A mixture of chemicals is flowing in a pipe \(\left(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\right.\), 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.

Step by step solution

01

Define the given variables

We are given the following variables: - Thermal conductivity of the pipe, \(k = 14 \frac{W}{m \cdot K}\) - Inner diameter of the pipe, \(D_i = 2.5 cm = 0.025 m\) - Outer diameter of the pipe, \(D_o = 3 cm = 0.03 m\) - Length of the pipe, \(L = 10 m\) - Mixture temperature, \(T_{mix} = 135 °C\) - Convection heat transfer coefficient of the mixture, \(h_{mix} = 150 \frac{W}{m^2 \cdot K}\) - Insulation thickness, \(t = 2.5 cm = 0.025 m\) - Ambient air temperature, \(T_{amb} = 20 °C\) - Convection heat transfer coefficient of ambient air, \(h_{amb} = 25 \frac{W}{m^2 \cdot K}\) - Desired surface temperature, \(T_s = 45 °C\)

02

Calculate the thermal resistances

We need to find the thermal resistance of the pipe's wall \(R_{wall}\), and the insulated part \(R_{insulation}\). For the pipe's wall: $$ R_{wall} = \frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} $$ For the insulation: $$ R_{insulation} = \frac{\ln(\frac{D_o+2t}{D_o})}{2 \pi k_{ins} L} $$

03

Calculate the total heat transferred for both situations

We will calculate the total heat transferred through the pipe before and after the installation of insulation. Without insulation: $$ Q_{before} = \frac{T_{mix} - T_{amb}}{R_{wall} + \frac{1}{h_{amb} \pi D_o L}} $$ With insulation: $$ Q_{after} = \frac{T_{mix} - T_{s}}{R_{wall} + R_{insulation} + \frac{1}{h_{amb} \pi (D_o + 2t) L}} $$

04

Evaluate the desired thermal conductivity of the insulation

Since we want to maintain the outside surface temperature at 45°C or lower, we need the heat transferred through the insulated pipe, \(Q_{after}\) to be equal or less than the heat transferred if the surface temperature were about 45°C. So: $$ Q_{after} \le Q_{before} $$ Using the expressions from steps 2 and 3, plug in the values and solve for \(k_{ins}\): $$ \frac{T_{mix} - T_{s}}{\frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} + \frac{\ln(\frac{D_o+2t}{D_o})}{2 \pi k_{ins} L} + \frac{1}{h_{amb} \pi (D_o + 2t) L}} \le \frac{T_{mix} - T_{amb}}{\frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} + \frac{1}{h_{amb} \pi D_o L}} $$ Solve for \(k_{ins}\): $$ k_{ins} \ge \frac{\ln(\frac{D_o+2t}{D_o})}{2 \pi L} \left(\frac{T_{mix} - T_{s}}{T_{mix} - T_{amb}} (\frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} + \frac{1}{h_{amb} \pi D_o L}) - \frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} - \frac{1}{h_{amb} \pi (D_o + 2t) L}\right) $$ Plug in the given values and find the appropriate insulation thermal conductivity, \(k_{ins}\).

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