Mixing Problem: How much particulate matter is in the tank when it overflows?

In summary, the problem involves a tank with a capacity of 1000 gallons that initially contains 500 gallons of polluted water with 50 pounds of particulate matter. Pure water is added at a rate of 20 gallons per minute while the mixed solution is drained off at a rate of 10 gallons per minute. The question is how much particulate matter is in the tank when it reaches the point of overflowing, which is when the amount of water exceeds 1000 gallons. To solve this, we need to use the rate of water flowing in and out, as well as a separate differential equation to find the time it takes for the water to reach over 1000
  • #1
johnq2k7
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Mixing Problem... need help work shown!

A tank with a 1000 gallon capacity intially contains 500 gallons of water that is polluted with 50 lb of particulate matter. At time (t=0), pure water is added at a rate of 20 gal/min and the mixed solution is drained off at a rate of 10 gal/min. How much particulate matter is in the the tank when it reaches the point of overflowing?

work shown below:

Volume of Tank Capacity= 1000 gal
Intial Volume of Water= 500 gal
Intial Polluted Matter= 50 lb

Pure water Rate IN= 20 gal/min
Mixed Solution Rate OUT= 10 gal/min

Overflowing= excess of 1000 galloons

change in rate= change in rate in - change in rate out
= 20 gal/min - 10 gal/min
= 10 gal/min

change in rate therefore equals 10 gal/min

intial total amount of material in tank is : 500 gallons + 50 lbs
since 1 galloon equals approx. 8.34 lb

therefore, there is intially 917 galloons of material in the tank

how do i use this information that I processed to find the particulate matter once it reaches the point of overflowing which is (>=1000gal)

Please help me with this problem
 
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  • #2


The rate of water flowing in is = rate in - rate out. or [tex]\frac{dA}{dt}= rate in - rate out[/tex]

oops I see you see that already. The rate you seem to be analyzing is that of the water. How about the particulate solution itself? How much is flowing in? How much is going out? So the rate of change of amount of solution must be negative because 0 - rate out. Also to examine the rate out. try using your units.

rate out is lb/min right? so... [tex]\frac{A(t) lb}{gal} \frac{gal}{min}[/tex] units will show you the answer.

where A(t) is amount in pounds (lb)
 
Last edited:
  • #3


Also I forgot to mention that you will need a separate DE to find out the time T it takes for the water to reach >1,000 gal then stick it into your particulate solution DE
 

1. What is a mixing problem?

A mixing problem is a type of mathematical problem that involves calculating the concentration of a solution when two or more substances are mixed together. This can be applied to various real-life scenarios, such as mixing chemicals in a lab or combining ingredients in a recipe.

2. How do you solve a mixing problem?

To solve a mixing problem, you need to set up an equation using the known quantities of the substances being mixed and the desired concentration of the final solution. This equation is then solved using algebraic methods to determine the unknown concentration.

3. What is the "work shown" method for solving mixing problems?

The "work shown" method involves breaking down the mixing problem into smaller, more manageable steps and showing all the work involved in each step. This allows for easier understanding and verification of the solution.

4. Why is it important to show work when solving mixing problems?

Showing work when solving mixing problems is important because it allows for a more thorough understanding of the process and helps to avoid mistakes. It also allows for easier troubleshooting if an error is made.

5. Can you provide an example of a mixing problem with work shown?

Yes, for example, if you have a 500 mL solution of 20% salt and you want to dilute it to a 10% salt solution, you would set up the equation: 0.2(500) = 0.1(x+500). Solving for x, you get x = 250 mL. Then, you would show the work for how much pure water and 20% salt solution to add to reach the desired 10% concentration.

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