Capacity and Losses
The kVA and amps that correction frees on your transformer or service, and the I²R losses it trims.
Apparent power is what your transformer and switchgear actually carry. At the same real load, a higher power factor means less of it: room for new load on equipment you already own, and less current in the conductors between your meter and the correction point, whose losses fall with the square of current.
Your numbers
Example values. Replace them with yours.
The peak kW through the transformer, service or panel you are rating: the bill’s demand line or a meter’s maximum. If you carried an average from a bill, replace it with the peak.
At the peak, if you have it; otherwise the bill’s or the Power Factor Check’s figure.
0.98 is what both manufacturers rate their equipment to reach (Accentz: 0.98–0.99; Trans Power: up to 0.98). The measured case on our results page reached 1.00.
Use the piece of equipment that limits you: the transformer’s nameplate kVA, or the main breaker or switchboard amps.
On the transformer nameplate. If the utility owns it, the rating is usually on the pad or in your service agreement.
Between phases: 480 for most US industrial services, 208 or 240 for many commercial buildings. Used for amps and for an amps-rated service.
Twelve months of kWh from the bills. Leave blank to skip the loss estimate.
Your blended energy rate: the bill’s energy charges divided by its kWh.
A working assumption for load losses in your own transformer, switchgear and feeders between the meter and the correction point. It is editable because it is an assumption; the engineering report replaces it with a measured value.
Results
96 kVA
115 A at 480 V
- Rating used
- 750 kVA
- Loading
- 585 kVA now (78%)490 kVA after (65% of rating)
- Current
- 704 A → 589 A16% less
- Headroom for new load
- 135 kW now255 kW after
- I²R losses upstream of the unit
- 30% lower
- Estimated loss energy saved
- 12,595 kWh a yearabout $1,259 (loss share 2.0%, $0.10/kWh)
At 0.82, 480 kW draws 585 kVA, 78% of the 750 kVA rating, and 704 A at 480 V. At 0.98 the same 480 kW draws 490 kVA (65%) and 589 A. That frees 96 kVA: room for about 255 kW of new load at 0.98, against 135 kW today. With 16% less current in the conductors and transformer between the service and the unit, their I²R losses fall about 30%. If 2.0% of your 2,100,000 kWh is lost that way, that is about 12,595 kWh, $1,259 a year at $0.10 per kWh.
Current reduction is what the rental measures: on one installed Reinigen unit’s display, the grid side drew 9% to 23% less current than the load side (see the Reinigen page). Accentz rates the unit to cut line current by up to 20%.
Released 96 kVA.704 A → 589 A at 480 V.
How this is calculated
Apparent power: kVA = kW ÷ PF, now and at the target; the difference is the capacity released. Loading is kVA over the rating; an amps-rated service is converted with kVA = √3 × volts × amps ÷ 1,000. Current: I = kVA × 1,000 ÷ (√3 × volts). Headroom is rating × PF − kW. Losses in a conductor are I²R, so at the same kW they scale with (PF now ÷ PF target)²; the kWh figure applies that to the share of your energy you assume is lost upstream of the unit.
Assumptions
- Nameplate arithmetic: no temperature, ageing or utility rating rules.
- The headroom at the target assumes the unit also corrects the reactive power of the new load, within its rating.
- The loss share is your assumption; it covers only conductors and transformers between the meter and the correction point, and nothing downstream of it.
- Harmonic current and its heating are not counted. Real kW is unchanged.
Estimates from your inputs and the formulas shown. They are not a ruling from your utility, a design or a guarantee. A survey and the six-month rental measure the real thing on your own meter.
Your numbers go with your request; nothing is sent until you submit the form.