effect of power factor improvement on energy consumption 4

[usamaegypt]

Hi,
I would like to ask whether enhancing the power factor by using capacitor banks could decrease energy consumption KWH
Best regards.

 

[ScottyUK]

No, other than reducing the I^2R losses in the conductors. PFC acts to reduce the kVArH consumption, not the KWH consumption. Reactive loading is normally billed separately by the utility if you are an industrial consumer, so it is often worth the investment to improve the load power factor.


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If we learn from our mistakes,
I'm getting a great education!

 

[jraef]

It saves on losses in the primary distribution system as well, that is why some utilities encourage PFC caps and penalize those with poor PF. Others, like mine (PG&E in California), do not even care because they feel the mitigation costs and risks are more than the transmission losses.

Still, it does little to nothing for the average user as far as kWH. If you are a steel ot textile mill or other large facility with a large sub-distribution system and lots of AC motors within your plant site, you may see some small amount of savings.

"Venditori de oleum-vipera non vigere excordis populi"

 

[aolalde]

I agree with ScottyUK and jraef.

The only improvement will be on the resistive loss (i^2*R) of the line and substation. Since improving the power factor will reduce the total line current, these losses will drop with the reduction ratio squared. But the resistance of the line an substation (including the transformers) is regularly too small, then the saving of kW is not that spectacular.
Assuming an hypothetic 480 Volts 3 phase line, driving 1000 amperes LOAD at 0.75 PF and with 0.0050 ohms per phase total resistance of line and transformers.
THE TOTAL LOAD POWER kVA= 1.732*.480*100 = 831.38
THE REAL POWER kW = 831.38 * .75 = 623.54
The LINE loss is 3* i^2*R/1000 = 15 kW

If the LOAD Power factor is improved to 0.90
THE REAL LOAD POWER REMAINS kW = 623.53
THE TOTAL POWER REDUCES TO kVA = 623.53/0.9 = 692.82
THE NEW LINE CURRENT = 692.82/(1.732*.480) =833.35
THE NEW LINE LOSS = 3 * 833^2*0.0050/1000 = 10.417 kW

THE SAVINGS WILL BE 15.00-10.417 = 4.593 kW

Low power factor means more reactive power demand and the utility’s Generator must increase the field excitation to provide it.

 

[mc5w]

The justifications for power facgor correction are:

1. Free up capacity of the system which can get you out of upgrading something. Savings can actually be substatial if you do not have to upgrade say a 1600 amp motor control center.

2. Getting the system to work off of a standby generator or cogeneration system. You benerally want improve power factor to around 85% to 95% inductive so that the generator field does not have to work so hard and so that the voltage regulator does not hit maximum field current.

3. If you are at the end of a long primary line improving power factor will improve reactive voltage drop in the line. One of the rules of thumb of reactive compensation is that you need to bring up substation power factor to 90% inductive to 95% capacitive.

4. Getting wires and busways to run cooler so that they live longer and have fewer other problems. This is related to reason number 1. Totally enclosed nonventilated busways run at about 100 degree Celsius temperature rise at full load - keeping a busway from running flat out actuallu does improve voltage regulation. At that much temperature rise you can only go about 100 feet before having to oversize the busway to control voltage drop.

 

[jraef]

All correct, but not what he asked about.

 

[weh3]

Correct, jraef.

Answer to original question: it depends on which kWh you are asking about. Load kWh or metered kWh. It makes no difference whether you put the question in terms of kWh or kW.

Increasing the power factor toward unity has no effect on the real power requirements of the load. There may be some secondary effects that reduce the kWh slightly, but there are no direct effects.

However, it is true, as others have pointed out, that all current, even reactive, contributes to copper loss behind the meter. You are heating your wires more, therefore paying more, with a low power factor than with a high pf.

William

 

[Airglider]

Hellow,

Installing quality VFD drives on motors will maximze motor efficiency and correct the power factor of the motor, under all load conditions to perfect.

Alan,

 

[ScottyUK]

Airglider,

Installing drives with an active power factor corrected front end will improve it to something resembling unity. Installing drives with a typical 6-pulse rectifier will introduce harmonics and give a non-unity power factor.

'Perfect' is not a term I associate with the real world. You can come close, but nothing is ever perfect that I know of.


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If we learn from our mistakes,
I'm getting a great education!