Erroneous voltage under load?

[Gdwats]

I've got a 24v 230AH fla bank powered by 9x 100W panels. It is at 26.6v and I turn on a 500W heater which gives PF of 0.97. I should have around 5000WH in the bank. After 10 min of running the heater the batteries read 24.2. There is no way they are at 50% charge. Discharging at C/10. It takes about another 10 min to read 100% charging at c/10.
Should I ignore the under load reading and run the heater for 30 min?

[NEOH]

We just had a thread regarding this exact issue.
Yes, voltage under load is different than RESTING VOLTAGE.

https://www.solarpaneltalk.com/forum/off-grid-solar/batteries-energy-storage/general-batteries/365013-discharging-agm

[Gdwats]

Well, I just read it and it shed no light on my issue really. I know what resting voltage is, and not to deplete below 50%max. But i don't know how to interpret the reading at c/10 discharge, short of turning everything off and waiting. calculating in W, and c is 5000W. Msybe thete is no short answer.

[AzRoute66]

There is no way they are at 50% charge.

There is your short answer.

Should I ignore the under load reading and run the heater for 30 min?

We already know that voltage is no indicator of state of charge unless the battery is at no load and rested (hours). So, if there is ANY concern that you are overdischarging, you need to go get the hydrometer wet. I think you'll find that the battery is fine. Installing an energy meter on the battery might give peace of mind in the long term.

[Sunking]

Bottom line your batteries can only handle a maximum 700 watt load. You need almost ten times more battery than you have, and ten times more panel wattage. You got exactly what you planned and asked for.

Batteries have Internal Resistance called Ri. What you are seeing is Current x Resistance voltage losses.

Voltage = Current x Resistance.

The resistance is a fixed variable, and current is the variable. As you increase current, you loose more voltage. FLA batteries vary a bit on how much current they can deliver without significant voltage sag or loss. Example a SLI (starting, lighting & ignition a car battery) is constructed with many very thin plates. This enables them to deliver very high current because the multiple thin plates have a lot of surface area and thus very low resistance. However a SLI battery is not worth a damn in a cycle application. If pressed into cycle service you only get 50 to 100 cycles or so.

True deep cycle batteries have heavy thick plates that can be cycled up to 1000 times. However those thick heavy plates do not have a lot of surface area, thus higher resistance than its SLI cousin. As a result they can only deliver roughly C/8 maximum charge and discharge currents.

So you have a 24 volt 230 AH battery. C/8 on a 230 AH battery = 230 AH / 8 Hours = roughly 30 amps. So 30 amps x 24 volts = 720 watts maximum load and you are trying to pull 5000 watts. Not going to work very well. You got exactly what you designed it to do. Fail. There is nothing wrong with your battery, it is doing exactly what it is suppose to do.

Think of it this way. I can put a 50 pound back pack on you, and you should be able to walk a few miles with it because the load is right for the vehicle carrying it (you). What you have done is put on a 350 pound backpack and you cannot even make it out the door. Why?

[AzRoute66]

So 30 amps x 24 volts = 720 watts maximum load and you are trying to pull 5000 watts. Not going to work very well. You got exactly what you designed it to do. Fail.

I don't see where he is trying to pull 5000 watts with his 500 watt heater.

[Sunking]

I don't see where he is trying to pull 5000 watts with his 500 watt heater.

You are right, my bad. I seen 5000 wh and overlooked the "h".

500 watts / 24 volts = 21 amps. No problem for a 230 AH battery. I was looking at 200 amps using 5000 watts, my bad.

OK having said that, I still do not see a problem, just Peukert Law.Voltage cannot be used to determine SOC on a working system. My question would be where is the measurement being taken?

[Gdwats]

Bottom line your batteries can only handle a maximum 700 watt load. You need almost ten times more battery than you have, and ten times more panel wattage. You got exactly what you planned and asked for.

Batteries have Internal Resistance called Ri. What you are seeing is Current x Resistance voltage losses.

Voltage = Current x Resistance.

The resistance is a fixed variable, and current is the variable. As you increase current, you loose more voltage. FLA batteries vary a bit on how much current they can deliver without significant voltage sag or loss. Example a SLI (starting, lighting & ignition a car battery) is constructed with many very thin plates. This enables them to deliver very high current because the multiple thin plates have a lot of surface area and thus very low resistance. However a SLI battery is not worth a damn in a cycle application. If pressed into cycle service you only get 50 to 100 cycles or so.

True deep cycle batteries have heavy thick plates that can be cycled up to 1000 times. However those thick heavy plates do not have a lot of surface area, thus higher resistance than its SLI cousin. As a result they can only deliver roughly C/8 maximum charge and discharge currents.

So you have a 24 volt 230 AH battery. C/8 on a 230 AH battery = 230 AH / 8 Hours = roughly 30 amps. So 30 amps x 24 volts = 720 watts maximum load and you are trying to pull 5000 watts. Not going to work very well. You got exactly what you designed it to do. Fail. There is nothing wrong with your battery, it is doing exactly what it is suppose to do.

Think of it this way. I can put a 50 pound back pack on you, and you should be able to walk a few miles with it because the load is right for the vehicle carrying it (you). What you have done is put on a 350 pound backpack and you cannot even make it out the door. Why?

Sorry no, I'm trying to pull just 500 W for half an hour, not 5000W. But thanks for the info, it is all good.

[Gdwats]

Your latest comment about SOC under load is quite true. I just don't feel comfortable uncapping batteries under charge/discharge to see the running SOC, so that is why I posed the question about the "sag" is it? I'll try putting a 5# backpack on a 250# man and see if he can walk for 20 miles in full sun tomorrow.

[Sunking]

Sorry no, I'm trying to pull just 500 W for half an hour, not 5000W. But thanks for the info, it is all good.

Sorry about that, it is on me. My bad. While under load the voltage you are most concerned with is 10.5 volts is 100% DOD dead or 1.75 vpc. What you are seeing is most likely normal as I explained current resistance losses.

Try this. Put the battery under a load and measure the voltage and current and call it VL and the current I or A. Voltage shall be taken on the battery Term Post and no where else is valid. Now shut the load off, wait about 15 seconds and measure the voltage again and call it OCV

Ri = [OCV - Vl] / Amps

Example lets say OCV = 25.2 volts Vl - 24 volts, and I = 30 amps.

RI = [25.2 - 24] / 30 amps = .04 Ohms.

Not the most accurate method, but will get you in the ball park. Just make sure you do the OCV record after you hit it with the load to minimize Peukert effect.

[Gdwats]

Oh Peuk. Ok, thanks. I will do exactly that. Is your 10.5V analogous to 21V for my 24V bank?


It looks like I am after the internal battery bank resistance here to be used in calculating P needed for devices? Like P needed is device power + P loss. And P loss is i^2(Ri + Rsys) or something similar?

oh, and I'm measuring V off the post via the CC readout.

[Sunking]

Oh Peuk. Ok, thanks. I will do exactly that. Is your 10.5V analogous to 21V for my 24V bank?

Yes Sir. 10.5, 21, and 42 volts

It looks like I am after the internal battery bank resistance here to be used in calculating P needed for devices?

No sir, power has nothing to do with it. Power is just the product. It is about Current and Voltage Sag under load. Just a sanity check in the design. As a rule of thumb, you limit discharge current to C/8 maximum. Where C = the battery Amp Hour Capacity, and divided by the 8 Hour discharge rate.

For example lets say I have an 80 AH Battery. That would mean I do not want to exceed 80 AH / 8 H = 10 amps. That will limit your voltage loss to roughly to 2 to 3% maximum. So say you have a 230 AH battery and your load is 100 Amps. Not going to work because C/8 = 28.75 amps. If the load is 100 Amps, you know the minimum capacity must be at least 800 AH or greater.

Surf around this forum and you will find hundreds of threads where people have a 100 AH 12 volt battery and wonder why their 2000 watt Inverter will not work right. I just love it when I tell them they only need 15 more batteries.

oh, and I'm measuring V off the post via the CC readout.

What did I say? Measure the voltage on the BATTERY TERM POST and no where else. Not only do batteries have resistance, wire resistance is even higher. If your system is designed properly, you will se as much as 1/4 volt difference between the Battery Term Post and the Inverter Input Terminals. That is 1/4 volt lower. Most DIY's have a lot more than 1/4volt. They failed to use large enough wire.

Voltage loss is additive. Here is where can go wrong. Your battery is fully charge and you hit it with and incur a 1-volt sag, plus another 1 volt or a total of 2 volts. Now if your battery was fully charged with no panel power available the battery resting voltage is 12.5 volts - 2 volts = 10.5 volts at the Inverter. Guess what is going to happen? Give you a hint. The Inverter low voltage disconnect is 10.5 volts and you are giving it 10.5 volts. What is going to happen very soon with your fully charged battery?

Your inverter goes DARK and quits making noise.