Optimizing Air Flow for CO2 and Heat Loss in Vans

[mess]

TL;DR

I am working on a campervan project that i intend to take far up north in extreme cold climates, as such ive sealed it very well with thick foam insulation. There is nearly 0 air flow between the outside of the van and the inside.

On one of my test trips it was very cold, i kept all the windows and vents closed, and by the morning i had a severe migrain.

My goal is to determine the ideal amount of air flow, in order to keep CO2 below 1000ppm and to minimize heat loss.

I calculated the following based off a similar post in this forum, and I am hoping this can be verified so I know that i am in the right ballpark and going in the right direction as far a solution.
Fresh air has about 0.04% (400ppm) co2, non ideal air has a high limit of 0.1% CO2 (<1000ppm). the difference is .06%.
in a van with typical sprinter dimensions that's about 500 cubic feet minus any things you might have added in your build, but ill say 500.
so 500 cuft * .06% = .3 cuft can be added in order to pass non ideal co2.
and 500 cuft * .26% = 1.3 cuft can be added before you start to get sick from co2. because 0.3% (co2 sickness concentration) - 0.04% (fresh) = .26% difference.
A single occupant adds about 12 cubic feet per day of co2, or about .5 cuft per hour of co2.
That means at best, a single occupant has less than an hour before CO2 levels become not ideal, and just under 3 hours before they might get headache, lack of concentration and sleepiness from too much co2.
now getting into fatal numbers.
500 cuft * 1% = 50 cuft can be added before you might die from co2. This will happen in 100 hours for a single occupant, that's about ~4 days, ~2 days for double, less than that with pets.

Next to figure out a solution through venting in air from the outside:

when the inside air reaches nearly 1000ppm (i have a sensor that controls a vent fan and valve that's attached to the roof of my van now), the vent will open and the fan will suck air from the outside, and mix it in with the return air going into my diesel heater.

Inside air:
500 cf @ 0.1% co2

air coming in from vent :
20cfm @ 0.4% co2

500 cf / 20 cfm = 25 minutes
in 25 minues the entire volume will have been diluted. with would bring it to 50% of the co2%, 500ppm.

therefore at 20cfm, .05% (500ppm) can be removed in 25 min

in one hour that's about ~1000ppm.

but one human is adding .5% (5000ppm) CO2 in one hour. Or 5x more than I am removing.

Therefore in order to maintain around 500ppm in the van i need a flow rate of 20cfm*5 or 100cfm.

And to maintain around 1000ppm CO2 I need a flow rate of about 50cfm? (I think i might have done this part wrong)

(pic of my 3d printed contraption)
Any help appreciated! thank you!

 

[BillTre]

Get masks for the people with an exhale valve plumbed to the outside.
You removed your CO₂ source.

 

[jrmichler]

Here's my calculation for 1000 PPM of $\rm{CO_2}$, where X = Cubic feet per hour of air (CFH). A quantity, X, of air at 0.04% $\rm{CO_2}$ enters, 0.5 CFH of $\rm{CO_2}$ is added, and the same quantity of air (plus the 0.5 CFH of $\rm{CO_2}$) leaves at 0.1% $\rm{CO_2}$:

$0.0004X + 0.5 = 0.001(X + 0.5)$, where $X$ is CFH of air.

Solve for X, and I get 833 CFH, or 14 CFM of air required.

 

[mess]

Here's my calculation for 1000 PPM of $\rm{CO_2}$, where X = Cubic feet per hour of air (CFH). A quantity, X, of air at 0.04% $\rm{CO_2}$ enters, 0.5 CFH of $\rm{CO_2}$ is added, and the same quantity of air (plus the 0.5 CFH of $\rm{CO_2}$) leaves at 0.1% $\rm{CO_2}$:

$0.0004X + 0.5 = 0.001(X + 0.5)$, where $X$ is CFH of air.

Solve for X, and I get 833 CFH, or 14 CFM of air required.

Thanks so much for this!

What was your process for arriving at that equation? I would like to learn :D

 

[jrmichler]

It's a simple flow balance, as illustrated in the sketch for the $\rm{CO_2}$:

The equation says that flow in is equal to flow out, with the flow being unknown $X$. Set up the equation, solve for $X$.

There is a tiny error because oxygen is taken from the incoming air stream and converted into $\rm{CO_2}$ that is not shown in the sketch or the equation. That error is so small as to be meaningless in the scope of this problem.

 

[mess]

It's a simple flow balance, as illustrated in the sketch for the $\rm{CO_2}$:
View attachment 273756
The equation says that flow in is equal to flow out, with the flow being unknown $X$. Set up the equation, solve for $X$.

There is a tiny error because oxygen is taken from the incoming air stream and converted into $\rm{CO_2}$ that is not shown in the sketch or the equation. That error is so small as to be meaningless in the scope of this problem.

Thanks for the explanation! I tried to do that on my own and after about 30 minutes I kind of got some fleeting partial clarity on the equation. What subjects should I study in order to develop equations like these modeling real world behavior?

 

[jrmichler]

What subjects should I study in order to develop equations like these modeling real world behavior?

The primary subject to study is physics. Of course, you need calculus to study physics, and algebra to study calculus. Other core subjects are statics and dynamics. Those subjects will not only give you a solid foundation, but get you about 1/3 of the way to a degree in either physics or engineering.

 

[Tom.G]

https://www.epa.gov/indoor-air-qual...-do-i-need-my-home-improve-indoor-air-quality

...homes receive 0.35 air changes per hour but not less than 15 cubic feet of air per minute (cfm) per person. as the minimum ventilation rates in residential buildings in order to provide IAQ that is acceptable to human occupants and that minimizes adverse health effects. ASHRAE also suggests intermittent exhaust capacities for kitchens and bathroom exhaust to help control pollutant levels and moisture in those rooms. ASHRAE also notes that "dwellings with tight enclosures may require supplemental ventilation supply for fuel-burning appliances, including fireplaces and mechanically exhausted appliances.

Cheers,
Tom