Question

Consider a 25-m-long thick-walled concrete duct $(k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ of square cross section. The outer dimensions of the duct are \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\), and the thickness of the duct wall is \(2 \mathrm{~cm}\). If the inner and outer surfaces of the duct are at \(100^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer through the walls of the duct.

Short answer

Question: Calculate the rate of heat transfer through the walls of a thick-walled concrete duct with a square cross-section, given the following information: - Length of the duct: \(25 \,\mathrm{m}\) - Thermal conductivity of concrete: \(k = 0.75 \,\mathrm{W/m\cdot K}\) - Outer dimensions of the duct: \(20 \,\mathrm{cm}\times 20 \,\mathrm{cm}\) - Thickness of the duct wall: \(t= 2 \,\mathrm{cm}\) - Inner and outer surface temperatures: \(T_i = 100 ^\circ\mathrm{C}\) and \(T_o = 30 ^\circ\mathrm{C}\) Answer: The rate of heat transfer through the walls of the duct is \(1.05\,\mathrm{W}\).

Step-by-step solution

Identify the given information

The given information in the exercise is as follows: - Length of the duct: \(L=25 \,\mathrm{m}\) - Thermal conductivity of concrete: \(k = 0.75 \,\mathrm{W/m\cdot K}\) - Outer dimensions of the duct: \(20 \,\mathrm{cm}\times 20 \,\mathrm{cm}\) - Thickness of the duct wall: \(t= 2 \,\mathrm{cm}\) - Inner and outer surface temperatures: \(T_i = 100 ^\circ\mathrm{C}\) and \(T_o = 30 ^\circ\mathrm{C}\)

Find the inner dimensions of the duct

First, we need to find the inner dimensions of the duct, which can be calculated by subtracting the thickness of the duct wall from the outer dimensions: Inner dimension: \(20\,\mathrm{cm} - 2\,\mathrm{cm} = 18\,\mathrm{cm}\)

Calculate the surface area of the duct walls

Now, we need to find the surface area of the duct walls. For the square duct, there are four wall surfaces with each having a height and width. The surface area of one of the duct walls can be calculated as: \(A = L \times w\) Surface area for outer surface: \(A_o = 25\,\mathrm{m} \times 0.2\,\mathrm{m} = 5\,\mathrm{m^2}\) Surface area for inner surface: \(A_i = 25\,\mathrm{m} \times 0.18\,\mathrm{m} = 4.5\,\mathrm{m^2}\)

Calculate the thermal resistance

The thermal resistance for a square duct is given by: \(R = \frac{L}{k} \times \frac{1}{A_o - A_i}\) Substituting the given values: \(R = \frac{25\,\mathrm{m}}{0.75\,\mathrm{W/m\cdot K}} \times \frac{1}{5\,\mathrm{m^2} - 4.5\,\mathrm{m^2}}\) \(R = \frac{25\,\mathrm{m}}{0.75\,\mathrm{W/m\cdot K}} \times \frac{1}{0.5\,\mathrm{m^2}}\) \(R = 66.67\,\mathrm{K/W}\)

Calculate the rate of heat transfer

Now, using Fourier's law of heat conduction: \(\dot{Q} = \frac{T_i - T_o}{R}\) \(\dot{Q} = \frac{100^\circ\mathrm{C} - 30^\circ\mathrm{C}}{66.67\,\mathrm{K/W}}\) \(\dot{Q} = \frac{70\,\mathrm{K}}{66.67\,\mathrm{K/W}}\) \(\dot{Q} = 1.05\,\mathrm{W}\) So, the rate of heat transfer through the walls of the duct is \(1.05\,\mathrm{W}\).