kvar energy savings unit power factor optimization Industrial application part 2

[shieldcracker]

Let me set the record straight power factor correction, raising the power factor as close to one saves energy. This video is misleading, lowering the amp draw on a load by raising the power factor does not make the load more efficient and thus saves energy. By lowering the amps the losses on the branch and circuit feeder are reduced and that is where the energy savings come from. The watts to the motor or any load stay essentially the same. POWER FACTOR CORRECTION HELPS WITH LINE LOSSES ONLY.

[lordandprotector]

If you have an old kwh meter that is really just an amp-hours meter, AND you have a low PF due to lightly loaded motors, this will make the power meter turn slower for the same or even higher actual power usage. It does NOT save any power/energy. And the descriptions given by these 2 guys are absurd.

 All it is is a capacitor, in parallel with the load, to balance the inductive portion of the load. As Nikola Tesla spins in his grave!

[Pyro4117]

gapind1, YOU need the study, as 5occerboy correctly pointed out, you left PF out of the calculation, which is where the savings are supposed to occur.

They don't show where the capacitors are installed, but if they aren't at the motor itself, there will be no I^R savings.

For commercial customers with a kVAR meter, PF improvement is valid. For residential customers it's a scam.

35+ years as a utility engineer and conservation program evaluator.

[ENERGYCONTROLPLUS]

Power Factor only works on Commercial facility that has a significant old motor loads because Utility companies will surcharge you for poor factor loss. Utilities companies do not charge for Poor power factor on Residential and will only charge you for true power.

The meter that you have reads loss of power factor but it does not matter as far as the residential side.

If what you are saying is true why isn't reconigze by EnergyStar? Please don't lie to your Customers.

[Gigivr6]

Show us the Watt consuming difference, I know the meter you have there can show it. stop hiding stuff from inosent people!!!

[taylortownmayor]

@Gigivr6

They DO show watts. His meter shows KW when it shows power factor.

At 1:44 you can see the pf is 0.822 and the power is 2.48 KW.

At 2:11 you can see the pf is now 0.957 and the power is now 2.50 KW.

At 2:32 you can see the pf is now 0.998 and the power is now 2.51 KW.

Rob and Rich ARE right. The meters DON'T LIE.. the power doesn't change!

Also efficiency by definition is the ratio of power out divided by power in. Since power use doesn't change, neither does efficiency!

[Gigivr6]

Read what you write, as the pf encrease so does the KW, so it will end up costing the costumer more!!!

[taylortownmayor]

Yeah, very slightly. I wouldn't call that minuscule amount much of an increase though, 20~30 watts isn't much, but yes will cost you more!

It is due to the capacitor supplying VARS to the system and raising voltage. The reduction in line current due to the capacitor will also reduce voltage drop.

The slightly higher applied voltage will raise the power just a bit.

The important observation is that the watts don't change all that much even though pf is changing dramatically.

[taylortownmayor]

Sorry though for ignoring that small of an amount. During the day I deal in megawatts so I might not pay attention at all to a difference of a few KW here and there.

[billfenn2]

The video actually shows that the KVAR unit consumes more watts. Homes are billed for the wattage that they use. Look at 1:44 in the video. It shows a power factor of .822 The display below it is KW or killowatts (what you pay for). The reading is 2.48KW. Move the video forward to 2:36 and you will see that the KW has INCREASED to 2.52 when the power factor has been corrected to .999. This video actually shows that the KVAR unit is consuming more energy.

[gottagobideo]

This only works for industrial buildings like a coca cola plant. 99.99 percent of business are billed for kw not kvar.

[Dougout2424]

Rich, I would like to talk with you. What is the name of your business?

Thanks!

[vacman1353]

5occerboy, shame on you for not doing your homework. What these units save are your (I squared R losses) and depending on your motor loads and distance from the transformer could be significant. On facilities that are billed with for Power factor penalties we turn out to be heros. Whats your power factor?

[5occerboy]

Vacman you are perpetrating a scam if you suggest residential customers are billed for power factor. They're not. I concede that improving PF on a motor reduces current a bit and so I2R losses are less. But in the real world the I2R losses are tiny compared to the work being done & the kWh consumed by the motor - so the savings are picayune unless these rare situations apply: 1) Motor is grossly oversized / mismatched for the task. 2) Conductors are either extremely long or way undersized.

[5occerboy]

These devices are scams. This mountebank doesn't understand electricity or power he has confused current draw with energy use (to do work). Residential consumers don't get billed for amps you get billed for the energy in kWh to do work! Also You're not billed for power factor unless you are a big industrial user. Shame on you.

[gapind1]

5occerboy,

I think you need to do some studing on electricity. Everything from the power company is charged by kWh. kWh comes from volts times amps which equals watts. The amount of wattage you use is what you pay for. So if you take 12 amps X 120v this=1440watts. If you lower the ampage to 9 amps x 120v =1080 watts. You will pay for more power @1440 watts then 1080 watts. Simple math! What do you think it means if you see a product with 50va? va stands for watts! Volts x Amps

[5occerboy]

Gapind you are a technical ignoramus! OK the utility charges you for energy in kWh or Watts over time. But your equation 12 amps X 120v =1440watts is WRONG for AC power. Correct equation is Watts=Volts x Amps x POWER FACTOR. I invite you to open a text book on AC power and do some study. I defy you to challenge my equation and tell me why you didn't include PF. Your equation is true for VA but not WATTS - you don't get the distinction between apparent power VA and true power WATTS do you?