It is based on Industry and made from my experience during Energy audit of Nepalese Industry and training.

1. Er. Govinda Neupane
Certified Energy Auditor
gchintaksusta@gmail.com

2. What is power factor?
 Power factor is the ratio of
active (real) power to
apparent power (total
power).
1. Active (or working) power
to perform the work
(motion) and
2. Reactive power to create
and maintain electro-
magnetic fields.
3. Apparent Power is vector
sum of the active power
and reactive power.

3. Power used in Industry

5. Example

6. From Above Example
 Demand Charge Saving
 200kVA*NRS.250=NRS. 50000 per month
Note : Average Demand charge taken by Nepal
Electricity authority is NRS.250 per kVA
From my 2 years experience in Energy sector average power
factor of Nepalese industry is 0.8

7. Cost benefits of Power Factor
improvement
Reduced Maximum demand (kVA) charges on utility bill
Reduced distribution losses(kWh) within the plant network
Better voltage i.e. motor terminal and improve performance of
motor.
A high power factor eliminates penalty charges imposed when
operating with low power factor.
Reduce heat loss of transformers and distribution equipment
Prolong the life of distribution equipment
Stabilizes voltage levels
Increase your system's capacity, etc.

8. How to determine the Rating of
capacitors required?
 Method-1
The Butwal steel mill has an average power factor of 0.8
with an average KW of 850. How much KVAR is required
to improve the power factor to 0.99 ?
Cos F1 = 0.8 , Tan F1 = 0.75
Cos F2 = 0.99 , Tan F2 = 0.142
KVAR required = kW ( Tanf1 - Tanf2 )
= 850 (0.75 – 0.142)
= 517 kVAr

9. Method-2
1. Locate 0.8 (original power factor) in column (1).Refer
table.
2. Read across desired power factor to 0.99 column. We
find .609 multiplier
3. Multiply 850 (average KW) by .609 = 518 KVAR.

11. Er. Govinda Neupane
Certified Energy Auditor
Energy Efficiency Unit
Butwal Chamber
Butwal, Nepal
gchintaksusta@gmail.com