Problem 32 - Calculus Single Variable

Question

A mixing tank contains \(200 \mathrm{~L}\) of water in which \(10 \mathrm{~kg}\) salt is dissolved. At time \(t=0,\) a valve is opened and water enters the tank at the rate of \(20 \mathrm{~L}\) per minute. An outlet pipe maintains the volume of fluid in the tank by allowing \(20 \mathrm{~L}\) of the thoroughly mixed solution to flow out each minute. What is the mass \(m(t)\) of salt in the tank at time \(t ?\) What is \(m_{\infty}=\lim _{t \rightarrow \infty} m(t) ?\)

Step-by-Step Solution

Answer

The mass of salt at time \(t\) is \(m(t) = 10e^{-\frac{1}{10}t}\). As \(t\) approaches infinity, \(m_{\infty} = 0\) kg.

Step 1: Understand initial conditions and input rates

The tank initially has 200 L of water with 10 kg of salt dissolved in it. Water is entering the tank at 20 L/min, and similarly, 20 L/min of the solution is exiting the tank to keep the volume constant. The salt is dissolved in the same incoming water.

Step 2: Express the rate of change of salt mass

Define the rate of change of the mass of salt in the tank, \(m(t)\), using the balance equation: \(\frac{dm}{dt} = \text{(rate of salt entering)} - \text{(rate of salt leaving)}\). Since incoming water has no salt, rate entering is 0. Rate leaving is \(\frac{20}{200}m(t)\) kg/min.

Step 3: Develop the differential equation

The differential equation for salt mass is \(\frac{dm}{dt} = - \frac{1}{10}m(t)\). This is derived because the fraction of salt in the solution that leaves the tank at a rate of 20 L/min is \(\frac{m(t)}{200}\). Multiply this fraction by 20 to find the rate of salt leaving.

Step 4: Solve the differential equation

The differential equation \(\frac{dm}{dt} = - \frac{1}{10}m(t)\) is separable. Separating variables, integrating both sides gives \(\ln(m) = -\frac{1}{10}t + C\). Solving for \(m(t)\), we find \(m(t) = Ce^{-\frac{1}{10}t}\). Use the initial condition \(m(0)=10\) kg to solve for \(C\), resulting in \(C=10\). So, \(m(t) = 10e^{-\frac{1}{10}t}\).

Step 5: Determine the limiting behavior

As \(t \to \infty\), evaluate \(m(t) = 10e^{-\frac{1}{10}t}\). The exponential term approaches zero, thus \(m_{\infty} = \lim_{t \to \infty} m(t) = 0\) kg.

Key Concepts

Separable Equations

Separable equations allow us to simplify complex differential equations into more manageable parts. This happens by separating the variables on either side of the equation. For our problem, the differential equation is \(\frac{dm}{dt} = - \frac{1}{10}m(t)\). Here, the variables \(m\) and \(t\) can be moved to different sides:

• On one side, you have \(\frac{dm}{m}\).
• On the other side, you incorporate \(dt\).

The equation becomes \(\frac{dm}{m} = -\frac{1}{10} dt\). By integrating both sides, you find a solution for \(m(t)\). This process transforms a differential equation into an integrable form, allowing us to solve it for functions like the mass of salt here.

Initial Value Problems

Initial value problems in differential equations specify the value of the unknown function at a specific point. In our chemical mixing problem, we know that at \(t = 0\), the mass of salt \(m(0)\) is 10 kg. This initial condition helps determine the constant of integration when solving the differential equation.
Once the function \(m(t) = Ce^{-\frac{1}{10}t}\) is derived, substitute \(t = 0\) and \(m(0) = 10\) to solve for \(C\). This yields \(C = 10\). Applying initial conditions ensures that the particular solution accurately reflects the real-world system described.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value, common in chemistry and physics. For our differential equation, \(m(t) = 10e^{-\frac{1}{10}t}\), this exponential function shows the mass of salt decreasing over time.

• This function is characterized by the negative exponent.
• It indicates that as time increases, the quantity reduces steadily.

The negative rate \(-\frac{1}{10}\) is the *decay constant*, showing how quickly the salt's mass dissipates from the tank. Exponential decay functions often signify natural decline, such as chemical concentration change, facilitating prediction of long-term behavior.

Chemical Mixing Problems

Chemical mixing problems explore changes in concentration and quantities of substances in solutions, often using differential equations. In our example, water flows into and out of the tank, with salt dissolving and leaving. These problems typically involve balancing:

• The rate at which substances enter a system.
• And the rate at which they exit.

Here, solving \(\frac{dm}{dt} = - \frac{1}{10}m(t)\) provides insights into how the salt amount changes over time. These models are crucial for predicting concentrations or output quality in engineering and environmental contexts. Careful setup and solution of differential equations ensure accurate predictions, aiding process optimization and control.