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 \(500-\) -tank is initially filled with pure water. A copper sulfate solution with a concentration of \(20 \mathrm{g} / \mathrm{L}\) flows into the tank at a rate of \(4 \mathrm{L} / \mathrm{min.}\). The thoroughly mixed solution is drained from the tank at a rate of \(4 \mathrm{L} / \mathrm{min}\).

Short answer

Answer: The equation for the mass of copper sulfate in the tank at any given time is M(t) = 10000 - 10000e^(4t).

Step-by-step solution

Writing the initial value problem for the mass of copper sulfate

Let \(M(t)\) represent the mass of copper sulfate in the tank at time \(t\). We are given that the concentration of copper sulfate in the inflow solution is \(20\text{ g/L}\), and the flow rate is \(4\text{ L/min}\). So, the rate at which copper sulfate enters the tank is \(20 * 4\text{ g/min}\). Since the tank is being drained simultaneously at a rate of \(4\text{ L/min}\), the concentration of copper sulfate in the outflow solution can be represented as \(\frac{M(t)}{500}\text{ g/L}\). Multiplying this by the outflow rate of \(4\text{ L/min}\), we get the rate at which copper sulfate leaves the tank, which is \(\frac{4M(t)}{500}\text{ g/min}\). Now we can set up the initial value problem for the mass of copper sulfate in the tank. The rate of change of mass concerning time can be represented as the difference between inflow rate and outflow rate, which is \(\frac{dM}{dt}=20*4-\frac{4M(t)}{500}\). The initial condition is that the tank is filled with pure water, so \(M(0)=0\).

Solving the initial value problem

We have the following separable differential equation: $$\frac{dM(t)}{dt} = 80-\frac{4M(t)}{500}$$ Multiplying both sides by \(500\), we have: $$500\frac{dM(t)}{dt} = 40000-4M(t)$$ Now, rearrange and make it a separable equation: $$\frac{500}{40000-4M(t)} dM(t) = dt$$ Integrate both sides: $$\int\frac{500}{40000-4M(t)} dM(t) = \int dt$$ Let \(u = 40000-4M(t)\), so \(-4dM(t) = du\) Integrating, we get $$\int\frac{-1}{4u}du = \int dt$$ $$-\frac{1}{4}\ln|u|=t+C_1$$ Now, replace \(u\) with \(40000-4M(t)\) and solve for \(M(t)\) $$-\frac{1}{4}\ln|40000-4M(t)|=t+C_1$$ We know that \(M(0) = 0\), so we can find \(C_1\) by plugging in the initial values. $$-\frac{1}{4}\ln|40000-4*0|=0+C_1$$ $$-\frac{1}{4}\ln|40000|=C_1$$ Now we can write the equation for \(M(t)\): $$\ln|40000-4M(t)| = 4t - \ln|40000|$$ Raise both expressions to the power of \(e\): $$40000-4M(t) = e^{4t}e^{-\ln|40000|}$$ $$40000-4M(t) = 40000e^{4t}$$ Now, solve for \(M(t)\): $$M(t) = 10000 - 10000e^{4t}$$ So, the equation for the mass of copper sulfate in the tank at any given time is $$M(t) = 10000 - 10000e^{4t}$$

Key concepts

Differential Equations

A differential equation is essentially an equation that relates a function to its derivatives. In calculus, they play a crucial role in modeling how physical quantities change over time or space. Differential equations can take many forms, and their complexity can range from simple, linear equations to highly non-linear and difficult to solve ones.

At their core, differential equations are mathematical representations of real-world processes that can't be expressed easily using traditional algebraic equations. These processes could involve rates of change such as velocity, acceleration, population growth, heat dissipation, and many other phenomena where one quantity changes in relation to another.

Separable Differential Equations

Separable differential equations, as the name implies, are a type of differential equations that can be separated into two parts: one that only involves the variable and its differential (e.g., dx) and another that only involves the function and its differential (e.g., dy). This special characteristic allows us to bring all terms involving the independent variable to one side of the equation and all the terms involving the dependent variable to the other side.

To solve a separable equation, we follow these steps: First, separate the variables, then integrate both sides of the equation. After integrating, we finish by solving for the function. Being able to recognize and efficiently solve separable differential equations is paramount when dealing with initial value problems in calculus, as these types of problems often result in such equations.

Mass-Balance Problems

Mass-balance problems in calculus are a practical application of differential equations. These problems involve calculating the quantities of substances (mass) within a system over time as they undergo various processes like dilution, chemical reactions, or flow dynamics.

In the example provided, the system in question is a tank with an inflow and outflow of liquid, and the substance is copper sulfate. We're interested in how the mass of copper sulfate inside the tank changes over time. To set up the mass-balance for the tank, we use the principle of conservation of mass: the rate of mass entering the tank minus the rate of mass leaving the tank equals the rate of change of mass within the tank. These problems are essential in engineering, environmental science, and many other fields where it is crucial to understand how substances move through and interact with systems.