caviglia
join:2004-10-18
Visalia, CA
| Figuring out draw in watts
Hey Guys,
Heres what I got. I've got a pop that has a 525ah battery bank. 5 12v batteries, each 105ah.
I'm running this pop off a battery back to determine if the battery bank is big enough and try to get an idea of what the usage is.
Every 24 hours of operation the batters drop .2v. So the batteries started out at 12.6 and after 4 days they are at 11.8
Is there a way to calculate my draw in watts? So I can figure out sizing on the solar panels...
Thanks,
Cavig |
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IntraLink
Premium,MVM
join:2002-08-14
Utah Valley
| Another way is to go buy a cheap watt meter device.
I think they come in the clamp over variety so you just hold it over the cable and it will tell you through inductance how much is going through the cable at the moment.
If you can calculate from your figures the wattage that would be better because it is taken over a larger time period and may more accurately reflect your draw average. |
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lutful
Premium
join:2005-06-16
Ottawa, ON
 ·TekSavvy Solutions..
3 edits | reply to caviglia
You could always use a "clip-on" automotive ammeter to check actual current draw from the 12V bus at different times of day.
Otherwise, you can only guesstimate using the above graphs where you will see the voltage drops depends on how much current you are drawing versus max capacity.
Edit: I think my wild guess of 360W) and public guess below of 30W are both wrong. C = 525Ah, so if average current draw is 5A, battery voltage with load is following C/100 curve. Please use a clip-on ammeter to check average current draw and determine the proper C/? curve for your application. |
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public
join:2002-01-19
Santa Clara, CA
·DSL EXTREME
| reply to caviglia
said by caviglia  :Heres what I got. I've got a pop that has a 525ah battery bank. 5 12v batteries, each 105ah. Every 24 hours of operation the batters drop .2v. So the batteries started out at 12.6 and after 4 days they are at 11.8 You are using about 10% battery capacity per day, or about 2.5A, 30W load. |
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thewildthang
join:2005-12-21
Winfield, AL
| reply to caviglia
Well, that's one way to end a thread, just give a guy the answer, short sweet and to the point. No logic, no formulas, no explanation. Sort of reminds me of a T-shirt that I saw the other day, it said "Because I'm your mother and I said so!". Ok, from your answer you state that he is using 10% of his battery each day. From this I deduct that your saying that his minimum allowable voltage is 12.42 volts. 13.8 - (13.8 * .10) = 12.42 . I'm not arguing or disagreeing with you, I'd just like to know more than the bottom line to the answer. |
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public
join:2002-01-19
Santa Clara, CA
·DSL EXTREME
| said by thewildthang :Well, that's one way to end a thread, just give a guy the answer, short sweet and to the point. No logic, no formulas, no explanation. Gee why do you need a formula?
Assuming that a lightly loaded battery has initial voltage 12.8V, and final voltage of 10.8V is reached in 10 days, it is evident that 525Ah battery is drained in 240 hours. |
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