Question

Brine containing 3 lbs./gal of salt enters a large tank at the rate of 2 gals / min and the mixture well stirred leaves at 1.5gal / min. If the tank contains initially 100 gal of water, with 4 lbs. of dissolved salt.

a ) Find the amount of salt in the tank at any time t in minutes .

b ) Find the amount of salt in the tank after 4 minutes.

Answer 1

Solution;

Let y represent the amount of salt in the tank at time t, where t is given in minutes;

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

$R_{in}=\frac{3lbs}{gal}×\frac{2gals}{min}=6lbs/min$

Volume of mixture at time time t;

$V(t)=100+(2-1.5)t$

$V(t)=100+0.5t$

$R_{out}=1.5×\frac{y}{100+0.5t}=\frac{1.5y}{100+0.5t}$

Hence;

$\frac{dy}{dt}=6-\frac{1.5y}{100+0.5t}$

Rewrite;

$\frac{dy}{dt}+\frac{1.5y}{100+0.5t}=6$

The general solution of the equation is;

$y(t)=e^{-\int(x)dx}\int e^{\int p(x)dx}Q(x)$

$y(t)=e^{-\int \frac{1.5}{100+0.5t}dt}×6\int e^{\frac{1.5}{100+0.5t}dt}$

$y(t)=-6(100+0.5t)^{-1.5}\int(100+0.5t)^{1.5}dt$

$y(t)=-6(100+0.5t)^{-1.5}\frac{(100+0.5t)^{2..5}}{2.5}+C$

$y(t)=\frac{100+0.5t}{2.5}+C$

The initial conditions are;

$t=0,y=4$

$y(0)=4=\frac{100}{2.5}+C$

$C=4-40=-36$

Amount at time t is;

$y(t)=\frac{100+0.5t}{2.5}-36$

Amount after t=4;

$y(4)=\frac{100+0.5(4)}{2.5}-36$

$y(4)=4.8lbs$