Question

Heat is generated steadily in a 3-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 maintaining 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: The optimal thickness of the insulation material is 2.0 cm.

Step-by-step solution

Calculate the rate of heat transfer from the ball surface to the environment

The rate of heat transfer from the ball surface to the environment can be calculated using the formula: \(q = hA \Delta T\) where: - \(q\) is the rate of heat transfer - \(h\) is the heat transfer coefficient, which is given as \(7.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - \(A\) is the surface area of the ball, which can be calculated as \(A = 4 \pi r_{1}^{2}\), where \(r_{1}\) is the radius of the ball. Given the diameter of the ball is \(3\ \mathrm{cm}\), the radius is \(r_1 = 1.5\ \mathrm{cm} = 0.015\ \mathrm{m}\). Therefore, \(A = 4 \pi (0.015\ \mathrm{m})^2 = 2.827 \times 10^{-3}\ \mathrm{m}^{2}\). - \(\Delta T\) is the temperature difference between the ball surface and the ambient air. The ball surface temperature is constant, and the ambient air temperature is given as \(26^{\circ} \mathrm{C}\). Now, we can calculate the rate of heat transfer, \(q\): \(q = 7.5 \frac{\mathrm{W}}{ \mathrm{m}^{2} \cdot \mathrm{K}} \times 2.827 \times 10^{-3}\ \mathrm{m}^{2} \times \Delta T = 0.02120\ \mathrm{W}\cdot \mathrm{K}\cdot \Delta T\)

Calculate the heat transfer through the insulation material

The heat transfer through the insulation material can be calculated using the Fourier's law of heat conduction: \(q = \frac{k \cdot A \cdot \Delta T}{t}\), where - \(q\) is the heat transfer - \(k\) is the thermal conductivity of the insulation material, which is given as \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - \(A\) is the surface area of the ball, which we have calculated in step 1 - \(t\) is the thickness of the insulation material - \(\Delta T\) is the temperature difference across the insulation layer Let's rewrite the equation from step 1 and substitute the value of \(q\): \(\frac{0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot A \cdot \Delta T}{t} = 0.02120\mathrm{W}\cdot\mathrm{K} \cdot \Delta T\)

Find the optimal thickness of the insulation material

Now, we will find the optimal thickness, \(t\), that maximizes heat generation while keeping the surface temperature constant. To do this, we will solve the equation obtained in step 2 for \(t\): \(t = \frac{0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot A \cdot \Delta T}{0.02120\mathrm{W}\cdot\mathrm{K} \cdot \Delta T}\) The temperature difference \(\Delta T\) gets canceled out in both the numerator and denominator: \(t = \frac{0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot A}{0.02120\mathrm{W}\cdot\mathrm{K}}\) Now, we can calculate the optimal thickness, \(t\), by substituting the values of \(k\) and \(A\): \(t = \frac{0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 2.827 \times 10^{-3}\ \mathrm{m}^{2}}{0.02120\mathrm{W}\cdot\mathrm{K}} = 0.020\ \mathrm{m} = 2\ \mathrm{cm}\)

Compare the calculated thickness with the given options

We have calculated the optimal thickness of the insulation material to be \(2\ \mathrm{cm}\). Comparing this value with the given options, we find that the correct answer is: (d) \(2.0\ \mathrm{cm}\)

Key concepts

Thermal Conductivity

Understanding thermal conductivity is essential when studying heat transfer. Thermal conductivity, denoted by the symbol \( k \), is a material property that indicates how well a material can conduct heat. This means it effectively measures how quickly heat passes through a material when exposed to a temperature difference. Thermal conductivity is expressed in units of \( \mathrm{W} / \mathrm{m} \cdot \mathrm{K} \). The larger the value of \( k \), the better the material conducts heat.

• Materials with high thermal conductivities, like metals, will transfer heat rapidly, making them good conductors.
• Materials with low thermal conductivities, like rubber, will transfer heat slowly, making them good insulators.

In the exercise, the ball was covered with a material with a thermal conductivity of \( 0.15 \, \mathrm{W} / \mathrm{m} \cdot \mathrm{K} \). This value indicates modest conductivity, allowing it to serve as an effective insulator without excessively high resistance to heat transfer.

Heat Transfer Coefficient

The heat transfer coefficient, denoted by \( h \), is a crucial factor in the study of convective heat transfer. It quantifies the heat transfer between a solid object and a fluid (like air or water) per unit surface area and per degree of temperature difference.

• This coefficient is expressed in units of \( \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \).
• A higher coefficient signifies that the material facilitates rapid heat transfer between its surface and the surrounding fluid.

In our given problem, the heat transfer coefficient is \( 7.5 \, \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \), which is common for surfaces exposed to air. This value is used to calculate how much heat is lost from the ball's surface to the ambient air. It is an important figure that helps determine how well the insulation material will need to perform to minimize this heat loss.

Insulation Thickness Calculation

Calculating the optimal insulation thickness involves balancing thermal resistance with adequate heat retention. The goal is to regulate heat emission without allowing the surface temperature to change drastically.
In our exercise, the thickness \( t \) is found by comparing the rate of heat transfer through the insulation and the heat loss to ambient air.

• Using Fourier's law of heat conduction, the formula \( q = \frac{k \cdot A \cdot \Delta T}{t} \) enables us to dissect how the material's thermal conductivity \( k \), the surface area \( A \), and thickness \( t \) affect heat transfer.
• Solving the equation gives us an optimal value for \( t \) which, in this case, was calculated to be \( 2 \, \mathrm{cm} \).

This computation ensures that the insulation is neither too thick nor too thin, optimizing the heat generation within the spherical ball while maintaining a consistent surface temperature.