Problem 48 - Calculus: Early Transcendentals

Question

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15L/min. How much salt is in the tank (a) after \( t \) minutes and (b) after one hour?

Step-by-Step Solution

Answer

After one hour, the tank has approximately 17.26 kg of salt.

Step 1: Define Variables

Let the amount of salt in the tank at time \( t \) minutes be \( S(t) \) kg. The volume of the solution in the tank remains constant at 1000 L because the inflow and outflow rates are equal (total inflow = 15 L/min and outflow = 15 L/min).

Step 2: Determine Salt Inflow

The first stream brings in 0.05 kg/L of salt at 5 L/min, so its contribution is \( 0.05 \times 5 = 0.25 \) kg/min. The second stream brings in 0.04 kg/L of salt at 10 L/min, so its contribution is \( 0.04 \times 10 = 0.4 \) kg/min. The total salt inflow rate is \( 0.25 + 0.4 = 0.65 \) kg/min.

Step 3: Determine Salt Outflow

The outflow rate of the solution is 15 L/min, and the concentration of salt in the tank at any time is \( \frac{S(t)}{1000} \) kg/L. Therefore, the salt outflow rate is \( 15 \times \frac{S(t)}{1000} = \frac{15S(t)}{1000} \) kg/min.

Step 4: Set Up the Differential Equation

We equate the rate of change of salt in the tank to the difference between inflow and outflow: \[ \frac{dS}{dt} = 0.65 - \frac{15S}{1000} \] This represents the rate of salt change in the tank over time.

Step 5: Solve the Differential Equation

Rearranging the equation: \[ \frac{dS}{dt} = 0.65 - 0.015S \]This is a separable differential equation: \[ \int \frac{1}{0.65 - 0.015S} \,dS = \int \, dt \]Solving the integrals, we find: \[ -\frac{1}{0.015} \ln|0.65 - 0.015S| = t + C \]Solving for \( S(t) \):\[ S(t) = \frac{65}{3} - Ke^{-0.015t} \] where \( K \) is a constant.

Step 6: Determine the Constant of Integration

Since initially \( S(0) = 0 \) (pure water), substitute into the equation:\[ 0 = \frac{65}{3} - K \rightarrow K = \frac{65}{3} \] Thus, \( S(t) = \frac{65}{3}(1 - e^{-0.015t}) \).

Step 7: Calculate Amount of Salt After One Hour

Substitute \( t = 60 \) (one hour) into the equation: \[ S(60) = \frac{65}{3}(1 - e^{-0.015 \times 60}) \]Calculate the value for an approximate solution: \[ S(60) \approx \frac{65}{3}(1 - e^{-0.9}) \]

Key Concepts

Brine

Brine is essentially a solution of salt (sodium chloride) in water. In many mathematical problems, especially those involving tanks, brine plays a key role. Here, brine with different salt concentrations is added to a tank. Studying such scenarios helps in understanding mixing problems and their dynamic behavior over time.

The key points regarding brine in this problem are:

• Two different streams of brine enter the tank, one with 0.05 kg/L at a rate of 5 L/minute and another with 0.04 kg/L at 10 L/minute.
• This means that brine is continuously mixing and changing the solution within the tank.
• Understanding brine addition emphasizes how various concentrations contribute to the overall salt balance in the tank.

This forms the foundation for solving how much salt accumulates over time.

Salt Concentration

Salt concentration is a measure of how much salt is dissolved in a particular volume of liquid, typically expressed in kg/L. It is crucial in problems involving solutions as it directly affects the rate at which substances flow in and out of a system.

In the given exercise, different brine solutions with varying salt concentrations are mixed. This affects the overall concentration of salt in the tank at any given time. The initial concentration is zero since the tank contains pure water. However, as brine enters, the concentration changes:

• The inflow brines introduce salt at a combined rate, calculated by their respective concentrations and flow rates, resulting in a total input of 0.65 kg/min.
• The concentration in the outflow depends on how the brine mixes in the tank, represented as \( \frac{S(t)}{1000} \).

Overall, the salt concentration is vital because it determines how quickly the solution inside the tank reaches equilibrium.

Separable Differential Equation

Separable differential equations are a category of differential equations that can be rearranged so that all terms involving the dependent variable are on one side, and all terms involving the independent variable are on the other. This separation allows us to integrate each side individually to find a solution.

In this tank problem, modelling the rate of salt change gives us a differential equation:

• The equation \( \frac{dS}{dt} = 0.65 - 0.015S \) represents how the amount of salt in the tank changes over time.
• This equation is separable, allowing it to be rearranged for easy integration.
• Solving such an equation involves integrating, leading toward an explicit solution for the salt amount \( S(t) \).

This powerful technique ultimately aids in predicting the salt content at any given time.

Rate of Change

The rate of change in this context describes how the quantity of salt in the tank varies over time. It is a critical component in setting up the differential equation.

• The problem involves calculating both inflow and outflow rates.
• The inflow rate is constant and derived from both streams adding up to 0.65 kg/min of salt.
• The outflow rate, however, is dependent on the salt concentration within the tank at any time \( \frac{15S(t)}{1000} \).

Understanding the rate of change is essential for predicting how systems evolve. By finding the balance between these rates, we understand how much salt the tank holds at any specific moment. This, ultimately, leads us towards long-term insights like equilibrium states.