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24 Νοε 2013 (πριν από 3 χρόνια και 6 μήνες)

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1

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

BASICS

OF

P
OWER
F
ACTOR
C
ORRECTION

2

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Harmonics

Reactive power

Unsymmetrical load

Flicker

GRID

LOAD

Loads create disturbances

3

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Caused by:

Reactive power

Inductive loads, Power Electronics

Harmonics

Power electronics, non
-
linear loads

Commutation

Converter and drives

Voltage sags an swells

Load variations, high inrush currents

Unsymetric grids

Unbalanced single phase loads

Radio frequencies

Ripple control

Voltage interruptions

Lightning, over load, switching operations

Different aspects of electrical power quality

4

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

What are the different types of loads?

Ohmic loads

Lighting bulbs

Iron

Resistive heating

Capacative loads

Capacitors

Underground cables

Over excited

synchronous

generators

Inductive loads

Electrical Motors

Transformers

Reactors/chokes

Overhead lines

Under excited

Synchronous

generators

Discharge lamps

Power electronic

GRID

5

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Three different types of loads: 1. OHMIC
-
LOADS

Ohmic loads

U and I in phase

Phase shift = 0

No penalty



In

resistive

circuits

the

voltage

and

current


waveforms

reach

their

peaks

and

troughs

as


well

as

the

electrical

zeros

at

the

same


instant

of

time
.





The voltage and current are said to be in


phase (


㴠=
°
⤠慮搠瑨d⁥瑩牥t楮灵琠灯w敲e楳


converted into active power. Thus, resistive


circuits have a
unity

power factor.




The ohmic resistance does not depend on


frequency.



U
-

Voltage

I
-

Current


=
0
°

6

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Three different types of loads: 2. INDUCTIVE
-
LOADS

Inductive loads

U is 90
°

ahead of I

90
°

phase shift

Penalty!



Most of the industrial loads are inductive in


nature e.g. motors, transformers etc. Due to


inductive reactance of the load, the current


drawn by the load lags behind the voltage


waveform electrically by an angle






The magnitude of


楳⁰牯灯牴楯r慬瑯⁴桥

††
楮摵捴楶攠牥r捴慮捥⸠卩c捥c瑨t⁣畲牥琠污杳

††
扥桩湤瑨t⁶潬瑡来Ⱐ楮摵捴楶攠汯慤猠慲a⁳慩

††
瑯⁨t癥v愠污杧楮朠灯w敲e晡捴潲.




Impedance
-
X
L
= 2 * 3.14 * f * L


U
-

Voltage

I
-

Current


=
90
°

7

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Three different types of loads: 2. INDUCTIVE
-
LOADS



Inductive loads cause


a phase shift between


current and voltage.



A positive as well as a


negative power can be


observed.

Phase shift
t
U, I and power
+ ve
+ ve
-ve
Power
Current
Voltage

8

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Three different types of loads: 3. CAPACITIVE
-
LOADS

Capacative loads

I is 90
°

ahead of U

90
°

phase shift

Over compensation

is risky!



Due to capacitive reactance of the load, the


current drawn by the load is ahead the


voltage by an angle






The magnitude of


楳⁰牯灯牴楯r慬瑯⁴桥

††
捡c慣a瑩癥t牥r捴慮捥c




Impedance

U
-

Voltage

I
-

Current


=
90
°

C
f
C
X
C







2
1
1
9

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Three different types of electrical power



S = Apparent Power



P = Active Power



Q = Reactive Power

P

Q
1

Q
C

S
1

Q
2

=

Q
1

-

Q
C



1

S
2

2



10

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Three different types of electrical power

Reactive Power (

kvar)

2

2

P

S

Q

-



Active Power

²

²

Q

S

P

-



[

KW

]

Apparent Power

²

²

Q

P

S

+



[

kVA

]

cos




= P/S




=

phase

displacement angle

sin




= Q/S

S

1


=

uncompensated

apparent

power

Q = S

sin



S

2


=

compensated

power

with

capacitors

for

compensation

Q = P

tan



Q

1

Q

C

Q

2



2



1

S

1

S

2



S = Apparent Power



P = Active Power



Q = Reactive Power

11

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

What is Active Power?

The

amount

of

input

power

which

is

converted

into

output

power,

is

termed

as

“active

power”

and

is

generally

indicated

by

P
.


The

active

Power

is

defined

by

the

following

formula
.






[W]



Ideally, entire input power i.e. apparent power should get converted
into the useful output, i.e. heating of an oven, movement of an motor,
light of an bulb.


cos
3




I
U
P
12

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

What is Reactive Power?

Electrical machines work on the principle of conversion of
electromagnetic energy.(e.g. electric motors, transformers). A
part of input energy is consumed for creating and maintaining
the magnetic field. This part of the input energy cannot be
converted into active energy and is returned to the electrical
network on removal of the magnetic field. This power is known
as “reactive‘‘ power
Q

and is defined as follows.






[VAr]



sin
3




I
U
Q
13

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

What is Apparent Power?

Applications

of

electrical

equipment

are

based

on

conversion

of

electrical

energy

into

some

other

form

of

energy
.

The

electrical

power

drawn

by

an

equipment

from

the

source

is

termed

as

Apparent

Power,

and

consists

of

active

and

reactive

power
.


The

current

measured

with

a

clamp

amp

indicates

the

apparent

power
.

It

is

defined

as

follows
:







[VA]

I
U
S



3
14

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

What is the power factor?

Power factor = cos


cos
-
phi = P (kW) / S (kVA)


Phase shift
t
U, I and power
+ ve
+ ve
-ve
Power
Current
Voltage

15

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Typical power factors in industries

Factory
Typical
uncompensated PF
Breweries
0,6..0,7
Butcher's
0,6..0,7
Cement plant
0,6..0,7
Compressor
0,7..0,8
Cranes
0,5..0,6
Drying-Plants
0,8..0,9
Machinery, big sized
0,5..0,6
Machinery, small sized
0,4..0,5
Plywood
0,6..0,7
Sawmill
0,6..0,7
Steel factory
0,6..0,7
Suggar
0,8..0,85
Tobacco
0,6..0,7
Water pumps
0,8..0,85
16

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Why to improve the power factor?


System kVA
-

release


Reduction of power bill (short pay back time: 6
-
18 month usually)


Reduction of ohmic losses


Power Quality improvement (harmonics, voltage sags..)


Higher kW loading of transmission and distribution equipment and/or
smaller dimensioning of this equipment (cable, transformer, bus bars,...)

17

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

How to improve the power factor?


PFC Capacitors



Over
-
excited synchronous generators


Active (real time) compensation


Reduce amount of inductive load


Usage of modern converter technology

18

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Principle of PFC

19

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Principle of PFC

Mechanical or

thermal work

Generation of

magnetic field

Active Energy

Reactive Energy

Capacitor

Supply

Corriente

0

Load

95

Current

Current

65

20

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC


Individual compensation


Group compensation


Centralised automatic compensation


Combined compensation


Active (real time, by means of Semiconductors) PFC

21

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 1. Individual (fixed) Compensation

Disadvantages



Many small capacitors are more expensive


than one single capacitor of total


equivalent rating



Low utilization factor of capacitors for


equipment not often in operation

In fixed compensation, capacitors are directly connected to the

terminals of the individual load (e.g. motor, transformer), and

switched by means of the load contactor or CB

together with the load.

Advantages



kvar produced on the spot



Reduction of line losses



Reduction of voltage drops



Saving of switch gear

22

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 1. Individual Compensation
-

lighting

23

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 1. Individual Compensation
-

motor

24

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 1. Individual Compensation
-

motor

For compensating of asynchronous motors the capacitor output should be

maximum 90 % of no load reactive power of the motor.


Higher kvar ratings lead to self excitation of the motor after disconnection

from the grid.


Risk of over voltage > 1,1 * U
nominal
!


Recommended kvar size ensures a PF of < 1 but > 0,9 in low load as well as

full load operation of the motor.


A thumb rule recommends: kvar = 35% of active power (kW) of a motor

Active power can be found on the rating plate of a motor.

25

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 1. Individual Compensation
-

transformer



PFC on LV bus bar



Compensation of no load


reactive power of the transformer



Voltage increase on LV side

26

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 1. Individual Compensation
-

transformers

For compensation of no
-
load reactive power of transformers the kvar output

of the capacitor is based on the reactive power consumption of the transformer

itself. The recommended values compensate the magnetizing power of a

not loaded transformer only.


The following approximation formula can be used:



Q
o

= S
o

= i
o

x S
N

/ 100



Q
o

= Transformer no
-
load reactive power in kvar


S
o

= Transformer no
-
load apparent power in kVA


i
o

= Transformer no
-
load current in % of the nominal current


S
N

= Transformer nominal power in kVA



28

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 2. Group Compensation

Disadvantages



No load reduction, loss reduction,


voltage drop reduction on individual


load lines

In group compensation, capacitors are connected to a group

of loads (e.g. motors), and switched by means of the

main load contactor or CB together with the load.


Advantages



Reduction of capital investment



Losses reduced in distribution lines



Voltage drops reduced in distribution lines



Higher utilization factor of capacitors

29

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 3. Centralized Compensation

Disadvantages



Load not lightened on distribution lines


within a factory

In factories with many loads of different output and operating times fixed

compensation is usually to costly and non
-
effective. The most economic solution

for complex applications is usually a centralized automatic capacitor bank,

controlled by a automatic PF controller. Point of connection is usually in the main

distribution panel close to the transformer.

Advantages



Best utilization of the capacitors



Most cost effective solution



Easier supervision



Automatic control

30

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 3. Centralized Compensation

31

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Frequently asked questions


What is the thumb rule for selection of kvar size for motor fixed
compensation?


How to find the active load of a motor for calculating the capacitor size?


In factories with many loads it is problematic to calculate the required
capacitor output during planning status.

1) Why?

2) How to select a suitable kvar size?


When to select:

A) Fixed

B) Group

C) Centralised

-

compensation

32

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 4. Active Compensation

Disadvantages



Requires high capital investment



High engineering efforts required



Higher losses of electronic switches

Typical applications
:



Cranes, Lifts



Spot welding, punching ... e.g. car industry



Paper mills, semiconductors, ....



All kind of short term loads

Advantages



Real time compensation



Reduction of reactive energy costs



Opens new fields of applications



Improvement of power quality

33

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Methods of PFC: 4. Active Compensation

34

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction





Caledonian Paper plc

Irvine, Scotland



Production of 325 tons deposited
paper per year year


Total load: 47 MVA


10MVA sensitive load


11kV (50 Hz) factory grid supplied
from 132 kV HV Scottish Power


Lowest voltage sag 34% of nominal
voltages


37 voltage sags per year

Methods of PFC: 4. Active PFC
-

Reference project

35

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction





LCL
-
Filter

IGBT
-
Converter

i
N


SIPCON DVR (LV)

u
a, b, c

i
st


C

i
L


GRID

Load

EMC filter

Methods of PFC: 5. Active harmonic filter

36

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

37

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction


Capacitor bank

Qc = 10 * 25 kVAr

HV Grid

Transformer

630 kVA, u
k

= 5 %

Current = 666 A

300 kW

cos


㴠=⸶.

M

3 ~

HV Grid

Transformer

630 kVA, u
k

= 5

%

Current = ???

300 kW

cos


㴠=⸶.

M

3 ~

Current reduction: ???

Example: Current reduction in main supply cable?

38

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Capacitor bank


Qc = ??? kvar

HV Grid

Transformer

630 kVA, u
k

= 5 %

Current = 666 A

300 kW

cos


㴠=⸶.

M

3 ~

HV Grid

Transformer

630 kVA, u
k

= 5

%

300 kW

cos


㴠=⸶.

M

3 ~

Example: Required kvars for target PF of 0.98?

40

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction


Question:



An electricity bill of a petrochemical factory shows a monthly demand of


720 000 kvarh reactive work




Monthly billing: 720 000 kvarh * x $




Daily operation, 24 hours a day




What remedial measures have do be carried out to reduce the


electricity bill?

Example for PFC: Kvar calculation based on electricity bill (kvarh)

41

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Example for PFC: Kvar calculation based on electricity bill (kvarh)

Answer:




Additional capacitor output has to be installed.



According the following formula the required kvar output can be calculated:


Q in kvar = W in kvarh / time in h



Time = 30 days * 24 hours



Q = 720 000 / 30 / 24 = 1000 kvar



By installing a 1000 kvar (2*50+9*100) capacitor bank the customer


will eliminate the kvarh consumption down to
ZERO.

42

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction




Capacitor bank

Qc = ?? kvar

HV Grid

Transformer

630 kVA, u
k

= 5

%

300 kW

cos


㴠〮㘵

M

3 ~

Question:



A textile factory with a total load of 300 kW shows


an actual power factor of 0.65 (phi=49,5
°
)



The local power utility asks for a target PF=0.95


(phi=18,2
°
)


What capacitor output is required to avoid

surcharges for low PF?

Example

43

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction

Solution:




Qc = P * (tang phi1
-

tang phi 2) = 300*(tan (49.5)
-

tan(18,2))
= 252 kvar



For a proper fine tuning of the target PF we recommend a capacitor bank


design: 25 + 25 + 50 + 50 + 50 + 50 kvar



Depending on types of loads, e.g. frequency converters, de
-
tuned capacitor


banks should be used

Example

44

March 2001

EPCOS FK PM PFC

Basics of Power Factor Correction





35 kV

110 kV

16 MVA

Station 10

1.6 MVA

4 %

320 kW

(Converter)

1x500 kW

2x250 kW

(Converter)

16 MVA

Station 15

1.6 MVA

4 %

5x300 kW

(Converter)

1.6 MVA

4 %

4x300 kW

(Converter)

n. o.

3x5 MVA

Power uility

S

k

"=310...360 MVA

Harmonic Filter

Single line diagram essential for system study