Question

A hot plane surface at \(100^{\circ} \mathrm{C}\) is exposed to air at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of $20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The heat loss from the surface is to be reduced by half by covering it with sufficient insulation with a thermal conductivity of \(0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming the heat transfer coefficient to remain constant, the required thickness of insulation is (a) \(0.1 \mathrm{~cm}\) (b) \(0.5 \mathrm{~cm}\) (c) \(1.0 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(5 \mathrm{~cm}\)

Short answer

Answer: 1 cm.

Step-by-step solution

Calculate initial heat loss

To calculate the initial heat loss (Q_initial), we can use the equation for heat loss by convection: Q_initial = hA(T_surface - T_ambient) where h is the heat transfer coefficient, A is the area of the surface, T_surface is the surface temperature, and T_ambient is the ambient temperature. Given h = 20 W/(m^2∙K), T_surface = 100°C, and T_ambient = 25°C, we can calculate Q_initial: Q_initial = 20 A (100 - 25) = 20A(75)

Calculate heat loss with insulation

To find the heat loss with insulation (Q_insulation), we simply divide the initial heat loss by 2: Q_insulation = Q_initial / 2 = (20A(75))/2

Use heat transfer by conduction to find insulation thickness

The equation for heat transfer by conduction is: Q_insulation = (kAΔT)/L where k is the thermal conductivity of insulation, ΔT is the temperature difference between the surface and ambient air, and L is the insulation thickness. Given k = 0.10 W/(m∙K), we can rearrange the equation to find L: L = (kAΔT)/Q_insulation L = (0.10A(75))/((20A(75))/2) We can see the terms A and 75 can be canceled out in both the numerator and the denominator: L = (0.10)/((20)/2) Now we can solve for the insulation thickness L: L = (0.10)/(10) = 0.01 m = 1 cm The required insulation thickness is 1 cm, which corresponds to option (c) in the multiple-choice question.