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Author Topic: Kundel Motor  (Read 22571 times)

gn0stik

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Re: Kundel Motor
« Reply #15 on: June 03, 2006, 07:49:49 PM »
Well if you read closely you can do some research, we can get a horsepower rating from rpms and torque. Horsepower can then be converted to other output types. I'll look into it. Of course, any results I get will be speculative.

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Re: Kundel Motor
« Reply #15 on: June 03, 2006, 07:49:49 PM »

gn0stik

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Re: Kundel Motor
« Reply #16 on: June 04, 2006, 03:37:55 AM »


Thanks for the correction jake.
« Last Edit: June 05, 2006, 03:59:37 AM by gn0stik »

jake

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Re: Kundel Motor
« Reply #17 on: June 04, 2006, 03:47:55 AM »
A 30w fan motor means 30 watts, right?

30/746 = 0.040 hp

0.040 hp @ 1800 rpm = 0.117 ft*lb or 1.4 in*lb.


 

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Re: Kundel Motor
« Reply #17 on: June 04, 2006, 03:47:55 AM »
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Offline Liberty

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Re: Kundel Motor
« Reply #18 on: June 04, 2006, 04:15:47 AM »
A 30w fan motor means 30 watts, right?

30/746 = 0.040 hp

0.040 hp @ 1800 rpm = 0.117 ft*lb or 1.4 in*lb.


 

On Stephen Kundel's web site, the highest voltage rating in AC volts is 4.72 volts and the highest current in AC is .91A.  The power calculation is P=IXE.  AC power is about 4.3 watts. 

Would this motor have the capability to self run with a 30 watt output?

Offline lancaIV

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Re: Kundel Motor
« Reply #19 on: June 04, 2006, 04:58:51 AM »
On kundel?s web-site I reed about the use of a 60W fan blade !

S
  dL

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Re: Kundel Motor
« Reply #19 on: June 04, 2006, 04:58:51 AM »
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gn0stik

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Re: Kundel Motor
« Reply #20 on: June 04, 2006, 06:57:27 AM »
A 30w fan motor means 30 watts, right?

30/746 = 0.040 hp

0.040 hp @ 1800 rpm = 0.117 ft*lb or 1.4 in*lb.


 

Yeah, that's what I thought too, but.... I checked to be sure, and that's what I found. HP*746 is supposed to equal Watts. So I figured I was missing something or remembering it wrong. Apparently not.

However, your math is wrong too. 1200-1500 rpm was what the motors use so to figure torque at 1800 (the conservative figure for kundel's motor would be over optimistic for figuring torque for HIS motor, since he's comparing his to theirs) we need to figure torque from the sample motor rpms, not his.


0.040 hp @ 1200rpm = 0.175 ft. lb
0.040 hp @ 1500rpm = 0.140 ft. lb

So we apply those torque figures to his motor at 1800(his) rpm.

(1800 rpm * 0.175)/5252 = 0.059hp         0.059*746 = 44.014 Watts output
(1800 rpm * 0.140)/5252 = 0.048hp         0.048*746 = 35.80  Watts output

(2500 rpm * 0.175)/5252 = 0.083hp         0.083*746 = 61.91 Watts output - hmm lancaIV? blade doesn't mean much though.
(2500 rpm * 0.140)/5252 = 0.067hp         0.067*746 = 49.98 Watts output

now remember input was in the 3-5 Watt range.

So before conversion to electricity we have the potential to have a minimum of around 700% efficiency, and a max of about 1500%(rough guesstimate) Pretty huge range.

Of course conversion to electricity from hp will not be 100% efficient, so I don't wan't to imply it's actually going to out put that. That's just it's gross efficiency, so to speak. I think the actual numbers would probably be in the 40-60% conversion efficiency range, but I'm just guessing.

So Using those very non-scientific guesses, we're looking at anywhere from 17(lowest output *.4 for 40% conversion efficiency) Watts to 30 (highest * .6 for 60% conversion efficiency)  Still pretty friggin impressive. 300 - 750% approximate efficiency after conversion losses, assuming conventional methods of conversion (EM).

Sorry for the mistake before... Please... correct these numbers too if I'm wrong.





« Last Edit: June 04, 2006, 07:43:46 AM by gn0stik »

Offline hartiberlin

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Re: Kundel Motor
« Reply #21 on: June 04, 2006, 05:02:15 PM »
Well, I think this motor MUST be measured with analog volt and amperemeters,
not with digital as there are fast pulses that can?t be measured by
this digital multimeter.
Also the torque is not very big and if it draws around 4 Watts of input power
the fan output power is probably just under 2 Watts.
Normal 30 Watts fan motors are also very inefficient and
as long as he does not measure the torque with a prony break under
a load, I guess this motor is pretty unefficient. It has very big magnet
airgaps and thus electrical to mechanical power conversion is low..

That the motor does not increase in input power under a load
just shows this inefficiency and it is just impedance matching in this
case and I want to see analog meter readings.

But the site is very nice made and also the motors and
the concept is clear understandable, that is very nice.
Also it might be usefull to use copper-graphite brushes in series
to drive the coils from a 12 Volts lead accid battery, this way the battery will
be reacharged due to the Newman sparking commutator effect and you could light up in
series an indescant lamp from the RF burst pulses...
and finally see, if more magnets on the motor need also more
or less input power..
We will see, at least this motor makes quite some noise
and might also have vibrational problems, so I don?t know how
long such a motor would last and might need much more
maintainance than a standard motor....
but interesting concept.

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Re: Kundel Motor
« Reply #21 on: June 04, 2006, 05:02:15 PM »
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Offline Liberty

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Re: Kundel Motor
« Reply #22 on: June 04, 2006, 05:13:36 PM »
A 30w fan motor means 30 watts, right?

30/746 = 0.040 hp

0.040 hp @ 1800 rpm = 0.117 ft*lb or 1.4 in*lb.


 

Yeah, that's what I thought too, but.... I checked to be sure, and that's what I found. HP*746 is supposed to equal Watts. So I figured I was missing something or remembering it wrong. Apparently not.

However, your math is wrong too. 1200-1500 rpm was what the motors use so to figure torque at 1800 (the conservative figure for kundel's motor would be over optimistic for figuring torque for HIS motor, since he's comparing his to theirs) we need to figure torque from the sample motor rpms, not his.


0.040 hp @ 1200rpm = 0.175 ft. lb
0.040 hp @ 1500rpm = 0.140 ft. lb

So we apply those torque figures to his motor at 1800(his) rpm.

(1800 rpm * 0.175)/5252 = 0.059hp         0.059*746 = 44.014 Watts output
(1800 rpm * 0.140)/5252 = 0.048hp         0.048*746 = 35.80  Watts output

(2500 rpm * 0.175)/5252 = 0.083hp         0.083*746 = 61.91 Watts output - hmm lancaIV? blade doesn't mean much though.
(2500 rpm * 0.140)/5252 = 0.067hp         0.067*746 = 49.98 Watts output

now remember input was in the 3-5 Watt range.

So before conversion to electricity we have the potential to have a minimum of around 700% efficiency, and a max of about 1500%(rough guesstimate) Pretty huge range.

Of course conversion to electricity from hp will not be 100% efficient, so I don't wan't to imply it's actually going to out put that. That's just it's gross efficiency, so to speak. I think the actual numbers would probably be in the 40-60% conversion efficiency range, but I'm just guessing.

So Using those very non-scientific guesses, we're looking at anywhere from 17(lowest output *.4 for 40% conversion efficiency) Watts to 30 (highest * .6 for 60% conversion efficiency)  Still pretty friggin impressive. 300 - 750% approximate efficiency after conversion losses, assuming conventional methods of conversion (EM).

Sorry for the mistake before... Please... correct these numbers too if I'm wrong.







These are all very good approximations, but to really find out under real operating conditions of the motor itself, a small alternator/generator should be attached to determine what power output can actually be obtained.  Then we will know without a doubt whether it can self power or not.  The numbers look very encouraging with such a wide window for the probability of more power out than in. 

However, the torque curve is unknown on this motor. (I suspect very low torque at high RPM range, and higher torque at lower RPM range, because of the way the motor slowed down when fan load was applied).  This could affect generator/alternator power production, that the numbers may not reveal to us in our calculations.  It will for this reason, be important to carefully match the alternator/generator used for test, to this motors' RPM and torque curve for a test that makes the most of this motor's capability, to be a fair test.

-----------

Well, I think this motor MUST be measured with analog volt and amperemeters,
not with digital as there are fast pulses that can?t be measured by
this digital multimeter.
Also the torque is not very big and if it draws around 4 Watts of input power
the fan output power is probably just under 2 Watts.
Normal 30 Watts fan motors are also very inefficient and
as long as he does not measure the torque with a prony break under
a load, I guess this motor is pretty unefficient. It has very big magnet
airgaps and thus electrical to mechanical power conversion is low..

That the motor does not increase in input power under a load
just shows this inefficiency and it is just impedance matching in this
case and I want to see analog meter readings.

But the site is very nice made and also the motors and
the concept is clear understandable, that is very nice.
Also it might be usefull to use copper-graphite brushes in series
to drive the coils from a 12 Volts lead accid battery, this way the battery will
be reacharged due to the Newman sparking commutator effect and you could light up in
series an indescant lamp from the RF burst pulses...
and finally see, if more magnets on the motor need also more
or less input power..
We will see, at least this motor makes quite some noise
and might also have vibrational problems, so I don?t know how
long such a motor would last and might need much more
maintainance than a standard motor....
but interesting concept.

The power consumption not going up with the load is more of a function of the design of the motor (taking advantage of the rotational power from magnets) than inefficency.  The voice coil is positioning the magnet for the next 1/2 cycle.  It contributes some to torque by pushing into the repelling field of the magnet, but because the magnets are trying to align themselves constantly, the rotational power largely comes from the magnets and therefore does not translate back directly into electrical load from a physical load being applied.

Liberty

« Last Edit: June 04, 2006, 05:25:27 PM by Liberty »

gn0stik

  • Guest
Re: Kundel Motor
« Reply #23 on: June 04, 2006, 07:50:03 PM »
Well, I think this motor MUST be measured with analog volt and amperemeters,
not with digital as there are fast pulses that can?t be measured by
this digital multimeter.
Also the torque is not very big and if it draws around 4 Watts of input power
the fan output power is probably just under 2 Watts.
Normal 30 Watts fan motors are also very inefficient and
as long as he does not measure the torque with a prony break under
a load, I guess this motor is pretty unefficient. It has very big magnet
airgaps and thus electrical to mechanical power conversion is low..

That the motor does not increase in input power under a load
just shows this inefficiency and it is just impedance matching in this
case and I want to see analog meter readings.

But the site is very nice made and also the motors and
the concept is clear understandable, that is very nice.
Also it might be usefull to use copper-graphite brushes in series
to drive the coils from a 12 Volts lead accid battery, this way the battery will
be reacharged due to the Newman sparking commutator effect and you could light up in
series an indescant lamp from the RF burst pulses...
and finally see, if more magnets on the motor need also more
or less input power..
We will see, at least this motor makes quite some noise
and might also have vibrational problems, so I don?t know how
long such a motor would last and might need much more
maintainance than a standard motor....
but interesting concept.

The figures we worked from are input power and what he estimated as torque. There were a lot of assumptions made in my calculations due to the lack of factual information we had to work with, but I believe that the math is sound. If our input power were more accurate, via use of an analog meter, that would be fine, but ultimately it wouldn't make that huge of a difference. To make this more accurate the analog meter on the input, a torque meter on the output, and a small alternator attached to the shaft with another analog meter on that are the only things that would give us real factual info.

This would give us true input power, true gross output power, and effective output power(based on the alternator used, this is variable). That is why the range on my calculations is so large, to allow for a lowest, and highest efficiency, based on the knowlege we have of the device.

Unless there is an error in the figures I'm sure it's output power is somehwere in the range I specified. If you can find a flaw, please let me know. I mean you must have a reason to believe the output power to be just under 2 watts, right?  Unless you meant the actual fan motor from which he got the torque comparison. which would actually be:

A typical fan motor is about 60% efficient, let's give it the benefit of the doubt and say 66% that gives us a nice round output number.

So, it can't be this... because it would have around 20W output. As is typical for a fairly efficient fan motor.

If Kundel's motor had the same efficiency his output with 4W input would be 2.64W, and the numbers just don't support that.

Let's work it backwards...

2.64W/746 = 0.0036 hp

(0.0036 hp * 5252) / 1800 rpm= 0.0123 ft. lb torque Which is less than 1/100th of what he estimated. It just doesn't figure...

Can you help me figure out how you came up with your numbers?







« Last Edit: June 04, 2006, 08:22:49 PM by gn0stik »

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Re: Kundel Motor
« Reply #23 on: June 04, 2006, 07:50:03 PM »
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Offline hartiberlin

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Re: Kundel Motor
« Reply #24 on: June 04, 2006, 08:12:49 PM »
Fans are very easy to turn,
you don?t need 20 Watts for this, you only have the low wind friction !
Until no prony brake measurements are made, these
above calculations are just speculations,
for instance if the normal 30 Watts fan motor
would only have a 20 to 30  % efficiency the
calculation would be quite different, also it is a question,
if the 30 Watt fan motor would also run at 30 Watts
input all the time or if the input is lower, but the motor
is just rated at maximum 30 Watts input....
questions over questions..

gn0stik

  • Guest
Re: Kundel Motor
« Reply #25 on: June 04, 2006, 08:28:23 PM »
Fans are very easy to turn,
you don?t need 20 Watts for this, you only have the low wind friction !
Until no prony brake measurements are made, these
above calculations are just speculations,
for instance if the normal 30 Watts fan motor
would only have a 20 to 30  % efficiency the
calculation would be quite different, also it is a question,
if the 30 Watt fan motor would also run at 30 Watts
input all the time or if the input is lower, but the motor
is just rated at maximum 30 Watts input....
questions over questions..

NO THEY ARE NOT.. The fan motor calcs are NOT speculation. IT'S HARD MATH. Just because a blade is easy to spin doesn't mean the motor is not capable of more work than that. Torque curves are based on RPM, PERIOD. More RPM less torque, and vice versa.. Efficiency losses are based on LOAD. Both motors calcs were figured with NO LOAD. And I even allowed for losses, and load in the effective output.

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Re: Kundel Motor
« Reply #25 on: June 04, 2006, 08:28:23 PM »
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Offline hartiberlin

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Re: Kundel Motor
« Reply #26 on: June 04, 2006, 10:15:05 PM »
Yes, but you don?t know, at what RPM the Kundel motor really is rotating
and what the windfriction loss power is at that RPM, probably less than 2 Watts...

gn0stik

  • Guest
Re: Kundel Motor
« Reply #27 on: June 04, 2006, 10:48:06 PM »
Yes, but you don?t know, at what RPM the Kundel motor really is rotating
and what the windfriction loss power is at that RPM, probably less than 2 Watts...

I got the figures for RPM from his forums, and yes, you are correct, I don't know what the wind friction, and rpm figures are under load, so that's why I did my calculations based on no load. Remember his motor consumes less power under load, due to the half duty cycle, and lower rpms, and hence slower switching speed/power consumption. This improves efficiency under load, as compared to a fan motor, rather than decrease it.

Again, lack of data forced me to do calculations with no load, and use known values of existing conversion methods estimate the effective output. I was very conservative in my calculations, and never gave his motor the benefit of any doubt. My effective output took 40-60% losses into consideration. Even if I halved that conversion efficiency and took values of 70-80% losses, it would still be 200-375% effective efficiency. Let's be extremely conservative, and say the conversion to electricity is only 10-15% efficient. The output would be 100 to 188% effective efficiency. That's after 85-90% losses during conversion.

The only thing that would make the calculations before conversion to electricity invalid would be his estimated torque.

Understand I'm approaching this very skeptically, but the math doesn't lie.

The only way to know for sure would be to build it, and close the loop. I have already said that. These calculations were simply to get an idea of whether or not it would be worth the effort. And the figures were very promising.

 

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