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Acidwrz · 12-19-2002, 11:17 PM

I think someone needs to clear this up to avoid further confusion. This thread has given some people ideas that do not follow the laws of thermodynamics so I will try to clear them up the best I can.

I will use these figures for the Potted HP-199-1.4-0.8 from TeDistributing

If a device is rated at 24.6 volts and running at 11.3 amps the device requires 277.98 watts of power to get it running at full power. But on the website it lists its deltaQ(entropy) as 172 watts. Doing the equation 172/277.98 gives us the actual efficency of the device not the iMA so to speak. This means that it is 61% effective. This 61% does not mean to only send 172watts of power to it by using a powersupply that is a lower voltage. you could figure this out by 172watts=V*I. If 277watts is sent into the system and only 172 of work is done by the power... Lesser gains would be found if you sent 172 watts into the system. Using the 61% constant for this device(if you do not believe me that the Volts and amps are constant look at the second graph of any device on tedist and prove me wrong) That means that only 104.92 watts would do work.

If any of this is wrong please reply and explain your reasonings on it.

12-19-2002, 11:17 PM	#10
Acidwrz Cooling Neophyte Join Date: Dec 2002 Location: Texas Posts: 4	I think someone needs to clear this up to avoid further confusion. This thread has given some people ideas that do not follow the laws of thermodynamics so I will try to clear them up the best I can. I will use these figures for the Potted HP-199-1.4-0.8 from TeDistributing If a device is rated at 24.6 volts and running at 11.3 amps the device requires 277.98 watts of power to get it running at full power. But on the website it lists its deltaQ(entropy) as 172 watts. Doing the equation 172/277.98 gives us the actual efficency of the device not the iMA so to speak. This means that it is 61% effective. This 61% does not mean to only send 172watts of power to it by using a powersupply that is a lower voltage. you could figure this out by 172watts=V*I. If 277watts is sent into the system and only 172 of work is done by the power... Lesser gains would be found if you sent 172 watts into the system. Using the 61% constant for this device(if you do not believe me that the Volts and amps are constant look at the second graph of any device on tedist and prove me wrong) That means that only 104.92 watts would do work. If any of this is wrong please reply and explain your reasonings on it.