Question

The tube in a heat exchanger has a 2 -in inner diameter and a 3 -in outer diameter. The thermal conductivity of the tube material is $0.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{F}$, while the inner surface heat transfer coefficient is $50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2},{ }^{\circ} \mathrm{F}$ and the outer surface heat transfer coefficient is $10 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$. Determine the overall heat transfer coefficients based on the outer and inner surfaces.

Short answer

Question: Calculate the overall heat transfer coefficients based on the inner and outer surfaces of a tube with an inner diameter of 2 inches, an outer diameter of 3 inches, a thermal conductivity of 0.5 Btu/hr-ft-°F, an inner surface heat transfer coefficient of 50 Btu/hr-ft²-°F, and an outer surface heat transfer coefficient of 10 Btu/hr-ft²-°F. Answer: The overall heat transfer coefficient based on the inner surface (U1) is approximately _____ Btu/hr-ft²-°F, while the overall heat transfer coefficient based on the outer surface (U2) is approximately _____ Btu/hr-ft²-°F.

Step-by-step solution

1. Calculate the inner and outer radii of the tube

First, convert the given inner and outer diameters of the tube from inches to feet to match the unit of the other given values. One foot is equal to 12 inches. Inner diameter = \(2 \thinspace in = \frac{2}{12} \thinspace ft = \frac{1}{6} \thinspace ft\) Outer diameter = \(3 \thinspace in = \frac{3}{12} \thinspace ft = \frac{1}{4} \thinspace ft\) Then, calculate the inner and outer radii of the tube by dividing the diameters by 2. Inner radius, \(r_1 = \frac{1}{12} \thinspace ft\) Outer radius, \(r_2 = \frac{1}{8} \thinspace ft\)

2. Compute the thermal resistance of each layer

Calculate the thermal resistance of each layer of the tube (conducting layer and convective layers at inner and outer surfaces). Thermal resistance of conductive layer: \(R_{cond} = \frac{\ln(\frac{r_2}{r_1})}{2 \pi k}\), where k is the thermal conductivity of the tube material. Thermal resistance of convective inner and outer layers: \(R_{conv,1} = \frac{1}{h_1 \cdot 2 \pi r_1}\) (inner surface) \(R_{conv,2} = \frac{1}{h_2 \cdot 2 \pi r_2}\) (outer surface) Plug in the values: \(R_{cond} = \frac{\ln(\frac{1/8}{1/12})}{2 \pi (0.5)}\) \(R_{conv,1} = \frac{1}{50 \cdot 2 \pi \frac{1}{12}}\) \(R_{conv,2} = \frac{1}{10 \cdot 2 \pi \frac{1}{8}}\)

3. Compute the overall heat transfer coefficient based on inner and outer surfaces

Calculate the overall heat transfer coefficient based on the inner and outer surfaces of the tube using the following formula: \(U_1 = \frac{1}{R_{conv,1} + R_{cond}}\) \(U_2 = \frac{1}{R_{conv,2} + R_{cond}}\) Plug in the values: \(U_1 = \frac{1}{R_{conv,1} + R_{cond}}\) \(U_2 = \frac{1}{R_{conv,2} + R_{cond}}\)

4. Calculate the final results

Compute the numerical values for \(U_1\) and \(U_2\): \(U_1\) = Overall heat transfer coefficient based on inner surface \(U_2\) = Overall heat transfer coefficient based on outer surface After calculating the values, present the final results.