Working out KWh if you have KVArh and PF

  • Thread starter Benjaminfross
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In summary: I don't pay the bills, so I don't know the details.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. I don't pay the bills, so I don't know the details.In summary, the conversation discusses a cost-sharing agreement between two companies for improving power factor (PF) in order to reduce electricity consumption. The equation for PF is explained and the discussion revolves around calculating kWh using PF and KVArh. However, it is clarified that improving PF does not directly reduce kWh, but only reduces KVArh. The conversation also mentions voltage optimization as a potential
  • #1
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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  • #2
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:
  • #3
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.
 
  • #4
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
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 

1. How do I calculate KWh from KVArh and PF?

To calculate KWh from KVArh and PF, you can use the following formula: KWh = KVArh x PF. This formula takes into account the reactive power (KVArh) and power factor (PF) to determine the actual energy used (KWh).

2. Can I use a different formula to calculate KWh?

Yes, there are other formulas that can be used to calculate KWh from KVArh and PF. One example is KWh = KVArh x cosφ, where cosφ is the power factor angle. However, the first formula (KWh = KVArh x PF) is the most commonly used and accurate formula.

3. What is the significance of PF in calculating KWh?

Power factor (PF) is a measure of how efficiently electrical power is being used. A higher PF means that there is less reactive power (KVArh) and more active power (KWh) being used. Therefore, PF is an important factor in accurately calculating KWh from KVArh.

4. Is it necessary to know the PF in order to calculate KWh from KVArh?

Yes, PF is a crucial factor in calculating KWh from KVArh. Without knowing the PF, the calculation will not be accurate. Therefore, it is important to have a measurement or estimation of PF in order to accurately determine KWh from KVArh.

5. Can KWh be calculated without using KVArh and PF?

No, KWh cannot be accurately calculated without using KVArh and PF. These two measurements are essential in determining the energy usage (KWh) as they take into account both the active and reactive power components of the electrical system.

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