Question

In the effort to find the best way to cool a smooth, thin-walled copper tube, an engineer decided to flow air either through the tube or across the outer tube surface. The tube has a diameter of \(5 \mathrm{~cm}\), and the surface temperature is held constant. Determine \((a)\) the convection heat transfer coefficient when air is flowing through its inside at $25 \mathrm{~m} / \mathrm{s}\( with a bulk mean temperature of \)50^{\circ} \mathrm{C}\( and \)(b)$ the convection heat transfer coefficient when air is flowing across its outer surface at \(25 \mathrm{~m} / \mathrm{s}\) with a film temperature of \(50^{\circ} \mathrm{C}\).

Short answer

Answer: The convection heat transfer coefficient (h) for both cases is \(332.12 \mathrm{~W/m^2K}\).

Step-by-step solution

Calculate the flow properties for both cases

To find the properties of air, we will use the given temperature values. For both cases, the given temperature is \(50^{\circ} \mathrm{C}\). 1. For the flow through the inside tube (case a), we use the mean bulk temperature of air = \(50^{\circ} \mathrm{C}\). 2. For the flow over the outside surface (case b), we use the film temperature of air = \(50^{\circ} \mathrm{C}\). We can use a table or online resources to get the following properties of air for both cases: - Density (\(\rho\)) = \(1.161 \mathrm{~kg/m^3}\) - Dynamic viscosity (\(\mu\)) = \(1.85 \times 10^{-5} \mathrm{~kg/ms}\) - Thermal conductivity (k) = \(0.0276 \mathrm{~W/mK}\) - Specific heat (Cp) = \(1006 \mathrm{~J/kgK}\) - Prandtl number (Pr) = \(0.7\)

Determine the Reynolds number

The Reynolds number (Re) for both cases: 1. For the flow through the inside tube (Re_D), $$Re_D = \frac{\rho V D}{\mu}$$ 2. For the flow across the outer surface (Re_x); NOTE: In this case, the characteristic length is a diameter, $$Re_x = \frac{\rho V D}{\mu}$$

Calculate Reynolds number values

Calculate the Reynolds number values for both cases: 1. \(Re_D = \frac{1.161 \times 25 \times 0.05}{1.85 \times 10^{-5}} = 78,265\) (Turbulent flow) 2. \(Re_x = \frac{1.161 \times 25 \times 0.05}{1.85 \times 10^{-5}} = 78,265\) (Turbulent flow)

Calculate heat transfer coefficients using appropriate correlations

Since the flow is turbulent for both cases, we will use the Dittus-Boelter equation to calculate the heat transfer coefficient: $$h = 0.023(Re_x)^{0.8} (Pr)^{n} \frac{k}{D}$$ Here, n = 0.4 for heating and Nusselt number (Nu) = \(\frac{h \cdot D}{k}\)

Calculate heat transfer coefficients for both cases

Using the Dittus-Boelter equation, determine the heat transfer coefficients for both cases: 1. Case a: Flow through the inside tube $$h_a = 0.023(78,265)^{0.8} (0.7)^{0.4} \frac{0.0276}{0.05} = 332.12 \mathrm{~W/m^2K}$$ 2. Case b: Flow across the outer surface $$h_b = 0.023(78,265)^{0.8} (0.7)^{0.4} \frac{0.0276}{0.05} = 332.12 \mathrm{~W/m^2K}$$

Final Answers:

a) When the air flows through the inside of the tube, the convection heat transfer coefficient (h) is \(332.12 \mathrm{~W/m^2K}\). b) When the air flows across the outer surface of the tube, the convection heat transfer coefficient (h) is \(332.12 \mathrm{~W/m^2K}\).