Question

A tank having a capacity of 1000 liters, initially contains 400 liters of sugar water having a concen-

tration of 0.2 Kg of sugar for each liter of water. At time zero, sugar water with a concentration of

50 gm of sugar per liter begins pumped into the tank at a rate of 2 liter per minute. Simultaneously,

a drain is opened at the bottom of the tank so that the volume of the sugar-water solution in the

tank reduces 1 liter per minute. Determine the following:

Answer 1

Solution;

(a) Amount if salt in the tank after a t minutes;

Define a function;

s(t)=kgs of salt in the tank at a time t(minutes)

And ;

s'(t)=rate of change of amount of salt in the tank (rate of salt going in -rate of salt going out)

Rate of salt going in;

$=\frac{0.5kg}{1litre}×\frac{2litres}{min}=1kg/min$

Initial volume is 400 litres ;

Volume after 1 minute us;

400+2-1

Volume after time t;

400+t

The rate of salt going out is;

$=\frac{s(t)kg}{(400+t)litres}×\frac{1litre}{min}=\frac{s(t)}{(400+t)min}$

Therefore the rate of change of salt in the tank is ;

$s'(t)=1-\frac{s(t)}{400+t}$

Rewrite;

$s'(t)+\frac{s(t)}{400+t}=1$

The equation is if the form s'(t)+p(t)s(t)=g(t) ,whose integration is obtained as;

$s(t)=\frac{1}{\mu(t)}\int \mu(t)g(t)dt+\frac{D}{\mu (t)}$

Where;

$\mu(t)=e^{\int p(t) dt}$ This yields;

$s(t)=\frac{800t+t^2}{2(400+t)}+\frac{D}{400+t}$

Initial conditions are;

At t=0,s(t)=0.2×400=80kg

$s(0)=80=\frac{D}{400}$

D=32000

Hence,the amount of salt in the tank after a time t is;

$s(t)=\frac{800t+t^2}{2(400+t)}+\frac{32000}{400+t}$

(b) Amount of salt in the tank after 20minutes

$s(t)=\frac{800×20+20^2}{2(400+20)}+\frac{32000}{400+20}=103.33kg$