Question

A tank containing 10 kilograms of salt dissolved in 1000 liters of water has two salt solutions pumped in. The first solution of \(0.2 \mathrm{~kg}\) salt \(/ L\) is pumped in at a rate of \(20 \mathrm{~L} / \mathrm{min}\) and the second solution of \(0.05 \mathrm{~kg}\) salt \(/ \mathrm{L}\) is pumped in at a rate of \(5 \mathrm{~L} / \mathrm{min}\). The tank drains at \(25 \mathrm{~L} / \mathrm{min}\). Assume the tank is well mixed. Solve for the salt concentration at time \(t\).

Short answer

The salt concentration at time \( t \) is \( c(t) = \frac{-170 + 180e^{0.025t}}{1000} \) kg/L.

Step-by-step solution

Define the Variables

Let \( x(t) \) be the amount of salt in kilograms in the tank at time \( t \) minutes. We know initially there are 10 kg of salt, so \( x(0) = 10 \). The volume of the water in the tank remains constant at 1000 L because the total inflow rate equals the outflow rate \((20 + 5 = 25)\).

Set Up the Differential Equation

The change in salt concentration over time equals the rate in minus the rate out. The rate salt enters from the first solution is \(0.2 \times 20 = 4 \text{ kg/min}\), and from the second solution is \(0.05 \times 5 = 0.25 \text{ kg/min}\). Thus, the total rate in is \(4.25 \text{ kg/min}\). The rate out depends on the concentration of salt in the tank at time \( t \) and is given by \(\frac{x(t)}{1000} \times 25\). Thus, the differential equation is:\[ \frac{dx}{dt} = 4.25 - \frac{25x(t)}{1000} \]

Simplify the Differential Equation

The differential equation is: \[ \frac{dx}{dt} = 4.25 - 0.025 x(t) \] This is a first-order linear differential equation.

Solve the Differential Equation

The differential equation \( \frac{dx}{dt} = 4.25 - 0.025x(t) \) can be solved using an integrating factor method. The integrating factor, \( \mu(t) \), is found by solving \( \mu(t) = e^{\int -0.025 dt} = e^{-0.025t} \). Multiply the entire differential equation by this integrating factor:\[ e^{-0.025t} \frac{dx}{dt} + 0.025e^{-0.025t} x = 4.25 e^{-0.025t} \]The left-hand side is the derivative of \( x e^{-0.025t} \), thus \[ \frac{d}{dt}(x e^{-0.025t}) = 4.25 e^{-0.025t} \] Integrate both sides with respect to \( t \):\[ x e^{-0.025t} = -170e^{-0.025t} + C \]

Solve for x(t) and Apply Initial Condition

To find \( x(t) \), divide by \( e^{-0.025t} \):\[ x(t) = -170 + Ce^{0.025t} \]Using the initial condition \( x(0) = 10 \), substitute \( t = 0 \):\[ 10 = -170 + C \]Thus, \( C = 180 \).The solution becomes \( x(t) = -170 + 180e^{0.025t} \).

Find the Salt Concentration

The salt concentration \( c(t) \) in kg/L is given by the amount of salt \( x(t) \) divided by the volume, \( 1000 \) L. Therefore:\[ c(t) = \frac{x(t)}{1000} = \frac{-170 + 180e^{0.025t}}{1000} \]

Key concepts

Salt Concentration

Salt concentration illustrates how much salt is dissolved in a solution. In the context of the problem, we are dealing with a tank containing water and salt being mixed continuously. To determine the concentration at any given moment, it's essential to understand the changes occurring within the tank over time.

In our problem, salt is added through two different solutions, each with a specific concentration and flow rate. The concentration of the first solution is \(0.2 \text{ kg/L}\) delivered at \(20 \text{ L/min}\), adding \(4 \text{ kg/min}\) to the tank. The second solution at \(0.05 \text{ kg/L}\) flows in at \(5 \text{ L/min}\), contributing another \(0.25 \text{ kg/min}\). Together, they add \(4.25 \text{ kg/min}\) into the tank.

However, as the tank drains at \(25 \text{ L/min}\) to maintain a constant volume, the concentration changes as salt exits. The salt concentration at any time can be expressed by dividing the amount of salt \(x(t)\) by the total tank volume of \(1000 \text{ L}\). This relationship helps us keep track of how salt concentration evolves due to inflow and outflow dynamics.

Integrating Factor

The integrating factor is a mathematical tool used to solve linear differential equations. It's particularly useful when dealing with first-order linear differential equations, like in our problem.

The key insight is that multiplying through by a cleverly chosen function transforms the left side of our differential equation into an exact derivative, making it much easier to solve. In our given equation, the integrating factor \( \mu(t) \) was determined as \( e^{-0.025t} \).

The reason for using the integrating factor involves ensuring that the equation's left side becomes the derivative of a product, in this case \(x(t) e^{-0.025t}\). After applying the integrating factor, integrating both sides with respect to \(t\) allows us to move ahead in solving for \(x(t)\). Remember, finding the correct integrating factor effectively simplifies and streamlines the process of unraveling a differential equation.

First-Order Linear Differential Equation

A first-order linear differential equation helps model systems where the rate of change of a quantity is proportional to the quantity itself, with an added constant rate. In our case, the equation models the rate of change of salt in the tank over time.

The general form of a first-order linear differential equation is \( \frac{dx}{dt} = ax + b \), where \(a\) and \(b\) are constants. Here, our equation was \( \frac{dx}{dt} = 4.25 - 0.025x(t) \).

Solving first-order linear differential equations generally involves:

• Setting up the equation based on the rates of input and output.
• Applying techniques like the integrating factor to simplify and solve the equation.
• Using initial conditions to find specific constants related to the particular solution.

Understanding and solving these equations are crucial in applications ranging from engineering to natural sciences—for any situation where there is proportional growth or decay plus a steady rate of change.