RoadtripsAndCampfires
Well-known member
DON'T RELY ON THIS INFO - YET
I am wondering if those who know about solar, (when you have time, this is long so I don't expect an instant answer) would read this and tell me where it is wrong. I've taken notes on videos, websites, etc. and put together this information before I go on to the next step. Learning all this stuff is hard but I am determined. The disclaimer is large so no one finds this post and uses it without it being verified. I know there are going to be options and choices not mentioned here but first I just want to see where I went wrong on the basics of learning it. Thank you for even taking the time to read it, it took me all day to write up these notes:
Things To Know About Solar
[font=Arial, sans-serif]Sun angle, intensity of the rays, weather, time of day and how many hours of sunlight will all affect the amount of charge the panels will receive so you'll want to be able to store enough power in the batteries to get you through night, clouds, etc. until you can charge again. Where you are on earth will affect all of these factors so there is not one answer for everyone's power needs. But there are charts to help you figure out what you can expect in a given area during a given month. One such chart, or solar map, is available at https://www.nrel.gov/gis/solar.html National Renewable Energy Laboratory.[/font]
[font=Arial, sans-serif]The system will include the necessary number of panels at the necessary wattage, a controller that will regulate the power being received and direct it to a battery to store it in, and the wires to connect it all. Mounting hardware is necessary if they will be mounted to a roof. Some people prefer to carry them in the RV and then lay them out in the sun at camp. This method will work if you can't (or don't want to) mount them on your roof but you do lose any charging one would get during driving time to and from your destination. There are kits that can be bought which will contain the panels, wiring and contoller and sometimes the mounting hardware, or items can be purchased separately.[/font]
[font=Arial, sans-serif]A Sun Hour is an hour of the sun hitting at full strength, generally noon and an hour or so before and after. So in the morning or evening it might take 3 hours to achieve one sun hour of power hitting your panels. And if you are in the middle of cloud or tree cover you may receive very little charge at all, even at noon. You may receive 3 sun hours between the hours of 11 and 1 in June in Arizona but only 1.5 sun hours or less between 11 and 1 in June someplace else.[/font]
[font=Arial, sans-serif]A 12-volt solar panel rated at 100 watts operating at ideal sun conditions will produce about 18 volts and a 5.5 amp current in one hour. Volts x amps = Watts (w):[/font]
[font=Arial, sans-serif]18 volts x 5.5 amps = 99 Watts (rounded to 100, or 100w).[/font]
[font=Arial, sans-serif]A watt is a unit of measurement of the power the device uses. A 100w light bulb requires 100 watts in an hour and a 40w light bulb only requires 40w. Your solar needs will be dependent upon how many watts you will use in a day. Will you power a small fridge or just an alarm clock, radio and phone? Don't figure that out yet but know that there are charts that exist that provide “averages” for things such as TVs, radios, etc. to help you estimate but you'll want to check your appliance ratings to determine the actual amount of power needed based on what you are using.[/font]
[font=Arial, sans-serif]Watt Hours is how much power is generated over a period of time:[/font]
[font=Arial, sans-serif]1 hr x 100W = 100 watt hours.[/font]
[font=Arial, sans-serif]2 hrs. x 100W = 200 watt hours.[/font]
[font=Arial, sans-serif]Using a Solar Map https://www.wholesalesolar.com/solar-information/sun-hours-us-map wholesale solar [/font]
[font=Arial, sans-serif]I charted sun hours six cities that represent areas I will likely travel in (H=high, L=low):[/font]
[font=Arial, sans-serif]THIS PART WAS CORRECTED WITH AN EDIT AS SOME GOT DROPPED:[/font]
[font=Arial, sans-serif]La Jolla, CA 5.24 H, 4.29 L[/font]
[font=Arial, sans-serif]L.A., CA 6.14 H, 5.03 L[/font]
[font=Arial, sans-serif]Grand Junction, CO 6.34 H, 5.23 L[/font]
[font=Arial, sans-serif]Albuquerque, NM 7.16 H, 6.21 L[/font]
[font=Arial, sans-serif]Corvallis, OR 6.34 H, 5.23 L[/font]
[font=Arial, sans-serif]Seattle, WA 4.83 h, 1.60 L[/font]
[font=Arial, sans-serif]If I were to base my below calculation of my needs on the lowest amount of daily sun hours, it would be 1.60 for Seattle. However, I will likely only be there once in the next several years and probably for only 2 days. For my purposes I plan to be sure my batteries are charged before I get there and put us on a low consumption plan during our stay rather than base my calculations for number of panels to install on the lowest amount of average Seattle sunlight. Likewise, La Jolla will be visited in the summer and only for a few days so I am tossing that one out. I am going to use 5 sun hours to figure out our initial solar panel setup for our needs.[/font]
[font=Arial, sans-serif]100w x 5.0 Sun Hours = 500 Watt Hours (Wh)[/font]
[font=Arial, sans-serif]Deductions[/font]
[font=Arial, sans-serif]The next part of the process is determining energy lost by traveling across wires, the amps used by the charge controller and any cell obstructes due to things such as dirt or snow. Deduct 1/3 of your total watts for these losses by multiplying .70:[/font]
[font=Arial, sans-serif]500Wh x .70 = 350Wh (represents 70% of the total power allowing for 30% loss).[/font]
[font=Arial, sans-serif]The charge controller sends the power generated by the solar panels to the battery which stores it for later use and it has an amp rating which varies according to the conroller.[/font]
[font=Arial, sans-serif]Determining Amps[/font]
[font=Arial, sans-serif]Watts divided by volts equals amps. We are using a 12v system so:[/font]
[font=Arial, sans-serif]350wh divided by 12 volts = 29 amp hours.[/font]
[font=Arial, sans-serif]Battery[/font]
[font=Arial, sans-serif]A standard car battery is meant to be rapidly discharged for starting the car and then recharged as the car is driven by the alternator. The amount of discharge at time of start is a short high-current charge and normally only discharges a small portion of their stored energy.[/font]
[font=Arial, sans-serif]A deep cell or deep cycle battery is designed to be drained of more power and is in use more often as your devices draw from them. It is a good idea to avoid discharge below 50% so if one chooses a battery system capable of twice the amp hours needed this avoids going below 50% of the full battery charge. [/font]
[font=Arial, sans-serif]29 x 2 = Ah x 2 = 58 Ah required.[/font]
[font=Arial, sans-serif]Temperature and De-rating[/font]
[font=Arial, sans-serif]Temperature affects the amount of power a battery can store. Manufacturers can provide you with information on how much to de-rate your amp hours based on temperatures. Here is an example of a battery de-rating:[/font]
The first number below is the temperature (F) followed by the percentage to de-rate:
80, 0%
70, 2%
60, 5%
50, 10%
40, 15%
30, 20%
20, 35%
[font=Arial, sans-serif]One would want to increase the battery by the percentage of derating according to the chart based on the temperature the battery will be stored at. If the temperature averages 60 degrees F multiply the Ah by the percent shown, in this case 5%:[/font]
[font=Arial, sans-serif]58 Ah x .05 = 2.9. [/font]
[font=Arial, sans-serif]Add the 2.9 to 58 and we round out to 60Ah.[/font]
[font=Arial, sans-serif]The Inverter – Another Deduction [/font]
[font=Arial, sans-serif]The inverter converts the DC power of the battery where you have stored your power into AC power so that it can be used by your AC appliances and devices. The inverter uses about 5% of power:[/font]
[font=Arial, sans-serif]60Ah divided by .95 = 63.1 so we'll call that 63Ah.[/font]
[font=Arial, sans-serif]Storing Power For 2 days without sun:[/font]
[font=Arial, sans-serif]I you can't charge the battery during two days of no sun you would need to multiply the amp hours by that many number of days and have it fully charged before you arrive at camp:[/font]
[font=Arial, sans-serif]63Ah x 2 = 126Ah.[/font]
[font=Arial, sans-serif]So one would need a 126Ah battery.[/font]
[font=Arial, sans-serif]- - - - - - - - -[/font]
[font=Arial, sans-serif]SO – IS THIS ALL CORRECT? Or?[/font]
I am wondering if those who know about solar, (when you have time, this is long so I don't expect an instant answer) would read this and tell me where it is wrong. I've taken notes on videos, websites, etc. and put together this information before I go on to the next step. Learning all this stuff is hard but I am determined. The disclaimer is large so no one finds this post and uses it without it being verified. I know there are going to be options and choices not mentioned here but first I just want to see where I went wrong on the basics of learning it. Thank you for even taking the time to read it, it took me all day to write up these notes:
Things To Know About Solar
[font=Arial, sans-serif]Sun angle, intensity of the rays, weather, time of day and how many hours of sunlight will all affect the amount of charge the panels will receive so you'll want to be able to store enough power in the batteries to get you through night, clouds, etc. until you can charge again. Where you are on earth will affect all of these factors so there is not one answer for everyone's power needs. But there are charts to help you figure out what you can expect in a given area during a given month. One such chart, or solar map, is available at https://www.nrel.gov/gis/solar.html National Renewable Energy Laboratory.[/font]
[font=Arial, sans-serif]The system will include the necessary number of panels at the necessary wattage, a controller that will regulate the power being received and direct it to a battery to store it in, and the wires to connect it all. Mounting hardware is necessary if they will be mounted to a roof. Some people prefer to carry them in the RV and then lay them out in the sun at camp. This method will work if you can't (or don't want to) mount them on your roof but you do lose any charging one would get during driving time to and from your destination. There are kits that can be bought which will contain the panels, wiring and contoller and sometimes the mounting hardware, or items can be purchased separately.[/font]
[font=Arial, sans-serif]A Sun Hour is an hour of the sun hitting at full strength, generally noon and an hour or so before and after. So in the morning or evening it might take 3 hours to achieve one sun hour of power hitting your panels. And if you are in the middle of cloud or tree cover you may receive very little charge at all, even at noon. You may receive 3 sun hours between the hours of 11 and 1 in June in Arizona but only 1.5 sun hours or less between 11 and 1 in June someplace else.[/font]
[font=Arial, sans-serif]A 12-volt solar panel rated at 100 watts operating at ideal sun conditions will produce about 18 volts and a 5.5 amp current in one hour. Volts x amps = Watts (w):[/font]
[font=Arial, sans-serif]18 volts x 5.5 amps = 99 Watts (rounded to 100, or 100w).[/font]
[font=Arial, sans-serif]A watt is a unit of measurement of the power the device uses. A 100w light bulb requires 100 watts in an hour and a 40w light bulb only requires 40w. Your solar needs will be dependent upon how many watts you will use in a day. Will you power a small fridge or just an alarm clock, radio and phone? Don't figure that out yet but know that there are charts that exist that provide “averages” for things such as TVs, radios, etc. to help you estimate but you'll want to check your appliance ratings to determine the actual amount of power needed based on what you are using.[/font]
[font=Arial, sans-serif]Watt Hours is how much power is generated over a period of time:[/font]
[font=Arial, sans-serif]1 hr x 100W = 100 watt hours.[/font]
[font=Arial, sans-serif]2 hrs. x 100W = 200 watt hours.[/font]
[font=Arial, sans-serif]Using a Solar Map https://www.wholesalesolar.com/solar-information/sun-hours-us-map wholesale solar [/font]
[font=Arial, sans-serif]I charted sun hours six cities that represent areas I will likely travel in (H=high, L=low):[/font]
[font=Arial, sans-serif]THIS PART WAS CORRECTED WITH AN EDIT AS SOME GOT DROPPED:[/font]
[font=Arial, sans-serif]La Jolla, CA 5.24 H, 4.29 L[/font]
[font=Arial, sans-serif]L.A., CA 6.14 H, 5.03 L[/font]
[font=Arial, sans-serif]Grand Junction, CO 6.34 H, 5.23 L[/font]
[font=Arial, sans-serif]Albuquerque, NM 7.16 H, 6.21 L[/font]
[font=Arial, sans-serif]Corvallis, OR 6.34 H, 5.23 L[/font]
[font=Arial, sans-serif]Seattle, WA 4.83 h, 1.60 L[/font]
[font=Arial, sans-serif]If I were to base my below calculation of my needs on the lowest amount of daily sun hours, it would be 1.60 for Seattle. However, I will likely only be there once in the next several years and probably for only 2 days. For my purposes I plan to be sure my batteries are charged before I get there and put us on a low consumption plan during our stay rather than base my calculations for number of panels to install on the lowest amount of average Seattle sunlight. Likewise, La Jolla will be visited in the summer and only for a few days so I am tossing that one out. I am going to use 5 sun hours to figure out our initial solar panel setup for our needs.[/font]
[font=Arial, sans-serif]100w x 5.0 Sun Hours = 500 Watt Hours (Wh)[/font]
[font=Arial, sans-serif]Deductions[/font]
[font=Arial, sans-serif]The next part of the process is determining energy lost by traveling across wires, the amps used by the charge controller and any cell obstructes due to things such as dirt or snow. Deduct 1/3 of your total watts for these losses by multiplying .70:[/font]
[font=Arial, sans-serif]500Wh x .70 = 350Wh (represents 70% of the total power allowing for 30% loss).[/font]
[font=Arial, sans-serif]The charge controller sends the power generated by the solar panels to the battery which stores it for later use and it has an amp rating which varies according to the conroller.[/font]
[font=Arial, sans-serif]Determining Amps[/font]
[font=Arial, sans-serif]Watts divided by volts equals amps. We are using a 12v system so:[/font]
[font=Arial, sans-serif]350wh divided by 12 volts = 29 amp hours.[/font]
[font=Arial, sans-serif]Battery[/font]
[font=Arial, sans-serif]A standard car battery is meant to be rapidly discharged for starting the car and then recharged as the car is driven by the alternator. The amount of discharge at time of start is a short high-current charge and normally only discharges a small portion of their stored energy.[/font]
[font=Arial, sans-serif]A deep cell or deep cycle battery is designed to be drained of more power and is in use more often as your devices draw from them. It is a good idea to avoid discharge below 50% so if one chooses a battery system capable of twice the amp hours needed this avoids going below 50% of the full battery charge. [/font]
[font=Arial, sans-serif]29 x 2 = Ah x 2 = 58 Ah required.[/font]
[font=Arial, sans-serif]Temperature and De-rating[/font]
[font=Arial, sans-serif]Temperature affects the amount of power a battery can store. Manufacturers can provide you with information on how much to de-rate your amp hours based on temperatures. Here is an example of a battery de-rating:[/font]
The first number below is the temperature (F) followed by the percentage to de-rate:
80, 0%
70, 2%
60, 5%
50, 10%
40, 15%
30, 20%
20, 35%
[font=Arial, sans-serif]One would want to increase the battery by the percentage of derating according to the chart based on the temperature the battery will be stored at. If the temperature averages 60 degrees F multiply the Ah by the percent shown, in this case 5%:[/font]
[font=Arial, sans-serif]58 Ah x .05 = 2.9. [/font]
[font=Arial, sans-serif]Add the 2.9 to 58 and we round out to 60Ah.[/font]
[font=Arial, sans-serif]The Inverter – Another Deduction [/font]
[font=Arial, sans-serif]The inverter converts the DC power of the battery where you have stored your power into AC power so that it can be used by your AC appliances and devices. The inverter uses about 5% of power:[/font]
[font=Arial, sans-serif]60Ah divided by .95 = 63.1 so we'll call that 63Ah.[/font]
[font=Arial, sans-serif]Storing Power For 2 days without sun:[/font]
[font=Arial, sans-serif]I you can't charge the battery during two days of no sun you would need to multiply the amp hours by that many number of days and have it fully charged before you arrive at camp:[/font]
[font=Arial, sans-serif]63Ah x 2 = 126Ah.[/font]
[font=Arial, sans-serif]So one would need a 126Ah battery.[/font]
[font=Arial, sans-serif]- - - - - - - - -[/font]
[font=Arial, sans-serif]SO – IS THIS ALL CORRECT? Or?[/font]