Mixing Problem with Constant Coefficient Differential Equation

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In summary, a tank initially contains 1160 L of pure water. A solution with a concentration of 0.03 kg of sugar per liter enters the tank at a rate of 7 L/min and is mixed and drained from the tank at the same rate. Using the rate in minus the rate out formula, an equation is formed with the amount of sugar, s(t), as a function of time, t. This equation can be rewritten in the form of a constant coefficient, linear, and separable differential equation. The constant coefficient method, linear method (using an integrating factor), and separation of variables method can be used to solve for the amount of sugar in the tank after t minutes and the concentration of sugar in the solution after
  • #1
CollegeStudent
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Homework Statement



A tank contains 1160 L of pure water. A solution that contains 0.03 kg of sugar per liter enters a tank at the rate 7 L/min The solution is mixed and drains from the tank at the same rate.

a.) How much sugar is in the tank initially?
b.) Find the amount of sugar in the tank after t minutes.
c.) Find the concentration of sugar in the solution in the tank after 72 minutes.

Homework Equations


The Attempt at a Solution



I made s(t) be the amount of salt at time 't'

ds/dt = rate in - rate out

Well the rate in would be the concentration times the rate... so

ds/dt = (.03kg)(7L/min) - (s/1160)(7L/min)

ds/dt = .21 - (7s/1160)

Kind of get confused from there...I figure I'll have to get the 's' on one side right? separable?Am I even on the right track here? Just starting out in the class.
 
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  • #2
CollegeStudent said:

Homework Statement



A tank contains 1160 L of pure water. A solution that contains 0.03 kg of sugar per liter enters a tank at the rate 7 L/min The solution is mixed and drains from the tank at the same rate.

a.) How much sugar is in the tank initially?
b.) Find the amount of sugar in the tank after t minutes.
c.) Find the concentration of sugar in the solution in the tank after 72 minutes.

Homework Equations





The Attempt at a Solution



I made s(t) be the amount of salt at time 't'

ds/dt = rate in - rate out

Well the rate in would be the concentration times the rate... so

ds/dt = (.03kg)(7L/min) - (s/1160)(7L/min)

ds/dt = .21 - (7s/1160)

Kind of get confused from there...I figure I'll have to get the 's' on one side right? separable?


Am I even on the right track here? Just starting out in the class.

Yes, you are on the right track. So far so good.

If you call ##a = \frac 7 {1160}## your equation can be rewritten$$
\frac{ds}{dt} +as = .21$$It is constant coefficient, linear, and separable, so any of these methods would work. I would suggest the constant coefficient method first, linear (integrating factor) second, and separation of variables last in order of preference.
 
  • #3
Constant coefficient? I wouldn't just integrate here? I'm looking all over the web for an explanation of "constant coefficient" but not seeing too much
 
  • #4
CollegeStudent said:
Constant coefficient? I wouldn't just integrate here? I'm looking all over the web for an explanation of "constant coefficient" but not seeing too much

That's "constant coefficient differential equation". If you haven't studied those yet use the integrating factor method for first order linear DE's.
 

1. What is a "Diff Eq's Mixing problem"?

The "Diff Eq's Mixing problem" refers to a type of differential equation that involves solving for the concentration of a substance in a mixture over time. This problem is commonly encountered in chemistry and chemical engineering.

2. How is the "Diff Eq's Mixing problem" different from other types of differential equations?

The "Diff Eq's Mixing problem" is unique because it involves multiple variables, including time and concentration, and takes into account the changing rates of the substance being mixed.

3. What are some real-world applications of the "Diff Eq's Mixing problem"?

The "Diff Eq's Mixing problem" is commonly used in industries such as pharmaceuticals, food and beverage, and environmental engineering to model the mixing of chemicals and substances in various processes. It is also used in atmospheric science to study the dispersion of air pollutants.

4. What are the key steps in solving the "Diff Eq's Mixing problem"?

The first step is to define the differential equation for the specific mixing scenario. Then, it is necessary to solve the equation using techniques such as separation of variables, substitution, or integrating factors. Finally, the solution is checked for accuracy and interpreted in the context of the problem.

5. Are there any challenges in solving the "Diff Eq's Mixing problem"?

Yes, there can be challenges in solving the "Diff Eq's Mixing problem" as it involves solving a complex differential equation and interpreting the results in the context of the specific mixing scenario. Additionally, the problem may involve multiple variables and parameters, making it more difficult to find an exact solution.

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