Answer to Question #238273 in Calculus for Anuj

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

Let s(t) = amount, in kg of sugar at time t. Then we have

$\frac{dt}{ds}$ = (rate of salt into tank) - (rate of salt out of tank)

$= (0.05 \frac{kg}{L} · 5 \frac{L}{min}) + (0.04 \frac{kg}{Min} · 10 \frac{L}{min}) − \frac{s kg}{1000 L} · 15 \frac{L}{min}\\ = (0.25 \frac{kg}{Min}) + (0.04 \frac{kg}{Min} ) − \frac{15s kg}{1000 min} \\$

So we get the differential equation

$\frac{ds}{ dt} = 0.65 − \frac{15s}{ 1000}\\ \frac{ds}{ dt} =\frac{ 130 − 3s}{ 200}$

We separate s and t to get

$\frac{1}{ 130 − 3} ds = \frac{1}{ 200} dt$

Integrate

$\int \frac{1}{130-30s}ds=-\frac{1}{30}\ln \left|130-30s\right|+C\\ \frac{1}{200}t+C\\ \implies s = \frac{130 − C_4e^{−3t/200}}{ 3}$

Since we begin with pure water, we have s(0) = 0. Substituting,

$C_4 = 130$

$s = \frac{130 − 130e^{−3t/200}}{ 3}\\ s(60) = \frac{130 − 130e^{−3t/200}}{ 3} = 25.7153$

Thus, after one hour there is about 25.72 kg of sugar in the tank.