Question

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem. A one-million-liter pond is contaminated by a chemical pollutant with a concentration of \(20 \mathrm{g} / \mathrm{L} .\) The source of the pollutant is removed, and pure water is allowed to flow into the pond at a rate of \(1200 \mathrm{L} / \mathrm{hr}\). Assuming the pond is thoroughly mixed and drained at a rate of \(1200 \mathrm{L} / \mathrm{hr}\), how long does it take to reduce the concentration of the solution in the pond to \(10 \%\) of the initial value?

Short answer

Answer: The time it takes to reduce the concentration of the pollutant to 10% of its initial value is approximately 1846.12 hours.

Step-by-step solution

Identify the variables and constants

Let \(M(t)\) represent the mass of the pollutant in the pond at any given time \((t)\) measured in grams. The initial mass of the pollutant is given by the product of the volume of the pond and the initial concentration of the pollutant. In this case, the pond has a volume of \(10^{6} \mathrm{L}\) and the initial concentration of the pollutant is \(20 \mathrm{g} / \mathrm{L}\). Initial mass, \(M(0) = (10^6 \mathrm{L}) \times (20 \mathrm{g} / \mathrm{L}) = 2 \times 10^7 \mathrm{g}\). The rate of water flowing in and out of the pond is \(1200 \mathrm{L} / \mathrm{hr}\). The volume of water in the pond remains constant.

Create the initial value problem

Now we have to determine the rate of change of the mass of pollutant in the pond, given by \(dM/dt\). Since the volume of the pond is constant, the rate of change of the mass is equal to -(the concentration of the pollutant × flow rate). Therefore, the initial value problem for the mass of the pollutant can be expressed as: $$\frac{dM}{dt} = -\frac{M(t)}{V} \times R,$$ where \(V\) is the volume of the pond (\(10^6 \mathrm{L}\)) and \(R\) is the flow rate (\(1200 \mathrm{L} / \mathrm{hr}\)).

Solve the initial value problem

To solve the initial value problem, we'll first separate the variables \(M\) and \(t\): $$\int \frac{dM}{M(t)} = -\frac{1200}{10^6} \int dt.$$ Now, integrate both sides: $$\ln|M(t)| = -\frac{1200}{10^6}t + C.$$ Next, find the value of \(C\) using the initial mass \(M(0) = 2 \times 10^7 \mathrm{g}\). By substituting the values, we get: $$\ln(2 \times 10^7) = C.$$ Finally, let's find an expression for \(M(t)\): $$M(t) = 2 \times 10^7 e^{-\frac{1200}{10^6}t}.$$ Now we need to find the time it takes to reduce the mass of the pollutant to \(10 \%\) of the initial value. This means that we need to find the value of \(t\) for which \(M(t) = 0.1 \times 2 \times 10^7\).

Find the time

To find the time \(t\), we set \(M(t) = 0.1 \times 2 \times 10^7\) and solve for \(t\): $$0.1 \times 2 \times 10^7 = 2 \times 10^7 e^{-\frac{1200}{10^6}t}.$$ Divide both sides by \(2 \times 10^7\): $$0.1 = e^{-\frac{1200}{10^6}t}.$$ To isolate \(t\), we need to take the natural logarithm of both sides: $$\ln(0.1) = -\frac{1200}{10^6}t.$$ Finally, solve for \(t\): $$t = -\frac{10^6\ln(0.1)}{1200} \approx 1846.12 \mathrm{hrs}.$$ So, it takes approximately \(1846.12\) hours to reduce the concentration of the pollutant in the pond to \(10 \%\) of the initial value.

Key concepts

Differential Equations

A differential equation is a mathematical tool used to describe the relationship between a function and its derivatives. In essence, it tracks how a quantity changes over time or space. For instance, when we talk about the mass of a pollutant in the pond changing due to water flow, we're delving into the realm of differential equations.

In our pond problem, we have an unknown function \(M(t)\), representing the mass of the pollutant at any given time \(t\). The differential equation we used is \(\frac{dM}{dt} = -\frac{M(t)}{V} \times R\). Here, \(\frac{dM}{dt}\) is the rate at which the mass changes over time.

Differential equations like these are powerful because they allow us to model real-world scenarios and predict future behavior by solving them with given initial conditions.

Mass Balance

Mass balance is a fundamental principle in chemical engineering and environmental science. It tells us that the total mass entering a system should equal the mass leaving the system, plus or minus any accumulation within the system.

In the context of our pond, we're applying mass balance to the pollutant. Initially, the pond is contaminated, and as pure water flows in and out at equal rates, it ensures the pond remains at a constant volume. This balance helps derive the differential equation governing the situation.

• The initial mass of pollutant is \(M(0) = 2 \times 10^7 \text{ g}\).
• The contamination decreases as water flows through, reducing pollutant concentration.

Understanding mass balance aids in setting up the equation that models how the pollutant's mass decreases over time.

Rate of Change

The concept of "rate of change" is critical when dealing with differential equations. It refers to how quickly or slowly a particular quantity changes. In our model, the rate of change is how fast the mass of the pollutant decreases due to the flushing action of the water flowing in and out.

The term \(\frac{dM}{dt}\) in our setup represents this rate of change for the pollutant mass. Each hour, some measurable portion of the pollutant is removed, and this continues over a period until only a desired lower concentration remains.

Grasping the rate of change allows us to predict how long it will take for the pond to reach a specific concentration goal, which in this case was to reduce to 10% of the initial concentration. The step-by-step solution determines this time using integration techniques.

Exponential Decay

Exponential decay refers to a process where quantities diminish at rates proportional to their current value. It is a common phenomenon in physics, chemistry, and other sciences. In this pond scenario, the pollutant undergoes exponential decay as the cleaning process continues.

The equation from our solution, \(M(t) = 2 \times 10^7 e^{-\frac{1200}{10^6}t}\), represents this decay. The pollutant decreases exponentially over time, characterized by an exponential function. This is typical when depicting resources being depleted or populations declining over time.

In simpler terms, the more the pollutant is present, the faster it is removed, which slows down over time as less pollutant remains in the pond. Recognizing exponential decay is key to accurately predicting the timeframe needed to achieve the desired purity level of the pond water.