Electrickery - I'm so confused.

Submitted: Wednesday, Apr 16, 2003 at 18:34
ThreadID: 4431 Views:1231 Replies:6 FollowUps:7
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The question is related to power useage when using an inverter.

OK I understand that if I want to run a 12V item that _Affordable_Storage_Drawers.aspx 1amp, then theoretically, a 30 amphere battery will run this for 30Hrs Iam I right?

This is is in the perfect world, I understand you have losses here and there and you need to apply a fudge factor.

OK so what if I run a 240V appliance which _Affordable_Storage_Drawers.aspx 1 amp at 240V, through and inverter. How long (again in the perfect world) will the 30 amphere batter last?

Am I making sense because the more I think about it, the more I get confused. To think I'm a professional pilot as well :-@
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Reply By: Rob - Wednesday, Apr 16, 2003 at 19:12

Wednesday, Apr 16, 2003 at 19:12
A 12V item drawing 1 amp would drain a 30A/H battery after 30
hours - yes. At these lower currents, this equation applies - however
it is unlikely that the same battery would supply 30 amps for 1 hour.
Probably more like 40 - 50 minutes. They usually aren't designed for
long periods of high discharge and, indeed, you have to apply a
fudge factor which will vary according to the purpose the battery
was designed for.

You are on the right track when you mention Power (as opposed
to voltage (volts) or current (amps)).

The power (in watts) you are using is calculated by multiplying the
voltage-drop across the appliance by the current flowing through it.

Thus your 12V item drawing 1 amp is consuming 12 watts of power.

Your 240V item drawing 1 amp is consuming 240 watts of power.
Thus an inverter would have to supply 240 watts at 240 volts.

In theory, the 30AH battery would need to supply 240 watts of power to the inverter at 12 volts

Working the equation backwards, 240 watts divided by 12 volts =
20 amps. Thus the battery would be supplying 20 amps at 12volts
into the inverter.

Again, another fudge factor. No inverter is 100% efficient - probably
70% - 90% depending on the design (and how much $$ you spent!)

You would probably be drawing around 25 amps (using the thickest hookup cable practical to minimise losses) from the battery.

I would hazzard a guess that you would get around 30 - 40 minutes
from this set up. The best way to find out is to test it. Remember
though that regularly cycling a 30 AH battery at a 25A discharge rate
will quickly destroy it unless it is purpose designed for this.

Hope this is clear & happy flying! Err, I trust that this is a, ahemm, 4WD
plane! :-)


Rob
AnswerID: 17795

Reply By: nissan4x4 - Wednesday, Apr 16, 2003 at 20:03

Wednesday, Apr 16, 2003 at 20:03
Some simple laws to remember about electrical power (Watts).
1. Power out = Power in; Power in = Power out
2. AC power does not equal DC power.
ie: 240 Watts AC is not the same as 240 Watts DC. The equivalent DC watts is about 70-77% (I can't remember the actual value) of the AC value. DC power is usually refered to as "RMS". If you see something with an RMS rating, then this is the DC representation. The best example of this is speakers. You will see values like "100W pmpo" (some combination of p m o), which basically means peak to peak music power for a given length of time. If you applie the 70% rule this comes down to 70Watts. Where as other speakers have a rating 100Watts rms. This means that they are really loud.

But the most important rule that will concern you will be Power Out = Power In. Nothing is for nothing.

Cheers.
AnswerID: 17796

Follow Up By: GaryInOz (Vic) - Wednesday, Apr 16, 2003 at 22:22

Wednesday, Apr 16, 2003 at 22:22
A few major corrections: Point 2. 240W DC does the same amount of work as 240W AC. As you said in Point 1., power out = power in.

The DC "equivalent" of AC Is called RMS (("square)root mean squared"). It is a mathematical way of averaging the waveform (sinewave) to derive the DC eqivalent. For sinewaves it is (.7071 x peak-to-peak volts)/2, other waveforms can be worked out as well.

PMPO is peak music power output and has no relation to RMS power whatsoever. It is mainly a marketing ploy by manufacturers and up until recently WAS ILLEGAL TO USE FOR SPECIFICATIONS OF DOMESTIC PRODUCTS, and is calculated by the maximum peak voltage multiplied by the maximum peak current and is typically about 8 times the value of the true RMS output (+ or - a bit with some smart electronic "fudges"). No "time" is given as it is only a transient measurement, and therefore NOT truly representitive of the real power output. (Hence the reason for the ban). Look at quality audio equipment and it will always be quoted with real RMS figures, if it isn't, choose another brand because you are being HAD!.

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FollowupID: 11163

Reply By: Jarrod - Wednesday, Apr 16, 2003 at 23:53

Wednesday, Apr 16, 2003 at 23:53
Dolfn,
The easiest way, for the average person to figure out exactly what what you want is fairly simple. Get an amp meter, (or ammeter as somone will soon correct me) Make sure it is capable of measuring DC amps more than the level you believe the inverter will draw, i.e., inverter _Affordable_Storage_Drawers.aspx 30 amps, ammeter needs to be able to test up to say 40 amps. put the ammeter in series with the positive side of the power supply (battery) and measure the exact current being drawn. Then apply your simple formula which you have correctly demonstrated above!! An auto elec will have an ammeter capable of measuring many hundreds of amps, but i would imagine your average dick smith multimeter would be fused at about 20 amps. Make sure your ammeter is connected in series, not parallel, ( like when you test voltage) An ammeter across battery terminals is virtually a direct short. (not good!)

A couple of things to remember,

1. standard wet cell lead acid batteries ( unless it is a deep cycle) do not like being flattened below 11.0 to 11.5 volts. Do this too many times and you severly reduce the operating life of the battery,

2. power in definitely equals power out, however power out is a combination of electrical energy and in the case of an inverter HEAT. so power in to give 200watts out may be in the vicinity of 250 watts. The other 50 Watts ?? -- put your hand on the inverter - its been converted to heat. ( still power out, just in another form.)


Hope this helps.
Jarrod.
AnswerID: 17822

Reply By: Bob - Thursday, Apr 17, 2003 at 12:41

Thursday, Apr 17, 2003 at 12:41
For those interested in calculating battery current drain when using inverters here is the answer from the tech dept at Prosine Canada
"For a 12 Vdc system, take the output wattage of the inverter and divide it by ten to approximate the current that the inverter will draw from the battery.
For a 1200 W load, the inverter will draw approximately 120 Adc
from the battery."

Makes it real simple, don't even need a calculator.

Regards
Bob
AnswerID: 17834

Follow Up By: Member - dolfn - Thursday, Apr 17, 2003 at 13:48

Thursday, Apr 17, 2003 at 13:48
What if the inverter is 100W, but you are only running a 40W appliance off it?
cheers
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FollowupID: 11187

Follow Up By: Bob - Thursday, Apr 17, 2003 at 15:30

Thursday, Apr 17, 2003 at 15:30
Same sums mate. 40W appliance = 4 Adc
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FollowupID: 11192

Reply By: David N. - Thursday, Apr 17, 2003 at 13:18

Thursday, Apr 17, 2003 at 13:18
As an (ex) electrical engineer I could add a bit here- but there's no point as it's been pretty well covered by the others.
If you have any further questions feel free to ask and I (and the others I am sure!) will happily try to answer your questions- in a simple and understandable manner.
Cheers
AnswerID: 17841

Follow Up By: paul - Thursday, Apr 17, 2003 at 16:46

Thursday, Apr 17, 2003 at 16:46
David, i have 48W by three 240v lights (therefore i presume still 144W at 12v or 240v) hooked up to a 300W inverter attached to a battery. Is it therefore correct as above that i am drawing 14.4amps at 12v dc from the battery when in operation ?

I have the 50cc christie setup, if i attach this at low current switch to the battery at the same time it will draw from the christie alternator set up rather than the battery - this sound right, the battery being only the medium for the current draw ? This i presume would be far better for my battery than running straight off my battery - even though it is an optima yellow top i understand drawing 14amps per hour for 2 hours would not be healthy for any battery - this sound right ?

thanks
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FollowupID: 11195

Follow Up By: David N. - Thursday, Apr 17, 2003 at 21:28

Thursday, Apr 17, 2003 at 21:28
I assume you mean three 48watt incandescent lights..... as flouros are a slightly different kettle of fish...
Yes, basically you are on the right track here.
(If your inverter was 100% efficient you would draw 12 Amps- there is no such thing as a 100% efficient- so depending on your inverter's efficiency approx 14 Amps would be a good ball-park figure.)

If you are running an alternator setup at the same time as the lights- any excess charge capability from the alternator will charge your battery- then once fully charged (and assume the alternator regulator is working properly) the battery would remain fully charged while the alternator supplies the load.

Drawing 14 Amps for two hours equates to roughly 28 AmpHours- not a problem for any normal "deep cycle" battery. eg: If your battery was say 100AmpHour capacity, you could regularly take out 60-70 AmpHours and still get good results from your battery. (If you try to never discharge your "deepcycle" battery by more than 60 to 70% you will get good life from it - especially if you then get charge back in to it soon ie:next day, not next week. A "starting battery" (normal car battery) should be discharged FAR less if you want reasonable life from it.

Just remember with a lead acid battery that the more deeply you discharge it the less cycles you will get from it- a standby battery on float charge will often last way over ten years.
A true "deep cycle" battery such as those used by Telstra and home power setups are not of any use to some one as a "portable" -as weight (and cost) is prohibitive. The "deep cycle" battery you buy for your car or boat is really only a warmed over starting battery ie: more suited to reasonably deep discharge than a starting battery- but NOT a true deep cycle battery.
The three worst enemies of the average mobile battery (4WD/ car/boat/van etc)
1. deep discharge (which leads to sulphation)
2. vibration
3. heat
If you minimise all of the above you should get good life from your battery.
Cheers
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FollowupID: 11203

Follow Up By: Flash - Sunday, Apr 20, 2003 at 22:07

Sunday, Apr 20, 2003 at 22:07
Thanks Dave- helped me too
realy helpful thanks a lot!!!
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FollowupID: 11287

Reply By: Member - Terry- Saturday, Apr 19, 2003 at 18:08

Saturday, Apr 19, 2003 at 18:08
Gees Dolfn, ask a simple question and get totally befuddled. I will have to read these replies again
terry


AnswerID: 17932

Follow Up By: Flash - Sunday, Apr 20, 2003 at 21:20

Sunday, Apr 20, 2003 at 21:20
I thought the question was a Good one!!!!!
happy easter
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FollowupID: 11277

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