Question

Benzene-free air at \(25^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) enters a \(5-\mathrm{cm}\)-diameter tube at an average velocity of $5 \mathrm{~m} / \mathrm{s}$. The inner surface of the 6-m-long tube is coated with a thin film of pure benzene at \(25^{\circ} \mathrm{C}\). The vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(25^{\circ} \mathrm{C}\) is $13 \mathrm{kPa}$, and the solubility of air in benzene is assumed to be negligible. Calculate \((a)\) the average mass transfer coefficient in \(\mathrm{m} / \mathrm{s}\), (b) the molar concentration of benzene in the outlet air, and \((c)\) the evaporation rate of benzene in $\mathrm{kg} / \mathrm{h}$.

Short answer

Answer: In this problem, the average mass transfer coefficient is approximately \(0.288\ \mathrm{m/s}\), the molar concentration of benzene in the outlet air is approximately \(0.193\ \mathrm{mol}\ \mathrm{C6H6}\ /\mathrm{mol}\ \mathrm{Air}\), and the evaporation rate of benzene is approximately \(2.14\ \mathrm{kg/h}\).

Step-by-step solution

Calculate Area and Volume Flow Rate of Air

First, we need to calculate the area and the volume flow rate of air. The area can be found using the diameter of the tube and the volume flow rate can be calculated using the average velocity of air. Area, \(A = \pi (\frac{D}{2})^2 = \pi(\frac{0.05}{2})^2 = 1.9635 \times 10^{-3} \mathrm{m}^2\) Volume flow rate, \(Q = A \times V = 1.9635 \times 10^{-3} \mathrm{m}^2 \times 5 \mathrm{m/s} = 9.8175 \times 10^{-3} \mathrm{m}^3/\mathrm{s}\)

Calculate Molar Flow Rate of Air using Ideal Gas Law

Next, we need to calculate the molar flow rate of air using the volume flow rate and the ideal gas law, PV = nRT. The molar flow rate can be found by: Molar flow rate = \(\frac{PV}{RT} = \frac{101.3\mathrm{kPa} \times 9.8175 \times 10^{-3}\mathrm{m^3/s}}{8.314\mathrm{J/(mol.K)} \times 298\mathrm{K}}\) Molar flow rate = \(0.0395\ \text{mol/s}\) (approx)

Calculate Average Mass Transfer Coefficient of Benzene

Assume that the rate of mass transfer of benzene to air is proportional to the difference in vapor pressures at the tube wall (\(13 \mathrm{kPa}\)) and the air stream (\(0 \mathrm{kPa}\)). Therefore, the mass transfer coefficient can be calculated by: Average mass transfer coefficient, \(k = \frac{QL}{PAT} = \frac{9.8175 \times 10^{-3}\mathrm{m^3/s}\times6\mathrm{m}}{101.3\mathrm{kPa}\times1.9635 \times 10^{-3} \mathrm{m}^2} = 0.288\ \mathrm{m/s}\) (approx)

Calculate Molar Concentration of Benzene at Outlet Air

Now, calculate the molar concentration of benzene in the outlet air. The mass transfer rate of benzene can be calculated using the average mass transfer coefficient: Mass transfer rate of benzene = \(k\times A \times \Delta P = 0.288\ \mathrm{m/s} \times 1.9635 \times 10^{-3} \mathrm{m}^2 \times 13 \mathrm{kPa} = 7.6303 \times 10^{-3}\ \mathrm{mol/s}\) Molar concentration of benzene in outlet air = \(\frac{\text{Mass transfer rate of benzene}}{\text{Molar flow rate of air}}=\frac{7.6303 \times 10^{-3}\ \mathrm{mol/s}}{0.0395\ \mathrm{mol/s}}=0.193\ \mathrm{mol}\ \ce{C6H6}\ \mathrm{/mol}\ \ce{Air}\)

Calculate Evaporation Rate of Benzene

Finally, we can calculate the evaporation rate of benzene in kg/h using the mass transfer rate of benzene: Evaporation rate of benzene = Mass transfer rate of benzene \(\times\) Molar mass of benzene \(\times 3600\) =\( 7.6303 \times 10^{-3}\ \mathrm{mol/s}\times78.11\ \mathrm{g/mol}\times\frac{1\ \mathrm{kg}}{1000\ \mathrm{g}}\times3600\ \mathrm{s/h} =2.14\ \mathrm{kg/h}\) (approx) So the answers are: (a) The average mass transfer coefficient is approximately \(0.288\ \mathrm{m/s}\) (b) The molar concentration of benzene in the outlet air is approximately \(0.193\ \mathrm{mol}\ \mathrm{C6H6}\ /\mathrm{mol}\ \mathrm{Air}\) (c) The evaporation rate of benzene is approximately \(2.14\ \mathrm{kg/h}\)