This document discusses power factor correction and capacitor banks. It begins with an overview and review of relevant power formulas. It then discusses how adding capacitance in parallel with an inductive load can improve the power factor by reducing the quadrature power. Graphs of power triangles are used to illustrate this improvement. The document also notes that improving power factor can reduce transmission losses and increase the capacity of distribution systems since less current is required.

1. Saxion University of Applied Science
Power systems analysis
power factor

2. 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

3. Saxion University of Applied Science
Power formula’s
Instantaneous power: )()()( titvtp 
Average power:
 
TT
dttitv
T
dttp
T
P
00
)()(
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(

4. 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 

5. 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

6. Saxion University of Applied Science
Additional formulas
RIP RMS
2
 XIQ RMS
2

X
U
Q
RMS
2

S
P
cos 1cossin 22
 
ZIS RMS
2


7. 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

8. 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/

9. 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

10. Saxion University of Applied Science
Industrial losses
Also see Ardeche paper for overview of machinery
Efficiency and P.F. for asynchronous machine.
Run industry at 1.05 pu.

11. 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
)(cos
2
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%.

12. 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