Roger Helman Writes:
> Perhaps I'm showing my ignorance, but doesn't the alternator produce
>the same number of amps per rev. So the more revs the more amps with a
>max 35 amps for the stock unit? So no matter what the amps being draw,
>the hp to drive the alternator remains the same? Or does the force (BHP)
>needed to spin the alternator increase with the amps being drawn? If the
>first case is true and there are enough amps to drive the Fan, then the
>running with a fan would NOT cause a decrease in available BHP from the
>engine.
Roger
In another post Iwrote about basic generator theory where the gernerator
(or alternator)
output is a voltage which naturally wants to be proportional to the engine
speed but the
voltage regulator compensates by changing the field voltage to maintain a
constant
voltage. Lets consider the generator/voltage regulator as a system with a
constant
voltage output.
That being the case the current delivered by the generator is strictly a
function of the
current requirements of the electrical system (ie the current needed to run
the lights,
the radio, the coil and, if applicable, the electric fan) and the battery
charging requiriments.
The battery is self regulating but that is a topic for another post.
In my previous post I talked about the voltage generated by moving a wire
through a
magnetic field but that was an idealized voltgage. The wires involved have
a resistance
and any current will cause voltage drops for which the voltage regulator
compensates by
raising the idealized voltage so that the Videal - Vdrop = Vout. The
regulator raises the
Videal by increasing the field voltage. This works well until the field
voltage needs to be
more than the battery voltage. This is where the current limits and it
limits at a lower
level at lower speeds. This is why a generator can charge a battery at
idle when the
lights are off but when the lights are on (and the fan, etc) the battery
will discharge.
An alternator has the same limitations but they are at a higher level and
can usually
keep up with typical LBC loads.
If the electrical system requires 20 amps then the load on the generator is
240 watts and
assuming an efficiency of 50% the load on the engine will be 480 watts or
0.74 HP. This
HP is constant over the speed range. That means the Ft-Lb load on the
engine at 1000 RPM
is 3.82 Ft-Lb of torque, twice the load at 2000 RPM which is 1.91 Ft-Lbs.
(If anyone wants to
see the math see me after class)
Hope that this wasn't to protracted. Just remember what Albert Einstein
said about the
conservation of energy. Energy can not be created or destroyed. It all
must be accounted
for somewhere.
Dave Massey, St. Louis MO
"That's not a leak, that's just Dave running off at the mouth again"
P.S. as I write this I am sitting in a Hotel in Red Stick, Louisiana
(that's Baton Rough to those of
you who Don't know French). When I come down for the Jazz Fest
I'll look you up so
you can show me what a car that has never seen snow looks like.
P.P.S. 1 HP = 540 Lb-Ft/sec 0.74 HP = 400 Lb-Ft/sec
1000 RPM = 16.67 RPS 16.67 RPS *2*PI = 104 radians per second
2000 RPM = 33.33 RPS 33.33 RPS *2*PI =208 radians per second
Lb-Ft = work Lb-Ft/sec = power (work per unit time)
Ft-Lb = torque torque * 2 * pi = work (angular to linear
conversion)
Therefore: torque * radians per second = power
and: torque = power/radians per second Now plug in the
numbers:
Torque = 400 Lb-ft/sec / 104 Rad/sec = 3.85 Ft-Lb
Torque = 400 Lb-ft/sec / 208 Rad/sec = 1.92 Ft-Lb
QED
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