Question

A tank contains 5,000 L of brine with 11 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. (a) How much salt is in the tank after t minutes? y = ___________kg (b) How much salt is in the tank after 20 minutes? () y = ________ kg

Answer 1

a

`y(t) = 11e^{-(t)/(100) }`

b

`y(20) = 9 \ kg`

Explanation:

From the question we are told that

The volume of brine in the tank is
`V_L = 5000 \ L`

The mass of the salt is
`y(0)= 11 \ kg`

The rate at which water enters that tank is
`\r V = 50 \ L / min`

Generally the rate at which salt enters the tank is mathematically represented as

`z = (0)/(V_L) * \r V`

=>
`z = (0)/(5000) * 50`

Here the zero implies that there is no salt entering the tank

Generally the rate at which salt leaves the tank is mathematically represented as

`u = (y(t))/(V_L) * \r V`

=>
`u = (y(t))/(5000) * 50`

=>
`u = (y(t))/(100)`

Generally the rate of salt flow in and out of the tank is

`(dy)/(dt) = z - u`

=>
`(dy)/(dt) = - (y(t))/(100)`

=>
`(dy)/(y) = -(dt)/(100)`

=>
`ln(y(t)) = -(t)/(100) + c`

at time t = 0 y(0) = 11

=>
`ln(11) = -(0)/(100) + c`

=>
`c = ln (11)`

So

`ln(y(t)) = -(t)/(100) + ln (11)`

taking exponent of both sides

`y(t)=e^{ -(t)/(100) + ln (11)}`

=>
`y(t) = e^{-(t)/(100) } * e^(ln(11))`

=>
`y(t) = 11e^{-(t)/(100) }`

Generally at t = 20 minutes

The amount of salt in the tank is mathematically represented as

`y(20) = 11e^{-(20 )/(100) }`

=>
`y(20) = 9 \ kg`