Alcohol Concentration Calculation In A Mixing Tank Problem

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Hey guys! Ever wondered how to calculate the changing alcohol concentration in a mixing tank? This is a classic problem in differential equations and has practical applications in various fields, from brewing to chemical engineering. Let's break down a common scenario and learn how to solve it. In this article, we will address the situation of calculating the dynamic concentration of alcohol within a tank as beer with a particular alcohol concentration is pumped into it, displacing a mixture of varying concentrations. This problem exemplifies the principles of mixture problems in differential equations, showcasing how to model changes in concentration over time. Let’s dive in and explore this interesting problem together!

Understanding the Problem Setup

Okay, so imagine we have a tank, initially containing a certain amount of beer with a specific alcohol percentage. Now, we're pumping in more beer with a higher alcohol percentage, and at the same time, we're letting some of the mixture out to keep the volume constant. The key question here is: How does the alcohol concentration in the tank change over time?

To make it more concrete, let's consider a specific example:

  • Initial condition: A tank initially holds 400 gallons of beer with a 3% alcohol concentration.
  • Inflow: Beer with a 6% alcohol concentration is pumped into the tank at a rate of 3 gallons per minute.
  • Outflow: The mixture is pumped out at the same rate (3 gallons per minute) to maintain a constant volume.

Our goal is to determine the alcohol concentration in the tank at any given time t. Understanding these initial conditions and flow rates is crucial for setting up the mathematical model.

This type of problem involves several key concepts. First, we need to track the amount of alcohol in the tank, not just the concentration. The amount of alcohol changes as we add more alcohol in the inflow and remove some in the outflow. Second, the concentration is simply the amount of alcohol divided by the total volume in the tank. Since the inflow and outflow rates are the same, the total volume remains constant in this specific problem, which simplifies things a bit. Lastly, we will need to set up a differential equation to describe the rate of change of the amount of alcohol in the tank, and then solve this differential equation to find the amount of alcohol as a function of time. From there, we can easily calculate the concentration.

Setting Up the Differential Equation

Alright, let's get down to the math! The core of solving this problem is setting up a differential equation that describes how the amount of alcohol in the tank changes over time. This is based on the principle of "rate in minus rate out."

Let's define some variables:

  • A(t): The amount of alcohol in the tank at time t (in gallons).
  • V: The volume of the beer in the tank, which remains constant at 400 gallons in this case.
  • r_in: The inflow rate (3 gallons/minute).
  • C_in: The alcohol concentration of the inflow (6% or 0.06).
  • r_out: The outflow rate (3 gallons/minute).
  • C_out(t): The alcohol concentration of the outflow at time t. This is equal to A(t) / V.

Now, let's break down the rates:

  • Rate in: The rate at which alcohol enters the tank is the inflow rate multiplied by the inflow concentration: r_in * C_in = 3 gallons/minute * 0.06 = 0.18 gallons/minute.
  • Rate out: The rate at which alcohol leaves the tank is the outflow rate multiplied by the outflow concentration: r_out * C_out(t) = 3 gallons/minute * (A(t) / 400 gallons) = 0.0075 * A(t) gallons/minute.

The rate of change of the amount of alcohol, dA/dt, is the difference between the rate in and the rate out. This gives us the following differential equation:

dA/dt = (rate in) - (rate out)

dA/dt = 0.18 - 0.0075 * A(t)

This is a first-order linear differential equation, which we can solve using various techniques. Recognizing the components that contribute to the differential equation—inflow rate, inflow concentration, outflow rate, and outflow concentration—is vital. The equation encapsulates the dynamics of how the alcohol amount changes, balancing the influx of alcohol against its efflux.

Solving the Differential Equation

Okay, we've got our differential equation: dA/dt = 0.18 - 0.0075 * A(t). Now, how do we solve it? There are a couple of common methods we can use: separation of variables or using an integrating factor. Let's go with separation of variables – it's often a straightforward approach for this type of equation.

Here’s how it works:

  1. Separate the variables: We want to get all the A terms on one side and the t terms on the other. To do this, we can rewrite the equation as:

    dA / (0.18 - 0.0075 * A) = dt

  2. Integrate both sides: Now, we integrate both sides of the equation:

    ∫ dA / (0.18 - 0.0075 * A) = ∫ dt

The integral on the left side can be solved using a u-substitution. Let u = 0.18 - 0.0075 * A, then du = -0.0075 dA, and dA = -du / 0.0075. Substituting these into the integral gives:

`∫ (-1 / 0.0075) du / u = ∫ dt`

`(-1 / 0.0075) * ln|u| = t + C1`

`-133.33 * ln|0.18 - 0.0075 * A| = t + C1`

The integral on the right side is simply t plus a constant of integration, C1.

  1. Solve for A(t): Now, we need to isolate A(t). Let's first multiply both sides by -0.0075:

    ln|0.18 - 0.0075 * A| = -0.0075t + C2 (where C2 = -0.0075 * C1)

Exponentiate both sides:

`|0.18 - 0.0075 * A| = e^(-0.0075t + C2)`

`0.18 - 0.0075 * A = C3 * e^(-0.0075t)` (where `C3 = ±e^(C2)`)

Now, solve for A:

`0.  0075 * A = 0.18 - C3 * e^(-0.0075t)`

`A(t) = (0.18 - C3 * e^(-0.0075t)) / 0.0075`

`A(t) = 24 - C * e^(-0.0075t)` (where `C = C3 / 0.0075`)

  1. Apply the initial condition: We know that initially (at t = 0), the tank contains 400 gallons of beer with a 3% alcohol concentration, so A(0) = 400 * 0.03 = 12 gallons. Let's plug this into our equation:

    12 = 24 - C * e^(0)

    12 = 24 - C

    C = 12

So, our final equation for the amount of alcohol in the tank is:

`A(t) = 24 - 12 * e^(-0.0075t)`

This solution represents how the amount of alcohol changes over time in the tank. Each step, from separating variables to applying the initial condition, plays a crucial role in obtaining this result. Grasping the mathematics behind solving differential equations is essential for mastering mixture problems.

Calculating the Alcohol Concentration

Great! We've found the equation for the amount of alcohol in the tank at any time t: A(t) = 24 - 12 * e^(-0.0075t). But what we really want to know is the alcohol concentration, which is simply the amount of alcohol divided by the total volume.

Since the volume in the tank remains constant at 400 gallons, the alcohol concentration C(t) is given by:

C(t) = A(t) / V

C(t) = (24 - 12 * e^(-0.0075t)) / 400

C(t) = 0.06 - 0.03 * e^(-0.0075t)

This equation tells us the alcohol concentration as a function of time. Let's analyze this result a bit:

  • As t approaches infinity: The term e^(-0.0075t) approaches 0. So, C(t) approaches 0.06, or 6%. This makes sense because we're continuously pumping in beer with a 6% alcohol concentration, so eventually, the mixture in the tank will also reach 6%.
  • At t = 0: We have C(0) = 0.06 - 0.03 * e^(0) = 0.06 - 0.03 = 0.03, which is 3%, our initial concentration. This confirms that our solution is consistent with the initial condition.

We can also use this equation to find the concentration at any specific time. For example, let's find the concentration after 1 hour (60 minutes):

C(60) = 0.06 - 0.03 * e^(-0.0075 * 60)

C(60) ≈ 0.06 - 0.03 * 0.6376

C(60) ≈ 0.04087

So, after 1 hour, the alcohol concentration in the tank is approximately 4.09%. The formula we derived enables us to predict the alcohol concentration at any given time, making it a powerful tool for understanding the mixing dynamics. This calculation highlights how the concentration gradually increases from the initial 3% towards the inflow concentration of 6%.

Key Takeaways and Real-World Applications

So, what have we learned? We've successfully solved a classic mixing problem using differential equations. We started with a real-world scenario, set up a differential equation based on the principle of "rate in minus rate out," solved the equation using separation of variables, and finally, calculated the alcohol concentration as a function of time. This process demonstrates a powerful application of calculus to a practical problem.

Here are some key takeaways:

  • Differential equations are a powerful tool for modeling dynamic systems, where quantities change over time.
  • Mixture problems often involve tracking the amount of a substance (like alcohol in our case) and its concentration.
  • The principle of "rate in minus rate out" is fundamental to setting up the differential equation.
  • Initial conditions are crucial for finding a unique solution to the differential equation.
  • Analyzing the solution helps us understand the long-term behavior of the system.

This type of problem isn't just limited to brewing beer! It has applications in many different fields, such as:

  • Chemical engineering: Calculating the concentration of reactants in a chemical reactor.
  • Environmental science: Modeling the dispersion of pollutants in a lake or river.
  • Pharmaceuticals: Determining drug concentrations in the bloodstream.
  • Food processing: Mixing ingredients in a food production process.

Understanding how to solve these types of problems gives you a valuable toolset for analyzing and predicting the behavior of various systems in the real world. The ability to apply mathematical models to practical situations is a highly sought-after skill in many industries. By understanding the underlying principles, we can adapt this approach to model and solve a wide range of mixture problems, showcasing the versatility and importance of this type of analysis. Whether it's ensuring the precise mixing of chemicals or understanding environmental contamination, these methods provide a solid foundation for tackling complex challenges.

Conclusion

I hope this step-by-step guide has helped you understand how to calculate alcohol concentration in a mixing tank! It might seem a bit daunting at first, but breaking it down into smaller steps makes it much more manageable. Remember, the key is to set up the differential equation correctly and then apply your calculus skills to solve it. So next time you're faced with a mixing problem, you'll be ready to tackle it head-on! If you have any questions, feel free to ask in the comments below. Keep learning, and keep exploring the fascinating world of math and its applications!