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

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The discussion focuses on a mixing problem involving a tank with a 1000-gallon capacity, initially containing 500 gallons of water polluted with 50 pounds of particulate matter. Pure water is added at a rate of 20 gallons per minute while the mixed solution is drained at 10 gallons per minute, leading to a net increase of 10 gallons per minute. To determine the amount of particulate matter when the tank overflows, users must analyze the rates of inflow and outflow of both water and particulate matter, applying differential equations to model the situation accurately.

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

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... \frac{A(t) lb}{gal} \frac{gal}{min} units will show you the answer.

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


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
 

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