Differential equations capacity tank problem (chemical solutions) mixture

In summary: Do you see how that is derived? And do you know how to solve that equation? In summary, a 300-gal capacity tank initially contains a solution of 200 gals of water and 50 lbs of salt. A solution containing 3 lbs of salt per gallon is allowed to flow into the tank at the rate of 4 gal/min. The mixture flows from the tank at the rate of 2 gal/min. The tank overflows when it reaches 300 gallons and the amount of salt in the tank at the end of 30 minutes is approximately 87.33 lbs. At the end of 60 minutes, the tank will have overflowed and the amount of salt in the tank will be approximately
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Homework Statement



A 300-gal capacity tank contains a solution of 200 gals of water and 50 lbs of salt. A solution containing 3 lbs of salt per gallon is allowed to flow into the tank at the rate of 4gal/min. The mixture flows from the tank at the rate of 4 gal/min. The mixture flows from the tank at the rate of 2 gal/min. How many pound of salt are in the tank at the end of 30 min? When does the tank start to overflow? How much salt is in the tank at the end of 60 min?

Homework Equations



[tex] \frac{dx}{dt} + \frac{F}{V_o}x = F_i C_i [/tex]

The Attempt at a Solution



I really don't know how to start the equation, i mean i don't know where to substitute the values on the right variable. So I just need help on where to assign the values given.
 
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  • #2
No, you need much more help than that. This is supposed to be an exercise in understanding what a differential equation says, not just in substituting numbers into a formula.

For one thing, the formula you have written makes no sense because you haven't said what "x", "t", etc. represent.

If x represents the amount of salt in the tank, in pounds, and t the time, in minutes, then dx/dt is the total rate at which salt flows into and out of the tank, in "pounds per minute". Now look at what you are given:
"A solution containing 3 lbs of salt per gallon is allowed to flow into the tank at the rate of 4gal/min." Okay, that is salt flowing into the tank at (3 lbs/gal)(4 gal/min)= 12 lbs/min. That rate is positive because it is coming in and so increasing the amount of salt in the tank.

You also have "The mixture flows from the tank at the rate of 4 gal/min." and " The mixture flows from the tank at the rate of 2 gal/min." Are you sure you have copied the problem correctly? There is no reason to give two different statements like that. I suspec that the correct statement is the second and that the "4 gal/min" in the first sentence were accidently copied from the mixture flowing in. If it were either "4 gal/min" or both, then there would be at least as much solution flowing out as in and the tank would never overflow. So I am going to assume just " The mixture flows from the tank at the rate of 2 gal/min."

The first thing that tells us is that solution is coming in at 4 gal per min and out at only 2 gal per minute so there is a net intake of 4- 2= 2 gal/min. Since the tank initially contained 200 gal, after t minutes, it will contain 200+ 2t gallons. (Until it reaches 300 gallons and overflows- when will that happen?) If there are x pounds of salt in the tank, then there are x/(200+ 2t) "pounds per gallon"- and that is going out at 2 gal/min so out at (x/(200+2t))(2)= x/(100+ t) pounds per minute. Since that is out, it is decreasing the amount of salt- it is negative.

Putting those together we have dx/dt= 12- x/(100+ t) with an initial value of x(0)= 50 pounds.
 

1. What is the differential equations capacity tank problem?

The differential equations capacity tank problem is a mathematical model used to determine the concentration of a chemical solution in a tank over time. It involves solving a system of differential equations that represent the flow of the solution into and out of the tank.

2. How is the capacity of the tank determined in this problem?

The capacity of the tank is typically given as a fixed value in the problem. It represents the maximum amount of solution that the tank can hold at any given time.

3. What are the main variables in the differential equations capacity tank problem?

The main variables in this problem are the concentration of the solution in the tank, the flow rate of solution into and out of the tank, and the volume of solution in the tank at any given time. These variables are used to create a system of differential equations that can be solved to determine the concentration of the solution at any point in time.

4. What are some real-life applications of the differential equations capacity tank problem?

This problem is commonly used in chemical engineering to model and control the concentration of chemicals in tanks used for industrial processes. It can also be applied to environmental engineering to study the movement and mixing of pollutants in bodies of water.

5. How are the differential equations in this problem solved?

There are various numerical methods used to solve the system of differential equations in this problem, such as Euler's method, Runge-Kutta methods, and the Finite Difference method. These methods involve breaking down the differential equations into smaller, simpler equations that can be solved iteratively to approximate the solution.

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