Question

Ambient air at \(20^{\circ} \mathrm{C}\) flows over a 30 -cm-diameter hot spherical object with a velocity of \(4.2 \mathrm{~m} / \mathrm{s}\). If the average surface temperature of the object is \(200^{\circ} \mathrm{C}\), the average convection heat transfer coefficient during this process is (a) \(8.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(15.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(18.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(21.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(32.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (For air, use $k=0.2514 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7309, \nu=1.516 \times\( \)\left.10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{\infty}=1.825 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{s}=2.577 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)$

Short answer

Based on the provided information and calculations, determine the closest option for the average convection heat transfer coefficient. Answer: (e) \(32.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Step-by-step solution

Calculate Reynolds number

To calculate Reynolds number, we use the formula: \(Re = \frac{\rho Vd}{\mu}\) where \(Re\) is Reynolds number, \(V\) is the velocity of air (\(4.2 \mathrm{~m} / \mathrm{s}\)), \(d\) is the diameter of the spherical object (\(0.3 \mathrm{~m}\)), \(\rho\) is the air density (we can calculate it using \(\rho=\frac{\mu}{\nu}\)), and \(\mu\) is the dynamic viscosity of air (\(1.825 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\)). First, we need to calculate the air density: Air Density (\(ρ\)) = \(\frac{\mu}{\nu} = \frac{1.825 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}}{1.516 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}} = 1.204~\mathrm{kg/m^3}\) Now, we can calculate Reynolds number: \(Re = \frac{(1.204~\mathrm{kg/m^3})(4.2~\mathrm{m/s})(0.3~\mathrm{m})}{1.825 \times 10^{-5}~\mathrm{kg/m\cdot s}} \approx 8381\)

Calculate Nusselt number

We use the following correlation for calculating Nusselt number for flow over a sphere: \(Nu = 2 + 0.43 Re^{1/2} Pr^{1/3}\) where \(Nu\) is the Nusselt number, \(Re\) is the Reynolds number (8381), and \(Pr\) is the Prandtl number (0.7309). Now, we can calculate Nusselt number: \(Nu = 2 + 0.43(8381)^{1/2}(0.7309)^{1/3} \approx 39.47\)

Calculate the average convection heat transfer coefficient

Now that we have the Nusselt number, we can find the average convection heat transfer coefficient using the formula: \(h = \frac{Nu \cdot k}{d}\) where \(h\) is the average convection heat transfer coefficient, \(Nu\) is the Nusselt number (39.47), \(k\) is the thermal conductivity of air (0.2514 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\)), and \(d\) is the diameter of the spherical object (0.3 \(\mathrm{~m}\)). Calculating the average convection heat transfer coefficient: \(h = \frac{(39.47)(0.2514 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})}{0.3 \mathrm{~m}} \approx 32.9~\mathrm{W/m\cdot K}\) Comparing the calculated value with the given options, the closest option is: (e) \(32.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)