Mixing Problem... please help.. work shown

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johnq2k7

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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
 
At time t there are \(\displaystyle 500+(20-10)t=500+10t\) gallons of, say, brine in the tank so:

\(\displaystyle \frac{dy}{dt}=0-\frac{y}{500+10t}(10)=\frac{-y}{50+t}, \;\ \frac{dy}{dt}+\frac{1}{50+t}y=0\) and \(\displaystyle y(0)=50\).

\(\displaystyle {\mu}=e^{\int \frac{1}{50+t}dt}=e^{ln(50+t)}=50+t, \;\ (50+t)y=C, \;\ y=\frac{C}{50+t}\)

\(\displaystyle 50=\frac{C}{50}, \;\ C=2500, \;\ y=\frac{2500}{50+t}\)

The tank reaches overflowing when 500+10t=1000, t=50 minutes so \(\displaystyle y=\frac{2500}{50+50}=25 \;\ lbs\)
 
galactus said:
At time t there are \(\displaystyle 500+(20-10)t=500+10t\) gallons of, say, brine in the tank so:

\(\displaystyle \frac{dy}{dt}=0-\frac{y}{500+10t}(10)=\frac{-y}{50+t}, \;\ \frac{dy}{dt}+\frac{1}{50+t}y=0\) and \(\displaystyle y(0)=50\).

\(\displaystyle {\mu}=e^{\int \frac{1}{50+t}dt}=e^{ln(50+t)}=50+t, \;\ (50+t)y=C, \;\ y=\frac{C}{50+t}\)

\(\displaystyle 50=\frac{C}{50}, \;\ C=2500, \;\ y=\frac{2500}{50+t}\)

The tank reaches overflowing when 500+10t=1000, t=50 minutes so \(\displaystyle y=\frac{2500}{50+50}=25 \;\ lbs\)
Click to expand...

Thanks for your help and workout process... however... don't I need to convert the pariticulate matter from lbs to galloons.. how come this conversion is not needed?... thanks for your help.. appreciate it
 
You can convert if you want, but it ain't necessary. Different ways to tackle such a problem.
 
In general, you cannot convert gallons(volume) to pounds(mass) - without knowing what it is (material - material property).

The number you quote - 1 gallon equals approx. 8.34 lb - is true for water. It will be different for different material.

In this problem there is an "unexpressed" assumption - the addition of particulates create negligible change in the volume of the mixture ( that is the volume of 500 gallons of water with 50 lb. particulate remains 500 gallons - but its mass changes from 4170 pounds to 4220 pounds).
 
thanks.. now I fully understand how the lbs to gallon conversion was neglected...thanks for explaining the details
 
To reiterate, that is a good explanation SK gave. Like he said, this problem just said a particulate. I just used brine as an example of something because that is commonly used. We could have used oil or oreo cookies. Doesn't matter in this case.
 
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