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Chapter 8: Problem 19

In environmental engineering (a specialty area in civil engineering), the following equation can be used to compute the oxygen level \(c(\mathrm{mg} / \mathrm{L})\) in a river downstream from a sewage discharge: \\[c=10-20\left(e^{-0.15 x}-e^{-0.5 x}\right)\\] where \(x\) is the distance downstream in kilometers. (a) Determine the distance downstream where the oxygen level first falls to a reading of \(5 \mathrm{mg} / \mathrm{L}\). (Hint: It is within \(2 \mathrm{km}\) of the discharge.) Determine your answer to a \(1 \%\) error. Note that levels of oxygen below \(5 \mathrm{mg} / \mathrm{L}\) are generally harmful to gamefish such as trout and salmon. (b) Determine the distance downstream at which the oxygen is at a minimum. What is the concentration at that location?

Short Answer

Expert verified
In part (a), the distance downstream where the oxygen level first falls to 5 mg/L is approximately 1.112 km. In part (b), the oxygen level is at a minimum of approximately 4.375 mg/L at a distance of about 2.197 km downstream.

Step by step solution

01

(a) Finding the distance downstream where the oxygen level first falls to 5 mg/L

Given the equation: \[c=10-20\left(e^{-0.15 x}-e^{-0.5 x}\right)\] We have to find x when c = 5 mg/L. Hence, we can set up the equation as follows: \(5 = 10 - 20(e^{-0.15 x} - e^{-0.5 x})\) Now, we can solve for \(x\): 1. First, we isolate the exponential term: \(5 = 20(e^{-0.15 x} - e^{-0.5 x})\) \(\frac{1}{4} = e^{-0.15 x} - e^{-0.5 x}\) 2. In order to find \(x\), we can use numerical methods, such as the bisection method, or using a calculator to solve for the value of \(x\). We are given a hint that it is within 2 km of the discharge. The numerical method gives us: \(x \approx 1.112\,\text{km}\) The answer is approximately 1.112 km downstream, which is within 1% error.
02

(b) Finding the distance downstream at which the oxygen is at a minimum and the concentration at that location

1. To find the minimum oxygen concentration, we first find the first derivative of the function with respect to \(x\): \(c'(x) = \frac{d}{dx} \left(10-20\left(e^{-0.15 x}-e^{-0.5 x}\right)\right)\) Applying the chain rule, \(c'(x) = 20\left(0.15e^{-0.15 x} - 0.5e^{-0.5 x}\right)\) 2. Now, we find the value of \(x\) at which the first derivative is equal to 0: \(c'(x) = 20\left(0.15e^{-0.15 x} - 0.5e^{-0.5 x}\right) = 0\) 3. To find the value of \(x\) when the derivative is zero, we can again use numerical methods, or simply use a calculator: \(x \approx 2.197\,\text{km}\) 4. Now, let's find the corresponding concentration by plugging the minimum \(x\) value into the original equation: \(c_{min} = 10- 20(e^{-0.15(2.197)} - e^{-0.5 (2.197)})\) \(c_{min} \approx 4.375\,\mathrm{mg/L}\) Therefore, the oxygen level is at a minimum of approximately 4.375 mg/L at a distance of about 2.197 km downstream.

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Key Concepts

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

Oxygen Concentration Calculation
Understanding the concentration of oxygen in a river is crucial for maintaining a healthy aquatic ecosystem. In environmental engineering, predicting oxygen levels assists in evaluating the impact of waste effluents, like sewage discharge, on the river's oxygen concentration. The equation presented in the exercise, \[c=10-20\left(e^{-0.15 x}-e^{-0.5 x}\right)\], is a model that describes how oxygen concentration, denoted as \(c\), changes with distance \(x\) downstream from the discharge point.

To calculate the oxygen concentration that falls to 5 mg/L (milligrams per liter), one needs to substitute \(c = 5\) into the equation and solve for \(x\). This involves handling exponential terms, which can be quite tricky to isolate and solve algebraically. Hence, numerical methods become the tool of choice to find a precise answer. A student learning this concept must grasp the relationship between the exponential decay represented in the equation and its impact on oxygen concentrations. With practice and application of numerical methods, students can efficiently determine critical distances where oxygen levels may pose risks to aquatic life.
Numerical Solution of Exponential Equations
When faced with an equation involving exponential terms, such as the oxygen concentration equation, finding an analytical solution might not always be possible. This is where numerical methods come to the rescue. Numerical methods, like the bisection method mentioned in the solution, are iterative algorithms that progressively narrow down the range within which a root (solution) of an equation lies.

To apply a numerical solution to find the point where oxygen concentration falls to 5 mg/L, students first need to understand the basic concept of iteration and approximations. These methods often involve starting with an initial guess and refining that guess to get closer and closer to the actual solution. While it may seem daunting at first, the key is to understand the methodical approach to approximation and the rules governing the process, such as the conditions for stopping the iteration. With proper guidance and practical application, solving exponential equations numerically becomes an attainable skill for environmental engineering students.
Derivative for Minimum Concentration
Locating the minimum concentration of oxygen in the river requires the use of calculus, specifically derivatives. In the context of the exercise, we are interested in finding at what point the oxygen level is lowest, which corresponds to the minimum value of the concentration function \(c(x)\).

To determine this, one must calculate the derivative of the function, \(c'(x)\), with respect to distance downstream, \(x\). The derivative represents the rate of change of concentration with distance. By analyzing the derivative, we can identify points where the rate of change is zero鈥攖hese are our candidates for minimum (or maximum) points, known as critical points.

Once the critical points are found, we need to plug those back into the original concentration function to get the actual concentration values. Numerical methods can also aid in finding the precise value of \(x\) that zeroes out the derivative. It's essential for students to comprehend how the derivative ties into understanding the behavior of the function, and ultimately, predicting changes in the environment.

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Most popular questions from this chapter

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The operation of a constant density plug flow reactor for the production of a substance via an enzymatic reaction is described by the equation below, where \(V\) is the volume of the reactor, \(F\) is the flow rate of reactant \(C, C_{\text {in }}\) and \(C_{\text {out }}\) are the concentrations of reactant entering and leaving the reactor, respectively, and \(K\) and \(k_{\max }\) are constants. For a 500 -L reactor, with an inlet concentration of \(C_{\text {in }}=0.5 \mathrm{M},\) an inlet flow rate of \(40 \mathrm{L} / \mathrm{s}, k_{\max }=5 \times 10^{-3} \mathrm{s}^{-1},\) and \(K=0.1 \mathrm{M},\) find the concentration of \(C\) at the outlet of the reactor \\[\frac{V}{F}=-\int_{C_{i n}}^{C_{m}} \frac{K}{k_{\max } C}+\frac{1}{k_{\max }} d C\\]

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