Question

A mixture of chemicals is flowing in a pipe $\left(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{\rho}=3 \mathrm{~cm}\right.\(, and \)L=10 \mathrm{~m}$ ). During the transport, the mixture undergoes an exothermic reaction having an average temperature of \(135^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of $150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. To prevent any incident of thermal burn, the pipe needs to be insulated. However, due to the vicinity of the pipe, there is only enough room to fit a \(2.5\)-cm-thick layer of insulation over the pipe. The pipe is situated in a plant where the average ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the insulation for the pipe such that the thermal conductivity of the insulation is sufficient to maintain the outside surface temperature at \(45^{\circ} \mathrm{C}\) or lower.

Short answer

Based on the given information and step by step solution, determine the thermal conductivity of the insulation material needed to maintain the outside surface temperature of the insulated pipe at or lower than 45°C. Answer: The insulation material should have a thermal conductivity of approximately 0.0347 W/m⋅K to maintain the outside surface temperature at or below 45°C.

Step-by-step solution

Write down the given information

We have the following information given in the problem: - Pipe thermal conductivity: \(k=14 \,\text{W}/\text{m} \cdot \text{K}\) - Pipe inner diameter: \(D_i=2.5\, \text{cm}\) - Insulation outer diameter: \(D_\rho=3\, \text{cm}\) - Pipe length: \(L=10\, \text{m}\) - Chemical mixture temperature: \(T_c = 135\,^{\circ} \mathrm{C}\) - Mixture convection heat transfer coefficient: \(h_c = 150\, \text{W} / \text{m}^{2} \cdot \text{K}\) - Ambient air temperature: \(T_A = 20\,^{\circ} \mathrm{C}\) - Ambient convection heat transfer coefficient: \(h_A = 25\, \text{W} / \text{m}^{2} \cdot \text{K}\) - Max outside surface temperature of insulated pipe: \(T_\rho \leq 45\,^{\circ} \mathrm{C}\) - Insulation thickness: \(t_\rho = 2.5\, \text{cm}\)

Determine the heat transfer rate per unit length of the pipe

First, we need to find the heat transfer rate per unit length through the pipe. Use the heat transfer equation for the chemical mixture: \(q_c = h_c (T_c - T_A)\) Plugging in the values, we get: \(q_c = 150 \frac{\text{W}}{\text{m}^{2} \cdot \text{K}} (135\,^{\circ} \mathrm{C} - 20\,^{\circ} \mathrm{C}) = 17250 \frac{\text{W}}{\text{m^2}}\).

Determine the heat transfer rate through the insulation

Now, we need to find the heat transfer rate through the insulation. We can use Fourier's law of heat conduction for this purpose: \(q_\rho = \frac{k_\rho A_\rho (T_c - T_\rho)}{t_\rho}\) However, since we know that the heat transfer through the pipe must equal the heat transfer through the insulation (conservation of energy), we can set the heat transfer rate per unit area through the insulation equal to that of the pipe: \(q_c = q_\rho\)

Calculate the thermal conductivity of the insulation material

Now we can rearrange the expression for \(q_\rho\) to solve for the thermal conductivity of the insulation material: \(k_\rho = \frac{q_c t_\rho}{A_\rho (T_c - T_\rho)}\) Here, \(A_\rho\) represents the area under the insulation layer, and since the insulation is cylindrical, its area can be calculated as: \(A_\rho = \pi(D_\rho - D_i)L\) To continue with the thermal conductivity calculation, first, calculate the area \(A_\rho\): \(A_\rho = \pi(3\, \text{cm}-2.5\, \text{cm})(10\, \text{m}) = 15.7\pi\, \text{m}^{2}\) (approximately) Now plug in the values to find the thermal conductivity \(k_\rho\): \(k_\rho = \frac{(17250 \frac{\text{W}}{\text{m^2}})(2.5\, \text{cm})}{(15.7\pi\,\text{m}^2)(135\,^{\circ} \mathrm{C} - 45\,^{\circ} \mathrm{C})}\) We can simplify and convert the units to get: \(k_\rho \approx 0.0347\, \text{W}/\text{m} \cdot \text{K}\) So, the insulation material should have a thermal conductivity of \(0.0347\, \text{W}/\text{m} \cdot \text{K}\) (approx.) to maintain the outside surface temperature at or below \(45\,^{\circ} \mathrm{C}\).