Question

Heat is generated steadily in a \(3-\mathrm{cm}\)-diameter spherical ball. The ball is exposed to ambient air at \(26^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(7.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The ball is to be covered with a material of thermal conductivity $0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The thickness of the covering material that will maximize heat generation within the ball while keeping ball surface temperature constant is (a) \(0.5 \mathrm{~cm}\) (b) \(1.0 \mathrm{~cm}\) (c) \(1.5 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(2.5 \mathrm{~cm}\)

Short answer

Answer: (b) 1.0 cm

Step-by-step solution

Calculate the heat transfer rate

The heat transfer rate between the sphere and ambient air can be calculated using Newton's Law of Cooling: \(q = hA(T_s - T_{\infty})\) , where \(q\) is the heat transfer rate, \(h\) is the heat transfer coefficient, \(A\) is the surface area of the sphere, \(T_s\) is the surface temperature of the sphere, and \(T_{\infty}\) is the ambient temperature. Since \(T_s\) is constant, we can rearrange the equation to express the heat transfer rate in terms of the surface area, heat transfer coefficient, and temperature difference: \(q = hA(T_s - 26)\).

Determine the central temperature

Calculate the central temperature of the sphere by applying the constant surface temperature condition, which dictates that the temperature drop from the center of the sphere to its surface must be the same: \(T_c - T_s = T_{\infty} - T_s\), where \(T_c\) is the central temperature of the sphere. Hence, the central temperature can be found as: \(T_c = 2T_s - 26\).

Calculate the temperature gradient in the covering

The temperature gradient \(\frac{dT}{dr}\) in the covering can be determined by dividing the temperature difference across the thickness of the covering by the thickness: \(\frac{dT}{dr} = \frac{T_s - T_{\infty}}{r_r - r_s}\), where \(r_s\) is the radius of the sphere, and \(r_r\) is the radius of the sphere including the covering material.

Find the heat generation rate and thickness

The heat generation rate \(Q\) can be calculated using Fourier's Law of Heat Conduction: \(Q = -kA\frac{dT}{dr}\), where \(k\) is the thermal conductivity of the covering material. We can then determine the thickness of the covering by rearranging the equation and relating it to the heat transfer rate calculated in Step 1: \(r_r - r_s = \frac{q}{hA} \times \frac{1}{k}\) or \(t = \frac{q}{hkA}\), where \(t\) is the thickness of the covering material.

Calculate the required thickness

Using the equations derived in the previous steps, the required thickness can be computed: \(t = \frac{7.5\times 4\pi\times 0.015^2}{0.15\times 4\pi\times 0.015}= 0.0075\,\mathrm{m} = 0.75\,\mathrm{cm}\).

Check the results against answer choices

Comparing this thickness to the given answer choices, we see that the closest option is (b) \(1.0\,\mathrm{cm}\). Note that although the calculations yielded \(0.75\,\mathrm{cm}\), there is no answer choice matching exactly, so we choose the closest available option to our calculated result: (b) \(1.0\,\mathrm{cm}\).