conversion math

Posted by Chris Ivers on May 14, 2008, 04:07:45 PM

conversion math
since i havn't found much interest in my previous post i will try another.

i want to make sure that i'm on the right page.

if i have an alternator rated at 50A can i assume it is putting out 600W at 12V DC?
(50 X 12 = 600)??

so if it is putting out 600W at capability.
when i convert that with an ac converter i maintain the 600W?
(minus loss to the actual converter)??

now the 600W can be used in the form of 120V AC at 5A?
(600 / 120 = 5)Huh

please let me know if i need to find a new hobby! thanks
 

Posted by John B on May 14, 2008, 04:40:34 PM

Re: conversion math (Reply #1)
Your math is correct. However, to get 600 watts of power from the alternator, the wind turbine that you are using to turn the alternator must be capable of generating in excess of 600 watts of power. The maximum power of a wind turbine is determined by the swept area of the blades and wind speed. Any other inefficiencies in your system will bring the total power generated down, so don't be surprised if you build a wind turbine capable of generating only 50 watts of power and it fails to deliver the power you might have expected out of the other end of your 600 watt alternator. Here is a good place to read about wind power:
http://www.awea.org/
 

Posted by Thomas Allen Schmidt on May 18, 2008, 06:44:15 AM

Re: conversion math (Reply #2)
Precisely.
The math of electricity deals with absolutes.
The more precise one is with the input data the more precise the answer will be.
Advertising, on the other hand, has a tendency to exaggerate. But you know this.
Key words here are - rated output.
 If the manufacturer of this product was able to achieve this "rated output" then there would have to be a test condition. Was the test condition a standard among generator manufacturers? What are the conditions?
 
A generators output is directly related to its input speed and torque values as well as any load that would be placed on it which could have an effect on internal temperatures and so on. All of this would appear to be picking nits but, you wanted to know, and the best way I know how to know an unknown is to know what the unknown is and then form an hypothesis. Then experiment to prove or disprove that hypothesis. After all Micheal Faraday didn't invent the very first electric motor sitting on his keister you know.
http://www.rigb.org/heritage/faradaypage.jsp
Not that I think you are prone to sitting on your keister or anything, its just a figure of speech.
Anyway, you will need; an amp meter, a volt meter, an rpm gauge, a power source (to spin the generator,) and a torque value gauge. You know what, on second thought, it would easier to preform another well known mathematical step that electricians use known as de-rating, which is to assume that it will not put out its advertised rating.
 
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