# Thread: Concentration in brine solution

1. ## Concentration in brine solution

A tank contains 100 litres of brine whose concentration is 3 grams/litres. Three litres of brine whose concentration is 2 grams/litre flow into the tank each minute and at the same time 3 litres of mixture flow out each minute.

Show that the quantity of salt, Q gram, in the tank at any time t is given by:
Q = 200 + 100e^(-0.03t)

i have managed to get Q = 200 - 100e^(-0.03t)

and cannot figure out why the minus is there, i have included the pdf of the working i have done so far, any help appreciated

thanks
jacs

2. Originally Posted by jacs
A tank contains 100 litres of brine whose concentration is 3 grams/litres. Three litres of brine whose concentration is 2 grams/litre flow into the tank each minute and at the same time 3 litres of mixture flow out each minute.

Show that the quantity of salt, Q gram, in the tank at any time t is given by:
Q = 200 + 100e^(-0.03t)

i have managed to get Q = 200 - 100e^(-0.03t)

and cannot figure out why the minus is there, i have included the pdf of the working i have done so far, any help appreciated

thanks
jacs

In your solution you get to a line:

$t=-\frac{100}{3}\ln(600-3Q)+c$.

But now at $t=0,\ Q=300$ putting this into the above equation
gives a term $\ln(-300)$ not $\ln(300)$ which you have.

To avoid this change your integral to:

$t=-\frac{100}{3} \int \frac{3}{3Q-600}dQ$,

or:

$-0.03t=\ln(3Q-600)+C$.

Which if I have done this right should now allow you to get the

RonL

3. I see, thank you I had that originally, but since it yeilded the negative ln result, I assumed I had somehow messed it up.

Of course, it is a simple matter to swap the sign to avoid that problem (now that you have pointed it out to me...lol.... )

I bow before you genius

thanks

jacs