Answer to Question #253916 in Differential Equations for DEE

Question

A 2000 L tank initially contains 40 kg of salt dissolved in 1000 L of water. A brine solution containing 0.02 kg/L of salt flows into the tank at a rate of 50 L/min. The solution is kept thoroughly mixed, and the mixture flows out at a rate of 25 L/min. (a) Find the quantity of salt in the tank at any time t > 0) prior to overflow. (b) Find the time of overflow.

Answer 1

Solution;

(a)

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

Then;

S(0)=0

Let V(t) be the volume in the tank at any time t

Then V(0)=1000

But ,net flow is 50L/min -25L/min.

Hence;

$V(t)=1000+25t$

Rate of salt going in;

$\frac{dS}{dt}_{in}=\frac{0.02kg}{L}×\frac{50L}{min}$ =$1\frac{kg}{min}$

Rate of salt going out;

$\frac{dS}{dt}_{out}=\frac{25S}{1000+25t}$

Hence the net rate of salt in tank is;

$\frac{dS}{dt}=1-\frac{25S}{1000+40}$

Rearrange as follows;

$\frac{dS}{dt}+\frac{25}{1000+25t}S=1$

The integrating factor;

$I.F=e^{\frac{25}{1000+25t}dt}$ =$t+40$

Hence;

$\frac{d}{dx}(40+t)S=(40+t)×1$

Integrating both sides;

$S(40+t)=\int(40+t)dt=40t+\frac{t^2}{2}+C$

Resolve as follows;

$S(t)=\frac{t^2+80t}{2(40+t)}+\frac{C}{40+t}$

But ;

$S(0)=40$

Applying this initial conditions;

$S(0)=40=0+\frac{C}{40}$

Hence ;

$C=40×40=1600$

Hence,the quantity of salt in the tank at t>0 prior to overflow is described as;

$S(t)=\frac{t^2+80t}{2(40+t)}+\frac{1600}{40+t}$

(b)

Time for overflow;

$V(t)=1000+25t$

At time of overflow,the tank is full hence;

$V(t)=2000L$

$2000=1000+25t$

$25t=1000$

$t=40$ minutes.