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#1
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![]() That's only applicable in some microgeneration programs. You are pulling out 1.8A through the pump and hence its yours. And more or less, the whole current is consumed by the motor other than some leakages.
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You wouldn't want to see my tank. I don't use fancy equipment and I am a noob ![]() |
#2
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![]() Quote:
here is a simple explanation of PF I found "Power factor is the percent of electrical power that does work. Resistive loads, such as lights and heater elements, always have unity (1.0) power factor; all power is used for work. Motors, because they are essentially large inductors, lag current and cause power factor issues. A motor with a .85 power factor uses 85 percent of the power for work. 15 percent is wasted. For example, a 480VAC, 10HP motor with a 1.0 power factor uses 10.6 amps to run at 10HP. Lowering the power factor to .8 requires the motor to consume 13.2 amps to produce that same 10 HP." I read up on this years and years ago and can talk about it all day (well maybe not all day) but it is hard to sit and type it out, hence the simple explanation I found. we used to be realy concerned with PF with lights to get the efficiency. you if you have a pump that is rated at 120 watts but has a power factor of .6 in reality it will use 200 watts to do the work. the extra 80watts are lost through ineficient circuts, heat, ect... so when the power company bills you you are billed for the 200 watts. this is why the hydro companies have rebates for big business to upgrade to high PF equipment, it lowers the demand on the grid. Steve
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![]() Some strive to be perfect.... I just strive. |
#3
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![]() Where are you getting your power factor numbers from?
In any case put your watt meter to the pump, that will tell you the watts. You're not billed kva in residential. And even if you were you need to look at what your whole system is drawing, not just one little part.
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-My 330g build thread |
#4
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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 04:11 PM. |
#5
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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 |
#6
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__________________
You wouldn't want to see my tank. I don't use fancy equipment and I am a noob ![]() |
#7
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![]() The RD is a DC motor ?
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#8
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#9
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![]() Exactly.
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#10
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![]() Quote:
__________________
-My 330g build thread |
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