Question

A 250-gal tank is initially filled with brine (salt-water mixture) containing 100 lb of salt. Brine containing 3 lb of salt per gallon enters the tank at the rate of 7 gal/s, and the well-mixed brine in the tank flows out at the rate of 9 gal/s. Find the amount x(t) of salt in the tank after 1 minute.

Answer 1

x(t) = 581.66 lb

Explanation:

From the given information:

Consider the salt quantity in the tank at time t be x(t) lb.

∴ the rate of change of salt content in the tank is
`(dx)/(dt)= Rate \ of \ \ inflow - rate \ of \ \ outflow`

where:

the rate of inflow = salt conc. × flow rate = 3 lb/gallon × 7 gallons

=21 lb/s

rate of outflow = salt conc. in the tank × flow rate

= x/250 × 9

= 9x/250 lb/s

∴

`(dx)/(dt) = 21 - (9x)/(250)`

`(dx)/(dt) + (9x)/(250)= 21`

This is a 1st order linear differentiation,

The integrating factor if
`e^{^{ \int (9)/(250)\ dt}} = e^{^{ (9t)/(250) }}`

∴

`e^{^{ (9t )/(250)}} \ \ x(t) = \int 21 e^{(9t)/(250)} \ dt + C`

`e^{^{ (9t )/(250)}} \ \ x(t) = 21 * (250)/(9)e^{(9t)/(250)} + C`

`x(t) = (1750)/(3)+Ce^{^{(-9t)/(250)}}` at t = (0) and x(0) = 100 lb

Hence;

`100 = (1750)/(3)+C`

`C = (1750)/(3) -100`

`C = - (1450)/(3)`

∴
`x(t) = (1750)/(3)-(1450)/(100)e^{^{(-9t)/(250)}}`

after time t = 1 minute i.e 60 s

`x(t) = (1750)/(3)-(1450)/(100)e^{^{(-9 * 60)/(250)}}`

`x(t) = (1750)/(3)-(1450)/(100)e^{^{(-540)/(250)}}`

x(t) = 581.66 lb