I need a lesson on basic formulas

Fe-Wood
Fe-Wood Solar Expert Posts: 96 ✭✭
I am trying to figure out how long my batteries will last using various draws of power. For some reason I can't get my head around the conversion from 24VDC to 120VAC with the added complication of Amp Hours.
Can someone supply a formula?
For instance- I want to run an appliance that takes 10 amps of 120 VAC. I will be running it for 20 min. My supply is 24VDC. How many amp hours will I use? Don't confuse me with loss factor and all that. Just the straight conversion for now. I can add the other stuff later.

Thanks-
A confused speculator...

Comments

  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    Re: I need a lesson on basic formulas

    Think in terms of Watts. Watts are Volts times Amps and come out the same regardless of Voltage.

    So 10 Amps @ 120 VAC is 1200 Watts. 1200 Watts on 24 Volts is (1200/24) 50 Amps.

    This does not include the power used by the inverter or its conversion efficiency, so it's a bit difficult to get that into actual Amp hours used. There is a further complication of the way batteries discharge too, as the Voltage goes down the current goes up to maintain the Watts and thus the Amp hours fall off faster the lower the battery's SOC. Yes, it's confusing because it's complex.

    The time factor works in the same way: 1200 Watts for 20 minutes is 400 Watt hours. 400 Watt hours on 24 Volts is (400/24) 16.7 Amp hours.
  • waynefromnscanada
    waynefromnscanada Solar Expert Posts: 3,009 ✭✭✭✭
    Re: I need a lesson on basic formulas

    The easiest, most simple way of looking at it (ignoring losses etc as you requested) is to go with watts.
    Since watts is Amps X Volts, you can consider your appliance that is drawing 10 amps @ 120 volts, the wattage will be 10 X 120 = 1200 watts.
    So now you know that 1200 watts will be sucked out of your 12 volt battery.
    To find the amps that will be drawn on your 12 volt battery, take that 1200 watts, and divide it by 12 volts, which will = 100 amps.
    To verify: 100 amps X 12 volts = 1200 watts.
    So there you have it. Now as you get more familiar with these calculations, you can begin to investigate losses, power factor etc, but for now this will put yu in the ball park :D
    Added later: OOPS, I went with 12 volts instead of your 24 volts, but actually this too should help you understand when you compare it with the 24 volt answer you got from Cariboocoot. And yes, you multiply the amps by the time that those amps are being drawn from your batteries, or any other source, to get the amp hours. AND whatever amp hours your appliance consumes on 120 volts, must be multiplied by 5 to get the amp hours from your 24 volt battery, or 10 to find the amp hours if drawing from a 12 volt battery.
  • Fe-Wood
    Fe-Wood Solar Expert Posts: 96 ✭✭
    Re: I need a lesson on basic formulas

    Thanks you guys-
    Still clear as mud but I think I have it!

    So lets say I want to run an appliance that is 11.9 amps at 120VAC, this become 1428 watts. 11.9X120=1428 Cool, got that!
    Now 1428 watts for (X).33hrs=471.24 watts (right?)
    Then, 471.24 watt from(/) a 24VAC supply =19.635 Amp Hrs ?
    Am I on the right track?
  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    Re: I need a lesson on basic formulas

    1428 Watts for 0.33 hours is 471.24 Watt hours. It's very important to use the right terms otherwise you can confuse yourself. :roll:

    And yes, divide that by the system Voltage (24) and you get Amp hours.

    Close enough for government work anyway. :p
  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: I need a lesson on basic formulas
    Fe-Wood wrote: »
    So lets say I want to run an appliance that is 11.9 amps at 120VAC, this become 1428 watts.
    • 11.9aX120v=1428w
    Cool, got that!
    • Now 1428 watts for (X).33hrs=471.24 watt*Hours (right?)
    • Then, 471.24 watt*Hours from(/) a 24VAC supply =19.635 Amp Hrs ?
    Am I on the right track?

    Just remember to carry the units through--They do not disappear.
    • 11.9 Amps * 120 VAC * 1/24 volt battery = 59.5 amps
    • 11.9a * 120v * 0.33 Hours = 471 Watt*Hours
    • 11.9a * 120v * 0.33 Hours * 1/24 volt batt = 19.635 Amp*Hours @ 24 volts
    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • Fe-Wood
    Fe-Wood Solar Expert Posts: 96 ✭✭
    Re: I need a lesson on basic formulas

    Yes! Follow through with the units. That may have been part of my confusion.
    I'm building an excel worksheet were all I have to do is input the units and the conversion formulas are written in.

    So, The next layer- Battery % of discharge. Seems everything I've read says don't go below 20-30% of capacity. To be conservative I'll go with a 70% discharge value. If my 20hr rating is 395 Amp Hrs. That means I can discharge 395*.07=276.5 is my usable amp hrs, (over 20 hrs?)- Correct? So how does that work with only 6 hrs of use?

    Sorry if this seems tedious. My mind is simple these days and I'm really trying to understand how this works out!

    Thanks
  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: I need a lesson on basic formulas

    There are several rules of thumb regarding batteries.
    • discharge to about 75% state of charge on a daily basis for longest life.
    • don't discharge too often to below 50% state of charge for longer battery life.
    • don't ever discharge below ~20% state of charge as a "weak cell" (or several) may go to zero SOC and actually begin to reverse charge--which will kill that (those) cell(s).
    • Do not let a battery set for days/weeks/months below ~75% state of charge--Lead Sulfate crystallizes and no longer participates in the charge/discharge process (permanent loss of battery capacity).
    And rules can be "broken"... For example, you can discharge below 50% state of charge for applications that are not daily use... For example, if a battery has 500 cycle life (very deep discharge) and you only use it on weekends--~50 times a year (trailer, cabin)... That is a 10 year cycle life--and frankly, most (cheap to moderately priced) batteries will not last much longer than 10 years anyways. So, deeper cycling is not the worth thing in the world there.

    The typical off-grid recommendation is 1-3 days of "no sun" and 50% maximum discharge. A good average is 2 days of no sun and 50% maximum discharge. So, 4x your daily load for sizing of the battery bank is a good place to start. Or, 25% discharge per day.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • Fe-Wood
    Fe-Wood Solar Expert Posts: 96 ✭✭
    Re: I need a lesson on basic formulas

    OK, I new it was going to get more confusing:grr
    If I want to run to a 50% rate of discharge. This would be the math. (seeing if my tables are correct)

    Battery bank = 24VDC 790 Amp Hrs @20 hr rate
    50% of charge = 395 Amp Hrs @20 hr rate
    That means (if my formula is correct) Excluding losses etc. I can run 2370 watts for 20 hours at 120VAC, Yes?

    IF this is right and I'm not convinced it is, what happens if I only want to run for 6 hrs? Meaning, does the battery have a different rate of discharge? Is it a straight line of discharge based on the 20 hrs rate?
  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: I need a lesson on basic formulas
    Fe-Wood wrote: »
    If I want to run to a 50% rate of discharge. This would be the math. (seeing if my tables are correct)
    If I understand you correctly--the correct term is depth of discharge (like using 10 gallons out of a 20 gallon fuel tank).
    • Battery bank = 24VDC 790 Amp Hrs @20 hr rate
    • 50% of charge = 395 Amp Hrs @20 hr rate
    • That means (if my formula is correct) Excluding losses etc. I can run 2370 watts for 20 hours at 120VAC, Yes?
    Let's see. Note that the 20 Hour rate is:
    • 790 Amp*Hours / 20 Hour rate = 39.5 amps for 20 hours for 100% to 0% state of charge
    So, to discharge the battery only 1/2 way at 20 hour rate (i.e., for 10 hours):
    • 395 AH / 10 hour rate = 39.5 Amps for 10 hours to 50% state of charge
    • 39.5 amps * 24 volts = 948 watts average load for 10 hours
    IF this is right and I'm not convinced it is, what happens if I only want to run for 6 hrs? Meaning, does the battery have a different rate of discharge? Is it a straight line of discharge based on the 20 hrs rate?

    The battery capacity at 6 hour rate appears to be smaller (more losses as battery is discharged quicker).

    Batteries are not perfect--So they are not linear and behave differently at different discharge/charge rates.

    So, rather than trying for 100% accurate numbers--it usually works "well enough" to just try and use "rules of thumbs"...

    For example, we use the 20 Hour Rate because that is fairly close to the recommended rate of discharge (5 hours a night for 4 nights would be 20 hours)... And similar charging currents during the day.

    But, there are some other rules of thumb too... For example (these are for flooded cell, but work fairly well for AGM's too):
    • Recommended rate of charge form solar 5% to 13%.
    • Recommended maximum rate of charge/discharge 13% or C/8 (batteries become significantly less efficient and can overheat above this charge/discharge current)
    • Recommended maximum surge current 0.4*C (or C/2.5). If you attempt to draw more than 0.4 times the Battery Amp*Hour capacity, the battery voltage collapse (such as starting a deep well pump).
    AGM's can perform much better--but, on average, it does not hurt to follow the above rules of thumb.

    And, when you start taking losses into account... For example assume the inverter is ~85% efficient:
    • 948 Watts * 0.85 inverter eff = 805.8 watts at the load
    And there are other other losses too; Solar Panel derating (less power when hot), charge controller losses, battery charge/discharge losses, etc...

    All together, for a flooded cell battery bank, the end to end losses (solar panel rating to AC output power) works out to around 52%... Yes, about 1/2 of the solar panel name plate rating gets to your AC load at night.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • ggunn
    ggunn Solar Expert Posts: 1,973 ✭✭✭
    Re: I need a lesson on basic formulas

    Rules of thumb are of course temporary stand-ins for the real calculations you need to do when designing a system, but figuring on 50% max DOD (depth of discharge) and a lumped efficiency of 75%, once you have an idea what your daily energy usage is, multiply it by 2.5-3 to get a rough idea how much battery capacity you will need if you can charge up every day.

    Then there are days of autonomy to consider, i.e., how many days of no sun you can tolerate at normal usage. If it's three days, then multiply by three again. A need for battery capacity of 8 - 10 times one's daily usage is not unreasonable, and it's why many folks who think they want to get off the grid think again when they see what it is going to cost them to do that.
  • icarus
    icarus Solar Expert Posts: 5,436 ✭✭✭✭
    Re: I need a lesson on basic formulas

    In general, the kiss principle applies.

    Forget about battery am RATES, just use a total AH capacity of the battery under some averge conditions, shoot for a25% total discharge rate and that will put you close enough for simple calcs.

    So to keep. It simple, my batteries are ~450 ah (12 vdc) or ~ 5400 watt hours of available power, 25% discharge would be ~ 1300 WH. Plug in a 13 watt cfl, and I could go 100 hours or so to be in the target range. If you are concerned about inverter efficiency etc, then factor that in from some known spec. Then in the real world, design around your loading specs, and you will surely be close enough. In the real world close is probably close enough, they don't have to be gnats ass calcs.

    Just my opinion,

    Tony
  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    Re: I need a lesson on basic formulas

    I agree with Tony; forget about the discharge rate. It does vary, but so does the Voltage. When you calculate Amp hours @ Voltage using the nominal the rate pretty much takes care of itself because that nominal becomes the minimum Voltage, leaving a bit of fudge factor on the Amps.

    For example, a "12 Volt" battery fully charged at rest may be 12.75 Volts. At 12 Volts resting it is considered to be 50% discharged, and you're not going to go below that. So you're calculating the Amp hours based on the ending or minimum Voltage on the curve.

    Also do not get Depth of Discharge confused with State of Charge: they are reciprocals. In one post you've said "I'll go with a 70% discharge value." That would mean 70% DOD which is 30% SOC which is a completely dead battery that likely won't recharge.

    The reason why 25% DOD (75% SOC) is so often recommended is because battery life shortens with greater depth of discharge. There's a nice little graph of cycle life here from Trojan on page 2: http://www.trojanbatteryre.com/PDF/datasheets/T105RE_Trojan_Data_Sheets.pdf

    Working out where and which way the "fudge factors" go in to formulas is half the battle. ;)
  • System2
    System2 Posts: 6,290 admin
    Re: I need a lesson on basic formulas

    The way I handle days of autonomy is, add 1 to the number of days of autonomy, subtract the fraction of a day that the sun is shining (equivalent peak hours), and multiply by that. So if I need three days of autonomy, and I have 4 h/d of peak sun, I multiply by 3+5/6. Suppose you have full sun on Sunday and the previous week, then Monday through Wednesday are stormy all day, then Thursday is sunny. You have to last from 14:00 on Sunday to 10:00 on Thursday without exceeding your limit on depth of discharge.

    You may find the average insolation instead of the hours of peak sun. Peak insolation is close to 1 kW/m², so the average insolation in kW·h/m²·d is close to the number of hours of peak sun. Where I'm going to build, it's about 3.75 kW·h/m²·d.

    By the way, unit symbols are case-sensitive. s is second; S is siemens (reciprocal of ohm).
    V - volt
    A - ampere
    W - watt
    J - joule
    h - hour
    H - henry (unit of inductance)
    d - day.
  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    Re: I need a lesson on basic formulas

    The typical method of handling "days of autonomy" is to calculate the battery capacity need for one day and multiply it by the number of days without good charging. There is no sense in trying to predict how much charging you'll get on a "partially lit day" as current drops off drastically with a lessening of available light.

    The standard off-grid method: one day's worth if 25% discharge. The second day you can take the batteries down to 50%. The third day you start the generator.

    It really is not a good idea to try and devise a battery bank that can handle many days worth of "no sun". This inevitably means you will be spending a lot of money on both batteries and recharge capacity that will not be utilized most of the time. It also means the batteries stand a greater chance of spending too much time below 75% SOC, which accelerates sulphation of the plates. Plan for what you can expect on average and don't be afraid of a good generator for when the unexpected happens.
  • Fe-Wood
    Fe-Wood Solar Expert Posts: 96 ✭✭
    Re: I need a lesson on basic formulas
    So, to discharge the battery only 1/2 way at 20 hour rate (i.e., for 10 hours):
    • 395 AH / 10 hour rate = 39.5 Amps for 10 hours to 50% state of charge
    • 39.5 amps * 24 volts = 948 watts average load for 10 hours



    -Bill

    I'm not getting why the duration of discharge has to change (be cut in half) along with the state of charge. It seems to my thick head you have cut it in half and half again
  • waynefromnscanada
    waynefromnscanada Solar Expert Posts: 3,009 ✭✭✭✭
    Re: I need a lesson on basic formulas
    Fe-Wood wrote: »
    I'm not getting why the duration of discharge has to change (be cut in half) along with the state of charge. It seems to my thick head you have cut it in half and half again

    Not right sure what you mean, but there are two things you should know, or learn.
    1) If you want your batteries to last, do not take them below 50% state of charge.
    2) The faster you discharge your batteries, the less power they will give you before they reach that 50% point.
    You can think of it this way - - - slowly pouring water into a container from a bigger one will conserve water because little will be spilled and wasted. But if you pour the water faster, some, or a lot is spilled and wasted, so the useful water you'll get will be greatly reduced.
    I know this is not the same as discharging a battery, but it might help you understand. Discharge your battery slowly and you'll get it's greatest capacity. Discharge it quickly, and it's as if much is wasted, and you'll get far less than full capacity.
  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    Re: I need a lesson on basic formulas

    Slightly more technical than what Wayne said: if you discharge (or charge) faster, that means more Amps flow. Inevitably that increases the amount of energy that goes to heat rather than to work. Even though you may not notice the heat in the conductors, it's still there. That's the "water spilling out" in his analogy.
  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: I need a lesson on basic formulas

    Nothing fancy--If you have a 20 Hour Rate to discharge the battery from 100% to 0% state of charge... Then discharging the battery (at the same rate, current level, wattage, etc.) to 50% state of charge will only take 10 hours.

    Remember the 20 hour rate is C/20... If you have a battery with 100 amp*hour capacity at 20 Hour rate... Then if you discharged it at 5 amps for 20 hours, it would be dead. Or 5 amps for 10 hours to 50% capacity. Both are using 50 Amp*Hours worth of battery capacity.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • ggunn
    ggunn Solar Expert Posts: 1,973 ✭✭✭
    Re: I need a lesson on basic formulas
    BB. wrote: »
    Nothing fancy--If you have a 20 Hour Rate to discharge the battery from 100% to 0% state of charge... Then discharging the battery (at the same rate, current level, wattage, etc.) to 50% state of charge will only take 10 hours.

    Remember the 20 hour rate is C/20... If you have a battery with 100 amp*hour capacity at 20 Hour rate... Then if you discharged it at 5 amps for 20 hours, it would be dead. Or 5 amps for 10 hours to 50% capacity. Both are using 50 Amp*Hours worth of battery capacity.

    -Bill
    Huh? 5A for 10 hours = 50A-h, but 5A for 20 hours = 100 A-h. Maybe that's not what you meant by "both".
  • icarus
    icarus Solar Expert Posts: 5,436 ✭✭✭✭
    Re: I need a lesson on basic formulas

    As for autonomy. (I pad auto spell checked that to "auto mommy! Got to watch the screen now and again!)

    I figure, as ' Coot says, 1 days discharge to 25% (I prefer 20, but the priciple is the same) the PV is sized in an an deal world to provide enough charge capacity to recharge from ~ 3 days of zero sun, so in fct, the are few days when thee is no sun, few still when there is zero sun the second or third day, so at the end of day three, one might be down a max of 60% but more likely somewhere in the 30-40% range, at ich point it is time to fire the genny.

    In practice, I tend to do it when my tri metric says I haven't been near 100% for more than 48 hours.

    I think you can ad infinitum many of these issues to death. It may be a fine exercise, but in the real world, simple rules apply, and work pretty well.

    Tony
  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: I need a lesson on basic formulas

    Hmmm... I guess I am not being clear here...

    The 20 hour rate for a 100 AH battery bank is:
    • 100 AH / 20 H = 5 Amps
    In 20 Hours, that will take a 100 AH battery (at the 20 Hour Rate) dead in 20 hours (100% to 0% state of charge).

    Since we only recommend people discharge normal deep cycle batteries to 50% state of charge (1/2 of their capacity), the above example would be:
    • 20 Hour discharge rate is = 5 amps
    • 10 Hours @ 5 amps would discharge the battery by ~50%
    • 10 hours * 5 amps = 50 Amp*Hours used to power loads
    I was concerned that the original poster was perhaps getting rates and ratings confused...

    For example, when we suggest using 50% of the bank capacity, that is not the same as:
    • 100 AH * 0.50 capacity * 1/20 Hour rate = 2.5 Amp rate of discharge
    First, the 2.5 Amp discharge rate would be ~40 Hour discharge on the full bank. Second, the 100 AH battery bank would be rated slightly higher (maybe 110 AH) at the 40 Hour Discharge rate...

    I am going to stop typing here--Obviously I am bringing more confusion rather than clarity to the situation.

    -Bill :blush:
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    Re: I need a lesson on basic formulas

    A little editorial re-work:

    A 100 Amp hour (at the "20 hour rate") battery with a 5 Amp steady load will be discharged completely in 20 hours. Hence the term "20 hour rate".

    Since you don't want to take it below 50% State of Charge, you can discharge it at 5 Amps for 10 hours.

    Take a look at the Concorde chart here: http://www.solar-electric.com/cosuagmba.html
    It shows different Amp hour capacities for the same batteries depending on the rate of discharge used (2 hour, 8 hour, 24 hour, 48 hour, and 120 hour). Notice how the battery's capacity goes up as the discharge is stretched out over longer time (because the discharge rate is lessened).
  • Fe-Wood
    Fe-Wood Solar Expert Posts: 96 ✭✭
    Re: I need a lesson on basic formulas
    You can think of it this way - - - slowly pouring water into a container from a bigger one will conserve water because little will be spilled and wasted. But if you pour the water faster, some, or a lot is spilled and wasted, so the useful water you'll get will be greatly reduced.
    I know this is not the same as discharging a battery, but it might help you understand. Discharge your battery slowly and you'll get it's greatest capacity. Discharge it quickly, and it's as if much is wasted, and you'll get far less than full capacity.

    This makes sense to me. I had water used to as a metaphor for simple house wiring once.
    The math part still eludes me, but then I've never been good at this kind of math. I'm still trying to work it out in my head.

    Thanks for all the help so far...
  • Fe-Wood
    Fe-Wood Solar Expert Posts: 96 ✭✭
    Re: I need a lesson on basic formulas

    Ok, I went back and re-read most of these posts and If I have it correctly this should be the result- Assuming no losses.

    Now, If I want to run a 23Watt 120VAC light for 6hrs I would use 5.75 DC Amp Hrs. Am I getting this or do I still have a calculation (read as understanding) problem? I'm probably making this more complicated on my end than it is.

    Regardless, I appreciate everyones continued help and input.
  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    Re: I need a lesson on basic formulas

    Let's see.
    23 Watts for 6 hours = 138 Watt hours.
    138 Watt hours / 24 Volts = 5.75 Amp hours.

    Yep; you've got it. :D

    Now you can go on to the higher math of inverter efficiency, power factor, and wiring losses. :cry:
  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: I need a lesson on basic formulas
    Fe-Wood wrote: »
    Now, If I want to run a 23Watt 120VAC light for 6hrs I would use 5.75 DC Amp Hrs. Am I getting this or do I still have a calculation (read as understanding) problem? I'm probably making this more complicated on my end than it is.
    We usually try to work in Watts because is it a "complete" representative of the power used (rate of power specifically).

    Amps and Amp*Hours are a Rate and an Amount (like gallons per minute is to watts, and gallons is to Watt*Hours; used respectively). However, we need the "volts" to do many of the calculations.

    So, assuming 24 volt nominal battery bank:
    • 23 watts * 6 hours * 1/24 volt = 5.75 Amp*Hours @ 24 volts
    So, you have that right... If you have an 85% efficient inverter, then the "fudge factor" would add:
    • 23 watts * 6 hours * 1/0.85 inverter eff * 1/24 volts = 7.96 Amp*Hours @ 24 volts including inverter losses
    So--while understanding the math is important for planning out your own system--The losses are very significant and cannot be ignored when you have multiple areas for losses to occur (solar panels, charge controller and wiring, battery bank, inverter, etc.)...

    For a typical flooded cell battery system, the losses from solar array to AC power is ~52% (typical).... So, if you have 5 hours of sun per day (average), to power your 23 watt lamp at night from solar panels charging during the day, the rough answer is:
    • 23w * 6h * 1/0.52 system eff * 1/5 hours of sun = 53 Watt minimum of solar panels
    And since you have variable sun/weather/loads--You should plan on ~1.33 to almost 2x as much solar panel to power your minimum required loads (those you need every day and cannot put off using for another day). This allows you to get a good charge on the battery (during bad weather and/or random heavy power usage days) and to add some more loads over time without impacting your present needs.

    Solar panels have dropped so far in price, that it using the 5% to 13% rule of thumb for battery bank rate of charge, it is better to get closer to 10% rather than the 5% that was all people could afford 5-10 years ago.

    It is almost never a good idea to add "extra batteries/larger battery bank" to a system for "more power"... A larger solar array is the only way (for most people) to get more useful power out of their solar system--plus it keeps maintenance issues lower (less batteries to take care, fewer to replace 5-10 years out when they go bad, etc.).

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • TheBackRoads
    TheBackRoads Solar Expert Posts: 274 ✭✭
    Re: I need a lesson on basic formulas

    Can we sticky this? Learned a lot :D
  • Fe-Wood
    Fe-Wood Solar Expert Posts: 96 ✭✭
    Re: I need a lesson on basic formulas

    BB and Cariboocoot-
    Thank you for sticking with me on this. I'm beginning to really get a handle on batteries (I think) its a balancing act, based on use, recharge rate and capacity. I get that finding the sweet spot is elusive. The variables are ever changing. Ones power use is never the same month over month and year over year. Through this exchange I realize I was going to be sizing my battery bank to small. Clearly, I didn't understand. I thought one could run the batteries to a 100% state of discharge everyday with no problems or diminished capacity. I didn't realize that as you go further into the discharge the curve gets steeper. I'm beginning to "see" it in my minds eye.

    A crucial bit of info that I got today is that batteries are 2.14 volt per cell. The amps can very but the volts are always the same. Of course, this is for a fully charged battery. I don't quite understand why they are 2.14 volts. I'm assuming it has to do with the electrolyte and lead relationship. What is stored is really Amps or Watts depending on how one wants to quantify it.

    Yes, now I can begin to think about and incorporate loss factors.
  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    Re: I need a lesson on basic formulas

    If you think batteries are tricky to understand, wait 'til you find out about photovoltaic panels. ;)
  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: I need a lesson on basic formulas

    The battery/cell working voltage is based on pure chemistry. Lead Acid around 2.14 volts. Alkaline around 1.5 volts, NiCad around 1.25 volts, Lithium primaries can be near 3 volts, etc.

    The energy is stored in a battery based on chemical potentials (the battery is basically corroding during discharge and "un-corroding" during charging.

    For our needs, both current (amp*hours) and voltage are being stored.

    For example, think of pumping water into a tank. Pump a 1,000 gallons into a tank/lake 10' higher than the home, it takes XXX Watt*Hours to do that. Pump the same amount of water 100' higher into a tank/lake, it takes 10x XXX Watt*Hours. And, if you run that water back through a water turbine (hydro power), that 100' water level will produce 10x as much power too...

    Same as a battery. Store 100 AH at 2.14 volts = 214 Watt Hours.

    Store that same 100 AH at 12.84 volts (6 cells in series) = 1,284 WH -- Or 6x as much energy.

    Different battery chemistries, and even different designs of lead acid batteries, have different pluses and minuses for the job at had.

    Lead Acid is relatively cheap, lasts a pretty long time, but is very heavy.

    AGM lead acid can supply much more surge current vs battery weight. Does not vent gases/electrolyte--so does not need added water or make a mess. However, they cost 2x as much. They call can be frozen without damage (usually--however, you cannot charge/discharge them while frozen).

    Forklift lead acid batteries can take 80% discharge and last several thousand cycles. However, they are not as efficient at energy storage and take a lot more distilled water.

    Lithium batteries (many different rechargeable chemistries--each with different characteristics) are, in general, lighter and can supply huge amounts of surge currents for their size. But they are much more expensive and don't last as many years as a deep cycle lead acid battery.

    You can learn quite a bit about batteries for off-grid use in these two FAQs:

    Deep Cycle Battery FAQ
    www.batteryfaq.org

    Think of the batteries as the "heart" of your off-grid system. Size them for the amount of discharge (and costs/size/needs) to power your loads.

    Size the inverter to power your loads.

    Size the solar panels to keep up with the loads AND to properly recharge the battery bank (this second part is important and was "ignored" for many decades by off-grid system designers/owners. A much larger bank needs a much larger array to keep it properly recharged--even if the attached loads are not that high).

    And size the battery charger to the solar array.

    For a first approximation, work your system design in two parts. Design the battery system to support your loads, assuming all the power is drawn at night. Then design the charging system assuming are no DC loads (solar energy into battery bank). That is the "worst case" design case (in most cases).

    If you need power mostly during the day (say water pumping), and the loads are variable (need to pump during sunny weather, no need to pump during cloudy weather), then you can use a smaller battery bank (or even no battery bank) and you will have a bit less losses so you can use a smaller solar panel array. However, for the most part, other than water pumping and no battery system, most people use most of their power at night and recharge during the day.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset