Question

A tank initially holds 200 gallons of fresh water (no salt). A brine solution containing 2 pounds of salt per gallon is poured in at a rate of 3 gallons per minute. The tank is kept well-mixed and drained at the rate of 3 gallons per minute. Write and solve an initial value differential equation to model the amount of salt, A, in the tank at time, t. 2.

Answer 1

1.99 pounds per gallon of salt in t=2 in the tank.

Explanation:

First we consider the matter balance equation that contemplates the input and output; the generation and consumption equal to the acomulation.

In this case we have no Generation neither do Consumption so, if we consider Acomulation = A(t), the rate of change of A(t) in time is given by:

`(dA)/(dt)+R_(out)A(t)= CR_(s)` ---(1)

where C is th concentration, with the initial value statement that A(t=0) = 0 because there is no salt in the time cero in the tank, only water.

Given the integral factor ->
`u(t)= exp[R_(out)]` and multipying the entire (1) by it, we have:

`\int (d)/(dt)[A(t) \exp[R_(out)t]] \, dt = CR_(in) \int \exp[R_(out)t] \, dt`

Solving this integrals we obtain:

`A(t)=(CR_(in))/(R_(out))+Cte*\exp[-R_(out)t]`

So given the initial value condition A(t=0)=0 we have:

`Cte=- (CR_(in))/(R_(out))`,

and the solution is,

`A(t)=(CR_(in))/(R_(out))-(CR_(in))/(R_(out))\exp[-R_(out)t]`.

If we give the actual values we obtain then,

`A(t)=2(pounds)/(gallon)-2(pounds)/(gallon) \exp[-3t]`.

So in t= 2 we have
`A(t)=2(pounds)/(gallon)`.