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 the 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 few 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 Radio 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 would save enough to keep the engine running
4 gallons/hr = 512 oz (approx) /hr
, So 6 oz = 6/512 *60 minutes = 0.703 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 making 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 power to it, in which case you may not save any
fuel (over this short time period) 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
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)
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?
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