Question

Benzene-free air at \(25^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) enters a 5 -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 benezene 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

Question: Calculate the evaporation rate of benzene in kg/h based on the following solution steps: 1. Determine the total flow rate of air in the tube. 2. Convert flow rate to the molar flow rate. 3. Calculate the average mass transfer coefficient. 4. Compute benzene concentration in outlet air. 5. Calculate evaporation rate of benzene.

Step-by-step solution

Determine the total flow rate of air in the tube

Calculate the volumetric flow rate of air by using the given average velocity and tube dimensions. Volumetric flow rate (\(Q\)) can be calculated using the formula: \(Q = A * v\), where \(A\) is the cross-sectional area of the tube and \(v\) is the average velocity. The area can be calculated as: \(A = \pi * (D / 2)^2\), where \(D\) is the diameter of the tube. The diameter is \(5\,\text{cm}=0.05\,\text{m}\).

Convert flow rate to the molar flow rate

To find the molar flow rate, we will use the ideal gas equation: \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. The molar flow rate can be calculated as: \(n_m = \dfrac{Q*P_{air}}{R*T}\), where \(P_{air}\) is the partial pressure of air in the stream, \(R\) is the ideal gas constant \(8.314\,\text{J/mol K}\), and \(T\) is the temperature in Kelvin (which is \(25^{\circ}\text{C}+273.15=\,298.15\,\text{K}\)).

Calculate the average mass transfer coefficient (a)

We know that the mass transfer coefficient, \(K\), is related to the mass transfer flux by the following equation: \(K = \dfrac{J}{x}\), where \(J\) is the mass transfer flux and \(x\) is the distance traveled. Using Fick's law, we can calculate the flux, \(J\), as: \(J \approx \frac{\Delta C}{\Delta x}\), where \(\Delta C\) is the difference in concentrations and \(\Delta x\) is the distance over which diffusion occurs. Since the molar flow rate of air is much higher than the benzene flux (due to its low vapor pressure), we can assume that the air molar flow rate remains unchanged during the process. Since we already know the benzene vapor pressure at \(25^{\circ}\text{C}\), we can assume that the molar concentration of benzene in the air at the exit is equal to the vapor pressure (\(13\,\text{kPa}\)). The distance over which diffusion occurs can be assumed to be the length of the tube, which is \(6\,\mathrm{m}\). Hence, the average mass transfer coefficient can be calculated as: \(K = \dfrac{J}{x}\).

Compute benzene concentration in outlet air (b)

Since we know the vapor pressure of benzene at \(25^{\circ}\text{C}\) (\(13\,\text{kPa}\)), we can assume that the molar concentration of benzene in the air at the exit is equal to the vapor pressure. It is important to note that the unit of this molar concentration of benzene in the air is \(kPa\).

Calculate evaporation rate of benzene (c)

The evaporation rate of benzene can be calculated using the concentration at the outlet, found in step 4, and the molar flow rate of air found in step 2. The molar rate of evaporation can be calculated as: \(n_{C_6H_6} = \left(\dfrac{P_{benzene}}{P_{air}}\right)* n_{air}\), where \(P_{benzene}\) is the vapor pressure of benzene in the air, and \(n_{air}\) is the molar flow rate of air. Finally, convert the molar rate of evaporation to mass rate using the molar mass of benzene (\(78.11\,\text{g/mol}\)): \(Evap\_Rate = n_{C_6H_6} * M_{C_6H_6}\). Convert the evaporation rate to \(\mathrm{kg} / \mathrm{h}\). Follow the steps given above to find the values of the average mass transfer coefficient, molar concentration of benzene in the outlet air, and evaporation rate of benzene in \(\mathrm{kg} / \mathrm{h}\).

Key concepts

Benzene Evaporation

Benzene evaporation is a fundamental concept in mass transfer, especially when dealing with volatile organic compounds. Benzene is a colorless liquid with a distinctive odor, and it tends to evaporate quickly because of its high vapor pressure. At a given temperature, molecules of benzene escape from the liquid surface into the air, causing evaporation.
In this context, the vapor pressure of benzene is the key factor, which is the pressure at which benzene vapor coexists with its liquid. At \(25^{\circ}\mathrm{C}\), benzene has a vapor pressure of \(13\,\text{kPa}\). Since this exercise assumes a surface coated with benzene and benzene-free air entering the system, understanding the interplay of these components helps evaluate the benzene evaporation rate.

Mass Transfer Coefficient

The mass transfer coefficient \(K\) is critical for quantifying how fast a substance like benzene moves from one phase to another, such as from a liquid film into the air. This coefficient indicates how effectively benzene vapor leaves the liquid phase beneath it and diffuses into the air passing over the surface.
Calculating the mass transfer coefficient involves using the flux and the concentration difference. In this exercise, the mass transfer flux \(J\) can be estimated using Fick's Law, relating the mass transfer coefficient to the concentration gradient and the geometry of the problem, such as the length of the tube.

Fick's Law

Fick's Law is one of the cornerstones of mass transfer and is used to describe the process of diffusion. Fick's Law states that the rate of transfer of mass per unit area (flux) is proportional to the concentration gradient. Mathematically, this can be expressed as:
\[ J = -D \frac{dC}{dx} \]
where \(J\) is the mass flux, \(D\) is the diffusion coefficient, and \(\dfrac{dC}{dx}\) is the concentration gradient.
In the given exercise, since benzene is present and diffusing through a tube, Fick's Law helps estimate how the benzene vapor concentration changes as a function of position from the tube's entry to exit. Specifically, it aids in determining the average mass transfer coefficient across the entire length of the tube.

Volumetric Flow Rate

Volumetric flow rate is a pivotal aspect in mass transfer exercises, as it describes how much volume is flowing through a system per unit time. For instance, in this exercise, understanding the volumetric flow rate of air is essential to establish downstream conditions.
The volumetric flow rate \(Q\) can be calculated from the known cross-sectional area of the tube and the air velocity, using the formula:
\[ Q = A \times v \]
where \(A\) is the area and \(v\) is the mean velocity. This gives insight into the capacity of air flowing through the tube, which in turn affects how much benzene can evaporate and become entrained in the airstream.