Consider a tank which initially holds `V_0` gal of brine that contains a lb of salt. Another brine solution, containing b lb of salt per gallon, is poured into the tank at the rate of e gal/min while, simultaneously, the well-stirred solution leaves the tank at the rate of f gal/min. The problem is to find the amount of salt in the tank at any time t.
Let Q denote the amount (in pounds) of salt in the tank at any time t. The time rate of change of Q, `(dQ)/(dt)` , equals the rate at which salt enters the tank minus the rate at which salt leaves the tank. Salt enters the tank at the rate of be lb/min. To determine the rate at which salt leaves the tank, we first calculate the volume of brine in the tank at any time t, which is the initial volume `V_0` plus the volume of brine added et minus the volume of brine removed ft . Thus, the volume of brine at any time is `V_0+et-ft` . The concentration of salt in the tank at any time is `Q/(V_0+et-ft)` , from which it follows that salt leaves the tank at the rate of `f\ Q/(V_0+et-ft)` gal/min.
Thus, `(dQ)/(dt)=be-f\ Q/(V_0+et-ft)` or `(dQ)/(dt)+f/(V_0+(e-f)t) Q=be` .
Example 1. A tank initially holds 100 gal of a brine solution containing 20 Ib of salt. At t = 0, fresh water is poured into the tank at the rate of 5 gal/min, while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time t.
Here, `V_0=100` , `a=20` , `b=0` and `e=f=5` . So, equation is
`(dQ)/(dt)+1/20 Q=0` . Solution to this linear equation is `Q(t)=Ce^(-t/20)` .
At t=0, we are given that `Q=a=20` , so `20=Ce^(-0/20)` or `C=20` .
So, `Q(t)=20e^(-t/20)` .
Note that `Q->0` when `t->oo` as it should, since only fresh water is being added.
Example 2. A tank contains 40 l of solution containing 2 g of substance per liter. Salt water containing 3 g of this substance per liter runs in at the rate of 4 l/min and the well-stirred mixture runs out at the same rate. Find the amount of substance in the tank after 15 minutes.
Note that here we have problem in terms of liters and grams, not in terms of pounds and gallons. It doesn't matter. The main thing is that all variables were presented using same units.
Since initially there are 40 l of solution and each liter contains 2 grams of substance then initially there are 40*2=80 grams of substance.
So, `V_0=40` `a=80` , `b=3` , `e=f=4` .
And equation is written as `(dQ)/(dt)+1/10 Q=12` . This is linear equation. Its soltuion is `Q(t)=120+Ce^(-t/10)` .
At t=0, we are given that `Q=a=80` , so `80=120+Ce^(-0/10)` or `C=-40` .
Thus, `Q(t)=120-40e^(-t/10)` .
We need to find `Q(15)` : `Q(15)=120-40e^(-15/10)~~111.075` grams.