# Math Help - Air flow in a room. Differential equations

1. ## Air flow in a room. Differential equations

The air in a small 12 ft by 8 ft by 8 ft room is 3% carbon monoxide. Starting at t = 0, fresh air containing no carbon dioxide is blown into the room at a rate of 100 cubic feet per minute. If air in the room flows out through a vent at the same rate, when will be the air in the room be 0.01% carbon monoxide?

Thank you

2. volume of room ... $12 \cdot 8^2 = 768 \, ft^3$

let r = percentage of room air changed in 1 minute ...

$r = \frac{100}{768} \cdot 100 \approx 13.02$%

let p = percentage volume of CO in the room at any time t

$\frac{dp}{dt} = -rp$

$p = p_o e^{-rt}$

$p = 3e^{-rt}$

$.01 = 3e^{-rt}$

$t = \frac{\ln{300}}{r} \approx 43.8 \, min$

3. Here's another way using the integrating factor thing. More roundabout than the skeets though.

$\frac{dy}{dt}=\text{rate in-rate out}$

The rate in of the air is $\left(\frac{100 \;\ ft^{3}}{min}\right)$

$\text{rate out}=\left(\frac{y(t)}{768} \frac{ft^{3}}{ft^{3}}\right)\left(\frac{100 \;\ ft^{3}}{min}\right)=\frac{25y}{192}$

The equation is then:

$\frac{dy}{dt}-\frac{25y}{192}=100$

The integrating factor is $e^{\int \frac{25}{192}dt}=e^{\frac{25t}{192}}$

Use the IC to solve for C. The amount of air at t=0 is

768(.97)=744.96.

$ye^{\frac{25t}{192}}=768e^{\frac{25t}{192}}+C$

Using y(0)=744.96, we find that C=-23.04

We have $\boxed{y=768-23.04e^{\frac{-25t}{192}}}$

.01% CO means 99.99% air. .9999(768)=767.9232

$767.9232=768-23.04e^{\frac{25t}{192}}$

Solving or t we find $\boxed{t=43.805 \;\ min.}$

You can also do this in terms of the CO instead of air. Either way should give the same solution.