Date:         Wed, 20 Oct 2004 12:09:01 -0700
Reply-To:     Rocket J Squirrel <j.michael.elliott@ADELPHIA.NET>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Rocket J Squirrel <j.michael.elliott@ADELPHIA.NET>
Subject:      Re: Inverter usage (was "Dual Battery Installation Instructions
              on Ebay")
In-Reply-To:  <20041020180610.5690.qmail@web13602.mail.yahoo.com>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed

Reinhard: good job explaining the why trying to run a microwave oven off
a battery/inverter setup is a bad idea. Let's hope that others
contemplating such a setup will dip into the archives before asking again.

One note for those not electrically savvy: Reinhard writes that
inverters draw current even when not powering anything. The figure he
mentions for a 500W unit is 500mA -- that's 500 milliamps, or 1/2 an
amp. About 6 watts on a 12V system. So yeah -- turn the thing off when
you're not using it. I use mine to grind coffee beans and run the laptop
once in a while. Rest of the time it is off. Using one to power a
microwave is just silly.

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Here's a simplified way to compute approximate battery current draw for
an appliance -- it does not account for battery voltage sag, resistive
losses in your wires,  and other details, but will get you close:

Look at your 120V product and find the power consumption. Let's pick a
100W light bulb as an example. Products with motors (my coffee bean
grinder) can have peak draws three time higher.

Assume the inverter is 60 to 90% efficient. Let's pick 75%. So total
power needs are 100W for the bulb and 33W for the inverter = 133W.  That
33W share for the inverter just ends up as wasted power:  the inverter's
internal cooling fan just blows it out as warm air.

Battery Current (amps) = power / voltage.

Let's just say voltage is 13 volts, even though we have "12V" systems.

133W / 13V = 10 amps.

Leave that 100 watt bulb on for one hour and you have consumed 10
ampere-hours.

Products designed specifically for 12v operation are going to be more
efficient than trying to run a 120v product off an inverter. If you have
the choice of getting a 12V Dustbuster or a 120V one, get the 12V one.
At least you won't be losing power running the inverter.

And when it comes to heating anything, use propane. It's just so much
more efficient.

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Jeepers -- I rambled even more than Reinhard!

--
Mike "Rocket J Squirrel" Elliott
71 Type 2: the Wonderbus
84 Westphalia: "Mellow Yellow (The Electrical Banana)"
KG6RCR



Reinhard Vehring wrote:

>Jeffrey,
>not an easy question, because efficiency depends on
>the type of inverter and the system setup.
>Generally the inverter efficiency is around 90 % at
>full load, maybe 95% for more expensive models and
>larger inverters. The efficiency drops for smaller
>loads. To find out how much, you need an efficiency
>graph for your particular model. They are ususally not
>easily avaliable for the cheap models (for a reason).
>A way to estimate how well the inverter works at low
>loads is to look at the "no load current draw" For a
>500 W inverter it maybe around 500 mA.(So don't leave
>it switched on if you don't need it) Performance at no
>load and low load can be quite good, but you'll have
>to pay for it is my guess.
>I've seen an efficiency graph for a 400 W inverter
>that had 75 % at 50 W load and 60 % at 25 W load.
>
>Now overall efficiency is an entirely different
>matter: In the case of a microwave load this is quite
>interesting. There are basically two types of
>inverters around: True sine wave inverters ($$$) and
>"modified" sine wave inverters. The latter control the
>RMS output by changing the width of the rectangular
>pulses they produce. Most loads run fine on this type
>of waveform but it affects efficiency. Microwave
>cooking power depends on peak voltage not RMS voltage,
>so you only get the full cooking power if the inverter
>produces reasonably high peak voltage. It only does
>that if you feed it at about 14 to 14.5 V. Unless you
>charge the battery by running the engine when you
>cook, this is  not going to happen. In fact drawing a
>large current draw from a small battery will drop the
>voltage to maybe around 11V or less. At that DC
>voltage the peak output voltage of the inverter drops
>and the efficiency of the microwave is lousy.
>Basically you need to run it longer to get your stuff
>hot. Now the overall efficiency of your system maybe
>only 40 to 50 %. To get it up you'd need a larger
>battery that tolerates the current draw. This is why I
>don't like the cheap small battery - cheap inverter
>solution.
>BTW: The gas burner is the way to go: Just one energy
>conversion:fuel - heat. Microwave in a car needs 4
>conversions: fuel - mechanical (alternator) -
>electrical DC - electrical AC - heat.
>Hope this helps, sorry about the rambling
>Reinhard
>--- Jeffrey Earl <jefferrata@yahoo.com> wrote:
>
>
>
>>Reinhard,
>>
>>This brings up a question I've had since recently
>>installing 400W inverter on my aux. battery: is the
>>power consumption largely dependent on the total
>>amperage draw of the devices connected thru the
>>inverter? Or does the inverter itself draw enough to
>>soon drain the aux. battery?
>>
>>Appreciate any advice you can offer.
>>
>>Jeffrey Earl
>>1983 diesel Westfalia "Vanasazi"
>>http://www.vanthology.com/
>>
>>
>>
>>
>>
>>
>>
>>
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