Electrical Systems: Power Factor Correction
Resistive devices, like electric resistance heaters and incandescent lights transform all the
power supplied to the device into heat or useful energy. Inductive devices, like motors,
use some of the power supplied to the device to energize the inductive windings and
create a magnetic field. This power, called reactive power, is alternately stored and given
up by the windings, but is not used to do actual work. When this happens, the line
supplying power to the device now carries the actual power used by the device and the
reactive power created by the device.
Actual power used by the device is measured in kW, reactive power created by induction
devices is measured in kVAr, and the apparent power in the supply lines is measured in
kVA. The mathematical relationships between these types of power are described by the
“power triangle” shown below. For example,
Ps = (Pa2 + Pr2)1/2
Ps = supplied power (kVA)
Pr = reactive power (kVAr)

Pa = actual power (kW)
The ratio of the actual power consumed by equipment (Pa) to the power supplied to
equipment (Ps) is called the power factor.
PF = Pa / Ps = kW / kVA = cos 
Devices which generate/require large amounts of reactive power in relation to actual
power consumed have low power factors. Such devices include:



Motors
HID and fluorescent lights with low PF ballasts
Devices which convert AC power to DC power such as:
 DC drives
 Welding machines
 VFDs
 Induction furnaces
Fully loaded motors generally have a power factor of about 80%. However, if the motor
is under loaded, the fraction of reactive power (for the coil) to actual power (for
mechanical work) increases and the power factor decreases.
Two potential problems are associated with low power factor. First many utilities have
explicit or implicit charges for low power factor. Second, low power factor increases the
current, and hence losses, in transformers and the electrical distribution system. These
losses cost money and generate excess heat in the electrical distribution system, which
may shorten equipment lifetime or cause production shut downs. These potential
problems are discussed in the sections that follow.
Power Factor Charges
Many utilities charge for low power. To measure power factor, the most common type of
utility meter measures the total kVAr-hours and kVA-hours over the billing period and
calculates the average power factor as:
PF = Cos [ ArcSin (kVArh / kVAh) ]
The most common methods of charging for low power factor are:
1. Adding a demand penalty when the power factor dips below a set amount (usually
90%)
For example: the Cinergy DS rate specifies a demand penalty of:
 0.9 - PF 
kWactual
 when PF < 0.9
 PF 
If actual power was 100 kW, the power factor was 80%, and the avoided
cost of demand were $15.67 /kW, the monthly power factor charge would
be:
 0.9 - 0.8 
100 kW 
  12.5 kW
 0.8 
12.5 kW x $15.67 /kW = $196
2. Basing the demand charge on the supplied power Ps (kVA), rather than the actual
power used Pa (kW).
For example, assume billing demand is based on kVA rather than kW and
the demand charge is $15.00 /kVA-month. If actual demand was 100 kW,
the power factor was 80%, the implicit monthly power factor charge
would be:
kVAPF = 80% = kW / PF = 100 kW / 0.8 kW/kVA = 125 kVA
kVAPF = 100% = kW / PF = 100 kW / 1.0 kW/kVA = 100 kVA
Penalty = $15 /kVA-month x (KVAPF = 80% - KVAPF = 100%)
Penalty = $15 /kVA-month x (125 kVA – 100 kVa) = $375 /month
3. Basing part of the overall charge on the reactive power kVAr, which increases as
power factor decreases.
For example, the Dayton Power and Light General Service Primary Rate
specifies a charge of $0.30 per kVAr. The relationship between reactive
and actual power is:
Pr (kVAr) = Pa (kW) x Tan[Cos-1(PF)]
If the actual power was 100 kW and the power factor was 80%, then the
power factor charge would be:
Pr (kVAr) = 100 (kW) x Tan[Cos-1(0.8)] = 75 kVAr
75 kVAr x $0.30 /kVAr = $23
Power Losses and Excess Heat Generation
In addition to possible power factor charges, low power factor also results in excess
current in the electrical distribution system upstream from the device. The excess line
current results in increased resistive losses, and hence heat gain, in the wiring and
electrical distribution equipment. The quantity of line losses associated with low power
factor correction can be calculated as follows:
LL1 = Line loss before power factor correction
LL2 = Line loss after power factor correction
% Line Loss Savings = (LL1 – LL2) / LL1
LL1 = I21R1 = (kVA1/V1)2 R1 =[(kW1/PF1) / V1]2 R1 = [kW2 R / V2]1 / PF12
Thus:
LL2 = [kW2 R / V2]2 / PF22
Assuming everything remains constant except for the power factor:
[kW2 R / V2]1 = [kW2 R / V2]2 = [kW2 R / V2]
And,
% Line Loss Savings = (LL1 – LL2) / LL1
% Line Loss Savings = [(kW2 R / V2) / PF12 – (kW2 R / V2) / PF22] / (kW2 R / V2)1 / PF12
% Line Loss Savings = [1 / PF12 – 1 / PF22] / 1 / PF12
% Line Loss Savings = 1 – (PF1/ PF2)2
For example, if the power factor were improved from 80% to 90%, the percent line loss
savings would be:
% Line Loss Savings = 1 – (PF1/ PF2)2 = 1 – (80%/ 90%)2 = 21%
In addition, the heat generation in upstream electrical distribution equipment would be
reduced by 21%. This may or may not be significant. If the electrical circuits are fully
loaded and tripping due to excess current, then power factor correction could mitigate this
problem.
Although percent line loss savings are relatively high, total energy savings are typically
small since line losses are small. For example, if line losses are 2% of the total power
draw, the total power savings from correcting the power factor would be:
2% x 21% = 0.42%
Some manufactures of power factor correction equipment claim that actual losses are
much greater than those calculated here, but there is little documented evidence of this in
the open literature.
Because of these effects, it is generally in the client's interest to maintain a relatively high
power factor. To maintain a high power factor:


purchase equipment with high power factor ratings, such as high power factor
lighting ballasts
avoid or replace dramatically oversized motors, since under-loaded motors have
low power factors.
If power factor is still a problem, consider adding electrical capacitors.
Sizing Capacitors and Estimating Savings
Capacitors work by canceling reactive power and current on the primary or upstream side
of the capacitor. For example, if a motor operates at 70% power factor, installing a
capacitor in the power supply line to the motor would reduce reactive power and line
current on the primary side of the capacitor, but would not change the line current on the
secondary (motor) side of the capacitor. Thus, installing capacitors directly upstream
from low-power-factor loads reduces line current throughout the plant’s electrical
distribution system; whereas installing capacitors directly down stream of the utility
meter at the electrical service entrance to the plant, results in power factor correction for
utility billing purposes, but will not reduce line losses and overheating throughout the
plant.
Primary side of
capacitor
Secondary side
of capacitor
Capacitor
Motor
kVAprimary < kVAsecondary
PFprimary > PFsecondary
The two primary types of capacitors are oil-filled and gas-filled. Oil-filled capacitors
typically last about 60,000 hours and may introduce voltage transients to the downstream
equipment. In addition, the oil is potentially flammable and may contain toxins requiring
special disposal. Nitrogen and helium gas-filled capacitors last about 120,000 hours,
filter voltage transients, are non-flammable and non-toxic. However, the substantially
lower cost of oil-filled capacitors make them much more popular than gas-filled
capacitors.
Capacitors are sized by the amount of reactive power (kVAr) they can cancel. Simple
capacitors are sized to compensate for a fixed amount of power. “Stepped” capacitors
have internal controls that adjust the amount of reactive power compensation.
Adding too much capacitance can push the system from “lagging” to “leading”; for
example, adding too much capacitance may change the power factor from 95% to 105%.
Although leading power factor does not harm equipment, purchasing excess capacitors is
expensive and serves no useful purpose. In addition some meters may read a leading
105% power factor as 95%. If so, you would not get credit for the power factor
correction from 95% to 100%. (Other utility meters would read a power factor of 105%
as 100%). Because of these reasons, we recommend a conservative approach to power
factor correction in which we never overcorrect the power factor past 100%.
A simple method to size the amount of capacitor kVAr required is described in the steps
that follow:
1. Find kVAr for each month: Pr (kVAr) = Pa (kW) x Tan[Cos-1(PF)]
2. To increase PF as close to 1.0 as possible, recommend additional capacitance equal to
minimum monthly kVAr during the past 12 months. This approach minimizes the
possibility of adding too much capacitance.
3. Subtract the recommended capacitance (kVAr) from recorded (kVAr) for each
month. This difference represents the reactive power (kVAr) if the recommended
capacitance were added.
4. Recalculate PF, kVA or kVAr and electricity costs for each month, using the reactive
power calculated in the previous step. These costs represent the costs if the
recommended capacitance had been added.
5. Calculate savings as the difference between the actual costs and the costs calculated
in the previous step.
6. To estimate the implentation cost, we note that the installed cost of capacitors is about
about $20 /kVAr - $50 /kVAr, depending on control complexity and size.
Examples follow:
AR x: Consider Installing Capacitors to Improve Power Factor
Annual Savings
Project Cost
CO2
Resource
Dollars Capital Other
Total
(tonne)
Electrical Fees
None $12,305 $46,000 $46,000 $92,000
ARC: 2.3212.3
Simple
Payback
90 months
Analysis
Electrical equipment that generates inductive loads, such as motors, creates current that is
not in phase with the current supplied by the electric utility. Inductive loads cause the
current waveform to lag behind the voltage waveform. This causes a portion of the
energy to return to the source, hence leaving less usable power for the equipment. The
power associated with this unusable current is called reactive power (kVAr). Because
some energy is returned to the source, a higher supply power (kVA) must be generated by
the utility in order to meet equipment needs. The ratio of power consumed by equipment
(kW) to total power in the electrical lines (kVA) is called the power factor. Most utilities
charge for low power factor. Low power factor can be corrected by adding capacitors,
which are rated in terms of kVAr.
Power Factor of 1. All Power is usable.
Power factor of 0.71. Unusable power is created.
(Source: www.wikipedia.com)
The diagram below shows the relationship between the various types of power: supplied
power (kVA), reactive power (kVAr), and actual used power (kW). The quantity of each
type of power can be calculated using trigonometric relations defined by the power
triangle.
kVA (power
supplied
by utility)
kW
PF =
kVA
kVAr (measure of
unused power)
kW (actual power used by machines)
Power factor is ratio of the actual power, kW, and supplied power, kVA. The power
factor can be calculated using the following relationships:
PF = kW / kVA = cos (tan-1 (kVAr/kW))
(1)
Reactive power is a measure of the unused power in the lines. The reactive power can be
calculated using the following relationship:
Reactive Power (kVAr) = kW x tan (cos-1 (PF))
(2)
The supplied power (kVA) can be calculated from the reactive power (kVAr) and power
factor from the following relationship:
kVA = kVAr/sin (cos-1(PF))
(3)
According to your rate structure “Large General Service Rate”, the power factor charge is
about $0.48 per kVAr per month. Your utility company charges for all kVAr in excess of
10% of the billed demand. According to the utility bills, the current power factor
averages about 0.85. Savings would be achieved if the power factor were increased from
0.85 to 1.0.
Recommendation
We recommend considering adding an additional 2,300 kVAr of capacitance to improve
the average power factor.
Estimated Savings
The data listed below is extracted from the utility bills for the years 2006 and 2007. The
utility bills summary shows actual demand (kW) and your current lagging reactive
demand (kVAr). According to your rate structure, the power factor charge is about $0.48
/kVAr in excess of 10% of billed demand. As indicated in the table below, installing
2,300 kVAr of capacitance would increase your average annual power factor from 0.85
(kW/kVA) to 1.0 (kW/kVA). Savings are calculated as the product of power factor
charge ($/kVA) and the current billed reactive demand (kVA).
Meter
Reading
Date
8/27/07
9/26/07
10/26/07
11/27/07
12/27/07
1/25/08
2/26/08
3/27/08
4/25/08
5/27/08
6/25/08
7/25/08
Tot/Avg
Billed
Unit Power
Metered
Actual
Actual Lagging
Power
Proposed
Reactive
Factor
Consumption Demand
Reactive
Factor
Capacitance
Demand
Charge
(kWh/period) (kW)
Demand (kVAr) (kW/kVA)
(kVAr)
(kVAr)
($/kVA)
2,780,825
4,219
2,065
2,538
0.86
$0.48
2,300
2,623,534
4,238
2,057
2,532
0.86
$0.48
2,300
2,669,789
4,185
2,247
2,719
0.84
$0.48
2,300
2,959,728
4,240
2,114
2,590
0.85
$0.48
2,300
2,616,545
4,235
2,138
2,612
0.85
$0.48
2,300
2,634,931
4,235
2,138
2,612
0.85
$0.48
2,300
2,936,662
4,443
2,138
2,624
0.86
$0.48
2,300
2,653,735
4,235
2,138
2,612
0.85
$0.48
2,300
2,566,901
4,235
2,138
2,612
0.85
$0.48
2,300
2,584,217
4,081
2,081
2,540
0.85
$0.48
2,300
2,311,135
4,251
2,136
2,614
0.85
$0.48
2,300
0.84
2,145,290
4,217
2,262
2,739
$0.48
2,300
31,483,292
4,234
2,300
2,138
2,612
0.85
$0.48
New
Reactive
Demand
(kVAr)
238
232
419
290
312
312
324
312
312
240
314
439
312
New Billed
New Power
Reactive
Factor
Demand
(kW/kVA)
(kVAr)
0
1.00
0
1.00
1
1.00
0
1.00
0
1.00
0
1.00
0
1.00
0
1.00
0
1.00
0
1.00
0
1.00
17
0.99
2
1.00
Savings
($/Period)
$991
$987
$1,078
$1,015
$1,026
$1,026
$1,026
$1,026
$1,026
$999
$1,025
$1,078
$12,305
The table shows that 1,300 kVAr of additional capacitance would be sufficient to avoid
almost all power factor penalties, and the savings from installing additional capacitance
would be about $12,305 per year. Hence, we propose adding an additional 2,300 kVAr of
capacitance.
Estimated Implementation Cost
Based on previous experience, simple capacitors can be installed for about $40 per kVAr.
Therefore, the estimated implementation cost would be about:
(2,300 kVAr x $40 /kVAr) = $92,000
We estimate that about 50% of this is for materials and 50% for labor.
Estimated Simple Payback
($92,000 / $12,305 /year) x 12 months = 90 months
AR x: Install Capacitors to Improve Power Factor
Annual Savings
Project Cost
CO2
Resource
Dollars Capital Other
Total
(tonne)
Electrical Fees
None $26,537 $10,625 $10,625 $21,250
ARC: 2.3212.3
Simple
Payback
10 months
Analysis
Electrical equipment that generates inductive loads, such as motors, generates current that
is not in phase with the current supplied by the electric utility. The power associated with
this unusable current is called reactive power (kVAr). The total power in the line (kVA)
includes both the useable (kW) and the reactive power (kVAr). The ratio of power
consumed by equipment (kW) to total power in the electrical lines (kVA) is called the
power factor. Most utilities charge for low power factor. Low power factor can be
corrected by adding capacitors, which are rated in terms of kVAr.
The diagram below shows the relationship between the various types of power: total
power (kVA), reactive power (kVAr), and actual used power (kW). The quantity of each
type of power can be calculated using trigonometric relations defined by the power
triangle.
kVA (total
power)
kW
PF =
kVA
kVAr (reactive
unusable power)
kW (actual power used by machines)
Power factor is ratio of the actual power, kW, and total power, kVA. The power factor
can be calculated using the following relationships:
PF = kW / kVA = cos (tan-1 (kVAr/kW))
(1)
The reactive power can be calculated using the following relationship:
Reactive Power (kVAr) = kW x tan (cos-1 (PF))
(2)
The total power (kVA) can be calculated from the reactive power (kVAr) and power
factor from the following relationship:
kVA = kVAr/sin (cos-1(PF))
(3)
In the “Utility Analysis” section of this report, both on-peak and off-peak power factors
less than 0.9 increase the billed demand. The cost for power factor below 0.9 during onpeak is:
= (measured on-peak kW) / (on-peak PF) x 0.9 x (on-peak demand cost)
The cost for power factor below 0.9 during off-peak is:
= {[(max kW) / off-peak PF x 0.9] - [(measured on-peak kW) / (on-peak PF) x 0.9]} x
(off-peak demand cost)
Recommendation
We recommend considering adding an additional 425 kVAr to improve the average
power factor.
Estimated Savings
The table below shows on-peak and off-peak actual demand (kW), power factor
(kW/kVA), supplied power (kVA), reactive power (kVAr), and billing demand (kW).
The on-peak and off-peak power factor remains lower than 0.9 for the entire year. Using
the rate structure form the “Utility Analysis” section of this report, the cost for this low
power factor is about $26,537 per year.
Meter
Reading
Date
Days
On-peak
Demand
(kW)
Off-Peak
Demand
(kW)
9/11/06
10/10/06
11/8/06
12/11/06
1/12/07
2/12/07
3/12/07
4/12/07
5/11/07
6/12/07
7/12/07
8/10/07
Tot/Avg
32
31
29
33
32
31
28
31
29
32
30
29
367
864
862
842
838
810
842
804
814
823
827
806
845
825
860
864
840
838
812
827
806
827
817
827
793
845
830
On-peak
Power
Factor
(kW/kVA)
0.73
0.73
0.73
0.73
0.74
0.76
0.74
0.76
0.74
0.75
0.73
0.75
0.74
Off-Peak
Power
Factor
(kW/kVA)
0.74
0.73
0.73
0.74
0.75
0.74
0.75
0.75
0.73
0.74
0.73
0.74
0.74
On-peak
Supplied
Power
(kVA)
1,187
1,174
1,157
1,145
1,093
1,114
1,093
1,076
1,106
1,107
1,099
1,123
1,123
Off-peak
Supplied
Power
(kVA)
1,157
1,192
1,151
1,130
1,077
1,115
1,076
1,103
1,117
1,112
1,084
1,146
1,122
On-Peak
Reactive
Power
(kVAr)
814
797
793
780
734
729
741
703
739
736
748
740
755
Off-Peak
Reactive
Power
(kVAr)
774
821
787
757
708
747
713
730
762
743
740
775
755
On-Peak
Billing
Demand
(kW)
1,068
1,057
1,041
1,030
984
1,003
984
968
996
997
989
1,011
1,011
Off-Peak
Billing
Demand
(kW)
0
16
0
0
0
1
0
25
10
4
0
21
6
Calculated
Power Factor
Cost ($/period)
$2,603
$2,371
$2,402
$2,322
$2,098
$1,937
$2,177
$1,885
$2,094
$2,165
$2,340
$2,143
$26,537
The on-peak and off-peak reactive power (kVAr) measured at the meter will be reduced
by 425 kVAr per month by installing 425 kVAr of capacitance. This is depicted in the
table below. 425 kVAr of capacitance will raise the on-peak and off-peak power factor
above 0.9. Therefore, the annual power factor cost would be about $0 per year.
Meter
Reading
Date
Days
On-peak
Demand
(kW)
Off-Peak
Demand
(kW)
9/11/06
10/10/06
11/8/06
12/11/06
1/12/07
2/12/07
3/12/07
4/12/07
5/11/07
6/12/07
7/12/07
8/10/07
Tot/Avg
32
31
29
33
32
31
28
31
29
32
30
29
367
864
862
842
838
810
842
804
814
823
827
806
845
825
860
864
840
838
812
827
806
827
817
827
793
845
830
On-peak
Power
Factor
(kW/kVA)
0.91
0.92
0.92
0.92
0.93
0.94
0.93
0.95
0.93
0.94
0.93
0.94
0.93
Off-Peak
Power
Factor
(kW/kVA)
0.93
0.91
0.92
0.93
0.94
0.93
0.94
0.94
0.92
0.93
0.93
0.92
0.93
On-peak
Supplied
Power
(kVA)
947
939
919
910
867
896
864
860
881
884
868
902
895
Off-peak
Supplied
Power
(kVA)
928
950
915
902
860
888
856
882
883
886
853
914
893
On-Peak
Reactive
Power
(kVAr)
389
372
368
355
309
304
316
278
314
311
323
315
330
Off-Peak
Reactive
Power
(kVAr)
349
396
362
332
283
322
288
305
337
318
315
350
330
On-Peak
Billing
Demand
(kW)
864
862
842
838
810
842
804
814
823
827
806
845
831
Off-Peak
Billing
Demand
(kW)
0
0
0
0
0
0
0
0
0
0
0
0
0
Calculated
Power Factor
Cost ($/period)
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
The annual savings for installing 425 kVAr of capacitance would be about:
$26,537 /yr - $0 /yr = $26,537 /yr
Estimated Implementation Cost
Based on previous experience, simple capacitors can be installed for about $50 per kVAr.
Therefore, the estimated implementation cost for installing 425 kVAr of capacitance
would be about:
425 kVAr x $50 /kVAr = $21,250
We estimate that about 50% of this is for materials and 50% for labor.
Estimated Simple Payback
($21,250 / $26,537 /yr) x 12 months = 10 months
AR x: Install Capacitors to Improve Power Factor
Annual Savings
Project Cost
CO2
Resource
Dollars Capital Other
Total
(tonne)
Electrical Fees
None $44,362 $6,750 $6,750 $13,500
ARC: 2.3212.3
Simple
Payback
4 months
Analysis
Inductive loads, such as motors, generate a reactive current that is not in phase with the
current supplied by the utility, hence, it is unusable. The power associated with this
reactive current is called reactive power, and is measured in kVAr. The useful power
drawn by machines and supplied by the utility is measured in kW. The total power in the
lines is the sum of the useful and reactive power and is measured in kVA. The ratio of
power consumed by equipment (kW) to power in the lines (kVA) is called the power
factor. Most utilities charge for low power factor. Low power factor can be corrected by
adding capacitors, which are rated in terms of kVAr.
The diagram below shows the relationship between the various types of power. The
quantity of each type of power can be calculated using trigonometric relations defined by
the power triangle.
kVA (line
power)
kW
PF =
kVA
kVAr (reactive
power)
kW (actual power used by machines)
The power factor can be calculated using the following relationships:
PF = kW / kVA = cos (tan-1 (kVAr/kW))
(1)
Reactive power is a measure of the unused power in the lines. The reactive power can be
calculated using the following relationship:
Reactive Power (kVAr) = kW x tan (cos-1 (PF))
(2)
The line power (kVA) can be calculated from the reactive power (kVAr) and power
factor from the following relationship:
kVA = kVAr/sin (cos-1(PF))
(3)
In the utility analysis section of this report, the power factor charge is kVA x 0.9. The
utility rate for meter 1 and 3, “General Rate DS01 Service at Secondary Distribution
Voltage”, the monthly power factor charge for each meter with a power factor less than
0.9 and a billed demand less than 1,000 kW is:
Billed kW/mo x 300 kWh/kW x ($0.033076 - $0.020094) kWh + Billed kW x $15.1454
/kW
The monthly power factor charge for each meter with a power factor less than 0.9 and a
billed demand greater than 1,000 kW is:
Billed kW/mo x 300 kWh/kW x ($0.033076 - $0.020094) kWh + Billed kW x $15.1454
/kW
According to the utility bills, the current power factors for meter 1 & 3 are less than 0.9
for every month. Savings would be achieved if the power factor were increased to 0.90.
Recommendation
We recommend considering adding an additional 250 kVAr to Meter 1 and an additional
200 kVAr of capacitance to Meter 3 to improve the average power factor for each meter.
Estimated Savings
The table below shows meter 1 actual demand (kW) current reactive demand (kVAr),
current power factor and the total calculated electrical costs based upon the current rate
structure for the previous seven billing periods. The calculated electrical cost for meter 1
is about $290,089 for the last seven billing periods.
Meter 1
Meter
Reading
Date
Days
Metered
Consumption
(kWh/period)
Avg. Daily
Consumption
(kWh/day)
Actual
Demand (kW)
Supplied
Power
(kVA)
10/5/2007
11/5/2007
12/6/2007
1/9/2008
2/7/2008
3/7/2008
4/8/2008
29
31
31
34
29
29
32
240,064
508,236
596,096
678,600
724,944
418,725
482,823
8,278
16,395
19,229
19,959
24,998
14,439
15,088
803
976
1,062
1,256
1,317
853
979
1,005
1,192
1,336
1,549
1,586
1,100
1,243
Current
Reactive
Demand
(kVAr)
604
685
810
907
883
694
765
Tot/Avg
215
3,649,488
16,974
1,035
1,287
764
0.80
0.82
0.80
0.81
0.83
0.78
0.79
Power
Factor
Penality
(kW)
102
97
140
138
110
137
139
0.80
863.08
Power
Factor
(kW/KVA)
Billed
Demand
(kW)
Calculated Total
Cost ($/period)
905
1,073
1,202
1,394
1,427
990
1,118
$26,402
$39,694
$45,123
$51,166
$53,369
$34,840
$39,495
1,158
$290,089
The table below shows meter 1 actual demand (kW) new reactive demand (kVAr), new
power factor and the total calculated electrical costs based upon the current rate structure
for the previous seven billing periods. The calculated electrical cost for meter 1 by adding
250 kVAr of reactive power would be about $276,063 for the last seven billing periods.
Meter 1
Meter
Reading
Date
Days
Metered
Consumption
(kWh/period)
Avg. Daily
Consumption
(kWh/day)
Actual
Demand (kW)
10/5/2007
11/5/2007
12/6/2007
1/9/2008
2/7/2008
3/7/2008
4/8/2008
29
31
31
34
29
29
32
240,064
508,236
596,096
678,600
724,944
418,725
482,823
8,278
16,395
19,229
19,959
24,998
14,439
15,088
803
976
1062
1256
1317
853
979
New
Supplied
Power
(kVA)
878
1084
1180
1395
1463
947
1087
Tot/Avg
215
3,649,488
16,974
1,035
1,148
354
435
560
657
633
444
515
0.91
0.90
0.90
0.90
0.90
0.90
0.90
Power
Factor
Penality
(kW)
0.00
0.00
0.00
0.00
0.00
0.00
0.00
514
0.90
0.00
New Reactive
Power
Demand
Factor
(kVAr)
(kW/KVA)
Billed
Demand
(kW)
Calculated Total
Cost ($/period)
803
976
1,062
1,256
1,317
853
979
$24,469
$37,990
$42,863
$49,312
$52,116
$32,231
$37,082
1,035
$276,063
The table below shows meter 3 actual demand (kW) current reactive demand (kVAr),
current power factor and the total calculated electrical costs based upon the current rate
structure for the previous seven billing periods. The calculated electrical cost for meter 3
is about $94,078 for the last seven billing periods.
Meter 3
Meter
Reading
Date
Days
Metered
Consumption
(kWh/period)
Avg. Daily
Consumption
(kWh/day)
Actual
Demand (kW)
Supplied
Power
(kVA)
10/5/2007
11/5/2007
12/6/2007
1/9/2008
2/7/2008
3/7/2008
4/8/2008
29
31
31
34
29
29
32
73,871
128,492
159,205
183,097
177,265
124,927
147,326
2,547
4,145
5,136
5,385
6,113
4,308
4,604
232
282
362
368
387
309
368
296.3
417.8
494.7
547
548
437
527
Current
Reactive
Demand
(kVAr)
184
309
338
405
387
309
378
Tot/Avg
215
994,183
4,624
330
467
330
0.78
0.67
0.73
0.67
0.71
0.71
0.70
Power
Factor
Penality
(kW)
34.67
94.42
83.60
124.39
105.64
84.23
106.57
0.71
633.52
Power
Factor
(kW/KVA)
Billed
Demand
(kW)
Calculated Total
Cost ($/period)
267
376
445
492
493
393
475
7,919
12,084
14,573
16,382
16,169
12,272
14,679
420
$94,078
The table below shows meter 3 actual demand (kW) new reactive demand (kVAr), new
power factor and the total calculated electrical costs based upon the current rate structure
for the previous seven billing periods. The calculated electrical cost for meter 3 by adding
200 kVAr of reactive power would be about $82,226 for the last seven billing periods.
Meter
Reading
Date
Days
Metered
Consumption
(kWh/period)
Avg. Daily
Consumption
(kWh/day)
Actual
Demand (kW)
Supplied
Power
(kVA)
10/5/2007
11/5/2007
12/6/2007
1/9/2008
2/7/2008
3/7/2008
4/8/2008
29
31
31
34
29
29
32
73,871
128,492
159,205
183,097
177,265
124,927
147,326
2,547
4,145
5,136
5,385
6,113
4,308
4,604
232
282
362
368
387
309
368
232.0
301.8
386.9
421.2
430.1
327.4
408.6
0
109
138
205
187
109
178
1.00
0.93
0.93
0.87
0.90
0.94
0.90
Power
Factor
Penality
(kW)
0.00
0.00
0.00
11.05
0.00
0.00
0.00
Tot/Avg
215
994,183
4,624
330
358
132
0.93
11.05
New Reactive
Power
Demand
Factor
(kVAr)
(kW/KVA)
Billed
Demand
(kW)
Calculated Total
Cost ($/period)
232
282
362
379
387
309
368
$7,259
$10,286
$12,981
$14,224
$14,157
$10,668
$12,650
331
$82,226
The annual savings for installing 250 kVAr of capacitance to meter 1 and 200 kVAr of
capacitance to meter 3 would be about:
($290,089 - $276,063) / 7 mo + ($94,078 - $82,226) / 7 mo = $25,878 / 7 mo
$25,878 / 7 mo x 12 mo/yr = $44,362 /yr
Estimated Implementation Cost
Based on previous experience, simple capacitors can be installed for about $30 per kVAr.
Therefore, the estimated implementation cost for installing 400 kVAr of capacitance
would be about:
(450 kVAr x $30 /kVAr) = $13,500
We estimate that about 50% of this is for materials and 50% for labor.
Estimated Simple Payback
($13,500/ $41,595/year) x 12 months = 4 months
AR I632: TAKE CORRECTIVE ACTIONS TO IMPROVE POWER FACTOR
Annual Savings
Project
Simple
Cost
Payback
Resource
CO2 (lb)
Dollars
Electrical Demand
437 kVA
$94,812
$8,000
1 month
Analysis and Recommendation
Motors and other inductance devices require more power in the lines than they actually
consume. Power factor is ratio of power used (kW) to the power that must be supplied in
the lines (kVA). The relationship between actual, supplied and reactive power is shown
in the figure below.
kVA
(power supplied
by utility)
PF =
kW
kVA
kVAr
(measure of
unused power)
kW (power used by machines)
According to the your electricity bills, the average plant power factor is about 68%.
Based on your electric rate structure, you can lower your demand charges by improving
power factor. However, you do not want to over correct the power factor to more than
100%. In March 2000, power factor peaked at 74%. Thus, we recommend adding only
enough capacitance to raise your power factor to 97% for that month. According to the
analysis below, this would be about 400 kVAr of capacitance.
Estimated Savings
Determining How Much Capacitance to Add
The highest power factor registered during the period for which we were supplied with
electricity data was for March 2000, when the plant power factor was 74% and the peak
power was 594 kW. If the power factor during this month were corrected to 97%, the
power factor during the rest of the year would never exceed 97%. Using standard
trigonometric relations, the angle theta in the power triangle shown above was about:
Cos = kW / kVA
 = Cos-1 (PF)
 = Cos-1 (74%) = 42.27 o
The reactive power was about:
Tan = kVAr / kW
kVAr = kW x tan
kVAr = 594 kW x tan (42.27) = 540 kVAr
If the power factor were increased to 97%, the reactive power would be about:
Cos = kW / kVA
 = Cos-1 (PF)
 = Cos-1 (97%) = 14.07 o
kVAr = kW x tan
kVAr = 594 kW x tan (14.07) = 149 kVAr
Thus, the amount of capacitance required to boost your power factor from 74% to 97%
during March 2000 would be about:
540 kVAr – 149 kVAr = 391 kVAr
We recommend adding about 400 kVAr.
Calculating Savings
According to your electricity bills, the annual average power factor is 68% and the annual
average demand is 579 kW. Using standard trigonometric relations, the angle theta in the
power triangle shown above is about:
Cos = kW / kVA
 = Cos-1 (PF)
 = Cos-1 (68%) = 47.156 o
The reactive power is about:
Tan = kVAr / kW
kVAr = kW x tan
kVAr = 579 kW x tan (47.156) = 624 kVAr
If 400 kVAr of capacitance were added, the average kVAr would be about:
624 kVAr – 400 kVAr = 224 kVAr
The angle theta would be about:
 = Tan-1 (kVAr/kW) = Tan-1 (224/579) = 21.150 o
The annual average power factor would be about:
PF = Cos = Cos (21.150) = 93%
The annual average kVA would be about:
kVA = kW / Cos = 579 / Cos(21.150 o) = 621 kVA
The billing kVA would be about:
621 kVA x [1 + (0.9 - 93%)] = 602 kVA
According to your electricity bills, the annual average power factor is 68% and the annual
average kVA is 852 kVA. Thus, without power factor correction the annual average
billing kVA would be about:
Actual kVA x [1 + (0.9 – PF)] = 852 kW x [1 + (0.9 – 68%)] = 1,039 kVA
Thus, the savings from correcting the average power factor to 93% would be about:
1,039 kVA – 602 kVA = 437 kVA
437 kVA x $18.08 /kVA-month x 12 months/yr = $94,812 /yr
Estimated Implementation Cost
According to quotes from other utilities, the total installed cost for simple capacitors is no
more than $20 per kVAr. If so, the total implementation cost for installing 400 kVAr of
capacitors would be about:
400 kVAr x $20 /kVAr = $8,000
Estimated Simple Payback
$8,000 / $94,812 /yr x 12 months/yr = 1 month
AR E061501: TAKE CORRECTIVE ACTIONS TO IMPROVE POWER FACTOR
Annual Savings
Project
Simple
Cost
Payback
Resource
CO2 (lb)
Dollars
Electric Demand
42 kVA
$3,669
$2,800
9 months
Analysis and Recommendation
Motors and other inductance devices require more power in the lines than they actually
consume. Power factor is ratio of power used (kW) to the power that must be supplied in
the lines (kVA). The relationship between actual, supplied and reactive power is shown
in the figure below.
kVA
(power supplied
by utility)
kW
PF =
kVA
kVAr
(measure of
unused power)
kW (power used by machines)
According to the your electricity bills, the average plant power factor is about 92.4%.
Based on your electric rate structure, you can lower your demand charges by improving
power factor. However, you do not want to over correct the power factor to more than
100%. In February 2000, power factor peaked at 95%. Thus, we recommend adding
only enough capacitance to raise your power factor to 99% for that month. According to
the analysis below, this would be about 140 kVAr of capacitance.
Estimated Savings
During February 2000, the plant power factor was 95% and the peak power was 765 kW.
Using standard trigonometric relations, the angle theta in the power triangle shown above
was about:
Cos = kW / kVA
 = Cos-1 (PF)
 = Cos-1 (95%) = 18.195 o
The reactive power was about:
Tan = kVAr / kW
kVAr = kW x tan
kVAr = 765 kW x tan (18.195) = 251 kVAr
If the power factor were increased to 99%, the reactive power would be about:
Cos = kW / kVA
 = Cos-1 (PF)
 = Cos-1 (99%) = 8.11 o
kVAr = kW x tan
kVAr = 765 x tan (8.11) = 109 kVAr
Thus, the amount of capacitance required to boost your power factor from 95% to 99%
during February 2000 would be about:
251 kVAr – 109 kVAr = 142 kVAr
We recommend adding about 140 kVAr. According to the your electricity bills, the
average plant power factor is about 92.4%. The average power supplied is about 811
kVA and the average power consumed is about 750 kW. Using these numbers, the
average reactive power during the year is about:
Cos = kW / kVA
 = Cos-1 (PF)
 = Cos-1 (92.4%) = 22.482 o
Tan = KVAr / kW
kVAr = kW x tan
kVAr = 750 x tan (22.482) = 310 kVAr
If 140 kVAr of capacitance were added, the average reactive power would be about:
kVAr2 = 310 kVAr – 140 kVAr = 170 kVAr
The average power supplied would be about:
2 = Tan-1(kVAr / kW) = Tan-1(170 / 750) = 12.771 o
kVA = kW / Cos2 = 750 kW / Cos(12.771) = 769 kVA
The average reduction in kVA would be about:
811 kVA – 769 kVA = 42 kVA
Thus, the power factor savings would be about:
42 kVA x $7.28 /kVA–mo x 12 mo/yr = $3,669 /yr
Estimated Implementation Cost
According to quotes from other utilities, the total installed cost for capacitors about $20
/kVAr. If so, the total implementation cost for installing 140 kVAr of capacitors would
be about:
140 kVAr x $20 /kVAr = $2,800
Estimated Simple Payback
$2,800 / $3,669 /yr x 12 months/yr = 9 months
AR E062901: INSTALL 500 KVAR OF CAPACITANCE TO IMPROVE POWER
FACTOR
Annual Savings
Project
Simple
Cost
Payback
Resource CO2 (lb)
Dollars
Power factor charge
500 kVAr
$6,000
$7,500
15 months
Analysis
Power factor is the ratio of actual power consumed (kW) to the line power supplied
(kVA). The utility recorded an annual average power factor of 83% for the electricity
service to the plant. Most plants strive to achieve a power factor of 95% or more to
reduce demand costs, reduce line losses and allow electrical equipment to run cooler.
kVA (power
supplied
by utility)
kW
PF =
kVA
kVAr (measure of
unused power)
kW (power used by machines)
Recommendation
We recommend installing at least 500 kVAr of capacitance to raise the plant power factor
and save on electricity charges.
Estimated Savings
Monthly demand, reactive power and power factor from the last year of billing data is
shown in the table below.
KW
1,322
1,258
1,243
1,188
1,298
1,181
1,084
1,175
1,231
1,154
1,064
1,163
Tot/Avg/Max
KVAr
522
662
932
984
1,058
886
743
774
763
805
701
767
9,598
PF
93%
89%
80%
77%
78%
80%
83%
84%
85%
82%
84%
84%
82%
The monthly average reactive power varies from a low of 522 kVAr to a high of 1,058
kVAr. Thus, it would be highly unlikely that you would ever over-correct your power
factor by installing 500 kVAr of capacitance. If you did, the annual savings would be
about:
500 kVAr x $1 /kVAR-mo x 12 mo/yr = $6,000
Estimated Implementation Cost
The contract price for installation of simple capacitors varies from $10 to $20 per kVAr.
Assuming $15 /kVAr, the implementation cost would be about:
500 kVAr x $15 /kVAr = $7,500
Estimated Simple Payback
$7,500 / $6,000 /yr x 12 months/yr = 15 months
AR X: INSTALL CAPACITORS TO IMPROVE POWER FACTOR
PF charge
Implementation Cost
Simple Payback
Present
27,048 kVAr/yr
Recommended
3,048 kVAr/yr
Annual Savings
24,000 kVAr; $7,200
$60,000
8.3 years
Analysis
Many types of electrical equipment require that more power be supplied to the equipment
than is actually consumed by the equipment. The ratio of power consumed by equipment
to power supplied to equipment is called the power factor. Your utility charges you for
both how much power your plant actually uses (measured in kW) and for the excess
power that the utility was required to supply (a measure of which is kVAr). Thus, low
power factor results in higher utility costs. In addition, low power results in overheating
in wires, circuit boards and motors. This overheating may cause circuit breakers to shut
off the load and interrupt production, and always reduces the lifetime and increases
maintenance costs for electrical equipment. It is therefore in your interest to maintain a
power factor as close to 100% as is economically feasible.
kVA (power
supplied
by utility)
kW
PF = kVA
kVAr (measure of
unused power)
kW (power used by machines)
The utility recorded an annual average power factor of 79% for the electricity service to
the main plant. Most industrial facilities have a power factor of 90% or above. A 79%
power factor corresponds to an average of 2,254 kVAr per month, for which the utility
charges 30 cents per kVAr. Thus, your average power factor charge is about $676 per
month or about $8,114 per year. Installing 2,000 kVAr of capacitors would raise the
average monthly power factor to about 99% without ever switching from lagging to
leading. The annual utility savings would be about $7,200.
Additional savings from longer equipment lifetimes, lower maintenance costs, and
avoided production stoppages are difficult to quantify, but are potentially even greater
than the utility savings. We think that this is especially critical in your plant where the
general difficulty of obtaining spare parts can put machines out of production for
extended periods. In addition, your past problem with circuit overheating was worsened
by your plant’s low power factor.
Recommendation
We recommend that you contact your utility about a power quality survey to identify the
circuits with the lowest power factors. You could then install corrective capacitance on
those circuits or on the main utility feed. In this way, you could raise the power factor
into the acceptable range of 90% or greater. This will increase the lifetime and efficiency
of your equipment and help minimize production shutdowns. It will also provide modest
savings on your utility bill.
Our recommendation of 2,000 kVAr represents the maximum capacitance you currently
need. Lesser amounts of capacitance would reduce your utility savings accordingly. We
estimate the cost of 2,000 kVAr to be approximately $60,000 with simple controls. It
could be as high as $100,000 with more complicated controls and features. Based on
utility savings alone, the simple payback would be 8.3 years; however, the payback
would be shorter if production and equipment maintenance savings were included.
Estimated Savings
2,000 kVAr/month x 12 months/year x $0.30 /KVAR = $7,200
Estimated Implementation Cost
2,000 KVAr x $30 /KVAr = $60,000
Estimated Simple Payback
$60,000
Simple Payback =
= 8.3 years
$7,200 / yr
AR X: INSTALL TWO 100-KVAR CAPACITOR BANKS
Present
Electric Demand (kW)
Estimated Implementation Cost:
Simple Payback:
Recommended
Annual Savings
476 kVA; $3,570
$1,000 for installation
4 months
Analysis
Your electric demand charge is assessed in terms of kVA, not in kW. This means that the
electrical demand charge directly depends on the overall power factor in your plant. A
high power factor results in several benefits: lower demand costs, lower voltage drop in
the circuit, lower overheating in wiring, and less maintenance. The power factor in the
last 12 months varied between 88.8% and 97.9%. Some capacitor banks were already
installed on bus bars high above the floor. You also have five 100-kVAR capacitor banks
in stock.
Recommendations
We tried to strike a balance between raising power factor and not to cross to “lead” power
factor. Our analysis indicated that you could install two 100-kVAR capacitor banks to
your circuits, preferrably at locations where some low power factor equipment items such
as welders, rectifiers, and HID lamps with low power factor ballasts, are located. In
addition, we have learned from the experience in other plants that an audio failure
warning system allows for instant detection of capacitor malfunction and thus helps avoid
excessive cost penalties if a capacitor bank fails. Without audio failure systems, it could
be many months before you discover a capacitor bank failure.
Estimated Cost Savings
Present
The billing demand varied between 829 kVA and 1,382 kVA, totalling 13,654 kVA for
the year (August 94 to July 95).
After adding two 100-kVAR capacitor banks
The annual total billing demand will be lowered from 13,654 kVA to 13,088 kVA for a
reduction of 476 kVA
Savings
476 kVA x $7.50/kVA = $3,570/year in annual demand cost savings
Estimated Implementation Cost
$1,000 for installing two 100-kVAR capacitor banks
Estimated Simple Payback
4 months
AR X: INSTALL CAPACITORS TO RAISE POWER FACTOR
Electric Demand (kW)
Estimated Implementation Cost:
Simple Payback:
Present
2,003 kW
After AR
0 kW
Annual Savings
2,003 kW; $25,819
$30,000
14 months
Analysis and Recommendation
Your monthly power factor during the past 12 months was consistently below the 90%
level required by your utility company. As a result, a total of 2,003 kW in accumulated
power factor related demand penalties were assessed resulting in $25,819 in cost
penalties. The power factor varied between 83.3% and 85.7%. This low power factor
problem could be corrected by adding capacitor banks at the source of the problem, the
electric main, or both.
When large electric motors are operated below 50% of their rated capacity, their power
factor decreases signficantly. The thermo-mechanical refiner was equipped with a huge
700 hp motor. Other large motors include a 500 hp motor at the hydropulper, 400 hp and
a 150 hp motors at wood refiner, and a 100 hp motor at primary screen. These are good
candidates to check for under loading. If so, capacitors should be installed at the motor.
500 kVArs of capacitors will be needed to raise the power factor to 90% level. You
should consult local heavy equipment vendors to check out motor loading problems and
to determine how to position the capacitors between the motors and the electric main.
Some capacitors have already been installed, but it is suspected that they were either
knocked out by lightning or power surges. If existing capacitors and fuses can be
repaired and put back into operation, the net need of new capacitors will be reduced
accordingly.
Estimated Savings
Based on last year's utility bills, we estimate that adding 500 kVAr would save you about
$25,819 per year in utility penalties. It would also reduce heat generation and voltage
drop, and allow you to decrease the size of wiring. The improvement of power factor
from 83% to 91% would result in an 8.8% reduction in electric current, an 8.8%
reduction in voltage drop across the circuit, and an 18% reduction in excess heat
generation.
Estimated Implementation Cost
$30,000 for 500 - 600 kVAR capacitors and controls
Capacitors installed at the electric main might require a sophisticated controls costing an
additional $10,000.
Simple Payback
14 months
AR X: INSTALL AUDIBLE CAPACITOR FAILURE ALARM
Electric Demand
Implementation
Cost:
Simple Payback:
Present
123 kW
Recommended
0
$150
Annual Savings
123 kW; $1731
$150
1 month
Analysis
Recently 200 kVAr of capacitance failed. The visual alarm which should signal when the
capacitor bank is not operational also failed. This dual failure caused the power factor of
the plant to drop below 90% and cost the company $1,731 in demand penalties.
Recommendation
Although a properly functioning visual alarm may alleviate such problems in the future,
we recommend installing an audible alarm. This type of alarm is almost impossible to
ignore and assures that you will be alerted to a capacitor bank failure.
Estimated Savings and Implementation Cost
Looking backward, we can say that a malfunctioning alarm caused the company to be
charged for 123 kW in demand penalties costing $1,731. A functioning audible alarm
would have avoided this cost. Audible alarms can be purchased and installed for about
$150.
Simple Payback
SP = $150 initial cost / $1,731 /yr savings x 12 months/yr = 1 month