Re: Mobile S-Rotor!

TomGee wrote: 
daestrom wrote:

"TomGee" <lvlus@xxxxxxxxxxx> wrote in message
news:1134257569.032605.78860@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
<snip>

Take it a bit further why don't you. 

SNIP

Are you still under the wrong impression that I claimed my device adds
no drag?


No.  And I never was.


Then why do your statements indicate that you are and have been?

But I do have the impression that you claim your
device will add drag that results in less fuel consumption when you
disconnect the alternator belt.


Yes, that's correct.  I'm saying that the engine will use less energy
to push the rotor through the slipstream than it uses to turn its
alternator.


Two completely different ways to develop torque, but the latter has
been completely overlooked in reference to its design and the result is
that the same principles that apply to the lift-types are also applied
to the drag-types, but wrongly, as I've said before.


Not overlooked.


Oh yes, overlooked.

You're just mis-informed about what has, or has not been
considered in designing vehicles.  Your device isn't already mentioned in
the literature because it is a less efficient way to drive an alternator
than a simple belt system.


No.  My device is mentioned in driving an alternator.  The S-rotor
works and is used by many to produce power from the wind -er,
slipstream.  But here again you show that you are under the wrong
impression that my device uses no simple belt system.


The belt system *I* was referring to was the conventional one driving an
alternator directly from the engine shaft.  I assume that your S-rotor
drives an alternator with no losses from rotor shaft to alternator shaft, to
give you the benefit of the doubt.  But it makes little difference in the
outcome.


I contend that "little difference", however, will look more attractive
to us in the long run.  I have designed a planetary gear system for my
device in case it is needed.  Gears are more efficient than belts,
aren't they?  Using a direct-drive alt. saves the space required for a
belt-driven one behind the rotor, but it adds some length to the
housing.

I hope to be
able to change its belt system to a direct-drive system, which is even
more efficient that the belt system it has now.


Nor does the literature mention driving
alternators on vehicles by using falling water or a friction wheel in
contact with the ground underneath the vehicle.  Many ideas are not
referenced in the literature simply because they are obviously
impractical
and need no further review.


What I'm reading here is your opinion that my device is obviously
impractical and needs no further review than yours, is that correct?


You can have it reviewed by anyone you like.


I know that - I don't feel restricted by your opinion at all.  I will
build it even if you could pass a law prohibiting any further research
based on your opinion of it.

You want to get power from
your S-rotor to drive an alternator and think that it will take less power
to push the vehicle against this additional drag, than it does to drive the
alternator with an engine-belt system.  Same amount of power delivered to
the alternator, just using an S-rotor instead of an engine belt.

Does that pretty much sum up your idea?  Oh, and you may replace a
belt-driven alternator with a direct-drive alternator at some point.


Yes, you got it.

Even if you completely ignore the exact method of developing torque, you
can
still analyze the 'system' from an energy in/out standpoint.

Your whole design is based on the premise that you can run an alternator
with a 'wind device'  using less engine fuel than an alternator driven
with
a conventional belt drive.  You seem to believe that the drag created by
your 'wind device' will not require as much power as a conventional
V-belt
drive.


No.  You have misinterpreted what you read.  My device will not require
as much energy from the vehicle engine to push it through the
slipstream as the vehicle alternator requires from the vehicle engine
to produce the electrical power needed by the vehicle.


So, you believe the S-rotor can develop the power needed to drive the
alternator, but add so little drag that the engine will not need to develop
as much power as it does with an engine-belt driven alternator.  Do you have
any evidence for this article of faith??


Yes.  The design is similar to a paddlewheel rotor type except in a
feature that is common to the lift-types where the winds strikes all
the blades simultaneously.  That combines the high-torque feature of a
drag-type blunt-surfaced paddlewheel with the multi-blade face area of
a lift-type propeller.  While the trend today is a maximum of three
blades for the lift-types, the S-rotor can have many more blades than
that to add up sufficient to produce the same or more torque at much
smaller sizes.

As far as impractical ideas go, can you see why it is better to use a
belt drive system on a car than to use a direct-drive system instead?
After all, direct-drive is more efficient that a belt 'n pulleys
system, no?


I'm not clear as to which belt drive you're talking about now.  The one you
might use from your S-rotor to drive an alternator, or the one currently
used on most engine-belt driven alternators?


What difference will that make?  My device can use belt-drive or
gearing.  I hope I can build an alternator, if I can't buy one, to
where I can drive it directly.  Hugh Piggiot's Dual Rotors Axial Flow
alternators seem ideal for my device.  I can minimize the weight and
add cooling fins to the rotors and perhaps get enough amps from them to
provide what's needed.

Anyway, the direct-drive is more efficient, but isn't always suitable.  For
example, to generate the needed voltage at idle speeds, the alternator
either needs a strong magnetic field, or a faster speed than the prime mover
shaft.  Historically, manufacturers have opted for a belt drive that allows
them to step-up the speed so the alternator spins fast enough to generate
useful voltage while the engine is at idle.


Yes, but not because it is more efficient that direct-drive or gearing.
 It is more easily applied to a vehicle engine using a belt-drive, so
the savings of gearing or direct drive do not exceed the costs of easy
adaptibility and maintenance of a belt-drive, thus we don't have the
more efficient direct-drive or geared-drives on vehicles today.

A redesigned alternator could produce the necessary voltage with a
lower/different speed.  Then a belt-drive to step-up the speed might not be
needed and direct drive could be used.


The amount of hp used by an ordinary alternator is not really relavent,
since your 'wind device' would have to supply the same amount of hp,
regardless of the exact value.


Certainly it's relevant, why else have you bothered to argue the point?
It's relevant to know how much hp is used by the alt. from the engine
so as to determine the amount, if any, of hp saved when disconnecting
the alternator from the veh. engine.  If my device saves on fuel usage
or not, how can we know that without knowing how much fuel the veh.
engine uses to turn its alternator?


The point is that any hp savings incurred by disconnecting the alternator,
will have to be generated by the S-rotor.  So if the alternator draws 2 hp,
the S-rotor will have to generate 2 hp to replace it.  If the alternator
draws 14 hp, the S-rotor will have to generate 14 hp to replace it.

And an S-rotor that generates 14 hp will incur more drag for the vehicle
than one that generates 2 hp.  So if you 'save' more hp when you disconnect
a larger alternator, you will incur more drag from the larger S-rotor needed
to replace said hp.


So you're saying that the amounts must be equal and so no savings of
fuel or emissions will occur, correct?  You're saying that the veh.
engine will use the same amount of fuel to push the rotor to get say 14
hp from it as it does to get that from its alternator?  You could be
right, but I hope not.  My calculations gave me 18.25 Newtons, but I
was advised the drag coefficient figure I used was too high.  The
helper estimated roughly a Cd of 0.3 at 60 mph and got 33 N, which is
just over 1 hp.  The lowest hp estimate I have found of the hp required
by a car alternator is  4 hp.

My device is open halfway its height along its length, where the rotors
turn in reaction to the momentum of the slipstream.  Thus, since my
device does not present a solid bluff surface to the slipstream, it
should use less than the estimated 1 hp, no?

You haven't proposed a new, more efficient
way of converting mechanical power to electrical (whether the mechanical
power comes from engine shaft, or S-rotor shaft doesn't change the
alternator).


Yes, I have.  My improvements to the S-rotor make it much more
efficient than the generic model.  Enclosing it allows the slipstream
to be funneled into it and compressed to increase its momentum.
Downsizing it allows for the combining of several rotors for smoother
and more powerful applications, and makes it possible to be used in
mobile applications.  Specially designed blades take advantage of the
lift-effect and more efficiently impact the second blades.


No, read what I said.  "You haven't proposed a new, more efficeint way of
converting mechanical power to electrical."  That means you're still using
the same alternator technology.  Not the S-rotor part, the alternator.  So
if the alternator draws 4 hp from the engine, it will need to draw 4 hp from
your S-rotor shaft.


So, *if* a particular, large alternator requires 27.3435222 hp from the
belt
drive off an engine to drive it, your S-rotor would have to develop the
same
27.3435222 hp to drive it.  How big would your 'wind device' need to be
to
develop 27.3435222 hp??  How much increased drag would such a device
induce
on the vehicle? How much more hp would be needed from the main engine to
push this extra drag through still air? (I'll give you a hint, it's more
than 27.3435222 hp).


Not according to the figures I provide above.  It will take less than 1
hp to push it through the slipstream at 60 mph.  The blades will not
increase that figure no matter how much current the alt. has to produce
because they are resisting the air less than a stationary wall does.

So according to you, it will take more hp to push my device through the
slipstream than it takes for the veh. engine to turn its own
alternator, correct?


Yep.  Because your S-rotor device has to develop the same hp that the engine
supplied to the engine-belt driven alternator, and to develop that much hp,
the engine will have to expend more hp.


Well, show your calculations about that and we can compare them with
those I've given you.

Energy has to be expended to push the vehicle through the air, to
push
the
turbine through the air.  As the turbine is less than 100%
efficient,
it
will always take more HP to push the turbine through the air, than
can
be
extracted from the turbine.



That's fine so long as the hp required is still less than the
vehicle
engines requires to turn it own alternator.  My calculations show
that
it takes about 1 to 2 hp for the vehicle engine to push my rotor box
than it takes for it to turn its own alternator, which is from 4 to
30
hp.


But if your box only adds 1 to 2 hp to the vehicle engine demand, it
cannot
develop more than 1/2 to 1 hp output.  So it couldn't drive even a 100
amp
alternator.  Can't get out more than you put in.


I cannot figure out what you mean by your statement above.  Are you
saying that my device can only get 1/2 hp of what it takes the veh.
engine to push it?


Yep.

How do you figure that? 

With some math ;-)  The Betz limit guarantees you won't get more than 59% of
the power that it takes to push your device through the air.  So if it takes
1 extra hp to push the added drag, then that amount of added drag cannot
possibly generate more than 0.59 hp.


You are confusing two
distinct and non-equivalent principles as being applicable to the same
situation.


Really?  Which two?


The Betz equations and the first law of thermaldynamics.

Your device would convert engine power in the form of drag, to
mechanical
power at no better than the Betz limit (and probably a lot less), so
you're
already down to 59% of the power from the engine.  A belt driving an
ordinary alternator is much higher than that.



The Betz limit does not apply to drag-type rotors, only to the
lift-types, because it has to do with the area swept by the blades
which is used to provide the lift to turn the blades.


Actually, the Betz limit *does* apply.


No, it does not. 

Does to...
Does not...
Does to...

Rather tiresome.


Yes, it is, but you started it by not providing support for your
opinion and ignoring the detailed reasons I provided to support my
opinion.

You don't seem to really understand the derivation of the
Betz limit.  It is *not* a function of airfoil, or 'bluff blades', or 'swept
area' or whatever.  It is derived from the conservation of
mass/energy/momentum in a slipstream where the pressure/elevation/density
some distance upstream of the device in question and
pressure/elevation/density some distance downstream of the device are the
same.  Doesn't have to be right *at* the device, just some place
upstream/downstream.


A function is a variable quantity whose value depends upon those of
other quantities.  One of the variable quantities depended upon by the
Betz equations is the area swept by the blades of a lift-type rotor.
The Betz limit is a derivative of the function of available energy
passing through an area swept by airfoiled blades.  The blades can only
use a certain amount of energy from all of the wind passing through the
swept area because some of it passes through unused.  The only way for
the total energy in the wind to be harnessed is explained by the 3rd
law of motion which explains the conservation of momentum when objects
collide.

There are no collisions as such in a lift-type blade because it does
not use momentum as it's motive power but uses instead the phenomenom
of the lift effect.

If you look at the fluid flow in/out
of *any* device with the same pressure/elevation/density between the
inlet
and outlet, the power calculations still yield no better than the Betz
limit.


So how do you figure that about my device? 

Because the Betz limit applies to *any* device in a fluid stream with inlet
pressure/elevation/density are the same as outlet
pressure/elevation/density.


So what you're saying is that the Betz limit applies to any device
because the power calculation still yield no better than the Betz
limit.  See any circular logic in that?

I don't have to know anything about the
internal workings of the device to know that extracting more energy from
each pound of air will slow down the flow rate of the fluid resulting in a
power drop, or that allowing the fluid to flow through the device faster
means less energy is extracted from each pound of fluid and the power will
again drop.


You can only get above 59% power extraction if you have a difference
in pressure/elevation/density between inlet/outlet.


Okay, let's take the inlet of an airfoiled wind propeller as its swept
area and its outlet as the area behind the blades.  The
behind-the-blades area has a lower pressure as the result of having
passed by the blades.  By your argument, the Betz limit doesn't exist!


Nope.  One simply draws the 'system' surface somewhat upstream of the
device, and downstream of the device some distance to where the
pressure/elevation/density are the same again.  The inlet area is actually
smaller than the outlet (has to be for the mass flow rate to be
'conserved').  As the air moves from my invisible 'inlet' face towards the
device, the flow area widens and the pressure rises.  After it passes
through the device, the pressure drops (the dp across the device does work).
As the flow continues on past the device, the flow area widens some more,
and the pressure rises back to atmospheric at the invisible 'outlet' face.

When the system includes the space in front of, and behind the device
extending out to the invisible 'faces', the pressure/elevation/density at
the inlet and outlet faces are equal, Betz applies.


No, all that's technobabble related to something other than the Betz
limit.

In fact, the derivation
of Betz doesn't even care about the nature of the device used.


The derivation of it doesn't, but that's not what we're talking about.
We're not talking about ratios, we're talking about appropriate
applications of concepts.

Betz wrote it wrt airfoiled propellers and their relationship to the
slipstream area through which they sweep.  If he had been asked if it
applied to drag-type rotors I'm sure he would have answered no.


Have you any evidence for this article of faith?  The Betz limit applies to
all 'turbo-machinery' which includes S-rotors, Savanious (sp), paddle
wheels, and axial flow wind turbines.


Sure, the reference to "swept area" is what assigns it specifically to
rotors having a swept area, which can only refer to a rotor having
airfoiled blades spaced apart through which the wind passes where the
blades are set at angles to the wind and not perpendicular to it.

It could be
any sort of 'black box' device with any manner of mechanism inside it and
the derivation is still the same.  It just happens that wind turbines is
the
most common application/reference on the internet of the limit.


It just happens that people have little understanding of the scope of
the Betz limit.


Well, there we agree.  Your refusal to accept that it applies to
non-axial-flow wind devices is proof of that.   But I suspect your meaning
was to imply that I am the one with little understanding of its scope.


A paddlewheel in an open flowing stream, with no change in head is
limited
in the same way (no head difference means no pressure change, and the
density of water is unchanged).  Only when a dam is built to add a
difference in head can more power be extracted.


Assuming by head you mean elevation of the source, what that has to do
with wind energy is unknown to me.


If a fluid flows up or down hill (against/with gravity), then that
complicates the issue.  Granted 'wind' very seldom does this, but it is part
of the Betz limit derivation so I felt it needed to be included for
completeness.


How can a paddlewheel have no
change in head as it moves in and out of the water?


The head I'm talking about is the elevation of the fluid, not the parts of
the device.


For the paddlewheel, the elevation of the fluid changes as it moves up
and down through the water as it rotates.  A paddlewheel will have some
paddles submerged part of the time as it turns.  The head for the
paddlewheel constantly varies up and down.

The pressure must
change for it as its elevation in the stream goes up and down.


Again, the pressure of the fluid (water in this case) is what isn't
changing, not the pressure exerted on the device.


Yes, but you refer to the device being affected by the water pressure
on it, as on a dam, and that means you refer to the pressure exerted on
the device.

The
elevation of a dam is incidental to the principle of the force of water
turning rotors.  It is the elevation of the water, among other factors,
that determines the force of water and how much power can be gained
from it.


And hence since the water is not at the same elevation at the inlet as at
the outlet, the Betz limit does not apply to overshot water wheels, or water
turbines fed from the penstock of a dam.


Wrong "hence".  It does not apply because it cannot apply to
blunt-surfaced rotors that use the 3rd law of motion as their motive
principle.

Your 'wind device' doesn't have a pressure difference
upstream/downstream,
and the air is not significantly change in density, so the Betz limit
still
applies.



Wrong again.  The air exiting the device has lost some of its energy in
turning rotor, thus it has lower pressure coming out than it had going
in, so the Betz limit does not apply to it, by your own words.


Nope.


Yep.

Yes, the air exiting the device has lost some energy, but that doesn't mean
it remains at some lower pressure forever.


Now you're adding on to your qualifications as to what your
understanding of the Betz equations are all about.  You can go on
forever that way, apparently, but it's still as illogical the farther
you go as it is at the beginning.

If it were at lower pressure,
how does it 'leave' the area behind the device?
Answer: It's remaining
kinetic energy is partially converted to pressure as it slows flowing
through a wider and wider flow field.  Once the pressure in this air is
equal to that of the downwind pressure of the flow stream, then it can flow
away from the area behind the device, making room for more air to flow
through the device.


That's totally silly.  The correct answer is that it is pushed out by
the incoming wind.

This is why it is important that the air leaving the
device still have some kinetic energy, so it can be converted to pressure
and restore the air pressure to the same level as the surroundings.


You know, all that would be hilarious if I did not think you were being
serious.

And
leaving some kinetic energy in the air means you're not getting all the
power out of the wind, hence the Betz limit.


But then the Betz limit does not apply when you get all the power out
of the wind or water by using a paddlewheel or an S-rotor where the
momentum of all the wind or water pushes against the blades.

 > The energy it lost was kinetic energy, so that means it is moving
slower.

And if it's moving slower, for mass flow rate to be conserved, the flow area
must widen.


But the mass flow rate is already slowed, and I see no reason why it
must be conserved.

Looking at imaginary flowlines through/past the device, they
diverge away from each other behind the device as the slower air needs a
wider area to carry away the same mass as that which flowed into the device.


But the slower air is the same mass that flowed into the device, it is
only its energy that has changed and that's why it has slowed.  Seems
to me like you're making up your own physics here like you claim I'm
doing.  You have different ideas about a number of things but most of
them appear to be wrong ideas.


my god, you are stubborn.

You are proposing a PPM.

If your device actually worked, every car would have a sufficient number mounted such that no gas was actually used.

j.
.

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