Benzene \((M=78.11 \mathrm{~kg} / \mathrm{kmol})\) is a carcinogen, and exposure to benzene increases the risk of cancer and other illnesses in humans. A truck transporting liquid benzene was involved in an accident that spilled the liquid on a flat highway. The liquid benzene forms a pool of approximately \(10 \mathrm{~m}\) in diameter on the highway. In this particular windy day at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) with an average wind velocity of \(10 \mathrm{~m} / \mathrm{s}\), the liquid benzene surface is experiencing mass transfer to air by convection. Nearby at the downstream of the wind is a residential area that could be affected by the benzene vapor. Local health officials have assessed that if the benzene level in the air reaches \(500 \mathrm{~kg}\) within the hour of the spillage, residents should be evacuated from the area. If the benzene vapor pressure is \(10 \mathrm{kPa}\), estimate the mass transfer rate of benzene being convected to the air, and determine whether the residents should be evacuated or not.

Step by step solution

01

Estimate the mass transfer coefficient

To estimate the mass transfer rate, we first need to determine the mass transfer coefficient. We can use the following simplified equation for the mass transfer coefficient in windy conditions, considering the vapor pressure of benzene: \(k = 0.01 \mathrm{~m} / \mathrm{s}\) This equation gives a rough estimate of the mass transfer coefficient, which is affected by wind. The higher the wind velocity, the higher the mass transfer coefficient.

02

Calculate the area of the liquid pool

Next, we calculate the area of the liquid benzene pool on the highway. The pool is assumed to be circular with a diameter of \(10 \mathrm{~m}\). To find the area, use the formula for the area of a circle: \(A = \pi r^{2}\) where \(A\) is the area, and \(r\) is the radius. The radius of the pool is half of its diameter, which is \(10/2 = 5 \mathrm{~m}\). \(A = \pi (5)^{2} = 25\pi \approx 78.54 \mathrm{~m}^2\)

03

Calculate the evaporative flux

Now we can find the evaporative flux of benzene at the surface of the pool. The evaporative flux, \(N_{Benzene}\), is equal to the product of the mass transfer coefficient (\(k\)) and the concentration difference, which is the vapor pressure of the benzene (\(P_{Benzene}\)) divided by the constant gas (\(R\)) and the absolute temperature in Kelvin (\(T\)): \(N_{Benzene} = k \cdot \frac{P_{Benzene}}{RT}\) Given \(P_{Benzene} = 10\mathrm{kPa}\), and using \(R = 8.314 \mathrm{~J} / \mathrm{(kmol \cdot K)}\) and the temperature in Kelvin (which is \(25^{\circ}\mathrm{C} + 273.15 = 298.15\) K ), we can find the evaporative flux: \(N_{Benzene} = 0.01 \cdot\frac{10\cdot10^{3}}{8.314 \cdot 298.15} \approx 4.02 \mathrm{~kmol} / \mathrm{m}^2\mathrm{s}\)

04

Calculate the mass transfer rate

Now we can calculate the mass transfer rate by multiplying the evaporative flux by the area of the benzene pool: \(Mass\ Transfer\ Rate = N_{Benzene} \cdot A \cdot M\) \(Mass\ Transfer\ Rate = 4.02 \cdot 78.54 \cdot 78.11\approx 24905.50 \mathrm{~kg} / \mathrm{kmol}\cdot\mathrm{s}\)

05

Determine if evacuation is necessary

Finally, we need to find out how much benzene will evaporate in one hour. We convert the mass transfer rate to kilograms per hour: \(Mass\ Transfer\ Rate\ (hourly) = 24905.50 \cdot 3600 \approx 896596.80 \mathrm{~kg} / \mathrm{hr}\) Since the mass transfer rate of benzene vapor per hour (\(896596.80\mathrm{~kg}\)) is greater than the considered threshold of \(500\mathrm{~kg}\), the residents should be evacuated from the area to ensure their safety.

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