Working out KWh if you have KVArh and PF

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SUMMARY

The discussion centers on calculating kilowatt-hours (KWh) from kilovolt-ampere reactive hours (KVArh) and power factor (PF) in the context of utility billing. The equation PF = KWh/√(KWh² + KVArh²) is highlighted, with users clarifying that utilities typically measure kW and KWh directly, negating the need for retrospective calculations post-power factor correction. Participants emphasize that improving PF primarily reduces KVArh, not KWh, and that billing practices vary by region, with some utilities adjusting charges based on PF thresholds.

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  • Understanding of Power Factor (PF) calculations
  • Familiarity with kilowatt (kW) and kilovolt-ampere reactive (KVAr) concepts
  • Knowledge of utility billing practices related to power factor
  • Basic mathematical skills for manipulating equations
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  • Research "Power Factor Correction techniques" to improve efficiency
  • Learn about "Utility Billing Structures" and how they vary by region
  • Explore "KVA vs. KW billing" to understand implications for large industries
  • Study "Voltage Optimization" and its impact on energy costs
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Electrical engineers, energy managers, and facility operators looking to optimize power factor and reduce energy costs in industrial settings.

Benjaminfross
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Hello,

Complete novice here. About to enter into a cost sharing agreement with another company, part of the cost sharing is improving the PF.

I've understood the equation for PF = KWh/sq(KWh2 + KVArh2)

They only benefit from PF improvement and not any other savings in reducing electricity consumption that we might do.

My thinking is record what PF is today - then after a period see what new PF is (get this on bill) work it back using the new reactive power (KVArh) and see what the Kwh would have been compared with what it is now - then stick a Price per KWh and split the saving?

I've forgotten my Maths GSCE - struggling to rework the formula.

say PF is 0.9 and KVArh is 2772 what's KWh?

Appreciate any help.
Thanks
Ben
 
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Benjaminfross said:
say PF is 0.9 and KVArh is 2772 what's KWh?
PF = kW / √(kW2 + kVAr2 ) →

PF*√(kW2 + kVAr2 ) = kW →

PF2*(kW2 + kVAr2 ) = kW2

. . . . . Some work for you . . . . .

kW = kVAr / √( ( 1 - PF2) / PF2 )

Remember "k" for "kilo". "K" means something like "Kelvin".
 
Last edited:
Welcome to PF!

What you are describing doesn't match my experience with utility power factor issues. Not sure if you have a different issue from what I've seen or if you have something wrong in your understanding, but I'll explain what I've seen and you can tell me if it seems right:

Utilities measure kW and kWh directly, so there is no bill correction needing to be done from before/after pf correction, because the kW and kWh haven't changed (aside from very minor efficiency improvements). Power factor is penalized directly on the bill and should be stated explicitly: For example, for a 70% pf under an 80% requirement, measured demand = 100 kW, billed demand = 100*(.8/.7)= 114 kW.

Both the measured and billed kW are shown, so there shouldn't need to calculate them.

Unless you are saying you want to calculate what you would have been billed for if the pf correction hadn't been done. Then you flip-over the calculation: 114*(.7/.8)=100 kW

Also, I'm not sure why/how you are dealing in kVar: usually if you know anything besides kW, it is usually total kva, since that's volts times amps.
 
russ_watters said:
Welcome to PF!

What you are describing doesn't match my experience with utility power factor issues. Not sure if you have a different issue from what I've seen or if you have something wrong in your understanding, but I'll explain what I've seen and you can tell me if it seems right:

Utilities measure kW and kWh directly, so there is no bill correction needing to be done from before/after pf correction, because the kW and kWh haven't changed (aside from very minor efficiency improvements). Power factor is penalized directly on the bill and should be stated explicitly: For example, for a 70% pf under an 80% requirement, measured demand = 100 kW, billed demand = 100*(.8/.7)= 114 kW.

Both the measured and billed kW are shown, so there shouldn't need to calculate them.

Unless you are saying you want to calculate what you would have been billed for if the pf correction hadn't been done. Then you flip-over the calculation: 114*(.7/.8)=100 kW

Also, I'm not sure why/how you are dealing in kVar: usually if you know anything besides kW, it is usually total kva, since that's volts times amps.

That's for the reply both of you - Yes your right - I thought an increased power factor will reduce kWh but just reduces kWArh - Its only £90 on the bill so called the customer and its voltage optimisation there looking to do.

Thanks for the help
 
russ_watters said:
Utilities measure kW and kWh directly, so there is no bill correction needing to be done from before/after pf correction, because the kW and kWh haven't changed
That's not my experience ( in Denmark ):

Big industries will pay less as per kWh, but will be billed as per kVA as well, because the kVA's involve losses in the transmission lines. Thus big industries are motivated to compensate/decrease the amount of kVA's.
 
Hesch said:
That's not my experience ( in Denmark ):

Big industries will pay less as per kWh, but will be billed as per kVA as well, because the kVA's involve losses in the transmission lines. Thus big industries are motivated to compensate/decrease the amount of kVA's.
Ok, it certainly could be different in different countries or even per utility. For mine, in philadelphia, USA, they measure peak kw, kWh and power factor. As long as the PF is above a certain threshold, it is ignored. Below the threshold, kw and kWh are adjusted.
 
russ_watters said:
As long as the PF is above a certain threshold, it is ignored. Below the threshold, kw and kWh are adjusted.
It's a little different in DK:

The minimum PF, sustained for 30 minutes ( within a quarter ) is measured. Then a billing factor as for the kWh's ( within that quarter ) is found in a look-up table.

Something like that.
 

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