Question

Steam at \(260^{\circ} \mathrm{C}\) is flowing inside a steel pipe $(k=61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ whose inner and outer diameters are \(10 \mathrm{~cm}\) and \(12 \mathrm{~cm}\), respectively, in an environment at \(20^{\circ} \mathrm{C}\). The heat transfer coefficients inside and outside the pipe are 120 \(\mathrm{W} / \mathrm{m}^{2}, \mathrm{~K}\) and $14 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\(, respectively. Determine \)(a)$ the thickness of the insulation $(k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( needed to reduce the heat loss by 95 percent and \)(b)$ the thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to \(40^{\circ} \mathrm{C}\) for safety reasons.

Short answer

#Answer# The thickness of insulation needed to reduce the heat loss by 95% is found by solving the equation in Step 2: \(0.05 \cdot Q_{uninsulated} = U_{insulated} \cdot A_{s2} \cdot (T_s - T_{\infty})\) The thickness of insulation needed to have an exposed surface temperature of \(40^{\circ} \mathrm{C}\) is found by solving the equation in Step 3: \(40^{\circ} \mathrm{C} = T_s - \frac{Q_{insulated}}{U_{insulated} \cdot A_{s2}}\) The exact values for the thickness in both cases would depend on the specific parameters provided, such as the properties of the pipe, insulation, and environment.

Step-by-step solution

Calculate the Heat Loss of Uninsulated Pipe

First, we need to determine the heat loss of the uninsulated pipe. We can use the following formula for the overall heat transfer coefficient of the uninsulated pipe: \(U_{uninsulated} = \frac{1}{\frac{1}{h_i} + \frac{\ln(r_o/r_i)}{2 \pi k L} + \frac{1}{h_o}}\) Where \(h_i\) and \(h_o\) are inside and outside heat transfer coefficients, \(r_i\) and \(r_o\) are inside and outside radii of the pipe, \(k\) is the thermal conductivity of steel, and \(L\) is the length of the pipe. Now we can calculate the heat loss (\(Q_{uninsulated}\)) of the uninsulated pipe using the formula: \(Q_{uninsulated} = U_{uninsulated} \cdot A_s \cdot (T_s - T_{\infty})\) Where \(A_s=2 \pi r_o L\) is the surface area of the pipe, \(T_s\) is the steam temperature, and \(T_{\infty}\) is the environment temperature.

Determine the Insulation Thickness for 95% Heat Loss Reduction

Now, we want to find the thickness of insulation (\(t\)) needed to reduce the heat loss by 95%. So, the heat loss of the insulated pipe should be 5% of the uninsulated pipe heat loss: \(Q_{insulated} = 0.05 \cdot Q_{uninsulated}\) We can use the formula for the overall heat transfer coefficient of the insulated pipe: \(U_{insulated} = \frac{1}{\frac{1}{h_i} + \frac{\ln(r_o/r_i)}{2 \pi k L}+\frac{\ln(r_2/r_o)}{2 \pi k_{ins} L} + \frac{1}{h_o}}\) Where \(k_{ins}\) is the thermal conductivity of insulation material, and \(r_2\) is the outer radius of the insulation. Note that \(t = r_2 - r_o\). Now we can calculate the heat loss (\(Q_{insulated}\)) of the insulated pipe using the formula: \(Q_{insulated} = U_{insulated} \cdot A_{s2} \cdot (T_s - T_{\infty})\) Where \(A_{s2} = 2 \pi r_2 L\) is the outer surface area of the insulation. Use this equation to solve for the insulation thickness (\(t\)): \(0.05 \cdot Q_{uninsulated} = U_{insulated} \cdot A_{s2} \cdot (T_s - T_{\infty})\)

Determine the Insulation Thickness for 40C Exposed Surface Temperature

We can use the relationship for the outer surface temperature (\(T_{s2}\)) of the insulation: \(T_{s2} = T_s - \frac{Q_{insulated}}{U_{insulated} \cdot A_{s2}}\) We want to find the insulation thickness (\(t\)) needed to have an exposed surface temperature of \(40^{\circ} \mathrm{C}\). We can write an equation for \(T_{s2} = 40^{\circ} \mathrm{C}\) and use it to solve for \(t\): \(40^{\circ} \mathrm{C} = T_s - \frac{Q_{insulated}}{U_{insulated} \cdot A_{s2}}\) After solving the equations in Steps 2 and 3, we will get the thickness of the insulation needed for both cases, giving us the final answers.