jeca86 said:
the air in a room with volume 180m^3 contains 0.15% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2m^3/min and the mixed air flows out at the same rate. Find the percentage of carbon dioxide in the room as a function of time. What happens in the long run?
Let Y(t) be the amount of CO2 at time t.
Initial Value given: Y(0) = 0.15
y' = amount of c02 flowing in - amount of co2 flowing out
The amount of co2 flowing in will be the amount of air flowing in times the percentage that is co2. So amount in = 0.0005 * 2m^3/min
The amount of co2 flowing out is (y(t))/(180) * 2m^3/min. This is because the room is imagined to have a uniform mixture, so the percentage of c02 per m^3 in the room will be the same as the percentage of co2 per m^3 flowing out.
So y' = 0.0005 * 2m^3/min - (y(t))/(180) * 2m^3/min
y' = 0.001 m^3/min - y/90 m^3/min
Thus, y' = 1/1000 - y/90 = (9-100y)/9000
(y')/(9-100y) = 1/9000
U substitution time.... Let u = 9-100y, du = -100dy
Then y = (9-u)/100, dy = -1/100 du
(y')/(9-100y) = (-du/100)/(u) = -1/100 (du/u)
So, -1/100 (du/u) = 1/9000
Integrate:
(-1/100) Ln(u) = t/9000 + C
Ln(u) = -t/90 + C
But u = 9-100y,so
Ln(9-100y) = -t/90 + C
9-100y = e^(-t/90) * e^C
y = -(Ce^(-t/90) - 9)/100
Use your initial condition now... Y(0) = 0.15 = 15/100
y(0) = -(C-9)/100 = 15/100
-(C-9) = 15
C = -6.
Therefore y = 0.06e^(-t/90) + 0.09
To see what happens in the "Long Run" Take the limit of y as t approaches infinity. Does it approach a specific value?
