Question

A large tank holds 400 gallons of salt water. A salt water solution with concentration 2 lb/gal is being pumped into the tank at a rate of 3 gal/min. The tank is simultaneously being emptied at a rate of 1 gal/min. If 100 pounds of salt was dissolved in the tank initially, how much salt is in the tank after 250 minutes?

Answer 1

So the tank contains 1866.67 lbs of salt after 250 minutes.

Explanation:

The tank salt concentration at any moment is given as

`SC_(tank)=(Q)/(current \,amount\, of\, water\, solution) =(Q)/(400 + 2t)`

Now the rate of change of concentration is given as

`(dQ)/(dt) = rate_(in) - rate_(out) =SC_(in) * FR_(in) - SC_(tank) * FR_(out)`

Here

SC_in is given as 2 lb/gal

FR_in is given as 3 gal/min

SC_tank is given in above equation

FR_out is given as 1 gal/min

Putting all in the equation gives

`(dQ)/(dt) = SC_(in) * FR_(in) - SC_(tank) * FR_(out)\\(dQ)/(dt) =2 * 3- (Q)/(400+2t)* 1\\(dQ)/(dt) =6- (Q)/(400+2t)* 1\\(dQ)/(dt) +(Q)/(400+2t)=6`

Multiplying by the integrating factor of
`√(400+2t)` on both sides

`(√(400+2t))(dQ)/(dt) +(√(400+2t))(Q)/(400+2t)=6(√(400+2t)\\(√(400+2t))(dQ)/(dt) +\frac{{Q}}{√(400+2t)}=6(√(400+2t))\\(d)/(dt)(Q(√(400+2t)))=6(√(400+2t))`

Integrating both sides with respect to t gives

`Q\left(t\right)=\frac{\left(2\left(2t+400\right)^{(3)/(2)}+c_1\right)√(400+2t)}{400+2t}`

For the initial condition Q(0)=100 lbs so the equation is given as

`100=((0+c_1)√(400))/(400+0)\\c_1=((400 * 100))/(√(400))=2000`

So the equation is

`Q\left(t\right)=\frac{\left(2\left(2t+400\right)^{(3)/(2)}+2000\right)√(400+2t)}{400+2t}`

Solving for the t=250

`Q\left(250\right)=\frac{\left(2\left(2(250)+400\right)^{(3)/(2)}+2000\right)√(400+2(250))}{400+2(250)}\\Q\left(250\right)=1866.67 lbs`