Methanol ( \(\rho=791 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\left.M=32 \mathrm{~kg} / \mathrm{kmol}\right)\) undergoes evaporation in a vertical tube with a uniform cross-sectional area of \(0.8 \mathrm{~cm}^{2}\). At the top of the tube, the methanol concentration is zero, and its surface is \(30 \mathrm{~cm}\) from the top of the tube (Fig. P14-104). The methanol vapor pressure is \(17 \mathrm{kPa}\), with a mass diffusivity of \(D_{A B}=0.162 \mathrm{~cm}^{2} / \mathrm{s}\) in air. The evaporation process is operated at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). (a) Determine the evaporation rate of the methanol in \(\mathrm{kg} / \mathrm{h}\) and \((b)\) plot the mole fraction of methanol vapor as a function of the tube height, from the methanol surface \((x=0)\) to the top of the tube \((x=L)\).

(First, we need to find the molar concentration \(C\) of methanol vapor at the methanol surface and then at the top of the tube using the Ideal Gas Law: \(C = \frac{P}{R T}\), where \(R = 8.314 \mathrm{m^{3} Pa /(mol K)}\).) $$ C = \frac{P}{R T} = \frac{17,000 \mathrm{Pa}}{(8.314 \mathrm{m^{3} Pa /(mol K)})({273 + 25})\mathrm{K}} = 2.05 \cdot 10^{4} \mathrm{mol/m^3} $$

(Use Fick's Law of Diffusion to find the mass flow rate of methanol vapor through the tube: \(N_{A} = -D_{AB} A \frac{d C_{A}}{d x}\), where \(D_{AB}\) is the mass diffusivity, \(A\) is the cross-sectional area of the tube, and \(d C_{A}/d x\) is the concentration gradient.) $$ N_A = -(0.162 \cdot 10^{-4} \mathrm{m^2/s})(0.8 \cdot 10^{-4} \mathrm{m^2})\frac{(2.05 \cdot 10^4 \mathrm{mol/m^3})}{(0.3 \mathrm{m})} $$

(Now we can find the mass flow rate of methanol \(m_A\) by multiplying the molar flow rate \(N_A\) by the molar mass \(M_A\): \(m_A = N_A M_A\).) $$ m_{A} = N_{A} M_{A} = (1.12 \cdot 10^{-7} \mathrm{mol/s})(32\,\mathrm{kg/kmol}) = 3.58 \cdot 10^{-6} \mathrm{kg/s} $$

(Finally, convert the mass flow rate of methanol to kg/h.) $$ \text{Evaporation Rate} = 3.58 \cdot 10^{-6} \mathrm{kg/s} \cdot \left(\frac{3600 \mathrm{s}}{1 \mathrm{h}}\right) = 0.0129 \mathrm{kg/h} $$ The evaporation rate of methanol is 0.0129 kg/h.

(Use the mole fraction profile equation \(x_A(x) = x_{A,0} + (N_A/AD_{AB}) \cdot x\), where \(x_A(x)\) is the mole fraction at a given height \(x\) and \(x_{A,0}\) is the mole fraction at the methanol surface (\(x=0\)). We know that the mole fraction at the top of the tube (\(x=L\)) is zero.) First, find the mole fraction at the methanol surface by solving for \(x_{A,0}\): $$ x_{A,0} = \frac{C}{C + C_{air}}. $$ Since we are given the entire system is at 1 atm (101325 Pa) and methanol vapor pressure is 17 kPa (17000 Pa), we can find the concentration of air: $$ C_{air} = \frac{101325-17000}{RT} = 9.67 \cdot 10^{4} \mathrm{mol/m^3}. $$ Now, find the mole fraction at the methanol surface: $$ x_{A,0} = \frac{2.05 \cdot 10^4}{2.05 \cdot 10^4 + 9.67 \cdot 10^4} = 0.175. $$ The mole fraction profile equation is given as: $$ x_A(x) = 0.175 + \frac{-1.12 \cdot10^{-7}}{(0.8 \cdot 10^{-4})(0.162 \cdot 10^{-4})}x. $$ Now, one can plot the mole fraction \(x_A(x)\) as a function of the tube height \(x\) from \(0\) to \(L=0.3\) m. To do this, create a plot with the x-axis representing the tube height \(x\) (ranging from 0 to 0.3 m) and the y-axis representing the mole fraction \(x_A(x)\). Use the equation above to calculate the mole fraction at different heights in the tube to generate points for the plot. The plot will be linear descending from \(x_A(0) = 0.175\) to \(x_A(0.3) = 0\).