There was a dude who did the Iditarod and wrote a book. There was a section of the trail where the dogs had to go balls out b/c the temp was so low (-somewhere in the minus 1 or 2 hundreds), if you stopped you would die.
I saw your pic, but not the calcs. For example, your water loss rate is listed as 0.034 kg/h, whereas I had 0.0242 kg/h. How’d you get your number?
If I use your water flow rate, my water vaporizing power becomes 21.3 watts instead of 15.17 watts. That’s still not enough to account for the difference between your 44 watts total and my 31 watts total.
How is your 44 watts distributed? How much is going into heating the air, and how much is going into heating the water?
@Machine_Elf - I will go through the details when I get some time later. but here are somethings to check :
I think that density is incorrect. I get a density of around 1.66 grams per liter. Can you please check ?
Also the latent heat of vaporization is a function of temperature.
I used the density of air at STP (20C), and meant to write 1.225 grams per liter, not 0.001225. Pretty sure I used 1.225 g/l in my calcs. Having said that, I should have used the density at 37C, which would be even less. I don’t think you’re ever going to have super-cold air in your lungs - it gets heated on the way in - so I think 1.66 g/l is not an appropriate choice.
For the water, we’re starting with water at 37C, so the heat of vaporization should be the value for that temperature. This is not terribly different from the value at STP. I took the 2260 kj/kg value from the top of this google search result, but if I go with this chart instead, it looks more like 2425 kj/kg, in which case my water-vaporizing power becomes 16.28 watts. Still not enough to match with your results.
You are estimating heat loss by the human body, whether the heat is lost in the lungs or the passageways, wouldn’t it still count ?
My point is that you’re not going to have air existing in the lungs or passages at 1.66 g/l; it’s going to warm up on the way in, resulting in a density corresponding to ~37C, so using such 1.66 g/l to estimate mass flow rate from tidal volume will give an implausibly high mass flow rate.
I was going with the calculation basis originally posted in your post.