Question

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 \((x=L)\).

Short answer

Answer: The evaporation rate of methanol is 0.0129 kg/h. The mole fraction of methanol vapor decreases linearly as a function of tube height, ranging from 0.175 at the methanol surface to 0 at the top of the tube.

Step-by-step solution

(Step 1: Calculate the molar concentration of methanol vapor)

(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 \(R = 8.314 \mathrm{m^{3} Pa /(mol K)}\).) $$ C = \frac{P}{R T} = \frac{17,000 \mathrm{Pa}}{(8.314 \mathrm{m^{3} Pa /(mol K)})({273 + 25})\mathrm{K}} = 2.05 \cdot 10^{4} \mathrm{mol/m^3} $$

(Step 2: Calculate the mass flow rate of methanol vapor)

(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.) $$ N_A = -(0.162 \cdot 10^{-4} \mathrm{m^2/s})(0.8 \cdot 10^{-4} \mathrm{m^2})\frac{(2.05 \cdot 10^4 \mathrm{mol/m^3})}{(0.3 \mathrm{m})} $$

(Step 3: Determine the mass flow rate of methanol)

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

(Part a: Calculate the evaporation rate in kg/h)

(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.

(Part b: Plot the mole fraction as a function of tube height)

(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 \(x_A(0.3) = 0\).

Key concepts

Methanol Evaporation

Evaporation of methanol is a process where liquid methanol turns into vapor. This happens in environments like a vertical tube as described in our exercise. Methanol has a unique combination of properties, with a density \(\rho = 791 \, \mathrm{kg/m^3}\) and a molar mass \(M = 32 \, \mathrm{kg/kmol}\). These are crucial for understanding how quickly methanol changes state from liquid to vapor.

Evaporation occurs because methanol molecules at the liquid surface gain enough energy to become gas molecules, escaping into the air. In our example, the methanol is in a tube, which limits the escape area and forms a vapor pressure layer of \(17 \, \mathrm{kPa}\).

Evaporation is affected by factors like temperature, pressure, and area, as these touch on the energy available to molecules and the amount of methanol that can evaporate at once. The calculated evaporation rate of methanol is \(0.0129 \, \mathrm{kg/h}\) based on the environmental and methanol-specific parameters.

Fick's Law of Diffusion

Fick's Law of Diffusion is essential when dealing with the movement of particles in a medium. It is expressed as the equation \(N_A = -D_{AB} A \frac{dC_A}{dx}\), where:

• \(N_A\) is the molar flow rate.
• \(D_{AB}\) is the mass diffusivity.
• \(A\) is the cross-sectional area.
• \(\frac{dC_A}{dx}\) is the concentration gradient.

This law helps calculate the mass flow rate of methanol vapor in the evaporative process. It considers how methanol molecules spread through their environment from high to low concentration areas. The concentration gradient in particular is vital as it drives the diffusion process.

In our problem, methanol's mass diffusivity in air is \(0.162 \, \mathrm{cm^2/s}\), guiding us to determine how quickly methanol vapor can diffuse. Finding the concentration changes along the tube lets us predict the pattern of methanol vapor spread. The calculated molar flow rate is a part of estimating how well the methanol evaporation process proceeds.

Mole Fraction

The mole fraction, denoted as \(x_A\), represents the ratio of the number of moles of a component to the total number of moles in a mixture. It's expressed by the formula:

• \(x_A = \frac{C}{C + C_{air}}\),

where \(C\) is the concentration of methanol and \(C_{air}\) is the concentration of air.

Understanding mole fractions helps quantify exactly how much methanol vapor is present compared to air. In a mixed environment like our tube, knowing how methanol's concentration changes with height is crucial. It helps visualize how methanol spreads from the bottom of the tube to the top, eventually vanishing.

For instance, calculating the mole fraction at the methanol surface yields \(x_{A,0} = 0.175\). This number decreases as you look higher in the tube, showing the distribution and gradual evaporation of methanol.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in understanding gases' behavior. It is represented as \(C = \frac{P}{RT}\), where:

• \(C\) is the molar concentration.
• \(P\) is the pressure of the gas.
• \(R\) is the universal gas constant \(8.314 \, \mathrm{m^3 \,Pa/(mol \, K)}\).
• \(T\) is the temperature in Kelvin.

This law is used prominently to calculate methanol's concentration at the start of the evaporation process, converting methanol's vapor pressure into a functional concentration.

In our specific case, the conversion gives us a concentration \(C = 2.05 \cdot 10^{4} \, \mathrm{mol/m^3}\).

Understanding the Ideal Gas Law simplifies finding out how gas molecules like methanol react to changes in pressure and temperature. This is especially helpful in calculating how evaporation impacts the methanol in our tube scenario.