The average heat transfer coefficient for air flow over an odd-shaped body is to be determined by mass transfer measurements and using the Chilton-Colburn analogy between heat and mass transfer. The experiment is conducted by blowing dry air at \(1 \mathrm{~atm}\) at a free stream velocity of \(2 \mathrm{~m} / \mathrm{s}\) over a body covered with a layer of naphthalene. The surface area of the body is \(0.75 \mathrm{~m}^{2}\), and it is observed that \(100 \mathrm{~g}\) of naphthalene has sublimated in \(45 \mathrm{~min}\). During the experiment, both the body and the air were kept at \(25^{\circ} \mathrm{C}\), at which the vapor pressure and mass diffusivity of naphthalene are \(11 \mathrm{~Pa}\) and \(D_{A B}=0.61 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. Determine the heat transfer coefficient under the same flow conditions over the same geometry.

Question: Determine the average heat transfer coefficient for air flow over an odd-shaped body using the Chilton-Colburn analogy. The sublimation rate of naphthalene is 0.003 kg/s, the surface area is 2 m², time is 3600 s, vapor pressure is 500 Pa, and mass diffusivity 0.25 × 10⁻⁵ m²/s. Assume air properties at 1 atm pressure and 300 K temperature, with R = 287 J/kg·K, and Cp = 1005 J/kg·K. Solution: Step 1: Calculate mass flux at the surface Calculate the mass flux (ṁ) using the given data: $$ \dot{m}=\frac{m_{\text{naphthalene}}}{A t} = \frac{0.003 kg/s}{2 m^2 \cdot 3600s} = 4.17 \times 10^{-7}\, kg/m^2·s $$ Step 2: Calculate mass transfer coefficient First, calculate the molar concentration (C) of the air using the ideal gas law: $$ C=\frac{P}{R T} = \frac{101325 \,Pa}{287 \,J/kg \cdot K \cdot 300 \,K} = 1.18 \, kg/m^3 $$ Then, calculate the mass transfer coefficient (k_M) using the given equation: $$ k_{M}=\frac{\dot{m}}{C D_{A B}\left(1-\frac{P_{A}}{P}\right)} = \frac{4.17 \times 10^{-7} \,kg/m^2·s}{1.18 \,kg/m^3 \cdot 0.25 \times 10^{-5} \, m^2/s \cdot \left(1- \frac{500 \,Pa}{101325 \,Pa}\right)} = 1.19 \times 10^{-2} \,m/s $$ Step 3: Use the Chilton-Colburn analogy to calculate the heat transfer coefficient To use the Chilton-Colburn analogy, we assume Re and Pr are constant. Based on this assumption, we can rewrite the analogy as follows: $$ h=k_{M} \cdot\left(\frac{Re_{D} \operatorname{Pr}}{Re_{D} \operatorname{Sc}}\right) = k_{M} \cdot \frac{Pr}{Sc} $$ Now, we can calculate the heat transfer coefficient (h) using the mass transfer coefficient (k_M) and Pr/Sc ratio. We will assume the Pr/Sc ratio to be 0.71, which is typical for air at atmospheric pressure and room temperature: $$ h = k_{M} \cdot \frac{Pr}{Sc} = 1.19 \times 10^{-2} \,m/s \cdot 0.71 = 8.45 \times 10^{-3} \,W/m^2·K $$ FINAL STEP: Report the heat transfer coefficient The average heat transfer coefficient for air flow over the odd-shaped body under the given conditions is 8.45 × 10⁻³ W/m²·K.