Question

A tank contains 200 liters of fresh water .Brine containing 2.5 N/liter of dissolved salt runs into the tank at the rate of 8 liters/min and the mixture kept uniform by stirring runs out at 4 liters per minute .Find the amount when the tank contains 240 liters of brine. The concentration of the salt in the tank after 25 minutes amounts to how much

Answer 1

Solution;

Let A(t) be the amount of salt in the tank at any time t.

$\frac{dA}{dt}=R_{in}-R_{out}$ .....(i)

From the given information;

$R_{in}=2.5×8=20N/min$

$R_{out}=\frac{4×A}{200+4t}$

$\frac{dA}{dt}=20-\frac{4A}{200+4t}$

Rewrite as;

$\frac{dA}{dt}+\frac{A}{50+t}=20$

This is a first order differential equation;

The integrating factor:

$I.F=e^{\frac{1}{50+t}dt}$ =$e^{log(50+t)}=50+t$

Solution of the equation is;

$A×I.F=\int (I.F×20)dt +c$

$A(50+t)=\int20(50+t)dt+c$

$A(50+t)=20(50t+\frac{t^2}{2})+c$

$A(50+t)=1000t+10t^2+c$

Initial condition;

A(0)=0

Apply;

$0(50+t)=0+0+c$

Hence ; c=0

And;

$A=\frac{1000t+10t^2}{50+t}$ ....(2)

The tank contains 240ltrs of brine after;

$\frac{240-200}{4}=10 minutes$

After 10 minutes;

$A(10)=\frac{(1000×10)+(10×10^2)}{50+10}$

$A(10)=188.83N$

Now concentration of salt in the tank after 25 minutes;

$A(25)=\frac{(1000×25)-(10×25^2)}{50+25}$

$A(25)=416.66N$