Mixed Separable problem HW

[shuss102]

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

 

[galactus]

At time t there are \(\displaystyle 500+(20-10)t=500+10t\) gallons.

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

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

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

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

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

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

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

\(\displaystyle t=50 \;\ min\)

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

 

[shuss102]

Thanks

Thank you that was very Helpful