Problem 27 - Advanced Engineering Mathematics

Question

A large tank is partially filled with 100 gallons of fluid in which 10 pounds of salt is dissolved. Brine containing \(\frac{1}{2}\) pound of salt per gallon is pumped into the tank at a rate of \(6 \mathrm{gal} / \mathrm{min}\). The well-mixed solution is then pumped out at a slower rate of \(4 \mathrm{gal} / \mathrm{min}\). Find the number of pounds of salt in the tank after 30 minutes.

Step-by-Step Solution

Answer

There are approximately 55.35 pounds of salt in the tank after 30 minutes.

Step 1: Differential Equation Setup

Let's denote the amount of salt in the tank at time \( t \), in pounds, as \( S(t) \). The rate of change of salt is given by the difference between the input rate and the output rate. The input is the brine entering with \( \frac{1}{2} \) pound of salt per gallon at \( 6 \) gallons per minute, thus contributing \( 3 \) pounds/minute of salt. The output depends on the concentration of salt in the tank: the total amount of salt \( S(t) \) divided by the total volume of liquid in the tank, \( V(t) \). The differential equation is then \[ \frac{dS}{dt} = 3 - \frac{S(t)}{V(t)} \cdot 4, \] where \( V(t) = 100 + 2t \) gallons, since the net gain in liquid is \( 2 \) gallons per minute.

Step 2: Solve the Differential Equation

We can rewrite the differential equation as \[ \frac{dS}{dt} = 3 - \frac{4S(t)}{100 + 2t}. \] This is a first-order linear differential equation and can be solved using an integrating factor. First, simplify and organize it in the form \( \frac{dS}{dt} + P(t)S = Q(t) \), where \( P(t) = \frac{4}{100 + 2t} \) and \( Q(t) = 3 \). The integrating factor \( \mu(t) \) is \( e^{\int P(t) \, dt} = (100 + 2t)^2. \)

Step 3: Integrate the Equation

Multiply the entire differential equation by the integrating factor: \[ (100 + 2t)^2 \frac{dS}{dt} + 4(100 + 2t)S = 3(100 + 2t)^2. \] This transforms the left side into the derivative of \((100 + 2t)^2 S(t)\). Integrate both sides with respect to \( t \): \[ \int d\left((100 + 2t)^2 S(t)\right) = \int 3(100 + 2t)^2 \, dt. \] The right side integrates to \( 3\left(\frac{(100 + 2t)^3}{3} - \frac{(100)^3}{3}\right) \). After integrating both sides and solving for \( S(t) \), apply initial conditions.

Step 4: Apply Initial Conditions

Given \( S(0) = 10 \), use it to find the constant of integration. Substitute \( t = 0 \) in the equation \( (100)^2 \cdot 10 = \left(\frac{(100 + 2t)^3}{3} - \frac{(100)^3}{3}\right) + C \). Solving for \( C \) gives \( C = 10 \,000 - \frac{(100)^3}{3} \). This first term simplifies the entire equation.

Step 5: Calculate Salt at 30 Minutes

Substitute \( t = 30 \) minutes back into the equation for \( S(t) \). Find \( V(30) = 100 + 2 \cdot 30 = 160 \, \text{gallons}. \) Use this to adjust the solution and solve for \( S(30) \). Substitute these values into the corrected solutions of \( S(t) \). The calculations yield \( S(30) \approx 55.3534 \) pounds.

Key Concepts

First-order linear differential equations

First-order linear differential equations are a fundamental type of differential equation that model various dynamic processes in nature and engineering. These equations have the form \( \frac{dy}{dt} + P(t)y = Q(t) \), where \( P(t) \) and \( Q(t) \) are functions of time. In the context of our mixing problem, the equation \( \frac{dS}{dt} = 3 - \frac{4S(t)}{100 + 2t} \) is a linear first-order differential equation that describes how the amount of salt, \( S(t) \), changes over time as brine is added and the solution is removed.
This equation is linear because the term containing \( S(t) \) is raised to the first power. Its structure allows for systematic methods like the integrating factor method to solve for \( S(t) \), which represents the amount of salt in the tank at any given time \( t \).
Understanding these equations is crucial, as they appear in various applications such as electrical circuits, fluid dynamics, and population growth models.

Integrating factor method

The integrating factor method is a powerful technique used to solve first-order linear differential equations. This method transforms the equation into a form that is easier to integrate.
In our example, we have the differential equation \( \frac{dS}{dt} + \frac{4}{100 + 2t} S = 3 \). We identify \( P(t) = \frac{4}{100 + 2t} \) and \( Q(t) = 3 \). The integrating factor, \( \mu(t) \), is calculated as \( e^{\int P(t) \, dt} = (100 + 2t)^{2} \).
Multiplying the entire differential equation by this integrating factor simplifies the problem. It turns the left side into the derivative of a product, which can then be integrated easily to find the solution.

• Calculate the integrating factor: \( \mu(t) = e^{\int \frac{4}{100 + 2t} \, dt} = (100 + 2t)^{2} \)
• Multiply the differential equation by \( \mu(t) \)
• Integrate both sides to solve for \( S(t) \)

By applying this method, we can efficiently handle problems involving dynamic changes, such as those found in mixing problems like the one discussed.

Mixing problems

Mixing problems are a fascinating example of how differential equations can model real-world scenarios. These problems typically involve a tank that contains a fluid with dissolved substances, such as salt. The process involves a fluid with a certain concentration entering and leaving the tank at given rates.
In our exercise, we have a tank initially containing 100 gallons of fluid with 10 pounds of salt. Brine with a half-pound of salt per gallon flows in at 6 gallons per minute, while the mixed solution leaves at 4 gallons per minute. Over time, the water level increases, affecting the concentration of salt.
The differential equation helps us track how much salt remains in the tank over time, allowing for applications in many fields.

• Understand the input and output rates (Brine in and solution out)
• Set up a differential equation to track change over time
• Solve using appropriate methods such as the integrating factor method

These problems highlight the crucial role of mathematics in predicting and controlling processes in industries like chemical engineering and water treatment.

Initial value problems

Initial value problems are a specific type of problem where the solution to a differential equation must satisfy a given initial condition. This means that at a particular initial time, the value of the dependent variable (like salt in our case) is specified.
For our tank problem, we are given that at time \( t = 0 \), the salt content is 10 pounds. This is crucial because solving the differential equation provides a general solution with a constant of integration.
This constant is determined by applying the initial condition, ensuring that the solution fits the real-world scenario from the beginning.

• Determine the initial condition: \( S(0) = 10 \) pounds
• Use the initial condition to find the integration constant
• Verify that the solution satisfies both the differential equation and the initial condition

Initial value problems are prevalent in physics and engineering, providing solutions that reflect actual initial states of systems and predicting their evolution over time.