Calculating Battery Current

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Canton

Senior Member
Location
Virginia
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Electrician
I have (32) 12 Volt batteries in series for a UPS for a voltage of 384 volts DC. The batteries are a Data Safe 12HX505. I now that with the batteries in series I have built up the voltage so the nominal current would be equal to one. I'm trying to figure out what the nominal current is? The info I have is they are 6 cell, the nominal Ah @ 8 hr rate to 1.75 volts/cell end voltage is 119 Ah and the Watts/Cell @ 15 min. rate to 1.67 volts/cell end voltage is 506 watts.

There are two of these strings in the cabinet each protected by a 400 amp breaker {(2) 400 amp breakers in cabinet}. So I have (2) strings of 32 batteries at 384 volts in each string. The name plate says 271 amps...is this 271 amps for one string or the whole cabinet. I trying to figure out the current in each string?:?
 
271A on the nameplate on what piece of equipment? Seems odd for a cabinet with two 400A breakers.

Your current will be controlled mostly by the load.

Short circuit current of each string at the breaker is the battery charged voltage (x12 in your case) divided by the internal resistance of the battery (x12 in your case) plus wire resistance.
 
271A on the nameplate on what piece of equipment? Seems odd for a cabinet with two 400A breakers.

Your current will be controlled mostly by the load.

Short circuit current of each string at the breaker is the battery charged voltage (x12 in your case) divided by the internal resistance of the battery (x12 in your case) plus wire resistance.

I have three battery cabinets and we are trying to size our wires. Each battery cabinet has (2) strings of batteries 32 batteries in series (64) total. There are (2) 400 amp breakers in each cabinet. One breaker for each string. I'm trying to figure out the nominal current that each string can provide. The cabinet says 271 amps, but I do not know if that is the string current or the cabinet current (both strings). Most of the information given on the name plate was per string. Is there some way to calculate the current with the information that I gave in post #1?

Thanks
 
I have three battery cabinets and we are trying to size our wires. Each battery cabinet has (2) strings of batteries 32 batteries in series (64) total. There are (2) 400 amp breakers in each cabinet. One breaker for each string. I'm trying to figure out the nominal current that each string can provide. The cabinet says 271 amps, but I do not know if that is the string current or the cabinet current (both strings). Most of the information given on the name plate was per string. Is there some way to calculate the current with the information that I gave in post #1?

Thanks
What you must know or find out is how much power the UPS can provide at full continuous load and at surge/short term load.
That will tell you how many total amps at 384 volts the UPS will draw.
You can then divide that by 3 to get the normal full load amps from each string. However the short circuit current available from each string is far higher than 400A, and so you really need to size the wire so that it is protected by that 400A breaker. Any lower ampacity in the wiring would make it a tap conductor and you would need to know whether there is any DC OCPD at the UPS input or not to see whether a reduced ampacity tap wire can be allowed.
Note also that if you turn off two of the three strings the UPS will be perfectly happy trying to draw its entire current requirement from the one remaining string.
 
What you must know or find out is how much power the UPS can provide at full continuous load and at surge/short term load.
That will tell you how many total amps at 384 volts the UPS will draw.
You can then divide that by 3 to get the normal full load amps from each string. However the short circuit current available from each string is far higher than 400A, and so you really need to size the wire so that it is protected by that 400A breaker. Any lower ampacity in the wiring would make it a tap conductor and you would need to know whether there is any DC OCPD at the UPS input or not to see whether a reduced ampacity tap wire can be allowed.
Note also that if you turn off two of the three strings the UPS will be perfectly happy trying to draw its entire current requirement from the one remaining string.

Thank you for the info. Is there some way to calculate the amount of current the battery can provide with the information I have? I guess I'm lookinf for the Ah of one battery but I don't know how to figure it with the info I have.
 
Thank you for the info. Is there some way to calculate the amount of current the battery can provide with the information I have? I guess I'm lookinf for the Ah of one battery but I don't know how to figure it with the info I have.

The AH of one battery, which is also equal to the AH of one cell and the AH of one string, is 119. This is at the eight hour rate, C8. An AGM battery suitable for use in a UPS will probably be capable of sustained operating current of 2-5 times C20. We can guesstimate that C20 for the battery would be about 140-150AH. So an allowable working current from one string could be between 300A and 750A, but the UPS is not going to try to pull that except under fault conditions.
Anyway, the batteries are quite capable of tripping the 400A breaker under a sustained overload, even more so under a short circuit.
I would size the wires to safely carry 400A, but I tend to be conservative.

The only reason I would look at the nominal current per string when sizing the wires would be to evaluate voltage drop. And at 384 volts that is not likely to be an issue.
 
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It seems if most of the information on the nameplate is per string, then the amp value would also be per string, especially considering there are two 400A breakers in the cabinet... but I can't say for sure. If the nameplate doesn't somehow indicate it is specifically per string, then only the manufacturer can answer this question with certainty.

You cannot determine the nominal current of just the battery string portion of the circuit. The connected load is determines for the most part what the nominal current is. Just the string portion of the circuit has only two current that can be determined without consideration for the entire circuit: 1) short-circuit current, as mentioned previously, and 2) no load condition, which is obviously zero.

If these are connected to a UPS, does your UPS units have a maximum current rating. That's what should be used to determine the maximum steady-state current for each circuit, and even this value is highly dependent on the load connected to the UPS. Unless you have a constant, continuous load connected to the UPS, nominal current cannot be determined much less have any meaning.
 
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I would size the wires to safely carry 400A, but I tend to be conservative.
I believe under the NEC, both the breaker supply and load wires must have an ampacity greater than 350A (next lower standard OCP rating), and the the calculated load must not be greater than the wires ampacity. Minimum size will be determined by the termination and equipment temperature limitations.
 
140621-1519 EDT

Canton:

The information in your first post is very confusing.

The info I have is they are 6 cell, the nominal Ah @ 8 hr rate to 1.75 volts/cell end voltage is 119 Ah and the Watts/Cell @ 15 min. rate to 1.67 volts/cell end voltage is 506 watts.
Assuming 6 cells at a nominal 2 V each and, thus, 12 V total for one battery, then for 119 ampere-hours at 12 V means the steady current would be 119/8 = 14.8 A. It is possible a somewhat lower voltage should be used because the final voltage for this rating is 1.75 V per cell.

Using the 15 minute rating and still assuming 2 V per cell, the current for a 12 V battery based on the wattage you provided is 506/2 = 253 A.

As a cross check I would expect the ratio of these two currents to be greater than 253/14.8 = 17 because the time ratio is 8/0.25 = 32. Fast discharge probably reduces the available load energy. So this difference may be reasonable.

Is the nameplate data really worded as you presented the data?

The battery short circuit current might be in the thousands of amperes.

.
 
140621-1519 EDT

Canton:

The information in your first post is very confusing.


Assuming 6 cells at a nominal 2 V each and, thus, 12 V total for one battery, then for 119 ampere-hours at 12 V means the steady current would be 119/8 = 14.8 A. It is possible a somewhat lower voltage should be used because the final voltage for this rating is 1.75 V per cell.

Using the 15 minute rating and still assuming 2 V per cell, the current for a 12 V battery based on the wattage you provided is 506/2 = 253 A.

As a cross check I would expect the ratio of these two currents to be greater than 253/14.8 = 17 because the time ratio is 8/0.25 = 32. Fast discharge probably reduces the available load energy. So this difference may be reasonable.

Is the nameplate data really worded as you presented the data?

The battery short circuit current might be in the thousands of amperes.

.

Yes, I double checked. The information is the same online @ Data Safe. In the same piece of paper I have here they list this battery's Short Circuit Current @ 4510 amps. Its also says the max discharge current (Amps-2 min rate) is 913 amps. The battery has an internal resistance of 2.8 ohms.

Just trying to figure out how they got 271 amps.....also for one stringer or the two stringers combined.
Thanks
 
...

Just trying to figure out how they got 271 amps.....also for one stringer or the two stringers combined.
Thanks
FWIW, I believe it is one string, but also for both. This equipment is designed with the intent to keep the standby system powered while performing maintenance on either of the two strings. Therefore the [max] 271 amp rating is established with the on-line maintenance condition in mind. Keep in mind the load determines if this rating is satisfied, as I mentioned previously. When both strings are enabled, the load current will be the same as when only one string is enabled.

If you have more than one of these paralleled to power the load, where the maximum steady-state current of the laod can exceed the 271A rating, there will have to be a safe switching procedure written for taking multiple strings off-line at the same time. I'd have to look it up, but it may even be required to have signage to the effect of this safe-switching procedure posted.
 
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where in Va?

where in Va?

913 amps. The battery has an internal resistance of 2.8 ohms.

Just trying to figure out how they got 271 amps.....also for one stringer or the two stringers combined.
Thanks

Canton, I suggest you call in a local electrical engineer to help you. Working with 4,500+ potential amps and confusing the data like you are sounds mighty dangerous.

PS: no battery capable of 4500a will have an internal resistance of 2.8 ohms......
 
140621-1938

Canton:

If I assume a linear drop in voltage over a 15 minute period from 2.00 to 1.67 V, then average voltage is 1.835 V over this time period. Now assume the given power at this point is the 506 W, and assume the current at this point is the desired average current, then current is 506/1.835 = 275.7 A. Close to your 271. If you use a voltage of 1.867 the the result is 271.02 A.

384 V with an internal resistance of 2.8 ohms calculates to a short circuit current of 384/2.8 = 137 A. This does not correlate with your stated short circuit current.

.
 
Thank you for the info. Is there some way to calculate the amount of current the battery can provide with the information I have? I guess I'm lookinf for the Ah of one battery but I don't know how to figure it with the info I have.

Forget the battery current. Just like any other circuit the conductors must be rated higher than the overcurrent device protecting them.
 
480.3 Wiring and Equipment Supplied from Batteries. Wiring and equipment supplied from storage batteries shall be subject to the applicable provisions of this Code apply-ing to wiring and equipment operating at the same voltage, unless otherwise permitted by 480.4.



A look at all of article 480 would seem to be a good idea.
 
Canton, I suggest you call in a local electrical engineer to help you. Working with 4,500+ potential amps and confusing the data like you are sounds mighty dangerous.

PS: no battery capable of 4500a will have an internal resistance of 2.8 ohms......

That is correct....it is 2.8 Mohms, I was typing fast. There is an electrical engineer, we are working with stamped drawings for a federal job. This job has tight time restraints and I cannot afford to wait for answers until this wed. and loose to more days.

There is no confusing data....all the information submitted is correct and can be found on Data Safes website for the 12HX505 batteries.
 
Forget the battery current. Just like any other circuit the conductors must be rated higher than the overcurrent device protecting them.


Yes...that is what I am going to do. I will just size my conductors for each string that is being provided with the 400 amp breakers. I'm just trying to figure out why (6) 400 amp breakers...how they came up that. There were a lot of changes with the batteries.
 
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