Question

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.

Short answer

Explain your answer. Answer: Yes, the residents should be evacuated from the area due to the benzene spill. The mass transfer rate of benzene vapor per hour was calculated to be 896596.80 kg/hr, which is greater than the considered threshold of 500 kg. This high amount of benzene vapor in the air can pose a significant risk to the residents' health and safety, making evacuation necessary.

Step-by-step solution

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.

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 \(10/2 = 5 \mathrm{~m}\). \(A = \pi (5)^{2} = 25\pi \approx 78.54 \mathrm{~m}^2\)

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: \(N_{Benzene} = 0.01 \cdot\frac{10\cdot10^{3}}{8.314 \cdot 298.15} \approx 4.02 \mathrm{~kmol} / \mathrm{m}^2\mathrm{s}\)

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: \(Mass\ Transfer\ Rate = N_{Benzene} \cdot A \cdot M\) \(Mass\ Transfer\ Rate = 4.02 \cdot 78.54 \cdot 78.11\approx 24905.50 \mathrm{~kg} / \mathrm{kmol}\cdot\mathrm{s}\)

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.

Key concepts

Evaporative Flux

Evaporative flux represents the rate at which a substance, such as benzene in this scenario, transitions from a liquid state to a gaseous state at the surface. When we talk about a scenario like a benzene spill, it is crucial to calculate the evaporative flux to understand how rapidly the liquid is turning into vapor and poses a potential risk to nearby areas.
The evaporative flux, denoted as \(N_{Benzene}\) for benzene, is calculated as a product of the mass transfer coefficient and the concentration gradient. This flux is a critical measure in our problem, as it provides the quantitative value indicating how much benzene is being transferred from the liquid pool into the air above. The formula is:

- \(N_{Benzene} = k \times \frac{P_{Benzene}}{RT}\)

In the formula, \(k\) is the mass transfer coefficient, \(P_{Benzene}\) is the vapor pressure of benzene, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
This approach simplifies the way to see the impact of evaporative flux, since higher flux values indicate a greater volume of benzene transitioning into vapor, increasing potential airborne risks. Understanding and managing this parameter is essential for safety assessments during chemical spills.

Mass Transfer Coefficient

The mass transfer coefficient \(k\) is a crucial factor when assessing how substances like benzene move from a liquid to a gaseous form due to the transfer to the surrounding air. In essence, this coefficient measures the efficiency at which molecules are swapped between phases.
For benzene evaporating from a pool, the mass transfer coefficient is influenced significantly by wind velocity, as seen on windy days. As the wind increases, so does the coefficient, hence amplifying the transfer rate of benzene into the air. In this problem, a given estimate for \(k\) is \(0.01\) m/s.
This value embodies a rough measure of mass transfer effectiveness. It's pivotal for performing correct calculations because it directly affects the evaporative flux. A small change here could lead to considerable variations in the mass flux estimates, which makes its determination a central part of the problem when looking at potential health risks from chemical exposure.

Vapor Pressure

Vapor pressure is crucial in understanding the volatility of a substance like benzene. It denotes the pressure exerted by the vapor in equilibrium with its liquid at a given temperature. This characteristic essentially tells us how much of the liquid will evaporate into the air under specific conditions.
In the benzene spill scenario, vapor pressure \(P_{Benzene}\) is necessary to calculate the evaporative flux. The given vapor pressure for benzene here is \(10\) kPa. This high value signifies benzene's propensity to convert into vapor readily, even in moderate conditions like the specified \(25^{\circ} \mathrm{C}\) temperature of the problem scenario.
Understanding vapor pressure aids in predicting exposure risks from volatile compounds. A high vapor pressure means that more of the substance will evaporate quickly, which could lead to increased airborne concentrations and requires detailed calculations to ensure safety when dealing with potential hazardous exposure, especially in residential zones near spill sites.