How to Calculate Maximum Demand Savings Achieved by Power Factor
Correction
When a power factor correction proposal is in the pipeline, a question often asked it “What will be the
cost savings / payback period? If you can answer this question the proposal will more than likely be
accepted but answering this question is not always easy. Some utilities are very secretive about how
they charge for poor power factor and refer to “special contracts”. Some utilities have power factor
penalty charges and are quite transparent on how these charges. Some Utilities build this into their
maximum demand (kVA) charges which is what is being considered here.
Generally, customers can have one year of meter data, averaged over half hour intervals, provided by
their energy supplier and this data along with electricity accounts can be used to predict potential cost
savings.
The potential cost savings are determined as follows.
(1) Take kVA calculated from meter data collected at ICP Meter
(2) Take kW calculated from meter data collected at ICP meter and divide by 0.96 (new power
factor) to determine new kVA value
(3) Subtract new maximum kVA from old maximum kVA to determine maximum demand tariff
(kVA) savings. This has to be done in accordance with how maximum demand is charged. For
example, if there is a monthly maximum demand (kVA) charge, on a month by month basis,
subtract new maximum kVA from old maximum kVA to determine maximum demand tariff
(kVA) savings for that month.
(4) Multiply largest kVA saving by $X.XXX being the charge made by electricity supplier for dollars
per kVA per month.
An example is to be considered. The site has very poor power factor. In September 2016, the site was
being charged monthly $7.149 per maximum kVA per half hour period by their Electricity Supplier,
Contact Energy. Due to the large data files, the example will only consider one month of data. However,
when doing this for real, the potential savings for each month need to be calculated and totaled.
Below is meter data from a revenue meter.
We are given kWH and kVArH. As the value of kWH in cell C2 is kW per hour used over half an hour, the
value of kW (kWH/H) is C2*2. Likewise, as the value of kVArH in cell D2 is kVArH per hour used over half
an hour, the value of kVAr (kVArH/H) is D2*2.
As kVA = √(kW2+KVAr2) kVA for row 2 = SQRT(((C2*2)^2)+((D2*2)^2))
Do this calculation in cell E2
Then highlight cell E2 and click into corner as indicated by red circle to fill the rest of column. Next we
calculate power factor. As the ratio of kWH to KVARH is identical to kW/KVAR…
cosΦ = kwH/kVAH = kWH / √(kwH2+kVARH2) – calculate in column F as shown below and fill in column as
done with column E.
Next we calculate kW in column G by multiplying our calculated value of kVA by our calculated value of
PF. Fill in column F as done with column E.
We have calculated existing kVA. Our proposal is going to guarantee a power factor of 0.96, so our new
kVA will be will be no greater than kW/0.96. Calculate new kVA in column H as shown below and fill in
column H.
Copy column E and “Paste Values” into column I
Copy column H and ‘Paste Values” into column J
Highlight Column I and click onto
and sort this column only Largest to smallest. Then,
Highlight Column J and click onto
and sort this column only Largest to smallest.
We now have existing maximum demand for the month and new maximum demand with power factor
correction.
Subtract new maximum demand from existing maximum demand to calculate savings in maximum
demand.
So savings for the month of September will be 43.66kVA. The cost per kVA is $7.149 per month.
Savings for the month will be 43.66*$7.149 = $312.
If this is typical usage, savings over two years would be $7491. Now you need to do this for the other
eleven months of the year to work this out with some accuracy. Good luck and happy number
crunching. The example above is a real one.
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