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#1
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Quote:
P = V * I * PF So PF = P/(V*I) = 82W/(115V*1.8A) = 0.4 Last edited by sphelps; 03-07-2013 at 05:11 PM. |
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#2
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I'm just restating what has been said but:
PF= True Watts/ Volts*Amperes True Watts= 82w Volts= 115v Amps= 1.80a PF= 82/207 = 0.396 You're pump output is only 40% of the power it is drawing. Not good, not efficient! haha |
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#3
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Quote:
You better change the motor if you want the consumption to cut to half! DC motors FYI!
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You wouldn't want to see my tank. I don't use fancy equipment and I am a noob ![]() |
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#4
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The RD is a DC motor ?
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#5
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RD does have DC motor pumps but I don't think 6.5 is DC motors. Can't seem to find the exact motor specs of 6.5 but with that power factor, its impossible to be a DC motors pumps. DC motors don't have any pf, the only thing that will have some PF is the converter to run the pump (hence the 0.97 in Steve's DC motor).
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You wouldn't want to see my tank. I don't use fancy equipment and I am a noob ![]() |
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#7
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Yeap very fancy, want to buy it?
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#8
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#9
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Exactly.
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