Question

A tank contains 10m^3 of free water, brine having a concentration of 15kgsalt/m^3 is sent into the tank at the rate of 200litres/min. The mixture is kept uniform by mixing and runs out at the rate of 100litres/min. What will be the exit brine concentration when the tank contains 15m^3 of brine

Answer 1

initial amount of salt = $10m^3$

rate of pumping in = 200L/min

Let A be amount of salt at any time t,

then 

$\frac{dA}{dt} = R_{in} - R_{out}$

R is rate of flow.

$R_{in} = \frac{1g}{1L} \frac{5L}{1 min} = 5 g/min$

$R_{out} = \frac{A}{200}*{5} = \frac{A}{40} g/min$

Hence equation of state will be, 

$\frac{dA}{dt} = 5-\frac{A}{40}$

solving equation,

$\frac{dA}{dt} + \frac{1}{40}A =5$

$e^{\frac{1}{40}t}A = \int 5e^{\frac{1}{40}t}dt$

$e^{\frac{1}{40}t}A = 200e^{\frac{1}{40}t} + C$

$A = 200 + Ce^{-\frac{1}{40}t}$

applying condition, A=20 at t=0

$20 = 200 + C \implies C = -180$

Hence desired equation of the state is given by,

$A = 200 - 180e^{-\frac{1}{40}t}$