Question

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flows into the tank at a rate of Rin=80 L/minutes (min). The fluid mixes instantaneously and is pumped out at a specified rate Rout. Let y(t) denote the quantity of salt in the tank at time t. Assume that Rout=40L/min. (a) Set up and solve the differential equation for y(t). (b) What is the salt concentration when the tank overflows?

Answer 1

a)
`y(t)=50000-49990e^{(-2t)/(25)}`

b)
`31690.7 g/L`

Explanation:

By definition, we have that the change rate of salt in the tank is
`(dy)/(dt)=R_(i)-R_(o)`, where
`R_(i)` is the rate of salt entering and
`R_(o)` is the rate of salt going outside.

Then we have,
`R_(i)=80(L)/(min)*50(g)/(L)=4000(g)/(min)`, and

`R_(o)=40(L)/(min)*(y)/(500) (g)/(L)=(2y)/(25)(g)/(min)`

So we obtain.
`(dy)/(dt)=4000-(2y)/(25)`, then

`(dy)/(dt)+(2y)/(25)=4000`, and using the integrating factor
`e^{\int {(2)/(25)} \, dt=e^{(2t)/(25)`, therefore
`((dy )/(dt)+(2y)/(25)}=4000)e^{(2t)/(25)`, we get
`(d)/(dt)(y*e^{(2t)/(25)})= 4000 e^{(2t)/(25)`, after integrating both sides
`y*e^{(2t)/(25)}= 50000 e^{(2t)/(25)}+C`, therefore
`y(t)= 50000 +Ce^{(-2t)/(25)}`, to find
`C` we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions
`y(0)=10`, so
`10= 50000+Ce^{(-0*2)/(25)}`

`10=50000+C\\C=10-50000=-49990`

Finally we can write an expression for the amount of salt in the tank at any time t, it is
`y(t)=50000-49990e^{(-2t)/(25)}`

b) The tank will overflow due Rin>Rout, at a rate of
`80 L/min-40L/min=40L/min`, due we have 500 L to overflow
`(500L)/(40L/min) =(25)/(2) min=t`, so we can evualuate the expression of a)
`y(25/2)=50000-49990e^{(-2)/(25)(25)/(2)}=50000-49990e^(-1)=31690.7`, is the salt concentration when the tank overflows