Open water in the coop in winter

[saysfaa]

I keep an open water bucket in my coop.

I often see people say not to do that in the winter because of the moisture it adds via evaporation.

I don't think it adds enough to matter (at least in a well ventilated coop), partly because I don't see evidence of it mattering in my coop and partly because I don't think the physics of it mattering make sense.

So, I tried to do the math to find out.

https://www.engineeringtoolbox.com/evaporation-water-surface-d_690.html

gives a formula that is the most applicable I found with a moderate amount of looking. See the link at the end of this post for more information on the difficulties of coming up with such a formula.

g h = Θ A (x s - x)

where

g s = amount of evaporated water per second (kg/s)

g h = amount of evaporated water per hour (kg/h)

Θ = ( 25 + 19 v ) = evaporation coefficient (kg/m 2 h)

v = velocity of air above the water surface (m/s)

A = water surface area (m 2 )

x s = maximum humidity ratio of saturated air at the same temperature as the water surface (kg/kg) (kg H 2 O in kg Dry Air)

x = humidity ratio air (kg/kg) (kg H 2 O in kg Dry Air)

---

For a swimming pool 50 meters by 20 meters

g s = (34.5 kg/m 2 h ) (1000 m 2 ) ((0.014659 kg/kg) - (0.0098kg/kg)) / 3600
Which calculates out as 0.047 kg/s

My math is weak so I experimented with their equation
0.014659 - 0.0098 = 0.004859
0.004859 x 1000 = 4.859
4.859 x 34.5 = 167.6355
167.6355 divided by 3600 = 0.04656542

Round to three decimal places = 0.047

So, to get the numbers easier for me to see... skip the divide by 3600 to get kg per hour instead of kg per second

167.6355 kg/hour

If the "swimming pool" is 1 sq meter instead of 1000 sq meters and the unit is g instead of kg then

A meter by meter surface would evaporate 167.6355 grams per hour (at these temperatures and this humidity).

A meter by meter = 100 cm by 100 cm
100 x 100 = 10,000 sq cm

167.6355 divided by 10,000 is 0.01676355 g per hour per sq cm

My water bucket is 16 cm in diameter.
area of a circle is pi r squared
3.16 x 8 x 8 = 202.24 sq cm

0.01676355 g per hour per sq cm X 202.24 = 3.39 g per hour.

That is a mostly meaningless number to me.

There are 240 grams of water in a cup of water. A chicken drinks 2 to 4 cups of water per day. So, 480 to 960 grams per day.

3.39g per hour x 24 hours = 81 grams per day.

My water bucket might produce as much as 1/5 of a chicken if all the moisture a chicken produces is released via their poop, breath, and such. I couldn't quickly find info on how much moisture a hen generates in a day but they certainly can't generate more than they take in.

My water bucket doesn't evaporate that much in the winter (and it only matters in the winter) because the formula was for water at 20C (68F), and air temperature of 25C (77F). I couldn't quickly find the numbers for different temperatures. Many sources state that there is much higher evaporation rates at higher temperatures without giving numbers. And I can see that it makes a lot of difference just in how long it takes spilled water to evaporate on a warm day vs a cold day.

Similarly, I didn't easily find the numbers for different humidity levels. I decided it didn't matter for this purpose, partly because the outside humidity level around here is very rarely lower than what was used in the swimming pool example. Higher humidity will slow the evaporation.

For a discussion about some of the difficulties of making a formula for the rate of evaporation of water see
https://van.physics.illinois.edu/ask/listing/1440
I noticed that most of the issues they discuss slow down the rate of evaporation in my context (for example, how much the body of water is mixed by stirring or current, how much wind moves across the surface).

Edit it fix grammar.

And add conclusion: It makes sense to continue not worrying about the open water.

 

[orelsi]

Never had problem with it. Good post!

 

[Sally PB]

I often see people say not to do that in the winter because of the moisture it adds via evaporation.

I'm more worried about water spilling. A horizontal nipple waterer would get around that, I think, but I don't have one.

Thanks for doing the research on this! Very interesting.

 

[aart]

I didn't follow all of your rabbit hole venture closely, but not seeing any temperature aspect?
Temperature could have an effect, especially if open waterer is heated to prevent freezing.
Spilled water, wattle dipping, feet immersion are also concerns/risks.

 

[NatJ]

I didn't follow all of your rabbit hole venture closely, but not seeing any temperature aspect?
Temperature could have an effect, especially if open waterer is heated to prevent freezing.

It's in there, hidden among the other bits:

My water bucket doesn't evaporate that much in the winter (and it only matters in the winter) because the formula was for water at 20C (68F), and air temperature of 25C (77F). I couldn't quickly find the numbers for different temperatures. Many sources state that there is much higher evaporation rates at higher temperatures without giving numbers. And I can see that it makes a lot of difference just in how long it takes spilled water to evaporate on a warm day vs a cold day.

Short version: the formula was for warm water in warmer-yet air. (Water 68F, air 77F) Cooling it down should make it evaporate slower, by an unknown amount.

 

[saysfaa]

Yes, open water would not be a good choice for every situation.

• One would want to choose a very stable bucket. Or weight it. Or tie it. Or brace it. Or something.
• And place the bucket in a corner. Or on a riser. Or some such place less likely to be stepped in. And not in a very crowded coop.
• It may not be appropriate if a rooster were part of the flock. Or, maybe a floppy-combed and -waddled breed. Or maybe when supplementing lights so they lay all winter so have bigger combs and waddles. Or so on.

Nothing is the best choice in every situation. Most of these aren't because of evaporation, though.

Heating the water would increase the evaporation. I skipped most of the discussion about that in these links as I don't heat the water and the math for one vs the other was so hard to find.

Now I'm not even sure whether the swimming pool equation was for a heated pool or not. If it was, then the math works closer for a heated waterer (if heated proportional to a swimming pool - who would heat that much?) and makes my bucket give off much less water than figured above.

The temperatures of water and of air are in the equation in the evaporation coefficient. They were given in the original example in the first link which said they came from experimentation.

 

[saysfaa]

It is really rough. And I don't really follow all the math either. Plugging in differences doesn't always work and may not here either. I don't see the mistakes yet but I make a lot of them.

And thank you for bringing up possible issues.

 

[saysfaa]

Oh. The evaporation was for a heated pool. The point of the example was to find how much heat needed to be added to maintain the temperature.

"...Most of the heat or energy required for the evaporation is taken from the water itself. To maintain the water temperature - heat must be supplied to the water..."