# Calculating watts from generator to batteries

beth
Solar Expert Posts:

**32**✭
Hi all I am trying to get an understanding of generator sizing.

If you have a battery bank of twelve 6v - 400 amp hour, so you have 6X400 =2400 watts

times twelve = 28800 watts in total.

Now lets say you plan on using half of this not to deplete your batteries more than 50%.

So when you have depleted your battery bank by 14400 watts and want to replace these, whats the correct math to see how different size generators will perform?

I know that the generator must supply the usage loads as well when charging. So for example if you had a 5 kw generator, I am sure there is some efficiency loss in its charging capability? But lets say from that 5kw generator you got 4kw per hour of out put, and while you are running your generator you are using 1000 watts in your house per hour, so you now have 3000 watts left for battery charging. Then it would take 14400 / 3000 or 4.8 hours to top up the battery bank.

Is my formula correct? What is the correct loss to calculate in for the generator charging efficiencies ?

I have read here that a battery bank requires a certain charge rate recommended of 5-13%.

How would this apply in the above example. Do you take 5-13% of the total battery bank, .05X28800 = 1440 watts, so you need at least that may watts going into the batteries to charge them at an appropriate rate?

Thanks for all replies!

If you have a battery bank of twelve 6v - 400 amp hour, so you have 6X400 =2400 watts

times twelve = 28800 watts in total.

Now lets say you plan on using half of this not to deplete your batteries more than 50%.

So when you have depleted your battery bank by 14400 watts and want to replace these, whats the correct math to see how different size generators will perform?

I know that the generator must supply the usage loads as well when charging. So for example if you had a 5 kw generator, I am sure there is some efficiency loss in its charging capability? But lets say from that 5kw generator you got 4kw per hour of out put, and while you are running your generator you are using 1000 watts in your house per hour, so you now have 3000 watts left for battery charging. Then it would take 14400 / 3000 or 4.8 hours to top up the battery bank.

Is my formula correct? What is the correct loss to calculate in for the generator charging efficiencies ?

I have read here that a battery bank requires a certain charge rate recommended of 5-13%.

How would this apply in the above example. Do you take 5-13% of the total battery bank, .05X28800 = 1440 watts, so you need at least that may watts going into the batteries to charge them at an appropriate rate?

Thanks for all replies!

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## Comments

5,400✭✭✭✭It takes ~20% more energy to charge the batteries than the batteries give you. Not a strict formula but pretty good rule of thumb.

So if you draw down 100 ah, you will need to generate ~120 ah to come even.

In round numbers, if you need to get the 120 ah, if you were charging at 20 amps, it would take five hours to fully recharge the batteries. That said, that too is not a strict linear relationship, as the batteries are charged faster for the first ~80, the final 20% takes more time with reduced currents.

I suggest that for more information you read the following links:

http://www.batteryfaq.org/

http://www.windsun.com/Batteries/Battery_FAQ.htm#Lifespan%20of%20Batteries

PS, you really need to learn to understand charging current in AMPS, not watts. Battery capacity is based on AMP/HOURS.

A 100 ah battery would then like to charge between 5-13 amps. Remember that any charging also needs to over come any loads that may be drawing on the battery at the same time.

Good luck and keep in touch,

Tony

17,615✭✭Let's start by recalculating your battery bank size. You have twelve 6V batteries wired how? If you connect batteries in series the Voltage goes up, but not the Amp hours. If you connect them in parallel the Amp hours go up but not the Voltage. I suspect you have a combination of series/parallel connections. For instance, if you have a "12 Volt" system then you have two 6V in series X 6 in parallel for (6*400) 2400 Amp hours. If it is "24 Volt", then you have four 6V in series X 3 in parallel for (3*400) 1200 Amp hours.

A 2400 Amp hour 12 Volt bank would want 240 Amps @ 14.2 Volts for charging, or 3408 Watts

A 1200 Amp hour 24 Volt bank would want 120 Amps @ 28.4 Volts for charging, or 3408 Watts

Same Watts, different Voltages at different Amperage. At 5% rate you can drop down to 1704 Watts.

Now, how long does it take to "replace used Amp hours"? Unfortunately it isn't as simple as "3400 Watts @ 28.4 Volts = 119 Amps * 5 hours = 600 Amp hours" because A). the last few Amp hours take longer to recharge (current rate drops as t battery comes up), . darn lead-acid batteries use about 20-25% more power to recharge than you get back out of them, C). not all chargers are power-factor corrected so aren't as efficient as they should be.

Your estimate of 4.8 hours was pretty close, but for the losses. In reality it might be closer to 7.

Now someone will tell me why I'm wrong.

32✭so cariboocoot, you are saying that at 10 % charge rate , 10% of a 1200 amp hour bank is 120 amps at a charging voltage of 28v, So if you put that into watts as you did = 3408 watts, then you can look at your generator rating of lets say 5k to see where your at. So it looks like a 5k gen would just begin to cover that with additional small loads ?

Thanks, and as well Icarus for those links

29,967adminUnfortunately, it gets more complicated...

You have the losses of charging the battery bank (~80% for flooded cell, ~90% efficient for AGM--Very rough and typically conservative numbers).

Then we have the actual charging voltage (28 or 29 volts for the 24 volt battery bank), we have losses in the battery charger (another 80% efficient or so)...

And we have another "loss" which is Power Factor... Basically the battery charger takes more current (not power) than it really transfers to the battery bank. So, the genset has to be ~1/0.6 PF = 1.66x larger that you would expect from all of the other losses... The only "good thing" about "bad" power factor (PF less than 1.0) is that it does not use (much) extra energy (fuel).

So, for a 120 amp battery charger, charging at 29 volts, running an 80% efficient battery charger, with a power factor of 0.6 would need a genset rated, at least:

- 120 Amp * 29 volts * 1/0.80 eff * 1/0.6 PF = 7,250 VA (or VAR) minimum genset

Now, note I said VA or VAR (volt amp reactive)--You can usually assume the VA rating of a genset is at least the Watt Rating (sometimes VA is > Watt rating--all depends on the manufacturer)... We usually use the equation for power as:- Power=Volts*Amps= Watts

Well, for AC (and other special cases), we use VA (or VAR or kVA or kVAR--VA and VAR are the same thing). Many electrical devices (electronics, motors, etc.) do not take the Current exactly as offered by the AC Grid or Off-Grid Inverter... So, with a lot of hand waving, it turns out these devices draw a lot more current than would be indicated by watts alone... So, the "real" power equation is--And the VA rating is:- Power=Volts*Amps*Power Factor
- VA=VAR=Volts*Amps

Basically, from the genset and inverter wiring point of view, a device with a "poor" power factor of 0.6 takes 1.67x ( or 1/0.6=1.67) as much current as its "Watts" Name plate would suggestion... For example, two 120 watt computers, one with a "bad" power factor of 0.6 and other with power factor correction that has a PF of 1.0:- Power=Volts*Amps=120 Volts * 1.0 amps = 120 watts (not really)
- Power = Volts * Amps * Power Factor
- Amps = Power / (Volts * PF)
- Amps = 120 watts / (120 Volts * 1.0 PF) = 1.0 amps (PF = 1.0)
- Amps = 120 watts / (120 Volts * 0.6 PF) = 1.67 amps (PF = 0.6)

So, in one case with PF=1.0, a generator would only need to output 120 Watts or 120 Volt*Amps.In the other case, with a PF=0.6:

- Power = V*A*PF
- VA = Power / PF = 120 watts / 0.6 PF = 200 VA (or VAR)

So, from an electrical point of view--the generator to supply the PF=0.6 120 Watt power supply need to be rated to output 200 "Watt" (really 200 VA or 200 VAR) to operate without overheating/popping a circuit breaker.So--Sizing a genset for charging your battery bank is pretty complex--and it does not help that many vendors do not explicitly say what their Power Factor is...

And, there are now some vendors (like Xantrex) that are making battery chargers that are Power Factor Corrected which allows you to charge your battery bank with a smaller genset (less fuel wasted by running a bigger genset that you really needed).

Note that many gensets and pieces of equipment have both Watt and VA ratings. And, for a genset or inverter, many times the Watts=VA rating--but not always--Some have a VA rating > than their Watt Rating (and marketing will advertize that their 200 VA inverter is larger than their competitor's 120 Watt inverter--somewhat true, but also somewhat miss-leading).

Confused the heck out of you yet?

-Bill

1,372✭✭✭beth, one question. Will this battery bank be used in an emergency backup system, or will it have daily use? For a backup system its more important to get it through the bulk charge stage, from 50% to 80%, and then stop. As mentioned above, that is the fastest charge stage and gets you the most amps back into your battery for the fuel used in the generator. When your utility service returns then you can let it recharge the bank the rest of the way. If you get an outage longer than a few days you should charge up most of the way once in a while.

If the battery bank will be used on a daily basis, then its important to budget fuel use to get it charged completely as often as possible to prevent sulphation in the batteries.

You also want to plan on using as much of the available power as possible while the generator is running. Gensets run most efficiently at full output. My 12 kw genset uses 1.5 gallons of fuel per hour at half load, and 2 gallons at full load. So I get an extra 6 kw for just 1/2 gallon more. My inverter could handle loads like my well pump and electric water heater, but to get the most from my fuel I wait to use those things when the generator is on to recharge my battery bank.

32✭It's for a non grid tied system Techntrek.

Yes Bill you sure did, I appreciate the info though.

I am still wondering With a 4kw trace sw4024 and 3 strings of L16's-

1200amp hour bank, How's the honda 5kw generator going to do. I just started working with this system so I don't know if I am going to have trouble. I just had to set the inverter to only pull 20 amps from the generator instead of the 30 where it was set, to keep the breaker from blowing and that is with no additional loads from the house?

29,967adminBeth,

So I don't create more confusion... Exactly what batteries (brand/model, or Voltage and 20 Hour Amp*Hour rating). How many batteries per string (4x 6 volt batteries, or 12x 2 volt batteries per string, etc.). There is quite a range in capacity and output voltage for Trojan batteries (if Trojan). I want to get your bank capacity correct (it will change the calculations quite a bit if I guess wrong).

For the Honda, what model number? I assume you are using the 120 VAC outlet (what is the current rating at 120 VAC)? Or is it the 240 VAC outlet (the SW4024 is probably a 120 volt input unit?).

For charging the battery bank, you are using the SW4024?

-Bill

32✭Bill, I currently have a 24v battery bank, 3 strings of 4, six v batteries, now they are Trojan L16s. I am about to replace them with something similar. The bank configuration will stay the same and the amp hour of the new batts. will be some where in the 350 - 400 AH. at the c20 rate. The Honda is the EM5000sx, I use the 240 outlet going through a Trace transformer, then through the trace inverter. Specs. of the gen. are rated current - 37.5/18.8 A., rated output - 4.5 kva

Thanks

29,967adminSo, your battery bank AH rating is 3x 400 AH = 1,200 AH at 24 volts.

The SW4024 appears to have a good power factor (assume 1.0). Assume 80% efficient charger at maximum charging current of 120 amps. And 30 amps 120 VAC maximum AC charging current.

I am confused about the popping a circuit breaker when the SW4024 was set to 30 amps. Which circuit breaker was it popping (inverter or another breaker elsewhere in your system)? Was it the Generator 240 VAC breaker or a 120 VAC breaker on the output of the step-down transformer? And what was its current rating? Was it opening right away, or after a period of time? Any pattern?

This would indicate that the SW has a "poor" Power Factor:

- 20 amps max programed / 30 max RMS draw by SW = 0.66 (very rough estimate)

Although--It is not impossible that there is a bad circuit breaker or some other issue (such as over heated CB panel) that is causing them to open at too low of current rating.In general, it would seem that the maximum charging current for your setup would be:

- 4,500 Watts * 0.80 inverter efficiency * 0.95 transformer eff * 1/29 volts battery charging = 117 Amps (120 amps max rating for SW)

However, because you have derated the system to 20 amps of 120 VAC:- 120 VAC * 20 amps * * 0.80 inverter efficiency * 1/29 volts battery charging = 66 Amps

Anyway--The ratio of charging current to battery bank capacity:- 66 amps / 1,200 AH = 0.055 or 5.5%

It is not high (you can go to 10-15% of Bank AH rating) without any problems.You are running your genset at:

- 20 amps / 37.5 amps maximum (at 120 VAC) = 0.53 or 53% of rated load

Which is perfectly OK--It would be nice to get your genset to operate at closer to 80% rated load (30 amps)--but you have the breaker issue.From reading the SW manual--their AC input current readings are a bit miss-leading... Trace chose to read only the "Real" in-phase current. So if there is a "Bad Power Factor" somewhere (inductive load leading to PF~0.6 instead of >0.9), their meter will not show that.

To (easily) measure the "True" (believe it or not, you can call these the vector sum of the "Real" and "Imaginary" components of the current--If you remember your real and imaginary numbers from math oh so many years ago) AC and DC currents in your system--A clamp on AC / DC rated amp meter (~$100 for a "cheap" one) may be needed.

But--if your system operate OK as is--The only reason to try and get more current (Power) from the genset would be to reduce run-time and noise (by running the genset at 80% load instead of 55% loading and getting more charging current for faster charging).

I looked through the SW Menu system in the manual--I did not see one that displayed the Battery Current... Is there a menu where you can monitor the battery DC current? And what is its value when charging (and what is the AC input current).

Lastly, the SW will automatically manage the AC input current from the genset so that it does not draw more than the present setting... You an add AC loads to your SW system and it will first cut back on the amount of charging current to the battery bank... And second, if the loads go very high (larger than 20 Amps AC of your present max AC2 input current setting), it will actually take power from the battery bank and "assist" the generator in supplying the AC loads too.

-Bill

17,615✭✭Beth;

You have quite a large battery bank! It will want 3500 Watts of power available to it for charging. Then add in the losses for the charging system (power factor, et cetera) and add on loads ... Even a 5kW is going to be working. The generator's breaker was probably tripping trying to supply more than its capacity.

I've been there, done that, reprogrammed the Outback! I can even charge mine off the 1000 gen if I remember to cut the charge rate in half and not turn anything on! Your battery bank is about 4X the size of mine.

What's worse: the Trojan L16's want even more current than 10%. They're 'happier' with 13%, due to the difficulties in getting 'tall case' batteries to remix properly. But for occasional charging this isn't an issue.

I suspect you're actually losing some capacity by going from the gen's 240 output through a transformer to feed the Trace @ 120.

32✭Thanks again guys,

Bill in answer to:

"I am confused about the popping a circuit breaker when the SW4024 was set to 30 amps. Which circuit breaker was it popping (inverter or another breaker elsewhere in your system)? Was it the Generator 240 VAC breaker or a 120 VAC breaker on the output of the step-down transformer? And what was its current rating? Was it opening right away, or after a period of time? Any pattern?"

It is the breaker on the generator, there is only one there, so I guess it functions for both 120/240. This was a new issue that just started occurring. The error light on the trace sw would start to flash after about 10 min. ( I read in the Manual that this means a generator issue) then in about another 5 min the breaker on the gen. will pop. I tried to turn off all power going to the house , but same thing. I know my batts. are old and one reads 4.6 v the others read 6v.

29,967adminHmmm... A 4.6 volt battery may be a huge issue (either it has a shorted cell, or a there is an open cell somewhere else).

I don't know, but it is possible the low voltage battery bank is causing issues (drawing too much out of phase, aka bad power factor, current) from the genset.

I hate to add to your costs/hassles... But you may want to get an AC/DC clamp on current meter like one of these: MA220 Extech AC/DC clamp meter (~$100)

The issue is that the SW current meters, on purpose, only shows you a portion of the total current in the AC wiring. The clamp meter will measure the RMS (Root Mean Square) current (the amount of current that is causing the breaker to pop). It also can measure the frequency of your genset (too high or too low of RPM?).

The Generator AC breaker that is popping is probably the 15 amp (or 20 amp) 240 VAC breaker--With your Trace transformer, you cut the voltage by 1/2 but double the current (power=Voltage*Current--So the equation still balances--1/2 the voltage and 2x the current is still the same amount of power).

-Bill|

32✭Thanks bill,

I realize that bad cell may be a problem, but what do you mean by "drawing too much out of phase" in you statement:

"I don't know, but it is possible the low voltage battery bank is causing issues (drawing too much out of phase, aka bad power factor, current) from the genset."

Also could you explain exactly what wires I would check with that meter and what kind of readings I would look for,

what other system checks would it be good for? Its probally a good tool to have?

17,615✭✭I shall try to explain in laymen's terms.

Really dead or shorted cell in one battery becomes lower resistance when charging. When this gets to the AC side of things, so to speak, it "flattens" the curve of the sine wave; bringing the Power Factor down so that the efficiency is even less than would normally be expected.

Always a good idea to have a DVM and learn how to use it. Checking for different Voltages between batteries is a good way of spotting problems, as well as checking for Voltage potential along wires. For instance, along a wire connecting (+) or one battery to (+) of another you should read '0' Volts between the two positive terminals. If it read '0.5' there is a potential bad wiring problem causing inequitable current sharing (one battery higher Voltage than the other, corroded or loose terminals, failing wire).

29,967adminHere is a fairly good discussion of AC power and Power Factor and non-linear AC wave forms...

Regarding electrical "Phase" between voltage and current...

Think about attaching a rope to a car to pull it up hill... If you stand right in front of the car and pull on the rope--100% of your pull goes into moving the car forward.

If you stand 45 degrees off to the side, only the Cosine of 45 degrees (0.7) of your force goes into pulling the car forward... The rest goes into pulling the car to the side... If you still want to pull 100 lbs forward, you need to pull ~140 lbs on the rope to pull 100lbs forward.

Voltage sine wave form and the current wave form--The current can either follow the voltage wave form or it can be "delayed" (for inductors) and the wire needs to carry more current for the equivalent amount of power...

Another thing that can happen is that the current only flows at the very peak of the voltage--This is sort of like standing in front of the car and pulling 200 lbs for 1 second and 0 lbs for 1 second... The average is still a 100 lb pull--but you need a rope rated for 200 lbs or it will break.

The Amp meter in the Inverter system only measures the current that pulls the car forward--This particular meter does not measure the total current being carried by the wire.

(Sorry, this is a very sort explanation of a long subject--please feel free to ask if you have more questions).

It is not difficult to use a Clamp on amp meter... But to make good use of it, you need to understand the basics of electricity to use it correctly.

I am going to use the old water analogies--It is not exact, but may help you understand....

Say you have a small water system with two storage tanks on the top of a hill... and the pipes run down to your home. Normally, the two pipes carry equal gallons of water to each tank--and the water flow is equal to the two tanks. However, one of the pipes/tank valves gets blocked... Now, from at the home, you are drawing water and replacing water in the tanks just like normal... But you now have 1/2 the storage capacity and you need to figure out what is wrong.

You use a Flow Meter (aka clamp on DC amp meter) around each pipe (wire) and check the flow (amps) going through each pipe. One will be carrying all of the water and the other will not be carrying any--So now you can find out what is going one.

On the other hand, say that one tank has a hole in it... The tank with the hole is now taking much more water than the other tank--so you are pumping water into the tanks, but find that your storage goes flat in a few hours (as the one tank drains water from itself an the other connected water tank)--The flow meter (clamp on amp meter) can again show how the flows are different between what should have been two identical water tanks... You can then use other tools to isolate the problem.

For testing your generator--You need to make a short extension cord (that you can plug between the generator and the inverter) with the three power wires (hot A, hot B, and neutral wires) split out from the cable (leave the insulation on the wires themselves). You can now clamp on the AC clamp meter and measure the total current through each of the wires. If the A/B (240 hot wires) are carrying 10 amps and your 20 amp breaker is popping--then probably you have a bad circuit breaker. (remember that 10 amps at 240 VAC is the same amount of power as 20 amps at 120 VAC).

The reason you have to "split" out the insulated wires is that for electrical current flow there is the power going out one wire and coming back on the second. If you put a current clamp around both wires--the net current flow is zero amps. You need to clamp on just one wire at a time to measure the current in that one wire.

Here is an OK video on using a clamp meter.

If your AC current is too high, then you can look at the Inverter/Charger/Battery bank for problems.

If you have somebody nearby who has some electrical experience, they can spend an hour with you and train you have to check out your system with a current meter.

It is a very handy tool (just like a DMM is)--but it is not quite as easy to understand as a volt meter is.

Also, note that I am saying get an AC and DC capable current clamp meter. You have both AC power and a DC battery bank.

The AC clamp meter are just fine and even quite a bit less expensive--But they cannot measure DC amps (they will just read zero amps if you try it).

-Bill

32✭Dear Bill and Cariboocoot,

Thanks for this test meter advice! I have a a question about each of the tests you recommend.

Cariboocoot

How exactly would I use my digital volt meter to do this test? Which meter probes (+ , - ) do I touch where.

Bill

So this test will show me the output of my gen. and if this is less than the rating of the breaker, and the breaker keeps popping, its bad. But why else would the breaker pop if it is designed to handle the max output from the gen? A short or problem somewhere between the gen and or in the inverters battery charger?

Thanks guys!

I

17,615✭✭A good digital meter reads either positive or negative Voltage, so it doesn't matter which probe goes where when checking this. Set on DC Volts; scale large enough to handle the full battery capacity (just in case). If you get reading across what should be a "null" parallel connection, then something is causing one side to be higher than the other - like a loose wire. Sometimes it's a bad battery, though.

29,967adminNot a "short" in the terms you are think of (a piece of bare wire touching something metal).

This is more along the lines of a bad/shorted battery/cell causing the Inverter's Battery Charger to miss-behave and draw too too much current from the AC side of the inverter...

Is this what is happening?--I really don't know. But without the tools and a knowledgeable person on site, we may never find where the problem is without changing a couple parts at a time... A $50 circuit breaker or $X,000 worth of batteries to "fix" the problem? And still may have the problem.

With your current battery bank--Perhaps identifying/isolating the weak/bad battery (ies) and either (for example) reshuffling 4 strings of batteries into three good strings (perhaps with a spare or too) or purchasing an new/used replacement to get by for another year or so.

I hate to tell you what to try / purchase next... I have not worked with any of your equipment and the problem (I don't think) is a common one (popping circuit breaker on generator when inverter max current is set below generator's output rating).

I think that somebody with a DC/AC current clamp can help identify where the problems may be--However, if you have demonstratably "dead batteries" in your current battery bank--pulling those and doing something else (rearange/replae a partial or full bank) first before looking for design issues in the inverter/charger is probably the first step I would take.

You do run the real risk that there is a hardware problem with the inverter/charger (for example, the AC current sense coil may be bad and the inverter is not actually monitoring the input current)--And it is old enough that it may need to be replaced vs repaired. Then you find the current battery bank does not support the "new inverter/charger" and you have to think about replacing the battery bank (that you may have just purchased a couple batteries to limp by on)...

And you can see where this ends up... And older system that is nickle and dime-ing you to death, vs a new system that empties your bank account all at once...

Maybe some other folks have some words of wisdom for you too... There is only so much I can do from a keyboard.

-Bill