Saxion University of Applied Science

Power systems analysis power factor

Saxion University of Applied Science

Overview op presentation Review of the theory

Power factor correction Capacitor banks at customers

Industrial losses

Benefits of power factor correction

Saxion University of Applied Science

Power formula’s

Instantaneous power: )()()( titvtp

Average power:

TT

dttitvT

dttpT

P00

)()(1

)(1

For single phase AC circuit: Average power: cosIVP

Quadrature power: sinIVQ

Apparent power: 22 QPSIVS

Complex power:

IVS (I* Complex conjugate)

SSSQSP )Im()Re(

Saxion University of Applied Science

Power in 1 phase single phase system

Apparent power: LLNph IVS 1

Average power: cos1 LLNph IVP

Quadrature power: sinLLN IVQ

Saxion University of Applied Science

Power factor correction 0- Power Triangle

Q

(inductive)

P

S

θ

Q

(capacitive)

P

S

By definition:

Positive angle, inductive, Lagging power factor

Negative angle, capacitive, Leading power factor

Saxion University of Applied Science

Additional formulas

RIP RMS

2 XIQ RMS

2

X

UQ

RMS

2

S

Pcos 1cossin 22

ZIS RMS

2

Saxion University of Applied Science

Power factor improvement 1- Adding capacitance

AC

Zcap

Z1

Capacitor bank

I network

Q1 inductive load

P1=P2

S1

New Q2 improved pf

θ1 θ2

Q Capacitor bank

S2

Power factor improvement

Saxion University of Applied Science

Power factor improvement 2

Overview of the Power factor reader.

Question: A inductive resistive load draws an average power of 20kW

with a PF of 0.6. The Voltage is 380V RMS / 50Hz. What capacitor has to be placed in parallel in order to bring the

to PF to 0.9?

References:

Powerfactor reader part of commercial material of ARTECHE, downloaded10 oct 2012 from http://www.arteche.com/

Saxion University of Applied Science

Adding power triangles

Summing Average Power, summing Quadrature Power to find power triangle

of the whole network

AC Z1 Z2 Z3

Q2 ind.

P1

S1

Q1 cap.

P2

P3

S2

S3 Q3 ind. Stotal

Qtotal

Ptotal

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Industrial losses

Also see Ardeche paper for overview of machinery

Efficiency and P.F. for asynchronous machine. Run industry at 1.05 pu.

Saxion University of Applied Science

Power factor improvement 3

How much does capacity of distribution system

increase by improving the power factor?

Transported active power in 3 phase system

cos33 LLLph IVP

If the PFold (cos ) is improved to PFimp (cos imp) the ration of powers becomes:

)(cos

)(cos2

2

improved

old

old

impoved

Ploss

Ploss

Improving PF from 0.8 to 0.95 this gives a ratio of 0.71,

which means a reduction of losses by 29%.

Saxion University of Applied Science

Power factor improvement 3 continue

Current ratios:

improved

old

improved

old

old

impoved

PF

PF

I

I

)cos(

)cos(

For improving PF from 0.8 to 0.95 this gives a ratio of 0.64, which a reduction of the required current by 16%. The capacity of the line seems to have increased