benifits of power factor improvement for industry

11
Benefits of Powerfactor improvement for Industry Er. Govinda Neupane Certified Energy Auditor [email protected]

Upload: govinda-neupane

Post on 23-Feb-2017

132 views

Category:

Engineering


5 download

TRANSCRIPT

Page 1: Benifits of power factor improvement for industry

Benefits of Powerfactor improvement for Industry

Er. Govinda NeupaneCertified Energy Auditor

[email protected]

Page 2: Benifits of power factor improvement for industry

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.

Page 3: Benifits of power factor improvement for industry

Power used in Industry

Page 4: Benifits of power factor improvement for industry
Page 5: Benifits of power factor improvement for industry

Example

Page 6: Benifits of power factor improvement for industry

From Above ExampleDemand 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

Page 7: Benifits of power factor improvement for industry

Cost benefits of Power Factor improvement

Reduced Maximum demand (kVA) charges on utility billReduced distribution losses(kWh) within the plant networkBetter 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 equipmentProlong the life of distribution equipmentStabilizes voltage levels Increase your system's capacity, etc.

 

Page 8: Benifits of power factor improvement for industry

How to determine the Rating of capacitors required?Method-1The 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 = 0.75Cos F2 = 0.99 , Tan F2 = 0.142

KVAR required = kW ( Tanf1 - Tanf2 )

= 850 (0.75 – 0.142) = 517 kVAr

Page 9: Benifits of power factor improvement for industry

Method-21. Locate 0.8 (original power factor) in column (1).Refer table.2. Read across desired power factor to 0.99 column. We find .609 multiplier3. Multiply 850 (average KW) by .609 = 518 KVAR.

Page 10: Benifits of power factor improvement for industry
Page 11: Benifits of power factor improvement for industry

Thank You.....!

Er. Govinda NeupaneCertified Energy AuditorEnergy Efficiency UnitButwal ChamberButwal, [email protected]