Question

A tank contains 3 kilograms of salt dissolved in 75 liters of water. A salt solution of \(0.4 \mathrm{~kg}\) salt \(/ L\) is pumped into the tank at a rate of \(6 \mathrm{~L} / \mathrm{min}\) and is drained at the same rate. Solve for the salt concentration at time \(t\). Assume the tank is well mixed at all times.

Short answer

The salt concentration at time \( t \) is \( \frac{30 - 27e^{-\frac{2t}{25}}}{75} \) kg/L.

Step-by-step solution

Define the Problem and Known Variables

We have a tank initially containing 3 kg of salt in 75 liters of water. A salt solution with a concentration of 0.4 kg/L enters the tank at a rate of 6 L/min. The tank is also being drained at the same rate, keeping the volume constant. Let's assume the concentration of salt at time \( t \) is given by \( Q(t) \) in kilograms.

Set Up the Differential Equation

Since the tank is well-mixed, we can set up a differential equation to model the change in the amount of salt over time. The rate of salt entering the tank is \( 6 \times 0.4 = 2.4 \text{ kg/min} \). The rate of salt leaving the tank is given by \( \left(\frac{Q(t)}{75}\right) \times 6 \), because 6 liters contain \( \frac{Q(t)}{75} \) kg of salt. Therefore, the differential equation is: \[\frac{dQ}{dt} = 2.4 - \frac{6Q}{75}\]

Simplify and Solve the Equation

Simplify the equation:\[\frac{dQ}{dt} = 2.4 - \frac{2Q}{25}\]This is a first-order linear differential equation. The standard form is \( \frac{dQ}{dt} + pQ = g(t) \), where \( p = \frac{2}{25} \) and \( g(t) = 2.4 \).

Find the Integrating Factor

The integrating factor formula is \( e^{\int p \, dt} \). Here, it becomes:\[e^{\int \frac{2}{25} \, dt} = e^{\frac{2t}{25}}\]

Solve the Differential Equation

Multiply through the differential equation by the integrating factor:\[e^{\frac{2t}{25}}\frac{dQ}{dt} + \left(\frac{2}{25}e^{\frac{2t}{25}}\right)Q = 2.4e^{\frac{2t}{25}}\]Integrate both sides with respect to \( t \):\[Q(t) \cdot e^{\frac{2t}{25}} = \int 2.4 \, e^{\frac{2t}{25}} \, dt\]Calculate the right side:\[= \frac{2.4 \times 25}{2} e^{\frac{2t}{25}} + C = 30e^{\frac{2t}{25}} + C\]

Solve for the Constant C

Initially, \( Q(0) = 3 \), so:\[3 \cdot e^{0} = 30 + C\]Solving gives \( C = 3 - 30 = -27 \).

Solve for Q(t)

Substitute \( C \) back into the equation:\[Q(t) = 30 + Ce^{-\frac{2t}{25}}\]\[Q(t) = 30 - 27e^{-\frac{2t}{25}}\]

Determine the Salt Concentration

The concentration \( C(t) \) is given by \( \frac{Q(t)}{75} \). So:\[C(t) = \frac{30 - 27e^{-\frac{2t}{25}}}{75} \text{ kg/L}\]

Key concepts

Salt Tank Problem

The Salt Tank Problem is a classic example used in differential equations to understand how substances mix over time in a controlled environment. Imagine you have a tank filled with a liquid mixture, such as water containing dissolved salt. The task is to determine how the concentration of salt changes as new solutions are added or removed.
This type of problem is essential for those studying chemical processes, environmental systems, and various engineering disciplines. It illustrates how substances equilibrate in well-mixed systems.

• This problem involves a few critical considerations: the rate of inflow and outflow, the initial concentration of the solute, and the behavior of the solution over time.
• It demonstrates the principles of material balance and the dynamic nature of mixed solutions, assuming perfect mixing at all times.
• The problem results in a first-order linear differential equation, a fundamental piece in mathematical modeling of natural processes.

Mixing Problems

Mixing Problems are a subset of differential equations that involve the blending and distribution of different materials within a mixture. Such problems typically ask you to find out the concentration of a substance as it changes over time.
In our exercise, the salt solution flowing into and out of the tank at equal rates leads to mixing problems. These assume that the tank's content has a consistent concentration throughout, akin to a perfect stirring process.

• The dynamic entry and exit of solutions mean that rates of change contribute critically to the results.
• Here, conditions for rates are given: A solution enters the tank at 6 L/min with a concentration of 0.4 kg/L, and the same amount leaves the tank, keeping volume steady.
• Such setups help in creating a model through differential equations, facilitating the evaluation of the ongoing process within the tank.

First-Order Linear Differential Equations

First-Order Linear Differential Equations are foundational in solving dynamic problems like those found in mixing operations. They capture how a system's variables change over time, constrained by linear relationships.
In the Salt Tank Problem, we use a differential equation \[ \frac{dQ}{dt} = 2.4 - \frac{2Q}{25} \] to model salt concentration over time.
This differential equation features:

• The expression \( \frac{dQ}{dt} \) represents the rate of change of the amount of salt in the tank.
• The term 2.4 is the consistent rate at which salt enters the tank.
• \( -\frac{2Q}{25} \) shows the proportional exit rate based on the salt present.

To solve first-order linear differential equations, we employ integrating factors. These simplify the equation to allow straightforward integration, leading to solutions that predict how systems evolve over time. In our solution, we found the integrating factor \( e^{\frac{2t}{25}} \) and solved for the general and particular solutions, given initial conditions, to determine how the salt concentration changes in the tank.