Author Topic: improvement  (Read 6139 times)

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ghurd

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Re: improvement
« Reply #33 on: December 14, 2005, 11:40:29 PM »
Recheck the math. Had this link open, so this is what I'll use. See the chart at around 650RPMs.


http://www.windstuffnow.com/main/alt_from_scratch.htm


G-

« Last Edit: December 14, 2005, 11:40:29 PM by ghurd »
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Arno

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Re: improvement
« Reply #34 on: December 15, 2005, 05:34:18 AM »
    Greetings Nothing to lose,


Thank you for a positive idea finnially, but I have decided to chuck it all and move to Florida, where I will no doubt start this all over again with air conditioning!


MERRY CHRISTMAS EVERYONE


arno

« Last Edit: December 15, 2005, 05:34:18 AM by Arno »

dinges

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Re: improvement
« Reply #35 on: December 15, 2005, 06:05:22 AM »
I haven't read windstuffnow's site, nor re-checked Willi's calcs.


The way I figured it out:


In star, voltage is 1.73 times single coil one; current is the same, so


R=U/I=1.73*U/I --> 1.73 times that of a single coil


In delta, voltage is the same, but current 1.73 times as big:


R=U/I=U/(1.73*I) = 1/1.73 times as big (small)


So the total difference in resistance between star and delta is 1.73/(1/1.73) = 3 times.


I was surprised when I had made this calculation about a week ago, because I expected only a 1.73 factor difference. No argueing with the numbers though...


Peter,

The Netherlands.

« Last Edit: December 15, 2005, 06:05:22 AM by dinges »
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willib

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Re: improvement
« Reply #36 on: December 15, 2005, 08:35:49 AM »
the resistance at any two points of star configuration is twice the phase resistance

ie ( 2x ).

and 2x/(2X/3) = 3 ..

so the resistance of star vs. delta is three times the delta phase resistance .as you said..

and the resistance of delta vs. star is 1/3 .

« Last Edit: December 15, 2005, 08:35:49 AM by willib »
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richhagen

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Re: improvement
« Reply #37 on: December 18, 2005, 12:38:01 PM »
Sorry to have to burst your bubble NTL, but you'll still only end up with 1500 watts of heat.  From thermodynamics, when you use the steam to perform work, you change its enthalpy (internal energy plus pressure times volume energy), during the process, and you end up with less thermal energy in the steam in the process.  All known processes involving using steam to convert thermal or pv energy to work still obey the laws of thermodynamics, one of which is the conservation of energy.  It was a good try however, but there is no free lunch there.  Now if you could bring the water into your house, and then perform some process on it and kick it out of the house as cold ice with a lower internal energy, and keep all the electical power consumed as heat in your home in the process, then you would end up with more than the 1500 watts of heat in your home, but we are back to using a heat pump or something else to pull it off.  Rich Hagen
« Last Edit: December 18, 2005, 12:38:01 PM by richhagen »
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