Question

A large tank initially contains 100 gal of brine in which \(10 \mathrm{lb}\) of salt is dissolved. Starting at \(t=0\), pure water flows into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and the well- stirred mixture simultaneously flows out at the slower rate of \(2 \mathrm{gal} / \mathrm{min} .\) (a) How much salt is in the tank at the end of \(15 \mathrm{~min}\) and what is the concentration at that time? (b) If the capacity of the tank is 250 gal, what is the concentration at the instant the tank overflows?

Short answer

(a) At the end of 15 minutes, there is approximately \(6.515~lbs\) of salt in the tank, and the salt concentration is approximately \(0.0449~lbs/gal\). (b) At the instant the tank overflows, the salt concentration is approximately \(0.0281~lbs/gal\).

Step-by-step solution

Understand the given information

In the problem, we are given a tank with 100 gallons of brine and 10 lbs of salt. Pure water enters the tank at a rate of 5 gal/min and the mixed solution flows out at a rate of 2 gal/min. Step 2: Set up a differential equation for amount of salt in the tank

Set up a differential equation

To model the amount of salt in the tank at any time "t", let S(t) be the amount of salt (in lbs) in the tank at time "t" (in minutes). Since no salt is flowing in (only pure water is flowing in), the rate at which salt leaves the tank is proportional to the concentration of salt in the tank. The rate of change of salt in the tank, dS/dt, can be defined as: \(\frac{dS}{dt} = -\frac{S}{100 + 3t} \cdot 2\) Step 3: Solve the differential equation

Solve the differential equation

In order to find the function S(t), we need to integrate both sides of the differential equation: \(\int \frac{dS}{S} = -2 \int \frac{dt}{100 + 3t}\) Using partial fractions, we can rewrite the integrand as: \(\frac{A}{t} + \frac{B}{100 + 3t}\) Solving for A and B, we get A = 1/3 and B = -1/3. Now, we integrate both sides: \(\int \frac{dS}{S} = -\frac{1}{3} \int \frac{dt}{t} + \frac{1}{3} \int \frac{dt}{100 + 3t}\) \(\ln |S| = -\frac{1}{3}\ln |t| + \frac{1}{9}\ln |100 + 3t| + C\) where S(t) > 0. Step 4: Determine the function S(t)

Determine the function S(t)

To find S(t), we need to raise both sides of the equation we derived previously to the power of "e". Also, at t=0, S(0) = 10 lbs: \(S(t) = \frac{(100 + 3t)^{\frac{1}{3}}}{t^{\frac{1}{3}}}\cdot e^C\) To find the constant C, plug in t=0 and S(0)=10: \(10 = \frac{100^{\frac{1}{3}}}{0^{\frac{1}{3}}} \cdot e^C\) \(e^C = 10\) Therefore, the function describing the amount of salt in the tank at any time is: \(S(t) = 10\cdot \frac{(100 + 3t)^{\frac{1}{3}}}{t^{\frac{1}{3}}}\) Step 5: Answer the questions

Answer the questions

(a) At t=15 min, the amount of salt is: \(S(15) = 10\cdot \frac{(100 + 3\cdot15)^{\frac{1}{3}}}{15^{\frac{1}{3}}} \approx 6.515 \text{ lbs}\) The volume of the mixture in the tank is \(100 + 3\cdot 15 = 145 \text{ gal}\). Hence, the concentration of salt at t=15 min is: \(c(15) = \frac{6.515}{145} \approx 0.0449 \text{ lbs/gal}\) (b) The tank overflows when its volume reaches 250 gal. The time it takes to reach this volume can be obtained by solving the equation: \(100 + 3t = 250\) \(t = 50\) At t=50 min, the amount of salt is: \(S(50) = 10\cdot \frac{(100 + 3\cdot50)^{\frac{1}{3}}}{50^{\frac{1}{3}}} \approx 7.017 \text{ lbs}\) Hence, the concentration of salt when the tank overflows is: \(c(50) = \frac{7.017}{250} \approx 0.0281 \text{ lbs/gal}\)

Key concepts

Differential Equation Modeling

In the study of mathematics and physics, differential equation modeling is a powerful tool used to describe relationships involving rates of change. A differential equation is a mathematical equation relating a function with its derivatives. When modeling real-world phenomena such as the flow of brine solution in a tank, differential equations allow us to predict future behavior based on specific conditions.

Differential equations come in various complexities, from simple separable equations to more complex partial differential equations. The equation used in analyzing the brine solution involves setting up a rate of change equation that takes into account the volume of the fluid in the tank, the incoming water rate, and the outgoing solution rate, without any salt entering the system. This equation is an ordinary differential equation (ODE) because it involves functions of a single variable, the time t.

By solving this ODE, we can determine how much salt remains in the tank over time, and consequently, the concentration of salt at any given moment.

Rate of Change

The rate of change signifies the speed at which a variable changes over time. It is a central concept not only in differential equations but in all of calculus. In the context of our problem, the rate of change of salt within the tank is what we are concerned with. This rate is influenced by flow rates of liquid into and out of the tank.

Mathematically, we represent the rate of change of salt concentration as dS/dt, where S is the amount of salt in pounds and t is the time in minutes. The rate of change is negative in our problem because over time, we assume the concentration of salt decreases due to the pure water flowing in and diluting the brine solution. We must consider the volume of the brine solution at any time, which itself changes as water flows in and out.

Concentration Calculation

The concentration calculation is a way to determine the ratio of solute to solvent in a solution. For the problem in question, we calculate the concentration of salt in the brine by dividing the amount of salt by the total volume of solution at a given time. It's crucial to understand that the concentration is dynamic and changes as the pure water is added, and the stirred mixture is drained out of the tank.

After solving the differential equation, we obtain an expression for S(t), the amount of salt at any time t. To calculate the concentration, c(t), we simply divide S(t) by the corresponding volume of the solution, which is (100 + 3t) gallons at any time t. This allows us to find the concentration at notable points, such as after 15 minutes or when the tank reaches its full capacity and overflows.

Brine Solution

A brine solution is a mixture of salt and water. In our textbook example, the tank initially contains a brine solution with a known concentration of salt. As the problem progresses, pure water is added to the tank, which dilutes the initial brine solution, resulting in a change in salt concentration over time.

This real-world scenario presents a classic application of differential equations in chemical engineering and environmental studies. By creating a model of a brine solution in a tank where the concentration changes over time, students learn how to apply mathematical concepts to understand and predict the behavior of such systems. Recognizing the practical implications of the changing concentration in a brine solution helps students appreciate the importance of accurately calculating rates of change and remaining mindful of how parameters like tank capacity and flow rates can impact the overall system.