Help Off Grid Chicken Coop

[SunEagle]

I thought the light bulb would be placed under the water bowl (because heat rises), not above it.

Putting a heat source below an object will allow the heat to rise but what is rising is "heated air" and using such a small heat load (7 watt) will not heat enough air to keep the water from freezing.

A better heat source is one the emits or radiates infrared energy. That energy is absorbed by the object which then eventually heats up if more heat goes in than leaves.

Even then depending on what you are heating, how cold is it and how much heat energy you are sending are all variables to the solution (or problem). I won't get into the thermodynamics (too much math) but you can find out how much heat rejection a certain wattage lamp sends out in BTU's and then determine how many BTU"s you need to keep the volume of water from freezing.

One BTU will raise 1 pound of water 1 degree F. One BTU = 0.293 watt hours. So depending on how many degrees F you need to heat and how much water you want to heat will determine how many watt hours you need to use times a fudge factor due to losses between energy generated and the energy (BTU's) that get to the water.

[Shockah]

Putting a heat source below an object will allow the heat to rise but what is rising is "heated air" and using such a small heat load (7 watt) will not heat enough air to keep the water from freezing.

A better heat source is one the emits or radiates infrared energy. That energy is absorbed by the object which then eventually heats up if more heat goes in than leaves.

Even then depending on what you are heating, how cold is it and how much heat energy you are sending are all variables to the solution (or problem). I won't get into the thermodynamics (too much math) but you can find out how much heat rejection a certain wattage lamp sends out in BTU's and then determine how many BTU"s you need to keep the volume of water from freezing.

One BTU will raise 1 pound of water 1 degree F. One BTU = 0.293 watt hours. So depending on how many degrees F you need to heat and how much water you want to heat will determine how many watt hours you need to use times a fudge factor due to losses between energy generated and the energy (BTU's) that get to the water.

Aaahhh,

As you can see, Hawaiians don't have much experience with frozen Chicken Coop water...

[J.P.M.]

But fortunately for life as we know it, ice also rises (floats.)

And for the same reason air warmer than its surroundings rises in a gravity field - it's less dense. In the absence of gravity, I don't believe ice will float in water. Put a water/ice mixture in a centrifuge and the Ice will "gravitate" to the center.

[inetdog]

And for the same reason air warmer than its surroundings rises in a gravity field - it's less dense. In the absence of gravity, I don't believe ice will float in water. Put a water/ice mixture in a centrifuge and the Ice will "gravitate" to the center.

I still like the ancient Greek philosophers' idea that something floats because of the quantity of levity it has.
Oh, and helium balloons move to the inside of a curve when floating inside your car. Much more accessible than a centrifuge!

[J.P.M.]

Putting a heat source below an object will allow the heat to rise but what is rising is "heated air" and using such a small heat load (7 watt) will not heat enough air to keep the water from freezing.

A better heat source is one the emits or radiates infrared energy. That energy is absorbed by the object which then eventually heats up if more heat goes in than leaves.

Even then depending on what you are heating, how cold is it and how much heat energy you are sending are all variables to the solution (or problem). I won't get into the thermodynamics (too much math) but you can find out how much heat rejection a certain wattage lamp sends out in BTU's and then determine how many BTU"s you need to keep the volume of water from freezing.

One BTU will raise 1 pound of water 1 degree F. One BTU = 0.293 watt hours. So depending on how many degrees F you need to heat and how much water you want to heat will determine how many watt hours you need to use times a fudge factor due to losses between energy generated and the energy (BTU's) that get to the water.

At he risk of sounding like I'm trying to separate fly crap from pepper, the way I learned it, I believe what you are describing belongs more in the province of heat transfer than Thermodynamics.

[J.P.M.]

I still like the ancient Greek philosophers' idea that something floats because of the quantity of levity it has.
Oh, and helium balloons move to the inside of a curve when floating inside your car. Much more accessible than a centrifuge!

Or backward when you slam on the brakes (assuming the car is moving relative to its surroundings). Some folks in the 17th century thought combustible stuff burned because it contained something called Phlogiston, which was released during combustion. That seems a bit like levity to me - quaint and perhaps of historical value, but not accepted as accurate - at least not at this time. Also as for accessibility, a bowl with water/ice stirred with a spoon will do as well.

[inetdog]

That seems a bit like levity to me - quaint and perhaps of historical value, but not accepted as accurate - at least not at this time.

True. What is really hard for us to accept today, given the general level of knowledge, is that what made something fall (gravity) was totally unrelated to what made it float (levity).
I was aiming more for levity(2), namely humor.

[Robert1234]

let's get back to our chickens cause they are getting thirsty...

Using some southern, red neck, chicken farmer math:

Heat Flow = U * A * (T1 - T2), where
U = Heat Transfer Coefficient
A = Surface Area of Watering Can
T1 = Temperature of Water in Can
T2 = Outside Temperature

To keep water from freezing, T1 in the final design will be 33 F (we'll use that later on).
T2 will be your outside design temperature (how low can you go - we'll design for 0 F)

The product (U * A) can be considered to be constant (although you should realize that as wind blows across the can the heat transfer coefficient "U" can increase), so let's run a quick experiment to estimate it's value.

Put a known amount of water in the can (let's say 1 gallon) and measure it's initial temperature (T1 start)
Put the can outside and measure the outside temperature (T2 start)
Leave the can outside for a period of time (let say an hour).
After set time, measure the temperature outside (T2 End) and the temperature of the water in the can again (T1 end)

Let's assume we used 1 gallon of water and measured the values to be:
T1 Start = 54 F
T1 End = 48 F
T2 Start = 36 F
T2 End = 34 F
Elapsed Time = 1 Hour

Calculate the heat flow:
As SunEagle says, 1 BTU will raise 1 pound of water 1 F. Since we used 1 gallon of water, that's about 8.3 pounds. Heat Flow is therefore:

(54 F - 48 F) * (8.3 lbs) / (1 Hour) = 50 BTU/Hr

Now to estimate the coefficient product, UA:
T1 = (T1 Start + T1 End) / 2 = (54 + 48) / 2 = 51
T2 = (T2 Start + T2 End) / 2 = (36 + 34) / 2 = 35

UA = (50 BTU/Hr) / (51 - 35 F) = 3.1 BTU / F / Hr

To figure the amount of heat necessary to keep your can of water from freezing in 0 F weather...

Heat Needed = 3.1 * (33 - 0) = 100 BTU / Hr = ~ 30 watts

Assuming 50% efficiency / fudge factor, we'll supply about 60 watts to keep the chickens hydrated.


Gee.. Now wasn't that a lot of fun to go through

[inetdog]

I will add one more consideration, since I know that it really is a factor for indoor pools, etc. :

The heat loss from the water container will also be dependent on the relative humidity in the coop. If the air is dry, cooling of the water by evaporation will be a significant factor even without a lot of air movement across the surface.

[J.P.M.]

let's get back to our chickens cause they are getting thirsty...

Using some southern, red neck, chicken farmer math:

Heat Flow = U * A * (T1 - T2), where
U = Heat Transfer Coefficient
A = Surface Area of Watering Can
T1 = Temperature of Water in Can
T2 = Outside Temperature

To keep water from freezing, T1 in the final design will be 33 F (we'll use that later on).
T2 will be your outside design temperature (how low can you go - we'll design for 0 F)

The product (U * A) can be considered to be constant (although you should realize that as wind blows across the can the heat transfer coefficient "U" can increase), so let's run a quick experiment to estimate it's value.

Put a known amount of water in the can (let's say 1 gallon) and measure it's initial temperature (T1 start)
Put the can outside and measure the outside temperature (T2 start)
Leave the can outside for a period of time (let say an hour).
After set time, measure the temperature outside (T2 End) and the temperature of the water in the can again (T1 end)

Let's assume we used 1 gallon of water and measured the values to be:
T1 Start = 54 F
T1 End = 48 F
T2 Start = 36 F
T2 End = 34 F
Elapsed Time = 1 Hour

Calculate the heat flow:
As SunEagle says, 1 BTU will raise 1 pound of water 1 F. Since we used 1 gallon of water, that's about 8.3 pounds. Heat Flow is therefore:

(54 F - 48 F) * (8.3 lbs) / (1 Hour) = 50 BTU/Hr

Now to estimate the coefficient product, UA:
T1 = (T1 Start + T1 End) / 2 = (54 + 48) / 2 = 51
T2 = (T2 Start + T2 End) / 2 = (36 + 34) / 2 = 35

UA = (50 BTU/Hr) / (51 - 35 F) = 3.1 BTU / F / Hr

To figure the amount of heat necessary to keep your can of water from freezing in 0 F weather...

Heat Needed = 3.1 * (33 - 0) = 100 BTU / Hr = ~ 30 watts

Assuming 50% efficiency / fudge factor, we'll supply about 60 watts to keep the chickens hydrated.


Gee.. Now wasn't that a lot of fun to go through

One of the smartest heat transfer engineers I ever worked with was from Mississippi. Another was from Atlanta. a third was from Louisville. They knew one another, took pity on me as a dumb Yankee for some reason and tried to teach me some stuff. I usually had fun. You seem like them some. That U*A seems not too far off from my experience.

[SunEagle]

At he risk of sounding like I'm trying to separate fly crap from pepper, the way I learned it, I believe what you are describing belongs more in the province of heat transfer than Thermodynamics.

You are correct. I was thinking about energy and heat and Thermo came to mind but "heat transfer" is a better description.

He still needs more than 7 watts no matter how it is described.

[paulcheung]

If is me, I would just use one of these to keep the water from frozen total. it will keep the surrounding water from freeze.

Cheers

http://www.drsfostersmith.com/product/prod_display.cfm?c=3578+3743+11368&pcatid=11368

[SunEagle]

If is me, I would just use one of these to keep the water from frozen total. it will keep the surrounding water from freeze.

Cheers

http://www.drsfostersmith.com/produc...8&pcatid=11368

Those are nice heaters. Depending on how cold the coop gets one of them should keep the water from freezing.

Now back to what it will cost him in batteries and solar panels to run even that 50 watt heater will be a big surprise to him.

[Robert1234]

If is me, I would just use one of these to keep the water from frozen total. it will keep the surrounding water from freeze.

Cheers

http://www.drsfostersmith.com/produc...8&pcatid=11368

Perhaps, but this is what you are trying to keep from freezing...

http://www.globalindustrial.com/p/outdoor-grounds-maintenance/kennels-dog-houses-barns/buckets-feeders/doublewall-poultry-fount-9833-galvanized-steel-3-gal?infoParam.campaignId=T9A&gclid=CNnzhpDn_bsCFQbl7Aodri0Afw

Drill a hole in it anywhere that is not sealed and all water will just leak out - not too good. Of course, you can always get one of these....

http://www.amazon.com/Farm-Innovators-All-Seasons-Plastic-Fountain/dp/B001CCSJCQ/ref=sr_1_2?ie=UTF8&qid=1389707546&sr=8-2&keywords=heated+poultry+water

or one of these...

http://www.amazon.com/Farm-Innovators-Poultry-HP-125-125-Watt/dp/B000HHQDPM/ref=sr_1_1?ie=UTF8&qid=1389707546&sr=8-1&keywords=heated+poultry+water

but what's the fun in that

Additionally, these manufacturers are not worried that much about thermal loses (note lack of insulation anywhere) so you can do better building or modifying one yourself. My greenhouse was off-grid for about a year. You both go old school and get really innovative when faced with a limited supply of electrical energy.

Getting back to the original question, the bottom line is it's still gonna take 40+ watts to keep the water melted hanging the can outside so you're gonna have to design your system for that if that is what you are doing. Referring back to Derrick's original response for the electrical need, it's spot on based on the description of the problem - and probably pretty "kind" on what will really be required. As he so artfully put it... "Wait until you price it out."

and with that I shall quote yet another great philosopher...

"and that's all i have to say about that"



IMG_1044.JPG

[J.P.M.]

Perhaps, but this is what you are trying to keep from freezing...

http://www.globalindustrial.com/p/outdoor-grounds-maintenance/kennels-dog-houses-barns/buckets-feeders/doublewall-poultry-fount-9833-galvanized-steel-3-gal?infoParam.campaignId=T9A&gclid=CNnzhpDn_bsCFQbl7Aodri0Afw

Drill a hole in it anywhere that is not sealed and all water will just leak out - not too good. Of course, you can always get one of these....

http://www.amazon.com/Farm-Innovators-All-Seasons-Plastic-Fountain/dp/B001CCSJCQ/ref=sr_1_2?ie=UTF8&qid=1389707546&sr=8-2&keywords=heated+poultry+water

or one of these...

http://www.amazon.com/Farm-Innovators-Poultry-HP-125-125-Watt/dp/B000HHQDPM/ref=sr_1_1?ie=UTF8&qid=1389707546&sr=8-1&keywords=heated+poultry+water

but what's the fun in that

Additionally, these manufacturers are not worried that much about thermal loses (note lack of insulation anywhere) so you can do better building or modifying one yourself. My greenhouse was off-grid for about a year. You both go old school and get really innovative when faced with a limited supply of electrical energy.

Getting back to the original question, the bottom line is it's still gonna take 40+ watts to keep the water melted hanging the can outside so you're gonna have to design your system for that if that is what you are doing. Referring back to Derrick's original response for the electrical need, it's spot on based on the description of the problem - and probably pretty "kind" on what will really be required. As he so artfully put it... "Wait until you price it out."

and with that I shall quote yet another great philosopher...

"and that's all i have to say about that"



[ATTACH=CONFIG]3585[/ATTACH]

I was too ruminating about what Mr. Gump might say.