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Fick s first diffusion law states that $$\text { Mass flux }=-D \frac{d c}{d x}$$ where mass flux \(=\) the quantity of mass that passes across a unit area per unit time \(\left(\mathrm{g} / \mathrm{cm}^{2} / \mathrm{s}\right), D=\) a diffusion coefficient \(\left(\mathrm{cm}^{2} / \mathrm{s}\right)\) \(c=\) concentration, and \(x=\) distance \((\mathrm{cm}) .\) An environmental engineer measures the following concentration of a pollutant in the sediments underlying a lake \((x=0\) at the sediment-water interface and increases downward): $$\begin{array}{l|ccc} x, \mathrm{cm} & 0 & 1 & 3 \\ \hline c, 10^{-6} \mathrm{g} / \mathrm{cm}^{3} & 0.06 & 0.32 & 0.6 \end{array}$$ Use the best numerical differentiation technique available to estimate the derivative at \(x=0 .\) Employ this estimate in conjunction with Eq. (P24.6) to compute the mass flux of pollutant out of the sediments and into the overlying waters \(\left(D=1.52 \times 10^{-6} \mathrm{cm}^{2} / \mathrm{s}\right)\) For a lake with \(3.6 \times 10^{6} \mathrm{m}^{2}\) of sediments, how much pollutant would be transported into the lake over a year's time?

Short Answer

Expert verified
Using the forward difference method, the estimated derivative at x = 0 is \(0.26 \times 10^{-6} \frac{\text{g}}{\text{cm}^3\text{cm}}\). The mass flux of pollutant is approximately \(-4.072 \times 10^{-13} \frac{\text{g}}{\text{cm}^2\text{s}}\), indicating it's moving out of the sediments and into the overlying waters. The total mass of pollutant transported into the lake over a year is approximately 50.87 g.

Step by step solution

01

Estimate the derivative at x = 0 using a numerical differentiation technique

Since we have only three data points, let's use the forward difference method to estimate the derivative at x = 0. The forward difference formula is given by: $$\frac{d c}{dx} \approx \frac{c(x + h) - c(x)}{h}$$ where h is the step size in x. At x = 0, h = 1 cm, we can approximate the derivative as follows: $$\frac{d c}{d x}(0) \approx \frac{0.32 \times 10^{-6} - 0.06 \times 10^{-6}}{1} = 0.26 \times 10^{-6} \frac{\text{g}}{\text{cm}^3\text{cm}}$$
02

Calculate the mass flux using Fick's first diffusion law

Now, using Fick's first diffusion law, we can calculate the mass flux: $$\text{Mass flux} = -D \frac{d c}{d x}(0)$$ Substitute the values D = 1.52 脳 10鈦烩伓 cm虏/s and the estimated derivative: $$\text{Mass flux} = -\left(1.52 \times 10^{-6}\right) \left(0.26 \times 10^{-6}\right) = -4.072 \times 10^{-13} \frac{\text{g}}{\text{cm}^2\text{s}}$$ The negative sign indicates that the mass flux of pollutant is moving out of the sediments and into the overlying waters.
03

Calculate the total amount of pollutant transported into the lake over a year

With the mass flux value obtained in step 2, we can calculate the total mass of pollutant transported into the lake over a year: Total mass of pollutant = Mass flux 脳 Area 脳 Time Convert the sediment area from m虏 to cm虏, and time in seconds for one year: Area = 3.6 脳 \(10^{6}\) m虏 脳 \(\Big(\frac{100 \text{ cm}}{1 \text{ m}}\Big)^2\) = \(3.6 脳 10^{10}\) cm虏 Time = 1 year 脳 365 days/year 脳 24 hours/day 脳 60 minutes/hour 脳 60 seconds/minute = \(3.1536 脳 10^{7}\) s Total mass of pollutant = (-4.072 脳 \(10^{-13}\) g/cm虏s) 脳 (\(3.6 脳 10^{10}\) cm虏) 脳 (\(3.1536 脳 10^{7}\) s) 鈮 -50.87 g The negative sign indicates that the total mass of pollutant has moved out of the sediments and into the overlying waters. The total mass of pollutant transported into the lake over a year is approximately 50.87 g.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Numerical Differentiation
Numerical differentiation is a method used to estimate the derivative of a function using discrete data points. It is particularly useful when dealing with functions that are challenging to differentiate analytically. In this exercise, we have concentration measurements at different depths of sediment in a lake, and we use these to estimate the rate of change of concentration with respect to depth.

The technique chosen for this problem is the forward difference method. This is because we have three points, and we want to estimate the change right at the start, specifically at the sediment-water interface which is at depth, x = 0 cm. The forward difference formula is given by: \[\frac{dc}{dx} \approx \frac{c(x + h) - c(x)}{h}\]where \(h\) is the step size between the data points. This method is simple and provides a reasonable approximation when the points are closely spaced, allowing us to approximate the derivative using the change in concentration from the first point to the second.
Mass Flux Calculation
Mass flux is a measure of how much mass of a substance passes through a unit area over a specified period. Applying Fick's first law of diffusion, we can calculate the mass flux of the pollutant from the sediments into the overlying water using the formula: \[\text{Mass flux} = -D \frac{dc}{dx}\]Here, \(D\) is the diffusion coefficient, a constant that signifies how easily the pollutant diffuses through the sediments.

In the problem, we substitute the values of \(D\) and the numerically estimated derivative at \(x = 0\). The negative value of the mass flux indicates the direction of movement from higher concentration in the sediments to lower concentration in the water. Calculating this helps us understand the rate at which pollutants escape into the lake water, an important consideration for environmental monitoring.
Environmental Engineering
Environmental engineering focuses on designing systems and solutions to protect and improve the environment. This problem highlights a typical scenario in environmental engineering where engineers assess pollutant transport and its environmental impact.

The calculation of mass flux is crucial in modeling and predicting how pollutants move from one area to another, such as from the sediments to a lake's open water. This helps in designing interventions to control pollution and informing about the potential long-term impacts on the ecosystem. All these outcomes are vital in establishing regulatory policies and environmental management strategies. Understanding how pollutants travel can assist in mitigating adverse effects on water quality and aquatic life.
Pollutant Transport
Pollutant transport refers to the movement of contaminants through environmental media like water, air, or soil. In this case, we're focused on how contaminants move from sediments into lake water.

Fick's first diffusion law is particularly useful for describing this diffusion-driven process, where the difference in concentration at different depths causes the pollutants to migrate from the sediment into the overlying water. This transport mechanism is a critical component of environmental modeling, allowing us to estimate the rates at which pollutants enter aquatic ecosystems.

By calculating how much pollutant moves into the lake over a year, engineers can gauge the potential risk to the aquatic environment and consider solutions, like remediation technology or preventive measures, to reduce pollutant levels and protect water quality.

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