How long will it take to reach 1000ppm CO2 from 2000ppm

[mess]

TL;DR Summary

If a fan is pumping in fresh air (400ppm CO2) at 50 CFM into a space that is 500 CF , another fan is pumping out the air of the space at 50cfm, and the space is currently 2000ppm, how long will it take to reach 1000ppm CO2?

would this be a quadradic equation since its dealing with half life? I think in 500/50 = 10 minutes, the fan would have brought 500 CF of fresh air, but the out fan would have been taking out the same amount of air which includes the fresh air, so at 10 minutes, it should have less then half the average CO2 ppm between the two. so it should be less than (2000+400ppm)/2 = 1200 ppm, but how much exactly?

 

[mfb]

Normally one fan should lead to the specified air flow already, a second fan might increase it.

Your room is probably more than 1 foot tall, so 500 square feet will correspond to far more than 500 cubic feet of air.

Assuming the air mixes fast relative to the exchange rate: You get an exponential distribution above the baseline of 400 ppm. C(t) = 400ppm + 1600ppm*e^-ct. Here c is the air flow divided by the volume.

 

[Delta2]

In order to write down the ODE that governs this system:
For a time window $(t,t+dt)$ :

• How much mass of CO2 is added to the room?
• How much mass of CO2 is removed from the room?

Assume that the density of the CO2 remains constant and equal to $\rho(t)$ during this time window of length $dt$. So at time $t+dt$ what is the new density of CO2 $\rho(t+dt)$. If all goes well and you answer correctly my questions you ll be able to find $d\rho(t)=\rho(t+dt)-\rho(t)$ and from that find the differential equation that governs this system.

 

[mess]

In order to write down the ODE that governs this system:
For a time window $(t,t+dt)$ :

• How much mass of CO2 is added to the room?
• How much mass of CO2 is removed from the room?

Assume that the density of the CO2 remains constant and equal to $\rho(t)$ during this time window of length $dt$. So at time $t+dt$ what is the new density of CO2 $\rho(t+dt)$. If all goes well and you answer correctly my questions you ll be able to find $d\rho(t)=\rho(t+dt)-\rho(t)$ and from that find the differential equation that governs this system.

Thanks for the response! What is dt? is it delta time? and what is p(t)?

How much mass of CO2 is added to the room?

A human adds 5000ppm CO2 in one hour. So me being alone will add that to the room

How much mass of CO2 is removed from the room?

I don't know, there is no constant rate to this because the fresh air is being mixed in with the inside air, and that mixture is being extracted.

 

[Chestermiller]

Let V be the volume of the room and f be the volumetric flow rate into and out of the room. The rate of CO2 entering the room is $fc_{in}$ where $c_{in}$ is the concentration of CO2 in the inlet stream (400 ppm). If the room is well-mixed, the rate of CO2 exiting the room is fc, where c is the instantaneous concentration of CO2 in the room air at time t. Therefore, the rate of increase of CO2 in the room is $fc_{in}-fc=f(c_{in}-c)$. And this is also equal to $V\frac{dc}{dt}$. Thus, from a mass balance on the CO2, we have $$V\frac{dc}{dt}=f(c_{in}-c)$$This is subject to the initial condition c(0)=2000 ppm.

 

[Delta2]

Thanks for the response! What is dt? is it delta time? and what is p(t)?

I guess you are not familiar with Calculus in order to be asking this question. Yes $dt$ is an infinitesimal increase in time. And $\rho(t)$ is the instantaneous density of CO2 in the room at the time $t$.

Anyway @Chestermiller explains quite well how to setup the differential equation that governs this system's behavior at post #5, I guess you can ask him for more details, he is more knowledgeable than me and since he has enter this thread grab the chance to talk with him :D.

 

[russ_watters]

@mess you don't need the differential equation if you just want to calculate the concentration after a certain time; @mfb gave you the equation for it. Otherwise if you want to see the dilution progress you can plug the differential equation into a spreadsheet and graph concentration vs time.

One thing to watch out for is the purge rate depends on the ventilation effectiveness. A perfectly mixed ventilation/room will have a 1.0 ventilation effectiveness. If mixing isn't perfect, it will be lower, and the concentration will be different in different places in the room. And if ventilation is introduced on one side of the room and exhaust is on the other, you will get displacement, and a greater than 1.0 effectiveness.

Some more resources, with different calculations for differently framed problems:
https://en.wikipedia.org/wiki/Dilution_(equation)#Dilution_ventilation_equation
http://faculty.washington.edu/airion/ENVH557/Dilution_Vent_1.pdf

 

[russ_watters]

Your room is probably more than 1 foot tall, so 500 square feet will correspond to far more than 500 cubic feet of air.

It's a van.