Ben, there is no question that anything consuming (using) electrical power
is going to require that power to be produced (or stored) somewhere,
sometime, somehow.  Whether you pull it from alternator or battery.  

 Since your objective (in this scenario) is to conserve fuel (rather than
battery power),  the real question is how much fuel you will save vs. the
utility of some electrical component you might decide to turn off.  For
instance, you might save fuel by turning off your GPS and Radio - but if

GPS is crucial to you taking the shortest, surest route to the nearest
filling station - then keeping it ON is perhaps worth much more than the

oz of fuel you might (I repeat might) save.

 

Lets look at some numbers.

 

At 12VDC 1 amp of current draw = 12 watts of power or much less than the
ordinary 60 watt household light bulb, but probably enough for GPS and

receiver.

 

12 watts of power (assuming 100% efficiency which you, of course, will NOT
get) would require  0.01609227 HP.  Lets round it up and say 0.0170 HP

 

A frequently used power equation for the rotary is HP = lbm fuel/0.55.  So
reworking the equation slightly  we get Fuel = HP*0.55  = 0.0170*0.55 =
0.00935 lb of fuel/min

 

So lets say you need to fly 30 minutes to get to the nearest airport.
0.00935*30 = 0.2804 lb fuel required to power the 12 watt component for 30
minutes.

 

Since there is approx 6 lbs of fuel per gallon, you would use 0.2805 / 6 =
0.0467 gallons of fuel or approx 6 oz of fuel (three jigger size shot
classes worth - you might need that after this flight {:>)).

 

At a throttled back cruise rate of  4 gallon per hour of fuel burn (if you
can lean it back that far and stay airborne and get good mileage) you

save enough to keep the engine running   

4 gallons/hr =  512 oz (approx) /hr   , So 6 oz = 6/512 *60 minutes =

minute or  42 seconds longer of flight (very approximate)

 

 

So the bottom line is the 42 seconds of flight more valuable than having
your GPS and radio - it could be if that made the difference between

the airport or not (but you don't know that bit of important information),
but then if you lose the most direct route to your airfield because you
turned off your GPS, you could easily use much more fuel just doing a 360
turn or two.

 

On the other hand, if your battery was fully charged  then you might not
draw the battery down enough to trigger the voltage regulator to send

to it, in which case you may not save any fuel (over this short time

by giving up your 12 watt GPS /Radio.

 

 

Hypothetical scenarios are fun to discuss and give you a ball park
quantification of factors that might turn out to be useful in decision
making (should you every face the situation), but my view is it is seldom
you find yourself in a real world situation that other factors don't screw
up the decision that the hypothetical scenario would call for {:>)

 

 

Just my 0.02 worth

 

Ed

 

Ed Anderson

Rv-6A N494BW Rotary Powered

Matthews, NC

eanderson@carolina.rr.com

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http://www.rotaryaviation.com/Rotorhead%20Truth.htm
<http://www.dmack.net/mazda/index.html>   _____  From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On
Behalf Of Ben Baltrusaitis
Sent: Thursday, March 19, 2009 9:09 AM
To: Rotary motors in aircraft
Subject: [FlyRotary] Alternator (Off topic)

 

Since it's quiet:

 

When I was a kid a guy at the parts store demonstrated to my Dad that when
electrical power was needed, a generator put a load on the engine. After
that, my Dad was careful not to run lights, radio, heater fan, or other
non-essentials when he was trying to get good gas mileage.

 

I have continued that tradition, however, I have seen it stated that
electrical draw on an alternator doesn't increase the mechanical load.

 

When low on fuel will it help to turn off electrical components not needed
for flight?

 

Is it true of an alternator; an electrical power demand doesn't cause an
increased mechanical load?

 

Or, does keeping headlights on during the day decrease gas mileage?

 

Thanks!

Ben



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