mixing problem

[SUhottie10]

The air in a room with volume 190 cubic meters contains 0.15% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2 cubic meters/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?

I followed an example in the book but that example didnt have percentages so I think that might be what messed me up. Can someone please just check my work and tell me what I might have done wrong?

dy/dt = rate in - rate out

rate in = (.05)*(2) = .1
rate out = (y(t)/180)*(2) = y(t)/90

dy/dx = .1 - y(t)/90
dy/dx = [9-y(t)]/90
dy/[9-y(t)] = dt/90
i integrated both sides and got:
-ln|9-y| = t/90 + C

y(0) = .15 --> -ln|9-.15| = 0/90 + C
C = -ln|8.85|

-ln|9-y| = t/90 - ln 8.85
|9-y| = 8.85e^(-t/90)

y(t) = 9 - 8.85e^(-t/90)

 

[galactus]

Check my figures and make sure.

At t=0, there are 28.5 cubic meters of CO in the tank.

Let y(t) be the amount of CO at time t. We are interested in finding \(\displaystyle y(\infty)\). What happens eventually. What does the amount of CO approach as time approaches infinity?.

\(\displaystyle \L\\\frac{dy}{dt}=\text{rate in-rate out}\)

rate in=\(\displaystyle \L\\(\frac{1}{20}\frac{m^{3}}{min})\cdot(2\frac{m^{3}}{min})=\frac{1}{10}\frac{m^{3}}{min}\)

rate out=\(\displaystyle \L\\\left(\frac{y(t)}{190}\cdot\frac{m^{3}}{min}\right)\cdot\left(2\cdot\frac{m^{3}}{min}\right)=\frac{y(t)}{95}\cdot\frac{m^{3}}{min}\)

\(\displaystyle \L\\\frac{dy}{dt}=\frac{1}{10}-\frac{y}{95}\)

\(\displaystyle \L\\\frac{dy}{dt}+\frac{y}{95}=\frac{1}{10}\)

Integrating factor:

\(\displaystyle \L\\\frac{d}{dt}\left[e^{\frac{t}{95}}y\right]=\frac{1}{10}e^{\frac{t}{95}}\)

After integration:

\(\displaystyle \L\\ye^{\frac{t}{95}}=\frac{19}{2}e^{\frac{t}{95}}+C\)

\(\displaystyle \L\\y=\frac{19}{2}+Ce^{\frac{-t}{95}}\)

But, \(\displaystyle y(0)=\frac{57}{2}\)

\(\displaystyle \L\\\frac{57}{2}=\frac{19}{2}+Ce^{\frac{0}{95}}\Rightarrow{C=19}\)

Therefore,

\(\displaystyle \H\\y=\frac{19}{2}+19e^{\frac{-t}{95}}\)

So, \(\displaystyle \L\\\lim_{t\to\infty}\left(\frac{19}{2}+19e^{\frac{-t}{95}}\right)=\frac{19}{2}\)

Therefore, the amount of CO in the tank eventually is \(\displaystyle \L\\\frac{19}{2}\approx{9.5} \;\ m^{3}\)

\(\displaystyle \L\\\frac{\frac{19}{2}}{190}=5% \;\ CO\)

Check me out. Make sure. Easy to make a mistake.

 

[SUhottie10]

Do you mind explaining to me how you got 28.5 cubic centimeters of CO in the tank and the 1/20 cubic meters/min and 1/190. Thanks. Percentages always confuse me. Of course its always the easy stuff that I have problems with. ha

 

[SUhottie10]

oh i just noticed I made a typo in my problem and its volume 180 cubic meters not 190 so i get the y(t)/190 part now. Sorry about the mistake. now that changes most of your work. I will try and figure it out!