Mixed Separable problem HW

shuss102

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Oct 7, 2011
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A tank with a 1000 gal capacity initially contains 500 gal of water that is polluted with a 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 is in the tank when it reaches the point of overflowing?

With this question i attempted to fig out so far, that the rate in minus rate out, is 10 gal/per min, which means that the rate in is greater than the rate out, and the tank fills up at t = 50 min , i get dA/dt to -10/500 lbs/min, however that is the conc alone. how do i figure out how much particulate is in the tank at the point of overflow?

thanks for the help
 
At time t there are 500+(2010)t=500+10t\displaystyle 500+(20-10)t=500+10t gallons.

So, dydt=0y500+10t(10)=y50+t\displaystyle \frac{dy}{dt}=0-\frac{y}{500+10t}(10)=\frac{-y}{50+t}

dydt+150+ty=0\displaystyle \frac{dy}{dt}+\frac{1}{50+t}y=0 and y(0)=50\displaystyle y(0)=50

The integrating factor is e150+tdt=eln(50+t)\displaystyle e^{\int \frac{1}{50+t}dt}=e^{ln(50+t)}

(50+t)y=C\displaystyle (50+t)y=C

y=C50+t\displaystyle y=\frac{C}{50+t}

50=C50,   C=2500,   y=250050+t\displaystyle 50=\frac{C}{50}, \;\ C=2500, \;\ y=\frac{2500}{50+t}

The tank reaches overflowing when 500+10t=1000\displaystyle 500+10t=1000

t=50   min\displaystyle t=50 \;\ min

So, y=250050+50=25   lbs\displaystyle y=\frac{2500}{50+50}=25 \;\ lbs
 
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