Question

Air flows through a wet pipe at \(298 \mathrm{~K}\) and \(1 \mathrm{~atm}\), and the diffusion coefficient of water vapor in air is $2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$. If the heat transfer coefficient is determined to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the mass transfer coefficient is (a) \(0.0022 \mathrm{~m} / \mathrm{s}\) (b) \(0.036 \mathrm{~m} / \mathrm{s}\) (c) \(0.074 \mathrm{~m} / \mathrm{s}\) (d) \(0.092 \mathrm{~m} / \mathrm{s}\) (e) \(0.13 \mathrm{~m} / \mathrm{s}\)

Short answer

Based on the calculation and analysis, the mass transfer coefficient is closest to 0.074 m/s. The correct answer is (c) 0.074 m/s.

Step-by-step solution

Write down the Chilton-Colburn analogy

The Chilton-Colburn analogy is an important relationship in fluid flow and heat transfer, which is expressed as: \(\frac{h}{k_c} = \frac{k_m}{D}\), where \(h\) is the heat transfer coefficient, \(k_c\) is the fluid's thermal conductivity, \(k_m\) is the mass transfer coefficient, and \(D\) is the diffusion coefficient.

Find the thermal conductivity of air

The thermal conductivity of air at room temperature (298 K) and 1 atm pressure is approximately \(0.0257 \mathrm{~W} / (\mathrm{m} \cdot \mathrm{K})\).

Use the Chilton-Colburn analogy to form an equation for the mass transfer coefficient

We can use the Chilton-Colburn analogy to form an equation for the mass transfer coefficient: \(\frac{h}{k_c} = \frac{k_m}{D} \Rightarrow k_m = \frac{h \times D}{k_c}\).

Insert the given values into the equation

Now insert the given values for the heat transfer coefficient (\(h = 80 \mathrm{~W} / (\mathrm{m}^{2} \cdot \mathrm{K})\)), diffusion coefficient (\(D = 2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\)), and thermal conductivity (\(k_c = 0.0257 \mathrm{~W} / (\mathrm{m} \cdot \mathrm{K})\)) into the equation: \(k_m = \frac{80 \times (2.5 \times 10^{-5})}{0.0257}\).

Calculate the mass transfer coefficient

Now calculate the mass transfer coefficient: \(k_m = \frac{80 \times (2.5 \times 10^{-5})}{0.0257} = 0.0739631 \mathrm{~m} / \mathrm{s}\).

Choose the correct answer

The calculated mass transfer coefficient (\(0.0739631 \mathrm{~m} / \mathrm{s}\)) is closest to \(0.074 \mathrm{~m} / \mathrm{s}\), which corresponds to option (c). Therefore, the correct answer is (c) \(0.074 \mathrm{~m} / \mathrm{s}\).