Question

A double-pipe heat exchanger is constructed of a copper $(k=380 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( inner tube of internal diameter \)D_{i}=1.2 \mathrm{~cm}\( and external diameter \)D_{o}=1.6 \mathrm{~cm}$ and an outer tube of diameter \(3.0 \mathrm{~cm}\). The convection heat transfer coefficient is reported to be \(h_{i}=800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the inner surface of the tube and $h_{o}=240 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( on its outer surface. For a fouling factor \)R_{f, i}=0.0005 \mathrm{~m}^{2}, \mathrm{~K} / \mathrm{W}\( on the tube side and \)R_{f, 0}=0.0002 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}$ on the shell side, determine \((a)\) the thermal resistance of the heat exchanger per unit length and \((b)\) the overall heat transfer coefficients \(U_{i}\) and \(U_{o}\) based on the inner and outer surface areas of the tube, respectively.

Short answer

The thermal resistance per unit length is 0.00112 m²·K/W, the overall heat transfer coefficient based on the inner surface area is 357.14 W/m²·K, and the overall heat transfer coefficient based on the outer surface area is 892.86 W/m²·K.

Step-by-step solution

Calculate the thermal resistance of the inner tube wall

First, we need to find the thermal resistance per unit length of the inner copper tube wall. We know that the thermal conductivity of copper is \(k = 380 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and the internal and external diameters of the tube are \(D_i = 1.2 \mathrm{~cm}\) and \(D_o = 1.6 \mathrm{~cm}\). We can use the formula for the radial conduction resistance for a cylindrical wall: \(R_{cond} = \dfrac{\ln(D_o/D_i)}{2 \pi k}\) Substitute the given values and convert the diameters to meters: \(R_{cond} = \dfrac{\ln(0.016/0.012)}{2 \pi (380)} = 6.095 \times 10^{-5} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

Calculate the total thermal resistance

Now, we will calculate the total thermal resistance per unit length of the heat exchanger by considering the fouling factors and the convective resistances on the inner and outer surfaces: \(R_{total} = R_{cond} + R_{f, i} + \dfrac{1}{h_{i}} \cdot \dfrac{D_{i}}{D_{o}} + \dfrac{1}{h_{o}} + R_{f, o}\) Replace the given values: \(R_{total} = 6.095 \times 10^{-5} + 0.0005 + \dfrac{1}{800} \cdot \dfrac{0.012}{0.016} + \dfrac{1}{240} + 0.0002 = 0.00112 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) So, the thermal resistance of the heat exchanger per unit length is \(0.00112 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\).

Calculate the overall heat transfer coefficient \(U_i\)

Now, let's determine the overall heat transfer coefficient based on the inner surface area of the tube. We can use the formula: \(U_i = \dfrac{1}{R_{total}} \cdot \dfrac{D_{o}}{D_{i}}\) Substitute the values: \(U_i = \dfrac{1}{0.00112} \cdot \dfrac{0.016}{0.012} = 357.14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) So, the overall heat transfer coefficient based on the inner surface area of the tube is \(357.14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Calculate the overall heat transfer coefficient \(U_o\)

Finally, we will determine the overall heat transfer coefficient based on the outer surface area of the tube. We can use the formula: \(U_o = \dfrac{1}{R_{total}}\) Substitute the values: \(U_o = \dfrac{1}{0.00112} = 892.86 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) So, the overall heat transfer coefficient based on the outer surface area of the tube is \(892.86 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To summarize, the thermal resistance of the heat exchanger per unit length is \(0.00112 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), the overall heat transfer coefficient based on the inner surface area is \(357.14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the overall heat transfer coefficient based on the outer surface area is \(892.86 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).