Question

A tank initially contains 50 litres of pure water. Starting at time t = 0 a brine containing 2 grams of dissolved salt per litre flows into the tank at the rate of 3 litres per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time t > 0

Answer 1

Let x be the amount of salt in the tank at time t.

∴ `("d"x)/"dt"` = inflow rate – outflow rate ........(1)

Given 2 grams of dissolved salt per litre flows into the tank at the rate of 3 litres per minute.

(i.e) inflow rate contain = 6 gram salt

∵ For one litre = 2-gram salt

For 3 litre = 6 gram salt

Tank contain 50 litres of water

∴ Out flow of rate of salt = `(3x)/50`

Substitute inflow rate and outflow rate in equation (1), we get

`("d"x)/"dt" = 6 - (3x)/50`

= `(6 xx 50 - 3x)/50`

`("d"x)/"dt" = (300 - 3x)/50`

= `(3[100 - x])/50`

`("d"x)/"dt" = (- 3[x - 100])/50`

The equation can be written as

`("d"x)/(x - 100) = (-3)/50  "dt"`

Taking integration on both sides, we get

`int ("d"x)/(x - 100) = int (-3)/50  "dt"`

`int ("d"x)/(x - 100) = (-3)/50 int "dt"`

`log(x - 100) = (-3)/50  "t" + log "c"`

`log(x - 100) - log "c" = (-3)/50  "t"`

`log((x - 100)/c) = (-3)/50  "t"`

`(x - 100)/"c" = "e"^((-3)/50)  "t"`

`x - 100 = "ce"^((-3"t")/50)`

Initial condition:

When t = 0, x = 0

`0 - 100 = "ce"^((-3 xx 0)/50)`

`- 100 = "ce"^circ`

`- 100` = c

Substituting c value in equation (2), we get

`x - 100 = - 100  "e"^((-3"t")/50)`

x = `- 100  "e"^((-3"t")/50) + 100`

x = `100[1 - "e"^((-3"t")/50)]`

∴ The amount of salt present in the tank `100[1 - "e"^((-3"t")/50)]`