Question

Air at \(40^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over an 8-m-long wet plate with an average velocity of \(2.5 \mathrm{~m} / \mathrm{s}\) in order to dry the surface. Using the analogy between heat and mass transfer, determine the mass transfer coefficient on the plate.

Short answer

Question: Calculate the mass transfer coefficient on a wet plate that is 8 meters long with air flowing over it at a temperature of 40°C and a pressure of 1 atm. The average air velocity is 2.5 m/s. Answer: The mass transfer coefficient on the wet plate is approximately 0.086 m/s.

Step-by-step solution

Find the Reynolds number

In order to determine the Reynolds number for this problem, we need the following formula: \(Re = \frac{\rho u L}{\mu}\) where \(Re\) is the Reynolds number, \(\rho\) is the density of air, \(u\) is the average velocity of air, \(L\) is the length of the plate, and \(\mu\) is the dynamic viscosity of air. The density (\(\rho\)) and dynamic viscosity (\(\mu\)) of air at \(40^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) are approximately \(1.127 \mathrm{~kg}/\mathrm{m}^3\) and \(2.08 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\), respectively. Now, we can calculate the Reynolds number: \(Re = \frac{(1.127 \mathrm{~kg}/\mathrm{m}^3)(2.5 \mathrm{~m} / \mathrm{s})(8 \mathrm{~m})}{2.08 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}} \approx 1.08 \times 10^6\)

Use the heat and mass transfer analogy

The Chilton-Colburn analogy relates the heat transfer coefficient (h), mass transfer coefficient (k), and the Reynolds number (Re): \(\frac{h}{k} = \frac{k_c}{k_m} = \frac{C_p \rho}{\rho_m} = \text{Constant}\) where \(k_c\) is the convective mass transfer coefficient, \(k_m\) is the mole mass transfer coefficient, \(C_p\) is the specific heat at constant pressure, and \(\rho_m\) is the molar density. For air at the given conditions, \(C_p = 1.005 \mathrm{~kJ}/(\mathrm{kg} \cdot \mathrm{K}) = 1005 \mathrm{~J}/(\mathrm{kg} \cdot \mathrm{K})\) and \(\rho_m = 44.62 \mathrm{~kmol}/\mathrm{m}^3\). Now, we can calculate the constant in the Chilton-Colburn analogy: \(\text{Constant} = \frac{C_p \rho}{\rho_m} = \frac{(1005 \mathrm{~J}/(\mathrm{kg} \cdot \mathrm{K}))(1.127 \mathrm{~kg}/\mathrm{m}^3)}{44.62 \mathrm{~kmol}/\mathrm{m}^3} \approx 25.43\)

Calculate the mass transfer coefficient

Finally, we can find the mass transfer coefficient (\(k\)) using the following formula, which is derived from the heat transfer coefficient formula for the turbulent flow regime: \(k = \frac{h}{\text{Constant}} = \frac{0.0296 \cdot Re^{0.8} \cdot Pr^{1 / 3}}{25.43}\) Here, \(Pr\) is the Prandtl number, which is approximately 0.7 for air at the given conditions. Now, substitute the values and calculate the mass transfer coefficient: \(k = \frac{0.0296 \cdot (1.08 \times 10^6)^{0.8} \cdot (0.7)^{1 / 3}}{25.43} \approx 0.086 \mathrm{~m} / \mathrm{s}\) The mass transfer coefficient on the plate is approximately \(0.086 \mathrm{~m} / \mathrm{s}\).