platypus

platypus · August 2013

I am first timer to your site, it's a great source, I found lots of interesting and very important information re batteries among other thing. I am at the moment arguing with a battery supplier that the battery bank that was delivered is not what was ordered. He assures me that the battery bank comprised of 12 x 2v 1600Ahr at 100c rate is what we asked for and have received. The battery is a clean skin battery without any markings it is an Exide battery bank and each battery weight 45Kg approximately 90 lb. After exhaustive search we have determined that the code for the battery is 7epzs875 which to me is an 875 Ahr at c20 rate according to Exide website.
They are trying to tell me that this battery will deliver 1600A at c100 even thou it is only 875Ahr at c20 capacity. I am searching for power and current discharge rates and energy content ( in Watts), for this battery and I am hoping that someone can help me with this, the supplier and the manufacturer are not very helpful, basically they do not answer any of my requests. All this came about when the system caught fire and most of the installation burned down only 4 months after installation. We assumed we have received 1600 amp batteries and treated them as such, with charge rate as high as 160a from solar.
with Kind regards Platypus

vtmaps · August 2013

Re: platypus

platypus wrote: »

After exhaustive search we have determined that the code for the battery is 7epzs875 which to me is an 875 Ahr at c20 rate according to Exide website.
<snip>
I am searching for power and current discharge rates and energy content ( in Watts),

Welcome to the forum Platypus,

To get the wattHours that a battery stores, you multiply the ampHours by the Voltage. Thus 2 volts X 875 ampHours = 1750 wattHours. You have 12 of these batteries, so 12 X 1750 = 21,000 wattHours = 21 kilowattHours.

This number is only valid when you are discharging the battery bank at the C20 rate which is 43.75 amps at 12 volts = 525 watts.

Also, it will ruin your batteris if you draw them down much below 50% of their capacity, so as a practical matter you have 10.5 kwh of storage.

Tell us more about your system... what do you use it for? What other components does it contain? Obviously this system was poorly designed or poorly assembled or it wouldn't have burned up. Perhaps we can help with other aspects of the system,

--vtMaps

stephendv · August 2013

Re: platypus

platypus wrote: »

The battery is a clean skin battery without any markings it is an Exide battery bank and each battery weight 45Kg approximately 90 lb. After exhaustive search we have determined that the code for the battery is 7epzs875 which to me is an 875 Ahr at c20 rate according to Exide website.

EPzS means it's a forklift battery and these are usually measured at the C5 rate, not C20. See here: http://www.sue.ba/sue/files/file/classic/Classic_EPzS_int.pdf on that chart there's a 7 EPzS 805 cell that weighs 45kg, so it matches pretty closely what you have, and that cell is measured in C5.

Using SMA's rough estimation for converting from C5 to C100:
C10 = C5/0.88
C10 = C100/1.25

So C100 = 1.42 * C5, which in your case would be 1242 Ah C100

Using a chart for flat plate deep cycle batteries (yours are tubular plate and might have a higher specific gravity): Page 16 here: http://www.batterysupplies.be/sites/default/files/files/111748CataloogENG72dpi.pdf

They get C100 = 1.76 * C5, which would be 1547Ah C100

So your supplier is sort of in the right ballpark, the exact C100 number should come from the manufacturer, but you'll have a hard time finding it, because in the forklift world no one cares about a 100 hour discharge

platypus · August 2013

Re: platypus

just silly question, if the battery in question is 805 A 2 volt that is 1610 watts of energy at C5 rate , and if I place a load of 16.01 watts for 100 hours, or C100 rate at the end of the 100 hours I will still have (1547x2=3094-1610=1484) 1484 watts of energy left in the battery. this just doesnot add up. We should stop building powerstations and start making batteries. this is what I do not understand. I have charged up a 12 volt bank which is 875a untill I raised the voltage to 15.3 volts and charged it for a futher 2 hours after this. than I placed a load of 102 Watts and and after 50 hours I was down to 40% or 11.9 Volts that is about 5100 watts of energy that I used so I must have 5838 Wats left in the battery. how slowly do I have to discharge the battery to get 1547 A or 3094 w. it is confusing at best.

stephendv · August 2013

Re: platypus

platypus wrote: »

just silly question, if the battery in question is 805 A 2 volt that is 1610 watts of energy at C5 rate , and if I place a load of 16.01 watts for 100 hours, or C100 rate at the end of the 100 hours

Think you're misunderstanding the C capacity rating. C100 is the capacity of the battery if it discharges from 100% to 0% in 100 hours. So if the battery is 805Ah at C5, it means that you can draw 805Ah in 5hours = 161A. At 2V, that's a constant 322W draw.

platypus · August 2013

Re: platypus

we have no clues as to what caused the fire, the system has been powering whole house for 4 months without any difficulty but as the winter has set in toward the end of April we have started to charge the batteries every other day by the Inverter/charger. the charger unfortunately has no shunt or temperature sensor to compensate for battery temperature and we think that during the day solar has managed to charge the battery and during the night the charger, being on timer, has continued to charge.

we have built the system with 24 volt 1600A battery but we now know the battery was only 875A but we have set all setting for a 1600A battery. I have 24 x 230 watt solar panels and 2 x 80 mppt solar chargers they are fused on the solar output and on the inverter from battery also. The wires from chargers are 50 milimitre square section and the wires from inverter to battery ar 100 milimitre square section.The distance from battery to charges and inverter was no more than 2 metres.

BB. · August 2013

Re: platypus

I am very sorry to read that you lost the system in a fire--I hope that you did not lose anything else/nobody was injured.

Can you give us more information on what happened with the fire?

In general, batteries will only catch fire if they are "boiled dry"... If flooded cell batteries, that would be over charging/not checking water levels every 2-4 weeks. For SEALED/AGM/GEL cell batteries, they will vent if overcharged and there is no way of restoring the lost water.

If the batteries where not the source of the fire--Another suspect is wiring and wiring connections.

Wiring, if under sized (not heavy enough copper) can easily overheat. We have seen here where a Licensed Electrician has installed way to small of wiring for the DC cabling. He did not understand that (for example) the current on the 24 volt side is 10x the current as on the 230 VAC side of the system. And he used the same gauge wire to the input of the AC inverter as he used on the output.

And poor connections (dirty/lose/poorly made) are another place fires can start. You can (carefully) use your had to check for warm/hot electrical connections (especially on the DC side) when the battery is under heavy loading or charging (high current flow). You can also use an IR Thermometer (non-contact type thermometer) to scan for high temperatures too (and save burns--chemical and acid--on your fingers).

Another issue with a failed connection is "Arc Faults". DC power makes for very nice Arc Welders and sustained arcs. Much more likely to have a sustained arc on the DC side of your system vs AC side. Again poor/failed/damaged connections or Hot to Ground faults can easily Arc Fault (Solar arrays are very prone to Arc Faults because there is not enough current to trip a fuse/breaker if there is a short circuit from PV+ to ground).

And lastly, the equipment itself. We are used to AC circuits that can supply way more power than we use on average (a computer and a few lights or fridge on a circuit that can supply 5x as much power as we are using).

For Battery Banks / Off Grid power systems--The charging current is near 100% of the wiring and components (inverters, chargers, charge controllers) current capabilities for many hours on end. And in some cases, it is possible to overheat devices that are not exceeding their rated output because of the many hours at heavy loads (i.e., a "car type" battery charger recharging a huge battery bank). Even generators have a typical max rating and a Prime Mover rating where the Prime Mover rating is only 80% or so of the typical max power rating (i.e., emergency backup, portable generator use).

We are very concerned about safety here, and while fuses/breakers are an important part of safety--There are a lot more issues to making for a safer system.

-Bill

platypus · August 2013

Re: platypus

Nobody was hurt and the damage was isolated in small section of the Garage, the battery,inverter, fuses and their enclosures all of the wireing and part of the wall was distroyed, the soot penetrated the whole garage. there was no sine of Arcking on any of the 24 volt wireing even thou the plastic was burned of( fuses were not blown). The 240 volt AC side of things there was nothing left to inspect. including the Inverter was consumed only the metal box was left, main fuse for the inverter in the fuse board was tripped. I have inspected the electrolite only few days prior so H2O is not likely but possible, we have built the system arround 1600 A 24 V but since the incident we have disovered that the battery was only 875A. we think we may have seriously overchaged already charged battery during the night of the incident. we have a new system now up and running it is twice the size than we planned before, we are happy with it.
I am however confused with the C rating of the batteries.I would like to understand the logic behind the rating. I am told that 875A battery at C5 rate is 1600A at C100 rate. When I loaded this battery rated 875A at C5 rate I got only 400A out of it, if I loaded it at only C100 rate I am getting close to the 800A . I do not see the possibility getting any more energy out than what is in the battery in the first place. so I do not understand how can a 800 A battery be rated as 1600 A battery no matter how slowly it is discharged. If anyone has insight into it it would be apreciated.

vtmaps · August 2013

Re: platypus

platypus wrote: »

I am however confused with the C rating of the batteries.I would like to understand the logic behind the rating. I am told that 875A battery at C5 rate is 1600A at C100 rate. When I loaded this battery rated 875A at C5 rate I got only 400A out of it, if I loaded it at only C100 rate I am getting close to the 800A . I do not see the possibility getting any more energy out than what is in the battery in the first place. so I do not understand how can a 800 A battery be rated as 1600 A battery no matter how slowly it is discharged. If anyone has insight into it it would be apreciated.

I am having trouble figuring out what it is you don't understand because of the units you use. If you don't understand the difference between an amp and an amphour, you will get nowhere.

Amps are a rate that electrons are moving through a conductor (water analogy: its like gallons per minute). If you multiply 'gallons per minute' by 'minutes' you get 'gallons' which is an amount of water.

An Amp is about 6,000,000,000,000,000,000 electrons per second. If you multiply 'amps' by 'hours' you get an amount of electrons. Batteries are rated in 'amphours' which is the amount of charge (electrons) they contain.

Peukert's law says that the amount of charge you can get out of a battery depends on how fast you draw the charge out. The faster you draw it out (amps = rate), the less charge (amphours = amount) you can draw from the battery.

Water analogy: Imagine you have a tank of water. You draw the water out at 1 gallon per minute and the tank is empty in 100 minutes. You conclude that the tank holds 100 gallons. But now you refill the tank and draw the water out at 2 gallons per minute. It takes only 40 minutes to drain the tank. Therefore your tank is rated 100 gallons at the one-gallon-per-minute rate, but the same tank is rated 80 gallons at the two-gallon-per-minute rate.

Going back to batteries: 875 ampHours at the C5 rate means that you can draw 175 amps for 5 hours. If you draw only 16 amps, the same battery will last for 100 hours (1600 amphours at the C100 rate).

Does this help you understand?

--vtMaps

Photowhit · August 2013

Re: platypus

platypus wrote: »

... this is what I do not understand. I have charged up a 12 volt bank which is 875a untill I raised the voltage to 15.3 volts and charged it for a futher 2 hours after this.

Why are you doing this? Is this your normal charging parameters? Bulk until the voltage rises to 15.3, then absorb for 2 hours? This is an equalizing voltage for most batteries...

platypus wrote: »

than I placed a load of 102 Watts and and after 50 hours I was down to 40% or 11.9...

This is a known load? is this AC measured at the inverter, does this include the 30+ watt load of the inverter it's self? does this include the 85% efficiency of the inverter in producing your 102watt load? Was this measured with a clamp meter with the electric going out of the battery? With a 12volt system you also have greater wire losses to consider. Also when was the battery voltage measured? did you let the battery rest with no load for 3-4 hours (everything off disconnected from system?) Voltage is a poor way of testing battery capacity, testing Specific Gravity (SG) will give you a true picture of the battery state of charge.

platypus wrote: »

that is about 5100 watts of energy that I used so I must have 5838 Wats left in the battery. how slowly do I have to discharge the battery to get 1547 A or 3094 w. it is confusing at best.

Have you read the battery FAQ's?

NorthGuy · August 2013

Re: platypus

platypus wrote: »

just silly question, if the battery in question is 805 A 2 volt that is 1610 watts of energy at C5 rate , and if I place a load of 16.01 watts for 100 hours, or C100 rate at the end of the 100 hours I will still have (1547x2=3094-1610=1484) 1484 watts of energy left in the battery. this just doesnot add up. We should stop building powerstations and start making batteries. this is what I do not understand. I have charged up a 12 volt bank which is 875a untill I raised the voltage to 15.3 volts and charged it for a futher 2 hours after this. than I placed a load of 102 Watts and and after 50 hours I was down to 40% or 11.9 Volts that is about 5100 watts of energy that I used so I must have 5838 Wats left in the battery. how slowly do I have to discharge the battery to get 1547 A or 3094 w. it is confusing at best.

First of all, when you're thinking about batteries it's better to think about amper-hours not watt-hours. Amper-hours measure the electric charge stored in the battery. To get watt-hours, you need to multiply by discharge voltage. If you discharge fast, voltage will be lower, so when you discharge the same amount of amper-hours you get less watt-hours. Also, when battery is fully charged, voltage is higher, so at the beginning of the discharge you get more watt-hours per one amper-hour of discharge than at the end.

The difference between watt-hours put in and taken out is dissipated as a heat, so if you discharge fast you may overheat the battery.

The capacity rating assume you need to get some minimum voltage from the battery. 1.75V/cell seems to be the standard. In your example C100 of 1547 means that if you discharge battery evenly for 100 hours at such a rate of discharge that the voltage goes to 1.75V/cell at the end, you'll get 1547 amper-hours.

When you apply a load to the battery, the voltage drops. The bigger the load, the bigger the voltage drop. Because of this voltage drop, discharging with higher load will reach 1.75V sooner. In your example, at C5 rate, battery will reach 1.75V after you discharge only 805 amper-hours. You cannot use the battery at C5 rate any more because the voltage is too low. But, when you remove the load, the voltage will come back up. So, you will be able to discharge more with lower load. If you continue discharging with C100 rate, you will get the rest (1547-805)=742 amper-hours until the battery reaches 1.75V again.

In practical terms, you never discharge battery very deep, so voltage never should go to 1.75V, or the battery may be ruined.

YehoshuaAgapao · August 2013

Re: platypus

platypus wrote: »

I am first timer to your site, it's a great source, I found lots of interesting and very important information re batteries among other thing. I am at the moment arguing with a battery supplier that the battery bank that was delivered is not what was ordered. He assures me that the battery bank comprised of 12 x 2v 1600Ahr at 100c rate is what we asked for and have received. The battery is a clean skin battery without any markings it is an Exide battery bank and each battery weight 45Kg approximately 90 lb. After exhaustive search we have determined that the code for the battery is 7epzs875 which to me is an 875 Ahr at c20 rate according to Exide website.
They are trying to tell me that this battery will deliver 1600A at c100 even thou it is only 875Ahr at c20 capacity. I am searching for power and current discharge rates and energy content ( in Watts), for this battery and I am hoping that someone can help me with this, the supplier and the manufacturer are not very helpful, basically they do not answer any of my requests. All this came about when the system caught fire and most of the installation burned down only 4 months after installation. We assumed we have received 1600 amp batteries and treated them as such, with charge rate as high as 160a from solar.
with Kind regards Platypus

Did you ever water the batteries at all (looks like the were at least somewhat going by later in the thread)? Exposed plates and/or overdischarging can cause a battery to burn up. Must be a terribly designed (or configured) system too to allow the voltage in the batteries to draw down that much. Flooded batteries are much harder to burn up than sealed AGMs. Flooded batteries can heat up with charged or discharged very deeply (AGMs also). Flooded don't heat up that bad when charged unless charged very rapidly. AGMs can heat up on charging, especially if at gassing voltage - AGMs are easier to burnup do to thermal runaway because hydrogen/oxygen recombination has to occur inside the battery. Flooded don't have such an issue as the hydrogen/oxygen is vented out, but the ultra deep discharge can cause them to heat up alot (especially if discharging quickly).

You won't have a very successful attempt at suing the battery company as they tend to expect you to know that batteries lose capacity (due to internal resistance) as they are discharged faster and also that batteries will be hooked up to a properly configured cut-off of some kind to not allow them to be drawn down so much and they expect you to properly maintain the batteries too (minimum: watering, not chronically running them in a deficit state of charge, not overdischarging them, and occasional EQ cycles). The only thing they really did wrong was giving you the 100-hour rate capacity (may have told you it was the 100-hour rate too, may not have) when almost anyone knows batteries are never discharged at a 100-hour rate, but a properly configured charge controller will not allow this misinformation (or mis-comprehended information) to cause a fire due to overdischarge.

Another possible cause of fire is extreme hydrogen buildup in the battery boxes due to lack of (zero or near-zero) ventilation, especially with the aggressive charging parameters it appears you have set up. If hydrogen is not vented out and it builds up significantly, then the slightest arc from a corroded or loose connection or melted wire insulation can torch things up pretty good.

BB. · August 2013

Re: platypus

You have to understand what the specifications are "laid against"...

For example, our rules of thumb are usually based on C/20 battery bank AH capacity rating as a generic starting point. Different battery mfg. can use other C/XX ratings for their rules of thumb.

Find the manual/specification for the specific battery model/mfg... And go with their settings/requirements as first. Sometimes, the chargers use % of AH capacity or similar--So you may have to "cheat" the setting to match the number required by the battery.

Then watch battery voltage/specific gravity/water usage/temperature and adjust your settings based on what you see happening.

-Bill

platypus · August 2013

Re: platypus

Going back to batteries: 875 ampHours at the C5 rate means that you can draw 175 amps for 5 hours. If you draw only 16 amps, the same battery will last for 100 hours (1600 amphours at the C100 rate).

Does this help you understand?

--vtMaps

thanks for that. I was under the impression that this is as it should be, a 1600 A hr battery if discharged aggressively ( C5 rate) will deliver 875 A hr of energy. However the manager of the manufacturer assured me that the battery they supplied which turned to be 875A hr. battery, will actually deliver 1600A hr. of energy if the load is at C100 rate.

Well this proved to be a load of tish as I have tried. after charging the battery the electrolyte density reached 1.280 , after resting the battery I have used number of 12 volt globes with total rated wattage of 102 watts and proceeded to discharge the battery and after 72 hours I had about 20 % power left. previously the same battery loaded through the inverter at the C5 rate the battery was capable of delivering roughly 400 A hr. of energy.

I now know that this 875A hr battery will not deliver more power no matter how slowly it is discharged but the faster it is discharged less energy will be available . I tried to obtain the spec. sheet for this battery but I found it impossible , I was hoping that someone may know about this battery. " 7 EPZS875 "
thanks for taking time and info. P.

vtmaps · August 2013

Re: platypus

platypus wrote: »

.
after charging the battery the electrolite density reached 1.280 , after resting the battery I have used number of 12 volt globes with total rated wattage of 102 watts and proceeded to discharge the battery and after 72 hours I had about 20 % power left.

How do you know that you had 20% power left? --vtMaps

Photowhit · August 2013

Re: platypus

You can Whine all you want, if you want answers, you might interact in the forum. I asked some direct questions and got no answers in helping you solve your problem. Have any answers to;

platypus wrote: »

... this is what I do not understand. I have charged up a 12 volt bank which is 875a untill I raised the voltage to 15.3 volts and charged it for a futher 2 hours after this.

Why are you doing this? Is this your normal charging parameters? Bulk until the voltage rises to 15.3, then absorb for 2 hours? This is an equalizing voltage for most batteries...

platypus wrote: »

than I placed a load of 102 Watts and and after 50 hours I was down to 40% or 11.9...

This is a known load? is this AC measured at the inverter, does this include the 30+ watt load of the inverter it's self? does this include the 85% efficiency of the inverter in producing your 102watt load? Was this measured with a clamp meter with the electric going out of the battery? With a 12volt system you also have greater wire losses to consider. Also when was the battery voltage measured? did you let the battery rest with no load for 3-4 hours (everything off disconnected from system?) Voltage is a poor way of testing battery capacity, testing Specific Gravity (SG) will give you a true picture of the battery state of charge.

platypus wrote: »

that is about 5100 watts of energy that I used so I must have 5838 Wats left in the battery. how slowly do I have to discharge the battery to get 1547 A or 3094 w. it is confusing at best.

Have you read the battery FAQ's?

platypus · August 2013

Re: platypus

I am sorry for not answering earlier, (time difference, Australia) the load on the battery was 12V globes just to eliminate any other loads the distance about 3 Ft using 4 mm SQ wire( capable to conduct about 40 A). I have made sure the battery is charger as much as possible by equalizing it. and no I have not read battery FAQ's I wish some / any Info. came with the battery they are clean skin and I was assured that they are 1600Ahr battery when they are half of that.

platypus · August 2013

Re: platypus

how do I know that I have only 20% left, the Voltage (11.55V) of the battery is reasonable indicator of capacity of the battery I am not far from this figure. the battery I used was cycled about 20 times to this point they are still being run in. I did not want to go bellow this as discharging them to low is not doing them much good.

vtmaps · August 2013

Re: platypus

platypus wrote: »

how do I know that I have only 20% left, the Voltage (11.55V) of the battery is reasonable indicator of capacity of the battery I am not far from this figure. the battery I used was cycled about 20 times to this point they are still being run in. I did not want to go bellow this as discharging them to low is not doing them much good.

The voltage of the battery is NEVER a reasonable indicator of the CAPACITY of a battery. The voltage of a battery may be a reasonable indicator of the state of charge (SOC) of a battery, but not while you have a load on the battery. You must disconnect the battery and wait a few hours to measure the voltage. If you really want to know the SOC, use your hydrometer... that is the gold standard.

platypus wrote:

I have used number of 12 volt globes with total rated wattage of 102 watts

Don't trust the rated wattage. If you want to know what they are drawing you must measure the current being drawn. Your conclusions are no better than the data you base your conclusions on. Also, the current drawn by the globes probably varies with the battery voltage. You need an amphour counter (such as Doc Watson or Trimetric) if you really want to draw firm conclusions.

platypus wrote: »

the battery I used was cycled about 20 times

Many (most?) flooded lead acid batteries do not reach their maximum capacity until they have been cycled about 50 times.

--vtMaps

SandyP · August 2013

Re: platypus

platypus wrote: »

how do I know that I have only 20% left, the Voltage (11.55V) of the battery is reasonable indicator of capacity of the battery I am not far from this figure. the battery I used was cycled about 20 times to this point they are still being run in. I did not want to go bellow this as discharging them to low is not doing them much good.

I also wanted to know what the SoC was during discharge and went with a Victron BMV-600 to "see" the amps going out. Unfortunately the accuracy of SoC when charging is somewhat less accurate due to the charge efficiency of the batteries being somewhat variable.

Photowhit · August 2013

Re: platypus

platypus wrote: »

.... I was assured that they are 1600Ahr battery when they are half of that.

If you've made your conclusion you can't look objectively. Did you " ... let the battery rest with no load for 3-4 hours (everything off disconnected from system?) Voltage is a poor way of testing battery capacity, testing Specific Gravity (SG) will give you a true picture of the battery state of charge."

Battery FAQs

I, too, would suggest you data could be better. "a number of globes" would suggest multiple connections and wiring that could leach current. An in line gauge or a very good clamp meter would give better results. Having SG readings would be more accurate measure of capacity. We didn't ask what temperature the battery is in, we're in summer here, but your in the cold part of your seasons, capacity is effected by how hot or cold the battery is.

NorthGuy · August 2013

Re: platypus

You might have damaged the batteries by over-discharging them, hence the loss of capacty.

vtmaps · August 2013

Re: platypus

Photowhit wrote: »

We didn't ask what temperature the battery is in, we're in summer here, but your in the cold part of your seasons, capacity is effected by how hot or cold the battery is.

Very good point! Up here in VT my batteries only have their rated capacity for a few weeks each year. --vtMaps

platypus · August 2013

Re: platypus

After charging the battery I have allowed 3 Hours stand by and the density of the electrolyte was 1.28 and the temperature of the battery was about 12 deg. Centigrade not sure what that is in F. The load wattage was calculated from the specs. stamped on the globes and was reasonably reflected by true RMS clamp meter to approximately 8 A, this changed as the voltage from the battery dropped and the current reading was reflected by the measurements. This test was first time I have discharged this battery so low. When I use the hydrometer when the battery is less than 50% of discharge, it's not very descriptive, this hydrometer is excellent indicator for when it reaches 50% or more. I have not been able to find good quality Hydrometer, I am sure they are out there, I am more interested for the battery to be fully charged than discharged as I do not intend to discharge more than 20%. this Battery was disconnected from all other instruments like the 2x 80A MMPT solar chargers which have all the AHr meters and calculators with the shunt and temp. sensor.

platypus

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