Question

Methanol \(\left(\rho=791 \mathrm{~kg} / \mathrm{m}^{3}\right.\) 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. P14-111). 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 \((x=L)\).

Short answer

Answer: The evaporation rate of methanol from the tube is \(1.74 \times 10^{-2} \,\mathrm{kg} \, \mathrm{h}^{-1}\).

Step-by-step solution

Convert given units to SI units

To have a consistent unit system, we will convert all given quantities to the SI unit system: Area of the tube = \(0.8 \mathrm{cm}^{2} = 8 \times 10^{-5} \mathrm{m}^2\) Planar height of the tube = \(30 \mathrm{cm}=0.3 \mathrm{m}\) Mass diffusivity, \(D_{AB} =0.162 \mathrm{cm}^{2} / \mathrm{s}=1.62 \times 10^{-5} \mathrm{m}^{2} / \mathrm{s}\) Pressure of methanol = \(17 \mathrm{kPa} = 17000 \mathrm{Pa}\)

Apply Fick's Law to find the evaporation rate

Now, we will apply Fick's Law, which states that flux of A through B is given by: \(N_A = -D_{AB} \frac{dY_A}{dz}\) Since the methanol concentration at the top of the tube(\(z=0\)) is zero, we can integrate Fick's Law and write the molar flux of methanol as: \(N_A = D_{AB} \frac{P_A}{RT} \frac{dY_A}{dz}\) To find the gradient, we will use the ideal gas law relation: \(Y_A P = \rho_A RT\)

Calculate the evaporation rate of methanol

We need to find the molar fraction, Y_A, of methanol at the evaporating surface (\(z = L = 0.3\,\text{m}\)). We have: \(Y_A = \frac{P_A}{P} = \frac{17000}{101325} = 0.1679\) As we know the concentration of methanol at the top of the tube (z=L) and at z=0, we can determine the gradient \(dY_A/dz\) by: \(\frac{dY_A}{dz} = \frac{0.1679-0}{0.3}\) Using these values in Fick's Law equation: \(N_A = (1.62 \times 10^{-5}) \frac{17000}{(8.314 \times 298)} \frac{0.1679}{0.3}\) \(N_A = 1.891 \times 10^{-5} \mathrm{kmol} \, \mathrm{m}^{-2} \,\mathrm{s}^{-1}\) Now, we are able to determine the evaporation rate by multiplying the molar flux by the cross-sectional area of the tube and the molar mass of methanol: \(Evap. Rate = N_A \cdot A \cdot M = 1.891 \times 10^{-5} \cdot 8 \times 10^{-5} \cdot 32\) The evaporation rate is in \(\mathrm{kmol}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}\), to convert it to \(\mathrm{kg}\, \mathrm{h}^{-1}\), multiply by \(3600\,\text{s}\): \(Evap. Rate = 1.891 \times 10^{-5} \cdot 8 \times 10^{-5} \cdot 32 \cdot 3600 = 1.74 \times 10^{-2} \mathrm{kg} \, \mathrm{h}^{-1}\) The evaporation rate of the methanol is \(1.74 \times 10^{-2} \,\mathrm{kg} \, \mathrm{h}^{-1}\).

Calculate mole fraction of methanol vapor along the tube height

To determine the mole fraction of methanol vapor along the height, we will use Fick's Law: \(Y_A = \int -\frac{D_{AB}}{RT} \cdot N_A dz\) where we know the mole fraction \(Y_A(L) = 0.1679\). For different distances \((z)\) along the height, we can determine the mole fraction, \(Y_A =\int -\frac{D_{AB}}{RT} \cdot N_A dz\). Then, plot the mole fraction as a function of height from the methanol surface to the top of the tube. In conclusion, we found that the evaporation rate of methanol is \(1.74 \times 10^{-2} \,\mathrm{kg} \, \mathrm{h}^{-1}\). The plot of the mole fraction of methanol vapor along the height can be constructed by evaluating the integral \(Y_A =\int -\frac{D_{AB}}{RT} \cdot N_A dz\) for different heights.