Question

A 1-in-diameter Stefan tube is used to measure the binary diffusion coefficient of water vapor in air at \(80^{\circ} \mathrm{F}\) and \(13.8\) psia. The tube is partially filled with water with a distance from the water surface to the open end of the tube of \(10 \mathrm{in}\). Dry air is blown over the open end of the tube so that water vapor rising to the top is removed immediately and the concentration of vapor at the top of the tube is zero. During 10 days of continuous operation at constant pressure and temperature, the amount of water that has evaporated is measured to be $0.0025 \mathrm{lbm}$. Determine the diffusion coefficient of water vapor in air at \(80^{\circ} \mathrm{F}\) and \(13.8\) psia.

Short answer

Answer: The diffusion coefficient of water vapor in air at the given temperature and pressure is approximately \(0.00221\mathrm{m^2/s} \times \frac{1}{C_0}\), where \(C_0\) is the concentration of water vapor at the bottom of the tube.

Step-by-step solution

Convert units to SI (from imperial system)

Given the use of imperial units in the problem, we need to convert these values to SI units before proceeding. Temperature: \(80^{\circ} \mathrm{F}\), pressure: \(13.8\) psia, diameter: \(1\) in, distance: \(10\) in, mass evaporated: \(0.0025\) lbm. To convert the temperature from Fahrenheit to Celsius: $$ T(°C) = \frac{T(°F) - 32}{1.8} $$ To convert the pressure from psia to atm: $$ P(atm) = P(psia) \times \frac{1}{14.696} $$ To convert the length from inches to meters: $$ L(m) = L(in) \times 0.0254 $$ To convert the mass from lbm to kg: $$ m(kg) = m(lbm) \times 0.453592$$

Calculate temperature, pressure, length, and mass values on SI units

Use the conversion factors to calculate the values. $$ T = \frac{80 - 32}{1.8} = 26.67^{\circ} \mathrm{C} $$ $$ P = 13.8 \times \frac{1}{14.696} = 0.9395 \mathrm{atm} $$ $$ D = 1 \times 0.0254 = 0.0254 \mathrm{m} $$ $$ L = 10 \times 0.0254 = 0.254 \mathrm{m} $$ $$ m = 0.0025 \times 0.453592 = 0.001134 \mathrm{kg} $$

Identify the formula for Fick's Law

Fick's Law of diffusion is given by the expression: $$ J = -D\frac{dC}{dx} $$ where \(J\) is the molar flux, \(D\) is the diffusion coefficient, and \(\frac{dC}{dx}\) is the concentration gradient.

Re-write Fick's Law for evaporation

For the evaporation experiment, we can express the molar flux \(J\) as: $$ J = \frac{m}{At}$$ where \(m\) is the mass evaporated, \(A\) is the cross-sectional area of the tube, and \(t\) is the time. Also, we can express the concentration gradient as: $$\frac{dC}{dx} = \frac{C_0-C_1}{L}$$ where \(C_0\) and \(C_1\) are concentrations at the bottom and top of the tube, and \(L\) is the distance between them.

Combine expressions and solve for the diffusion coefficient

Combine the expressions obtained in steps 3 and 4, and solve for the diffusion coefficient \(D\): $$ D = - \frac{mL}{A t (C_0 - C_1)} $$

Calculate the cross-sectional area of the tube

To calculate the cross-sectional area \(A\) of the tube, we use the formula for the area of a circle: $$ A = \pi (\frac{D}{2})^2 $$ where \(D\) is the diameter of the tube. Plug in the value from step 2: $$ A = \pi (\frac{0.0254}{2})^2 = 5.067 \times 10^{-4} \mathrm{m^2} $$

Calculate the time in seconds

Since the experiment was conducted for 10 days, we will convert this time into seconds: $$ t = 10 \times 24 \times 60 \times 60 = 864,000 \mathrm{s} $$

Calculate the concentration difference

Given that dry air is blown over the open end of the tube, the concentration of vapor at the top of the tube is zero, so \(C_1 = 0\). Since the concentration of vapor at the bottom of the tube is not given, we will use \(C_0 - C_1 = C_0\) as the concentration difference.

Calculate the diffusion coefficient

With all values known, we can now plug them into the equation obtained in step 5 to find the diffusion coefficient \(D\): $$ D = - \frac{0.001134 \times 0.254}{5.067 \times 10^{-4} \times 864,000 \times C_0} $$ Since this exercise only asks you to determine the diffusion coefficient of water vapor in air at \(80^{\circ} \mathrm{F}\) and \(13.8\) psia, you can consider the result in terms of \(C_0\): $$ D = - \frac{0.001134 \times 0.254}{5.067 \times 10^{-4} \times 864,000} \times \frac{1}{C_0} \approx 0.00221\mathrm{m^2/s} \times \frac{1}{C_0} $$ Please note that a more accurate value of \(D\) could be calculated if the concentration \(C_0\) was provided.