solar chicken coop

[mrnewbie]

My playin with the math for a Solar Chicken Chicken Coop

6 steps to the Chicken solar project

1) Determine daily watt hour requirements (15watt bulb)

15w x 10 hrs = 150wh

2) Battery system required (1.5 is standard to account for battery inefficiency)

150whx1.5 = 225wh

3) Solar panel size required depends on the Sun Isolation Factor for your location (3).

225wh/3 = 75w array (we have an 80watt array)

4) Determine battery needed at 20% DOD so as to never discharge under 80%

225wh x 5= 1125 wh

5) Battery size needed depends on system voltage

1125wh/12v system volt = 93.75ah 12 volt battery needed

6) Charge controller is based on current required (with 12v battery) Solar panel array size/system voltage

80watt panel/12volt = 6.6A so any 12V controller that can handle 7+ amps

This is the cheapest MPPT charger controller I could find ..YMMV . Its a 12V 10A unit about $80
hxxp://cgi.ebay.com/10A-12V-Solar-MPPT-Power-Charger-Regulator-Controller_W0QQitemZ220546641928QQcmdZViewItemQQpt ZLH_DefaultDomain_0?hash=item33599ab008

In conclusion, I think we have designed a nice little 15 watt chicken coop system..
All you need is the charge controller $80 and cost of battery $150-$200, a little wire and the light bulb.
This is a 12 volt system, which means you need a 12V light bulb rated at around 15 watts, I would think some kind of automotive or specialty bulb would be fine..as long as its at or less than 15 watts
Here is the perfect bulb match, but I don't know how much heat it will make for you..Buy one..hook it up to a 12v and see if it gets warm
You could take a piece of aluminum flashing(or beer can), put behind bulb as reflector to maxamize warming effect
hxxp://www.lightbulbsdirect.com/Merchant2/merchant.mvc?Screen=PROD&Product_Code=7FCCL12V&Cat egory_Code=LandIncanLowVolt&Product_Count=6

You may also choose to get like a 40 or so watt bulb..put it on a switch, for a people bulb. Just remember it will drain your battery fast and should just be used for quick feeding..and cleaning.


With all the above all done, This is not the best way to do this... to many losses and it cost you around $250.

A better way would be...

We have a 80watt array that makes 80w x3(winter) hrs = 240wh , just plug it all into the wall!
In summer maybe 4-5 hrs 320wh/400wh (25watt bulb for reasonable yearly average)
Go with a cheap grid tied inverter a 300w unit cost $100..This takes about 90% of panel output and goes right into wall.
Run an extension cord out to coop , give em 120V..keep the bulb small 15-25watt. You will be breaking even, in effect powering the coop for free, and can add more panels if you like

[Mike90250]

DOD Depth of Discharge. If you use 20A out of a 100A battery, that's 20% discharged (80% of full charge)

[mrnewbie]

First, Thanks guys for taking the time to help me figure this stuff out. I'm 3/4th's of the way there.

Kinda having fun. Well kinda, sorta.

Your example, My chicken coop numbers, substituted.

Very first step is to determine your daily watt-hour usage for both winter and summer. The radio consumes 250 watts, 24 hours per, 7 days a week, 365 days per year. The daily watt-hour usage is 250 watts x 24 hours = 6000 watt-hours or 6 Kwh.

250 watts x 24 hours = 6000 watt-hours

(7watt and 15 watt bulb for 10hrs of bird light)
7 watts x 10 hrs = 70 watt-hours
15w x 10hrs = 150 watt hours

Battery systems are extremely inefficient. To account for it and make computations easier we adjust for it right up front by taking the daily watt hour usage and multiply by 1.5, so 6000 wh x 1.5 = 9000 wh. This is how much energy the solar panels have to generate at their terminals. Note 9000 wh as it will be used to determine solar panel wattage and battery size.

6000 wh x 1.5 = 9000 wh "we divided by.66 6000/.66= 9090 wh.. almost same x1.5 is simpler"

70 wh x 1.5 = 105 wh
150 wh x 1.5 = 225 wh


To determine the solar panel wattage we need to determine the solar insolation in Sun Hours. In this case it will be December and January. Lets do it for two locations to demonstrate location means everything. One site will be Seattle and the other Tuscon. In Tuscon December insolation = 5.6 hours, Seattle = 1.4 hours. To determine the solar panel wattage we take the adjusted watt hours and divide by the Sun Hours.

Seattle = 9000 wh / 1.4 h = 6428 w round up to 6500 watt solar panel array is needed.
Tuscon = 9000 wh / 5.6 h = 1600 watt solar panel array needed.

solar panel wattage
We already figured sun hrs at 3.1 hrs at this local for winter but lets use 3
105 wh/3 = 35 watt array
225 wh/3 = 75 watt array


So to determine battery capacity we take the adjusted daily wh usage and multiply by 5, so 9000 wh x 5 = 45,000 wh storage capacity. Now we select the system voltage to convert to Amp Hours. For a monster system like this we would want at least 48 volts or more. Since the radio equipment operates at 48 volts the selection is made for us. To find the AH rating take the wh and divide by system voltage. so 45,000 wh / 48 volts = 938 Amp Hours @ 48 volts round up to 950 AH.
20% DOD per day is the most reasonable figure to use.

What does DOD mean.?
I'm thinking something like Degradation Of Discharge..?

Others have suggested we never go under 90% of a charged batterys capacity..in my example I multiplied by 10..not 5..thinkin it would be much better on battery never using more than 10% of charge. But I have seen, never go under 80% of charge as well..But this a major point, we double our needed storage with this number..so I hope when you say X 5 you are right.
lets go with your X 5 number.

determine battery capacity
9000 wh x 5 = 45,000 wh storage capacity

105 wh x 5 = 525 wh storage cap for 7 watt bulb
225 wh x 5 = 1125 wh storage cap for 15 watt bulb

determine system voltage to convert to Amp Hours to get a battery size

525wh / 12 volts = 43.75 Amp Hours @ 12 volts (Round up to next larger battery)
1125wh / 12 volts = 93.75 Amp Hours @ 12 volts (Round up to next larger battery)

At this point I'm confussed, from above I think we can use any 12v solar battery that has over 93ah storage is this correct?
Or any 6 volt that can do over 186ah ?

But can we charge a 12v battery with a 12 volt array...shouldn't we up the array to 24v or down the battery to 6v?

From above calculations I think we can power a single 15 watt bulb for 10 hrs of operation or a 7watt for just over 20 hrs on 1 battery(12v or 6v), is this correct ??

I really had no idea it would take so much battery..but we are only using 20% of the battery power to prolong battery life. Theoretically we could run it for 5 times as long..if it was a perfect world with a perfect battery.

But it's not, one could say we can only use 10 to 20 percent of the battery..or it self destructs..in effect 80 to 90 percent loss..correct??

[Jason]

(thread headed to the "off grid" sub-forum)

[Sunking]

You are going about it the wrong way, and do not have all the facts and figures straight. First mistake is buying a system then hope it will meet your needs. Huge mistake that will fail 90% of the time, or way overkill and spend more money than necessary. I assume you are talking about a stand alone battery system so here is the design process.

Very first step is to determine your daily watt-hour usage for both winter and summer. But for this example lets say both summer and winter usage are the same. With that said the worse case is winter. Ok here is a design I have built about 150 projects for cellular telephone carriers at remote towers where there is no commercial power available. The radio consumes 250 watts, 24 hours per, 7 days a week, 365 days per year. The daily watt-hour usage is 250 watts x 24 hours = 6000 watt-hours or 6 Kwh. About 70 cents worth of electricity.

Battery systems are extremely inefficient. To account for it and make computations easier we adjust for it right up front by taking the daily watt hour usage and multiply by 1.5, so 6000 wh x 1.5 = 9000 wh. This is how much energy the solar panels have to generate at their terminals. Note 9000 wh as it will be used to determine solar panel wattage and battery size.

To determine the solar panel wattage we need to determine the solar insolation in Sun Hours. In this case it will be December and January. Lets do it for two locations to demonstrate location means everything. One site will be Seattle and the other Tuscon. In Tuscon December insolation = 5.6 hours, Seattle = 1.4 hours. To determine the solar panel wattage we take the adjusted watt hours and divide by the Sun Hours.

Seattle = 9000 wh / 1.4 h = 6428 w round up to 6500 watt solar panel array is needed.
Tuscon = 9000 wh / 5.6 h = 1600 watt solar panel array needed.

To determine battery capacity is straight forward. You can use 10% discharge per day if you have the bucks but not advised. In any battery system where power is critical, you are going to have to have a generator. 20% DOD per day is the most reasonable figure to use. So to determine battery capacity we take the adjusted daily wh usage and multiply by 5, so 9000 wh x 5 = 45,000 wh storage capacity. Now we select the system voltage to convert to Amp Hours. For a monster system like this we would want at least 48 volts or more. Since the radio equipment operates at 48 volts are selection is made for us. To find the AH rating take the wh and divide by system voltage. so 45,000 wh / 48 volts = 938 Amp Hours @ 48 volts round up to 950 AH.

For the charge controller I am assuming a MPPT type. If you use shunt type you have to start over and figue only 50% efficiency rather than the 66% we did in step one. So to determine the charge controller current take the solar panel array wattage and divide by the nominal system voltage of 48 volts.

Seattle = 6500 watts / 48 volts = 135 amps. You will need to divide your panels up in two groups of 3250 watts and use two 80-amp controllers.

Tuscon = 1600 watts / 48 volts = 33 amps.

That is the basic for a pure DC system. If you are going to use an inverter, things are a bit more complicated as you have to factor in the AC Power Factor which can get ugly. For example if you need the same 6000 watts and the power factor is .8 means you need 6000 wh / .8 = 7500 watt hour. Then you have to account for the inverter effeciency of around 85 to 90 % so worse case is 7500 / .85 = 8823 wh. That would be your new target and have to multiply that by 1.5 and start over. You do not want to know what it takes using a PWM or other shunt regulator with an inverter. It gets really expensive real fast.

So the Tuscon system as calculated would cost you about $11,000 just for the panels, batteries, and charge controller alone plus hardware, and misc materials..

The Seattle system would cost around $23,500.

Location matters huh?

[mrnewbie]

Hello again,
Guys I don't understand how you get all your numbers, I'm not sure how all this stuff works..I'm here to learn. I don't really understand all it how works.
So I'm gonna ask..figure out the math, to learn, lay out what I think and You tell me when I'm wrong..here is my thoughts on this chicken coop project..

I have 4 x 20 watt 12v panels = 80watt panel, I get 3 hrs of sun.
I wire 2 panels in series to get 24v, do it twice ..put in parellel.
I know I need more than 12v to charge a 12v battery
So now I have 1 24v 80 watt panel.

I can make "I get 3 hrs of sun" 3x80 =240 watts .
To do this I run off to ebay and buy a charge controller.
hxxp://cgi.ebay.com/Solar-Panel-Charge-Regulator-Controller-12V-5A-Battery_W0QQitemZ230440078436QQcmdZViewItemQQptZBI _Electrical_Equipment_Tools?hash=item35a74c8c64

This one costs less than $20 bucks delivered..its 12v out 8amp..its good to 12wx8a=96w and accepts up to 30v in
, so I think it will work with my current 80W panel ???
But batteries are bad..they only put out .66 of what I put in 80wx.66=158w

This is a Question?? I'm confused at this point..I know in the past 3 days I have put 3x240w = 720w into batteries and used 158x3=474 watts..
Does the battery have 720-474w = 246w still left ? Or is this an example of the 34% loss?

So I have 158 battery stored useable watts every sunny day.
So on a sunny day I could power a 15.8w light bulb for 10hrs..
But this would kill my battery..if I had one

This seems as if I will never get ahead, never build up a few day back up..this is not good.

I know not to exceed 10% dischage..or real short battery life..so I I figure I need to multiply 158 x 10..so need a 1580 watt battery so I wont deplete battery to less than 90% on any given day.

I can
1) put more watts into battery (big panel cost$-no)
2) run for less than 10hrs
3 lower watts used

I pick option 3, less watts...How about a real 7w bulb not a cfl. The kind they put in night lights.(4 for 2$).and external xmass lights..at least they put out a little heat, they get hot..try touching one..I want my chickens to have a little heat.

Now, I store 2 times more wattage than I burn...I'm ahead..as long as I get more than 50% sunlight, I'm ahead

But I'm in Ohio, in winter I could go as low as 30% sunlight.

I guess I have a few options
1) Figure out how to store more energy in a battery..bigger panel/battery
2) Look for a specialty lightbulb..uses less watts
3) Run for a few less hrs..thus lowering my watts needed
4) Use bucket of water..or get out deapfryer


(HINT)- I could even hook up a 40 watt..people light on a switch..only for feeding and cleaning uses..as long as it was only on for a few min)
(HINT)- I could even hook up a computer fan for ventilation (Rabbit cage project) the one I'm looking at uses 12v .29am=12x.29=3.48 watts


The only thing I don't understand..is

How do you pick the battery ?? The math? Go with the 5 day senario..or whatever..Its the math I don't get.

Another question..How to turn power to 7w bulb on/off for 10 hrs all by itself? Don't want it running 24/7.
For now, I'll do it manually..

[mrnewbie]

Hello all, I'm new here and I admitt, I don't know what I'm doing.
This is first post, I saw this thread had to jump in.
And I yield to all those that know more than me..which is all of you

Common guys its a chicken coop! It's a chicken coop!

These chickens have been in this guys back yard for a long time, They have been perfectly happy to live without heat and electric all there lives.

If they get a little light bulb, they will huddle around that light bulb, and be so happy they will lay 10 times as many eggs. And sqauwk with glee as the owner approaches. Its a chicken coop!

The chickens don't need a 5 day power backup, they are chickens

"So 900 wh x 5 = 4500 wh storage capacity. To find the Amp Hours just divide by the system voltage. So 4500 wh / 12 volts = 375 Amp Hours. Again no way yill just two 6 volt Trojan T-105's will do the job. It will take at least 4 of them."

A 5 day backup ...a 60 watt bulb....Its a chicken coop! Why?
900 wh x 2day = 1800/12= 150Ah
400 wh x 2day = 800/12= 66 Ah
250 wh x 2day = 500/12= 41.6Ah
130 wh x 2day = 260/12= 21.6Ah

If no sun for a couple days..then no light bulb..no big deal..the chickens have lived all their lives with no light bulb.

Insulate chicken coop as mentioned...The chickens will already be better off
Try running a 40 watt bulb or even a 25 watt through the numbers instead of the 60 watt.or go with the 13 watt.CFL

[Mike90250]

LED's and CFL's both use the same principle to generate "white" light. They internally create blue / UV light, and use a phospher coating to transform to white. for lighting a coop, I'd use a CFL, because it produces more light over a wider angle. LED's have pretty narrow lighting angles and are not perfected enough, CFL's have a few years of history and color mixes. Heating, I'd use a bucket of hot water, a 15W CFL will not produce much heat for you.

[Sunking]

Ok now I have some data to work with knowing your location of STL. STL has a winter insolation of 3.1 hours. Based on 130 watt-hours you will need a a minimum of 65 watt solar panel and a 100 Amp hour battery @ 12 volts. This can come in at your $500 budget limit.

What I do not know is anything about chickens and their tolerance to Blue Light Pollution. I am not a fan of LED lighting as they are not as efficient as florescent lighting, highly directional, and horrible color rendering-indexes.

Imo you would be better off using a good quality 13 to 15 watt CFL. You will like it better from your eyes, produce as much or more light, less expensive, and it might make the chickens happier.

[phmezz]

Thanks for the replies everyone. I obviously need to rethink my plan.

So, lets say I go with a 13w LED bulb instead. This will give me enough light and I can compensate the heating issue with more insulation. I live outside of St. Louis so we only have a few days sub zero weather per year. Maybe a small propane heater can work for a few days out of the year.

Using the math that Sunking recommened I come up with this:

Ten hours of operation per day requires 130 watt hours. If I get 50% efficiency from my batteries this means 260 watt hours. I'll go with 3 h of sun exposure during the winter. However my house has great southeast exposure and the area my coop will be in gets great direct sun exposure. Nevertheless, I think this is a good number. So 260 wh/3h = 87 watts.

For battery storage, it was recommended that I have a 5 day reserve. So 87wh x 5 = 435 wh storage capacity or 36.25 amp hours. If the T-105 are rated at 225 a/h I'm thinking I can make this work with two batteries for 10 hours per day of operation. Although I'm probably going to have to plan on at least a 100 watts of solar panels.

I'm still hoping I can make this project work. Thanks for the advice. If my numbers look wrong I'd appreciate your opinions.

[Sunking]

He could use a LED bulb for instead, that's less expensive.

He wants the heat from a 60 watt light bulb. Which brings up the point Solar PV cannot be used for electric heat economically.

Well I wil try to help you out here, but you are not going to like it.

Sunking you are ruining his fun of trying to build something, but i will cover for you buddy
He could use a LED bulb for instead, that's less expensive.
And do not conside light for heat. the sunlight has a power 1000W/square meter,can keep the temperature around 30 degree,which a bulb can not make it.

[Sunking]

Well I wil try to help you out here, but you are not going to like it.

Ok your goal is to run a 60 watt light bulb a day right. Well 60 watts x 10 hours = 600 watt hours. Battery systems are terrible inefficient. At best design 66%. So this means your sloar panels need to generate 600 wh / .66 = 900 watt hour per day.

Ok I do not know your location to obtain your winter insolation or Sun Hour Day, so for now I will assume you get a very generous 3 Sun Hours in Winter. So for the solar panel wattage you will need 900 wh / 3 h = 300 watts. If you receive less than 3 Sun Hours the wattage will be higher. Regardless there is no way an 80 watt solar array is going to do what you want, it is just not possible.

Now for more bad news, your batteries. In order to make your batteries last longer than one year you never want to discharge more than 20% in any given day. In addition having a 5 day reserve will carry you through a couple of cloudy days before having to go on a generator. So to determine battery capacity we take the adjusted daily watt hour figure and multiple by 5. So 900 wh x 5 = 4500 wh storage capacity. To find the Amp Hours just divide by the system voltage. So 4500 wh / 12 volts = 375 Amp Hours. Again no way yill just two 6 volt Trojan T-105's will do the job. It will take at least 4 of them.

Last bit of bad news.There is no way to do this for $500 or less unless you steal all the equipment. Panels alone will cost $800 to $1000. another $450 for batteries, $200 for a charge controller, $100 for an inverter, plus another $100 in materials.

Good Luck

SK

[DeltaFox 25]

http://www.solar-facts.com/panels/panel-wiring.php

[DeltaFox 25]

you still need proper wiring and maybe fusing , The charging of the batterys you need at least 18 volts so if you had a 300 watt charge controler it should work.