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A little help with the math?
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Brad White_194
Member Posts: 74
In the big scheme of things, that is an excellent idea and one could probably back that up with the real numbers in your experience. I learned something today, or at least something to try or at least consider.
Thanks, ps.
Brad
Thanks, ps.
Brad
0
Comments

I am trying to explain to the customer
why they need to get a knew frequency drive instead of leaving the bad one on bypass. Nothing like money to make it clear.
So 24 amps is the motors draw without the drive. Its 460 volt three phase. It averages about 9 amps with the drive running. So the math is 24ampsX460volts=11040watts multiplied by three legs =33120 Watts at full load per hour. As opposed to 12420 watts per hour with the drive on average. That makes the difference 45440 watts per hour. Or 45 KW per hour. Does that all sound right? Just don't want to put my foot in my mouth.0 
No.
Volt x Amps X 1.73 X .8 / 1000 = kW for 3 phase loads
Square root of 3 (1.73) instead of "3 legs"
Include power factor (induction motor = 0.8), unless power factor correcting capacitor (not starting cap.) or other technique is used.
Multiply by run time to get kWhr, or kWh.
Calculate for both amperage conditions.
Then multiply by utility rate (last kWh, not average) to get $ difference.0 
Thank you Bob I am writing that formula down. Next time I need it I will have no idea were I put it. But I thank you. This time it will prevent hoof in mouth with the customer.0 
Damn, is Bob good....
Excellent, Bob.
I would add to the equation, Will, that the number of hours at various loads should be factored in. As load decreases (or increases, egads), the HP changes by the cube or cube root. As a very rough explanatory rule of thumb, a 10% increase in capacity takes 33% more HP... When you reduce a load by 10% your HP drops by about 27%.
That first increment does a lot of the work from the starting point.
If in doubt though, picking a safe constant operating point will get you there and any savings on top of that is a bonus and will make you look even better!0 
That is an important point to stress, Brad. FWIW  I like to degrade that cube exponent to 2.7 to account for drive losses.0
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