Question

A large tank is filled to capacity with 700 gallons of pure water. Brine containing 3 pounds of salt per gallon is pumped into the tank at a rate of 7 gal/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of pounds of salt in the tank at time t.

Answer 1

The number A(t) of pounds of salt in the tank at time 't' is;

`{A(t)}= -2,100 * e^{(-t)/(100) } + 2,100`

Explanation:

In the question, we have;

The volume of pure water initially in the tank = 700 gal

The concentration of brine pumped into the tank = 3 pounds per gallon

The rate at which the brine is pumped into the tank, = 7 gal/min

The rate at which the well mixed solution is pumped out = The same 7 gal/min

The number of pounds of salt in the tank at time 't' is found as follows;

The rate of change in A(t) with time = The rate of salt input - The rate of salt output

`The \ rate \ of \ change \ in \ A(t) \ with \ time = (dA)/(dt)`

The rate of salt input = 7 gal/min × 3 lbs/gal = 21 lbs/min

The rate of salt output = (A(t)/700) lb/gal × 7 gal/min = (A(t)/100) lb/min

Therefore, we have;

`(dA)/(dt) = 21 - (A(t))/(100)`

Therefore;

`(dA)/(dt) + (A(t))/(100)= 21`

The integrating factor is
`e^{\int\limits {(1)/(100) } \, dx } = e^{(x)/(100) }`

`e^{(t)/(100) } * (dA)/(dt) + e^{(t)/(100) } *(A(t))/(100)= e^{(t)/(100) } *21`

`(d)/(dt) \left[ e^{(t)/(100) } *{A(t)}{} \right]= e^{(t)/(100) } *21`

Using an online tool, we get;

`{A(t)}= c_1 * e^{(-t)/(100) } + 2,100`

At time t = 0, A(t) = 0

We get;

`0= c_1 * e^{(-0)/(100) } + 2,100`

c₁ = -2,100

Therefore, the number A(t) of pounds of salt in the tank at time 't' is given as follows;

`{A(t)}= -2,100 * e^{(-t)/(100) } + 2,100`