Hello Frank,

The concept of adjusting the output voltage of an energy source by
controlling the loading applied to it by a high efficiency power converter
is the way Maximum Power Point Tracking (MPPT) works. This has been done on
PV arrays in non-terrestrial applications for decades! The approach will
also work in wind and hydro applications utilizing permanent magnet
alternators. The circuit simply adjusts the load voltage the alternator
"sees" until maximum power transfer occurs for that particular operating
condition. This maximum power transfer optimizes the efficiency of energy
flowing from the source to the load. The goal would be to design the
alternator to optimally load the rotor (over the wind speed range of
interest) and the load interface circuit (MPPT) to optimally load the
alternator. A byproduct of this overall effect keeps the rotor RPM and tip
speed ratio controlled where maximum efficiency occurs throughout the entire
system, from wind input to power output.

A very early example of an approximation of this technique was utilized in
the original Jacobs direct drive turbines. They had a flyweight governor
that changed the blade pitch above governing wind speed, yielding a flat
power output curve. The variable blade pitch governor became active at only
18 mph and provided a situation where the rotor optimally loaded the
alternator! The field current in this DC generator (before high current
rectifiers) was proportional to output current. This is a form of
regenerative positive feedback where near maximum power transfer occurs
without increasing rotor speed. To me, this is analogous to the current day
MPPT! Maybe someday MPPT, in conjunction with a diversion load controller,
will become the standard for the small turbine to battery interface:-)

Brent Peterson

----- Original Message -----
From: Frank R. Leslie <fleslie@...>
To: Tom Gray <tomgray@...>; <awea-wind-home@egroups.com>
Sent: Friday, July 16, 1999 3:24 PM
Subject: [a-w-h] Re: Another from Eric


>A neat analogy! For a given rotor size and generator or alternator, we
could
>measure the output power in watts, increase the load, and find the maximum
>power transfer. This would probably result in a nonstandard voltage, but
>that's where the power of dc-dc converters comes in. Convert to the voltage
>needed to drive the battery charger circuit, or even better, combine the
>electronics.
>
>Frank
>
>
>-
>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=--=-|
>| Frank R. Leslie | Pers. email: mailto:fleslie@... |
>| 1017 Glenham Drive, NE | Prof. email: mailto:f.leslie@... |
>| Palm Bay FL 32905-4855 | Home: 407-768-6629 | KD4EYQ | 990521 |
>| Work: retired | Fax: email only please | 28-01.3130N / 80-35.6136W |
>| --- |
>
>-----Original Message-----
>From: Tom Gray [mailto:tomgray@...]
>Sent: Friday, July 16, 1999 10:36 AM
>To: awea-wind-home@egroups.com
>Subject: [a-w-h] Another from Eric
>
>
>I should have omitted the calculus and described Betz limit by analogy:
>
>If you had a company making Widgets (for example), what price should you
>sell each Widget for? If you sold them for a low price with a very small
>profit
>on each Widget, you'll need to sell lots of them. (ie you don't slow the
>wind
>much) On the other hand, if you make a huge profit on each Widget, you
>wouldn't need to sell very many -- and nobody would buy them. (ie slowing
>the wind to a stop) There exists some combination of price and the
>associated
>quantity of Widgets sold that yields the maximum possible profit.
>
>If you knew the equation that described the connection between the price
>you charged, and the number of Widgets sold at that price; then you can
>solve for the right combination. Yes, you can use calculus and the answer
>falls right out. But you can also solve it by "guessing" at possible
>solution
>and then looking at the results.
>
>Using the equation for efficiency as a function of wind speed reduction:
>
>efficiency = [(V1 + V2)*(V1^2 - V2^2)]/(2*V1^3)
>
>V1 = Initial velocity
>V2 = Final velocity
>
>You can choose V1 = 1 and many possible fractions for V2, calculate the
>efficiency, and keep doing it until efficiency doesn't go any higher. The
>answer happens to be V2 = V1/3 with an efficiency of .593.
>
>The answer has nothing to do with the aerodynamics of the turbine, or even
>the
>fluid used in the flow! You could just as well speak of a rotor submerged
>in a river.
>
>Regards,
>
>Eric



------------------------------------------------------------------------
You have received this message because you are a subscriber to the American Wind
Energy Association's Home Energy Systems list. To view previous messages from
the list, subscribe to a daily digest of the list, or stop receiving the list by
e-mail (and read it on the Web), go to
<http://www.egroups.com/list/awea-wind-home>. To unsubscribe from the list,
send a blank message to <awea-wind-home-unsubscribe@egroups.com>