StephenSDH
Senior Member
- Location
- Allentown, PA
I have a customer who had a company (http://www.correctioncontrols.com/) come to their plant and claim they can reduce their KW consumption by over 6%. According to my customer they actually hooked up equipment to a machine to show him it works(I have a feeling it was just KVA reducing instead of KW reduction) There are NO incentives/fines by the power company for bad PF. My understanding of power factor correction is that it will only save you money on line losses(I^2R), and you generally do it because of fines, reduce equipment size, or reduce voltage drop. From my calculations the customer might be lucky to save 1 percent on their electric bill if the whole plant was PF corrected at the inductive loads. The company doesn't come out and say the are installing capacitors on the equipment they call them "SR Control System". I believe it is just capacitors with some magic beans tossed in. I have the feeling it is a Fly-by-night company. Before I put my foot in my mouth by telling the customer this, I wanted to check to see if anyone can vouch for a dramatic KW reduction by clipping a power correction device to to a load. Below are my calculations of a sample KVA reduction savings. Thanks ahead of time!
-Steve
Generic 3 Phase Power Savings Calculator
Voltage 480 V
Conductor Impedance 0.0265 ohm/1000ft (500 MCM)
Sets of Conductors 2 Sets (Common for 800A Service)
Feeder Distance 500 feet
Conductor Resistance 0.006625 3 Sets of 500MCM Copper
Power Factor 0.75
Current 600 A
Apparent Power 498.24 KVA
Real Power 373.68 KW
Reactive Power 329.55 KVAR
Current 600 A
Real Current 450 A
Reactive Current 396.8 A
Power Loss(I^2*R*3Ph) 7.155 KW (@ Noncorrected PF)
Power Loss(I(real)^2*R*3Ph) 4.0246875 KW (@ PF == 1)
Maximum Power Savings 3.1303125 KW
Percentage Power Savings 0.84% Percentage Savings
-Steve
Generic 3 Phase Power Savings Calculator
Voltage 480 V
Conductor Impedance 0.0265 ohm/1000ft (500 MCM)
Sets of Conductors 2 Sets (Common for 800A Service)
Feeder Distance 500 feet
Conductor Resistance 0.006625 3 Sets of 500MCM Copper
Power Factor 0.75
Current 600 A
Apparent Power 498.24 KVA
Real Power 373.68 KW
Reactive Power 329.55 KVAR
Current 600 A
Real Current 450 A
Reactive Current 396.8 A
Power Loss(I^2*R*3Ph) 7.155 KW (@ Noncorrected PF)
Power Loss(I(real)^2*R*3Ph) 4.0246875 KW (@ PF == 1)
Maximum Power Savings 3.1303125 KW
Percentage Power Savings 0.84% Percentage Savings