Question

A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 10 L/s. The mixture is kept stirred and is pumped out at a rate of 25 L/s. Find the amount of chlorine in the tank as a function of time. (Let y be the amount of chlorine in grams and t be the time in seconds.)

Answer 1

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let c (t ) denote the amount of chlorine (in grams) in the tank at time t .

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(c (t )/(1000 + (10 - 25)t ) g/L) * (25 L/s) = 5c (t ) /(200 - 3t ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after t s there will be (1000 + 10t ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time t is (1000 - 15t ) L. Then the concentration of chlorine per unit volume is c (t ) divided by this volume.

So the amount of chlorine in the tank changes according to

$\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}$

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5t )^(5/3), then integrate both sides to solve for c (t ):

$\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0$

$\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0$

$\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0$

$\dfrac{c(t)}{(200-3t)^{5/3}}=C$

$c(t)=C(200-3t)^{5/3}$

There are 20 g of chlorine at the start, so c (0) = 20. Use this to solve for C :

$20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}$

$\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}$