dimensional heat conduction through the container

The heat conduction equation for a steady one-dimensional heat conduction in a spherical coordinate system is given by: $$\frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{dT}{dr}\right) = 0$$ where \(T\) is the temperature and \(r\) is the radial distance.

02

Express the boundary conditions

There are two boundary conditions for this problem. The first one is at the inner surface of the container, with an inner radius of \(r_1 = 2\) m, and a temperature of \(0^{\circ} \mathrm{C}\): $$T(r_1) = 0^{\circ} \mathrm{C}$$ The second boundary condition is at the outer surface of the container, with an outer radius of \(r_2 = 2.1\) m. Here, the heat transfer by convection is given by \(q = hA(T - T_{\infty})\), where \(A\) is the surface area of the sphere. Since the heat transfer due to conduction through the material is equal to the heat gained by convection, we can write: $$-kA\frac{dT}{dr}\Big|_{r=r_2} = hA(T(r_2)-T_{\infty})$$

03

Solve the differential equation for the temperature

To find the temperature variation in the container, we first need to solve the heat conduction equation: $$\frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{dT}{dr}\right) = 0$$ Integrating with respect to \(r\), we get: $$r^2\frac{dT}{dr} = C_1$$ Dividing both sides by \(r^2\) and integrating again, we obtain: $$T(r) = -\frac{C_1}{r} + C_2$$

04

Apply the boundary conditions and find the constants

Now, we apply the boundary conditions to find the values of \(C_1\) and \(C_2\): For the first boundary condition, \(T(r_1) = 0^{\circ} \mathrm{C}\): $$0 = -\frac{C_1}{r_1} + C_2$$ For the second boundary condition, we have: $$-k \frac{dT}{dr}\Big|_{r=r_2} = h(T(r_2)-T_{\infty}) \Rightarrow -k\left(-\frac{C_1}{r_2^2}\right) = h\left(-\frac{C_1}{r_2} + C_2 - T_{\infty}\right)$$ Solve this system of two equations to find the values of \(C_1\) and \(C_2\) and substitute them back into the expression for the temperature: $$T(r) = -\frac{C_1}{r} + C_2$$

05

Evaluate the rate of heat gain to the iced water

Finally, the rate of heat gain to the iced water can be evaluated using the heat transfer coefficient and the temperature of the outer surface. We will use the relation: $$q = hA(T(r_2) - T_{\infty})$$ Use the obtained temperature function \(T(r)\) with the substituted values of \(C_1\) and \(C_2\), and plug in the values for \(h\), \(A\), and \(T_{\infty}\) to calculate the rate of heat gain to the iced water.

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