Question

A tank initially holds 500 gal of a brine solution having a concentration of \(0.1 \mathrm{lb}\) of salt per gallon. At some instant, fresh water begins to enter the tank at a rate of 10 \(\mathrm{gal} / \mathrm{min}\) and the well- stirred mixture leaves at the same rate. How long will it take before the concentration of salt is reduced to \(0.01 \mathrm{lb} / \mathrm{gal}\) ?

Short answer

Short answer: It will take approximately 115.13 minutes for the concentration of salt in the tank to be reduced to 0.01 lb/gal.

Step-by-step solution

Identify the given information and find the initial salt mass

We are given that the tank initially holds 500 gallons of brine solution with a concentration of 0.1 lb of salt per gallon. To find the initial mass of salt in the tank, multiply the volume of the solution by its concentration: Initial salt mass = Volume × Concentration = 500 gal × 0.1 lb/gal = 50 lb

Set up the differential equation to represent the rate of change of salt mass

Let x(t) represent the mass of salt at time t. We know that the rate of change of salt mass in the tank depends on the rate at which water enters and leaves the tank. The fresh water entering the tank does not have any salt, so it only dilutes the salt concentration. Since water leaves the tank at the same rate it enters (10 gal/min), the rate of change of salt mass with respect to time is: \(\displaystyle \frac{dx}{dt}=-\frac{10x}{500}\)

Solve the differential equation

We can solve this differential equation by separating the variables: \(\displaystyle \frac{1}{x} dx = -\frac{1}{50} dt\) Now, integrate both sides: \(\displaystyle \int \frac{1}{x} dx = -\frac{1}{50} \int dt\) We get: \(\ln (x)=-\frac{t}{50}+C_1\) To find the integration constant C1, we can use the initial condition x(0)=50: \(\ln(50)=-\frac{0}{50}+C_1\) So, \(C_1=\ln(50)\) Thus, the mass of salt at any time t is given by: \(x(t)=50e^{-\frac{t}{50}}\)

Find the time at which the concentration becomes 0.01 lb/gal

To find the time when the concentration of salt is reduced to 0.01 lb/gal, we can set up the following equation using the mass of salt and solution volume: Concentration = \(\frac{x(t)}{500}\) = 0.01 lb/gal Substitute the expression for x(t) from Step 3: \(\frac{50e^{-\frac{t}{50}}}{500}\)=0.01 Solving for t: \(e^{-\frac{t}{50}}\) = 0.01 × 10 Take the natural logarithm of both sides: \(-\frac{t}{50} =\ln(0.01\times10)\) \(t=-50\ln (0.01\times 10)\) \(t\approx 115.13\) minutes Hence, it will take approximately 115.13 minutes for the concentration of salt to be reduced to 0.01 lb/gal.

Key concepts

Brine Solution Problem

The Brine Solution Problem is a classic type of differential equation problem often encountered in chemical and engineering contexts. It involves understanding how a solute, like salt, mixes with a solvent, like water, over time. In this exercise, the tank initially contains 500 gallons of brine with a salt concentration of 0.1 lb per gallon, totaling 50 lbs of salt. Fresh water is introduced at a rate of 10 gallons per minute and an equal volume is simultaneously withdrawn. This scenario leads to a dynamic process consisting of continuous dilution of the salt solution as water enters and leaves the tank at the same rate. Solving the Brine Solution Problem typically involves determining how the concentration of salt changes over time until it reaches a desired level.

Rate of Change

The Rate of Change in this context describes how quickly the salt concentration in the tank decreases over time. It is formulated as a differential equation which mathematically expresses how the quantity of salt changes. The rate at which the salt leaves the tank is proportional to its current concentration, given by \(\frac{dx}{dt} = -\frac{10x}{500}\), where \(x(t)\) is the mass of salt at time \(t\). Here, the negative sign indicates a decrease in salt quantity since fresh water entering the tank contains no salt, thus diluting the salt concentration further. It encapsulates the idea that the decay and dilution process happens simultaneously at the same rate that water is added to and removed from the tank. Understanding this differential equation is crucial in predicting how long it takes for the salt concentration to reach a target level.

Separation of Variables

Separation of Variables is a useful technique for solving differential equations where variables can be separated on opposite sides of the equation for integration. In the brine solution problem, we transform the rate of change equation \(\frac{dx}{dt} = -\frac{10x}{500}\) into a more manageable form by separating \(x\) and \(t\): \(\frac{1}{x} dx = -\frac{1}{50} dt\). This setup allows for intuitive integration, where each side is integrated independently, resulting in two antiderivatives: \(\ln(x) = -\frac{t}{50} + C\). This method simplifies solving for \(x(t)\), which tells us the mass of salt as a function of time. Using initial conditions helps solve for constants, ultimately providing the specific solution to the problem.

Exponential Decay

Exponential Decay characterizes the reduction of salt concentration in the tank over time. It's described by the solution \(x(t) = 50e^{-\frac{t}{50}}\), derived from solving the differential equation. The function signifies that the mass of salt decreases exponentially at a constant decay rate. The variable \(t\) represents time, and the decay constant \(-\frac{1}{50}\) determines the rate at which the exponential decreases. As \(t\) increases, \(x(t)\) asymptotically approaches zero, illustrating that even after long periods, remnants of salt continue to exist but in miniscule quantities. This decay model is crucial in calculating how quickly the concentration reaches a specific threshold, such as the 0.01 lb/gal goal in the problem.