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It's hard to say if it does or not. With the very small amount of water in the air, you'd think conduction wouldn't be a big deal. Yet the same could be said for the evaporative effect. The difference in the amount of actual water from near zero to saturated at cold temperatures is small, but may not have much of an effect on the evaporation rate given a much warmer bird. The moisture will go into the relatively dry air anyways given that it is heated by the birds.
Rocket Dad--nice article! Very good explanation of winter ventilation, now I wonder how much heat has to be added to get this effect--e.g. can you get this effect from the heat of the chickens alone, with a solar heat sink?.....
It depends upon the stocking rate. At some point you reach a heat balance where the heat generated by the birds equals the amount of heat lost through the structure and ventilation to produce a given temperature in the barn. If you want it any warmer than that you have to add supplemental heat.
I have a stocking rate of around 1.75 sq ft per bird for 2500 birds.
4 lb layers put out around 40 BTU per bird. So I figure I have bird heat of 100,000 BTU per hour.
The barn is well insulated so I figure I only lose about 5 BTU per sq ft of wall and ceiling regardless of outside temperature (the loss does vary a little with temperature, 5 BTU is a good average). For my barn that equals about 30,000 BTU.
So I have a surplus of 70,000 BTU right? No, we still need to ventilate to remove moisture, carbon dioxide, ammonia and dust. If the ventilation were to ever fail though the birds would be in serious trouble just from overheating.
So, I found at an inside temperature of 65 F, that I needed about 1300 CFM of ventilation to keep the humidity at 80% with an outside air temperature of -10 F. (This ventilation rate increases rapidly with a rise in outside air temperature. It takes little of this very dry air to remove the moisture generated in the barn).
How much energy does it take to warm that air?
(1300 CFM * 60 Minutes * 75F rise in temp) / constant of 55 = 106,363 BTUs
So my heat load is 106,000 BTUS + 30,000 lost through the structure = 136,000 BTU
The birds only produce 100,000 BTU so to maintain 65 F I need an additional 36,000 BTU an hour.
I have an 80,000 BTU radiant heater to make up the difference. Propane gives 90,000 BTU per gallon. So I can figure 36,000 / 90,000 = .40 gallons of LP per hour.
That's the general math behind it. In those conditions I actually burnt less propane than that. The numbers involved are highly variable. I could have been taking in less air than I figured, the birds could have been producing more heat, the insulation may have less losses than I figured...
The biggest demand on the heat is the air for ventilation. I can burn obscene amounts of propane by setting the ventilation rate just a little too high. The whole point of heating this barn is to keep feed costs down. If they eat ad libitum in cold temperatures they can easily eat 25% more feed (they eat $1000 a week anyways). The money saved in feed buys a lot of propane which I can use to provide a better winter environment for the birds. There is a point of diminishing returns though.