Question

A tank contains 1 kilogram of salt dissolved in 100 liters of water. A salt solution of \(0.1 \mathrm{~kg}\) salt \(L\) is pumped into the tank at a rate of \(2 \mathrm{~L} / \mathrm{min}\) and is drained at the same rate. Solve for the salt concentration at time \(t\). Assume the tank is well mixed.

Short answer

The salt concentration at time \( t \) is \( C(t) = \frac{11e^{0.02t} - 10}{100} \) kg/L.

Step-by-step solution

Define the Variables

Let \( S(t) \) be the amount of salt in the tank at time \( t \) in kilograms. The initial condition is \( S(0) = 1 \) kg since the problem states there is initially 1 kg of salt dissolved. The concentration of the incoming solution is 0.1 kg/L.

Set Up Differential Equation

The rate of salt entering the tank is \( 0.1 \times 2 = 0.2 \) kg/min, because the solution is entering at 2 L/min each containing 0.1 kg of salt. The rate of salt leaving the tank is \( \frac{S(t)}{100} \times 2 \), since the tank drains at 2 L/min and \( \frac{S(t)}{100} \) is the concentration of the salt in the tank. The change of salt over time is given by: \( \frac{dS}{dt} = 0.2 - \frac{2S(t)}{100} \).

Simplify the Differential Equation

Rewrite the differential equation as \( \frac{dS}{dt} = 0.2 - 0.02S(t) \).

Solve the Differential Equation

This is a first-order linear differential equation. We can solve it using the method of integrating factors. The integrating factor \( \mu(t) \) is \( e^{\int -0.02 \, dt} = e^{-0.02t} \). Multiply the entire differential equation by the integrating factor to obtain \( e^{-0.02t} \frac{dS}{dt} + e^{-0.02t} \cdot 0.02S = 0.2e^{-0.02t} \).

Integrate Both Sides

The left side of the equation becomes the derivative of \( e^{-0.02t} S(t) \). Integrate both sides to get: \( e^{-0.02t} S(t) = -10e^{-0.02t} + C \), where \( C \) is the constant of integration.

Solve for the Constant of Integration

Using the initial condition \( S(0) = 1 \), substitute into the equation: \( e^0 \cdot 1 = -10e^0 + C \). Solving, we find that \( C = 11 \).

Find the General Solution

Substitute \( C = 11 \) back into the solution: \( e^{-0.02t} S(t) = -10e^{-0.02t} + 11 \). Multiply through by \( e^{0.02t} \) to isolate \( S(t) \): \( S(t) = 11e^{0.02t} - 10 \).

Express the Solution in Terms of Concentration

The concentration of salt at time \( t \) is \( \frac{S(t)}{100} \). Therefore, the concentration is \( C(t) = \frac{11e^{0.02t} - 10}{100} \) kg/L.

Key concepts

Salt Concentration

Salt concentration refers to the amount of salt present in a given volume of solution. In this exercise, we start with water inside a tank containing 1 kilogram of salt in 100 liters of water. As a saline solution flows into the tank, understanding the changes in salt concentration becomes crucial.

The exercise involves a salt solution being pumped in and out, impacting the concentration over time. To determine how concentration varies, we define \( S(t) \,\text{as the amount of salt in kilograms at time}\, t.\). Initially, \( S(0) = 1\) kg. The incoming solution has a salt concentration of 0.1 kg/L.

Finding the concentration of salt over time involves creating a mathematical model. We use an initial value problem where the starting point is known, and then execute a step-by-step approach to find the salt present at any time \( t\, \, \ ext{allowing us to calculate changes in concentration with time.}\).

• A consistent inflow and outflow ensure that the solution inside is well-mixed.
• The goal is to adjust the mathematical model until it reflects the changing concentration as accurately as possible.

Integrating Factor Method

The Integrating Factor Method is a common technique to solve first-order linear differential equations. In our scenario, the equation can be simplified to \( \frac{dS}{dt} = 0.2 - 0.02S(t)\). This form of equation is perfect for applying the integrating factor method, which helps in finding a function solution over time.

The method begins by identifying the integrating factor, \( \mu(t) = e^{\int -0.02 dt} = e^{-0.02t}\). By multiplying through by \( \mu(t),\, \ ext{we transform the original equation into a simpler form that allows integration.}\)

Upon integration, the left side simplifies to the derivative of \( e^{-0.02t} S(t)\), making the equation easier to solve:

• This equation, once integrated, provides a clear path to solving the general solution.
• The integrating factor technique is a powerful method as it transforms complex differential equations into understandable ones.

This method, though sometimes challenging, is critical for situations where variables change linearly with time.

First-order Linear Differential Equation

The differential equation presented in this problem, \( \frac{dS}{dt} = 0.2 - 0.02S(t)\), is a classic example of a first-order linear differential equation. These types of equations have wide applications in physics and engineering, making them vital in understanding dynamic systems.

To solve them:

• We initially express the differential equation in standard form so it becomes manageable.
• The goal is to predict different trajectories of a system's behavior over time, using only initial conditions and rates of change.

First-order linear differential equations are fundamental in translating real-world problems into mathematical models. They often include terms for both constant rates of change and those depending on the state of the system, such as the amount of salt in our tank.

Being linear allows these equations to be quite solvable compared to non-linear equations, often through techniques like the integrating factor method or separation of variables. Understanding these principles is essential for modeling how quantities evolve across different scenarios in everyday applications.