When Total DC Load Exceeds the Size of Batteries

How did you figure max amp load?

If your figures are correct, you must reduce the load.

when you say reduce the load, do you mean that if I have an electrical equipment required to be in the design, should I omit that equipment. I am not clear on this

the load, not the battery, what are you operating and for how long? 2 seconds, a day a week? what is your expected amp draw from this motor hear element, fan, whatever?

I am estimating number of hours which will be 24hrs because most of the loads might be used 24 hours except the light that no need them during day time, but only for 12 hours.

I am don't like to provide you the type of load, but this is for 24 hours load and all they are dc equipment. no ac involve. I am using 12v dc battery for this total load.

you're wasting our time, bye

Increase the efficiency of your load....

do you have any advice that helps me to keep this 12 dc battery enough for my whole dc load.?

No advice. Need additional batteries.

Please present some numbers.

ok, I will tell you what I did, and please correct me if I am wrong.

Total Amp Hour needed = (Total continues max amp load = 60 amp) + (10% of Largest intermittent amp load =1 amp)

then

20% of (total amp hour) =12.2 amp-hrs

12.2 + 61=74 amp-hrs

from this number I decide the batteries capacity. so I use 40% discharge as the limit

I have to choose the battery so it does not discharge below %50 of its capacity

so

74 amp-hrs/0.4 = 185 amp-hrs and from their I use a battery rated higher than this number

I think I am not clear above, but let me know because I am confused

That mostly makes sense. How did you arrive at the initial 60A of load?

60A is my total max load of all dc equipment that considered as continues only. manufacturer of that equipment provide a max power (divided Pmax by 12vdc) or provide max load on the device itself. Then I added all equipment's max load together.

That is reasonable also. However, one thing that may help you is that not all of this equipment might be demanding maximum load all the time. You should also consider non-concurrent loads, and diversity factors even for those loads that might be sometimes, but not always, active at once. Your local regulations may specify some of these numbers.

Assuming your rough estimate of 185 ah per hour to a 50% discharge point to be reasonable to run that for 24 hours you would need a 4440 AH battery bank or roughly 45 100ah 12 volt deep cycle batteries.

Now how do you plan to recharge them?

how are you arriving at a deep cycle battery at all with almost no parameters given at all?

I read the whole thread.

His load is at most 60 amps on a 12 volt system. He figures a 20% safe overhead on that and wants to run it for up to 24 hours at a time without discharging the batteries below 50%.

The rest was just simple math.

you are assuming n motor starting load or CB motor drive?.

more info

What type of info to provide you. I provided some numbers above

That "maniacally laughable" formula is pretty much the same as LMTD formulas used for heat exchange calculations. Some of the deep principles cross over multiple disciplines.

Excellent reply!

See[for instance]:

http://www.smartgauge.co.uk/peukert2.html

https://en.wikipedia.org/wiki/Peukert%27s_law

http://planetcalc.com/2283/

http://www.ajdesigner.com/phpbatterylife/battery_life_peukert_number_time.php

Battery capacity SHOULD be sized to deliver the reqd. load. It is NOT the otherway round because battery is REQUIRED to deliver the load.

In your case, if you are unable to reduce the load, then battery BACKUP TIME for the given load WILL REDUCE.

a 20Ah battery capacity is rated at 87.5% depletion. You need to up your battery capacity. Check the datasheets on your battery. It will show you a curve of what your battery voltage will be at the rated discharge time. At 87.5% for a sealed lead acid battery at a dishcarge rate of .2C you should be about 10.5 VDC. For NiCAD, NIMH or other dry cell rechargables it is around 10 VDC. At a higher discharge rate like .4C or .5C (C=battery capacity) your battery voltage gets worse on a variable scale.

Also, depleting your battery to 50% or lower will seriously decrease the life of your battery and will require longer charge times which, if they aren't met you will have trouble even getting a usable 12V our of your battery.

Conclusion, redo your calculations based on 87.5% battery capacity. Either increase your capacity or add batteries.