Question

Consider a tube for transporting steam that is not centered properly in a cylindrical insulation material \((k=0.73\) $\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\(. The tube diameter is \)D_{1}=20 \mathrm{~cm}$ and the insulation diameter is \(D_{2}=40 \mathrm{~cm}\). The distance between the center of the tube and the center of the insulation is \(z=5 \mathrm{~mm}\). If the surface of the tube maintains a temperature of \(100^{\circ} \mathrm{C}\) and the outer surface temperature of the insulation is constant at \(30^{\circ} \mathrm{C}\),

Short answer

Based on the given information and calculations, the heat transfer rate per unit length along the misaligned tube within the cylindrical insulation material is found to be 67.97 W/m. This solution assumes one-dimensional heat transfer and considers the maximum heat transfer rate through the insulation material.

Step-by-step solution

Convert Units

First, we should convert all dimensions to SI units, i.e., meters: \(D_{1} = 0.2\thinspace \mathrm{m}\) \(D_{2} = 0.4\thinspace \mathrm{m}\) \(z = 0.005\thinspace \mathrm{m}\)

Calculate Mean Radius and Thickness

Next, we need to find the mean radius \((R)\) and thickness \((\Delta R)\) of the cylindrical insulation material: \(R = \frac{D_{1}+D_{2}}{4} = \frac{0.2 + 0.4}{4} = 0.15\thinspace \mathrm{m}\) \(\Delta R = \frac{D_{2}-D_{1}}{2} = \frac{0.4 - 0.2}{2} = 0.1\thinspace \mathrm{m}\)

Find Resistance of Insulation Material

Now, we can find the resistance of the insulation material, \(R_{t}\), using the equation for the resistance of a cylindrical shell: \(R_{t} = \frac{\ln{\frac{R + \Delta R}{R}}}{2 \pi k}\) Since the tube is off-center, we need to adjust R according to the coordinate where max heat rate occurs which is when R = R + z: \(R_{t} = \frac{\ln{\frac{(R+z) + \Delta R}{(R+z)}}}{2 \pi k} = \frac{\ln{\frac{0.15+0.005+0.1}{0.15+0.005}}}{2\pi(0.73)} = 1.0293\thinspace \mathrm{K/W}\)

Calculate Heat Transfer Rate per Unit Length

We can now calculate the heat transfer rate per unit length \((Q)\) using the temperature difference and the resistance: \(Q = \frac{T_{1} - T_{2}}{R_{t}} = \frac{100^{\circ}\mathrm{C} - 30^{\circ}\mathrm{C}}{1.0293\thinspace \mathrm{K/W}} = 67.97\thinspace \mathrm{W/m}\) The heat transfer rate per unit length along the tube is \(67.97\thinspace \mathrm{W/m}\).