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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??
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??
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