Power savings 2

[ZWE1967]

Please help:

A 3-phase induction motor rated 150hp(110kw) has a constant running load of 104 Amps per phase at approximately 466 volts across phases.Norminal pf is about 0.87

We noticed about 10Amps drop per phase when the pf was corrected upwards to about 0.95 using third party equipment.

How best can I translate this drop of (10A)to KWh over a 24hr period to try and estimate the savings, here is what I have:

KW = V(phase)*I(phase) * sqrt(3)*pf where v=466, i=10, pf=0.95

Kwh=kw*24

Will this work?

 

[jraef]

How best can I translate this drop of (10A)to KWh over a 24hr period to try and estimate the savings...

You can't. There is no energy savings. The drop in current is at the corrected power factor because the capacitors are providing the VARs, but the line SOURCE must still supply all of the kW, which is all you are billed for (from an energy standpoint). Common misconception, heavily promoted by the PFC "energy saver" people as a scam.

Sorry to break the bad news.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
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[rcw retired EE]

I think there may be some confusion between amps and watts.

Calculate kw for the before condition
kw = V x I x1.732 x pf = 446 x 104 x1.732 x 0.87/1000 = 69.9 kW. The motor is drawing 70 kW (93.8 HP)
kVA = V x I X 1.732/1000 = 80.33 kVA
power factor = kw/kva = 69.9/80.33 = 0.87

For the new operating point with the power factor correction:
kVA = 446V x 94 A x 1.732/1000= 72.6 kVA
kw = 446V x 94 A x 0.95 x1.732/1000 = 69.0 kW
Power factor = 69/72.6 = 0.95

The motor draws the same kw. The only thing you are saving is the I^2R losses in the cables and transformers between the meter and the motor. There are no power savings in the motor.

Typical power system losses would be in the 0.25% to 2% range for an industrial system. Take the high figure, 2% of 80 kVA = 1.6kW say 2 kw. The equivalent R for 2 kw loss at 104 amps is = 2000 watt/(104 x 104) = 0.185 ohms. (Since power = I^2 x R)

The 10 amp reduction in your power system saved 10A x 10A x 0.185 ohms = 19 watts. Over 24 hours, that's 0.456 kW.

But there are other advantages to power factor correction.

 

[ZWE1967]

Thanks for your input,

 

[Walden]

You appear to be an industrial user if so; you may have a kVA meter rather than a kWh meter. If you have, then you will save on your electricity bill if you improve the power factor.

 

[roydm]

If you have a maximum demand meter you have shaved 8 kVA off your maximum demand. Check what your demand charges are (typically $5-10 I think) so that's 8 x 5=$40 or 8 x 10=$80 for each and every month that you are using the motor during the peak period.
BTW if you are on Max Demand it's well worth reading up on the subject, the savings can be quite staggering. A while back I was able to save one facility in excess of $20K per year just by re-scheduling their work day slightly. It's not unreasonable to save 5-10% on your utility bill.
Roy

 

[LionelHutz]

Look at your electric bills and see if there are penalties or other charges for the kVA usage. If so, you can reduce your bill by correcting the power factor and lowering the kVA. Still, just doing so on a single 150hp motor to save 8kVA doesn't make much difference.

 

[jraef]

You appear to be an industrial user if so; you may have a kVA meter rather than a kWh meter. If you have, then you will save on your electricity bill if you improve the power factor.

Even that depends upon your PoCo. PG&E for instance, the largest PoCo in Northern Calif., does not charge power factor penalties for the majority of their industrial customers. So improvement means nothing to the users (other than the minor reduction in losses mentioned above).


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[dpc]

Just a word of caution regarding power factor penalties and utility billing - there are a multitude of methods used by utilities for billing energy and demand usage. Some are based on kVA, some on kVAr, some on kVArh, etc.

Some utilities have "ratchet" charges based on the highest demand for the previous 12 months. The variations are endless.