EE207 Electrical Power - Lecture 2

EE207 Electrical Power

Lecture 2

Active, Reactive, and Apparent Power –Part2

Rajparthiban Kumar EE207 Electrical Power 2

Reactive Power

• Reactive Power: This power is responsible for the

production of the magnetic flux in electrical

systems, and it does not produce any useful work

or heat.

• Reactive power will surge back and forth between

the source and the load. Its unit is VAR and is

measured using a special instrument called

varmeter.

• Reactive power flows from/to a system due to the

presence of inductive or capacitive elements.


Reactive Power

• Capacitors and inductors both are reactive

elements. An inductor absorbs reactive power and

it appear as reactive load, whereas a capacitor

supplies reactive power and it appears as reactive

source.

• The current in an inductor lags the voltage across

it by 90o. Conversely, the current in a capacitor

leads the voltage across it by 90o.

• An inductive reactance is expressed as jXL, and a

capacitive reactance is expressed as –jXC .


Reactive Power

Example

A sinusoidal voltage source has a voltage peak

Vp= 162Vand frequency of 60Hz is applied to

the terminals of an inductor. The resulting

current has a peak of Ip= 7.5 A.

a) Calculate the value of the inductor in [H].

b) Plot the current, voltage and instantaneous power

waveforms.

c) Calculate the power absorbed by the reactor.

d) Plot the phasor diagram.


Reactive Power

a) The inductance L

b) Assume the voltage is a cosine signal thus the current

will be 90o lagging. i.e.

mH..

L.fLX

..I

VX

L

p

pL

7350602

6216212

62157

162

========⇒⇒⇒⇒========

ΩΩΩΩ============

ππππππππ

A..

I&V.V

.represphasorinor

)tcos(.IandtcosV

oooo

o

9035902

570551140

2

162

9057162

−−−−∠∠∠∠====−−−−∠∠∠∠====∠∠∠∠====∠∠∠∠====

−−−−======== ωωωωωωωω


Reactive Power

0 45 90 135 180 225 270 315 360 405 450 495 540 585 630 675 720

0

I =7.5(A)

V=162(V)

P Instant.


You can observe from the graph that

The instantaneous power has positive and

negative peaks at 607.5 var.

The instantaneous power pulsates at 120Hz

which is twice the supply frequency, and it is

average is zero.

c) Reactive power absorbed by the inductor:

Reactive Power

var..

IVQ effeffL 56072

57

2

162============


d) Phasor diagram

• Note that the reactive power is calculated as the

product of the effective values of the voltage

times the current component which is out of 90o.

Reactive Power

2

162====V

2

57.I ====


• Reactive power can be obtained from the product

of the out of phase components of the effective

values of the voltage and current.

• To understand this let’s consider the following

example: A sinusoidal AC voltage

is applied across the terminals of a network and

produced a current given as:

Reactive Power

)(cos)( γγγγωωωω ++++==== tVtv p

)cos()( θθθθωωωω ++++==== tIti p

v(t)

i(t)

Z


Active Power

• Note that the voltage and current are at different

phase angles (γ, and θ respectively) and the load is

given as Z (impedance). Let’s consider this

general example to find Q.

• To simplify the analysis, the phasor representation

will be used, and the cosine function (i.e. cosωt)

will be used as a reference.

• Thus the rms voltage and current phasors are

given as:

orms

orms

II

VV

θθθθ

γγγγ

∠∠∠∠====

∠∠∠∠====


• Since the phase difference between the voltage

and the current is then the active power is

given as:

• The above term can also be obtained by taking

the Imaginary part of the result of the

Active Power

jdcjIII

jbajVVV

rmso

rms

rmso

rms

++++====++++====∠∠∠∠====

++++====++++====∠∠∠∠====

)sin(cos

)sin(cos

θθθθθθθθθθθθ

γγγγγγγγγγγγ

)( θθθθγγγγ −−−−

adbc

)sincoscos(sinIV

)sin(IVP

rmsrms

rmsrms

−−−−====

−−−−====

−−−−====

θθθθγγγγθθθθγγγγ

θθθθγγγγ


Active power

multiplication of the voltage times the

complex conjugate of the current. i.e.

)adbc(

)adbc(j)bdac(Im

)jdc)(jba(Im

VIImQ

*

*

−−−−====

−−−−++++++++====

++++++++====

====


Active, Reactive Power

• Active and reactive powers function independently of each other and, consequently, they can be treated as separate quantities in electric circuits.

• Both place burden on the transmission line that carries them, but, whereas active power produces a tangible result(heat, mechanical work, light, etc.), reactive power only presents power that oscillates back and forth.

• Transformers, ballasts, induction motor absorb reactive power, and it produces magnetic field in these devices.



Example

An ac generator (G) is connected to a group of R,

L, C circuit elements. The respective elements

carry currents as show by the figure. Calculate the

active and reactive power associated with the

generator.

G

16.12A

2ΩΩΩΩ j3ΩΩΩΩ

14A4ΩΩΩΩ

-j3.5ΩΩΩΩ20A



Solution

• The two resistors absorb active power given by:

• The inductor absorbs reactive power given by:

• The capacitor generates reactive power given by:

• Thus the net reactive power in the circuit is:

W

).()(RIP

1304520784

21216414 222

====++++

====××××++++××××========

varXIQ LL 58831422 ====××××========

var.XIQ CC 1400532022 ====××××========



QT

• Note that QT is negative and that shows that the

load will generate reactive power, which means

that this power has to be absorbed by the

Generator. However, the power absorbed by the

resistor must be supplied by the Generator.

varQQQ LCT 8125881400 −−−−====++++−−−−====++++====


Apparent Power

• The apparent power (S) is a combination of the active

power and reactive power and is expressed as:

• The unit of the apparent power is VA (Volt-Ampere).

VI)sinVI()cosVI(S

IandV

then,currenttheandvoltagethebetween

angleanthatgminassu

QPVIS

QP

oo

====++++====

∠∠∠∠∠∠∠∠

++++========

4 34 214 34 2122

22

0

θθθθθθθθ

θθθθ

θθθθ


Apparent Power

Example

A wattmeter and a varmeter are connected into

a 120V single phase line that feeds an ac motor.

They respectively indicate 1800W and 960var.

a) Calculate the in phase and quadrature (90o out of

phase) components of the current.

b) The line current.

c) The apparent power

d) The phase angle between the voltage and the line

current.


Solution

a) The in phase current (Ip) corresponds to the Active

power measured, thus:

The quadrature component Iq corresponds to the reactive

power measured, thus:

Apparent Power

AV

PcosII p 15

120

1800================ θθθθ

AV

QsinIIq 8

120

960================ θθθθ

V=120V

Ip =15A

Iq=8Aθθθθ

I


b) From the phasor diagram, the current I can be found

as:

c) The apparent power S

d) The phase angle between V and I

Apparent Power

AI

IIjIII qpqp

1781522

22

====++++====

++++====++++====

VAQPVIS 20401712022 ====××××====++++========

o

p

q.)

P

Q(tan)

I

I(tan 12811 ============

−−−−−−−−θθθθ


Power Factor

• Power factor (p.f.) of an alternating current device or circuit is the ratio of the active power P to the apparent power S, and is given as:

• Power factor can never be greater than unity (100%) as P can never exceed S.

• The power factor is said to be lagging if the current lags behind the voltage. Conversely, the power factor is said to be leading if the current leads the voltage.

θθθθcosS

P.f.p ========


• The power factor of a resistive circuit is Unity

(100%) because the apparent power and the active

power are equal.

• Pure inductive or capacitive circuits have zero

power factor. (note θ=90o between V and I).

• A combination of an inductive and resistive

elements will have a lagging power factor.

• A combination of capacitive and resistive circuit

elements will have a leading power factor.

Power Factor


Power Triangle

• The relationship between S, P, and Q can be

shown graphically using a Power Triangle

according to the following rules:

– Active power P absorbed (delivered) by a circuit or

device is considered to be positive (negative) and is

drawn horizontally to the right (left).

– Reactive power Q absorbed (delivered) by a circuit or

device is considered to be positive (negative) and is

drawn vertically upwards (downwards).

Inductive &

resistive circuit

power triangle

+P

+Q

+P

-Q

Capacitive &

resistive circuit

power triangle

S

S


Example 7-8

A single phase motor draws a current of

5A from a 120V, 60 Hz. The power factor

is 65 percent. Calculate

a) The active power absorbed by the load

b) The reactive power supplied by the line.


a) Apparent power

b) The reactive power

WSP

S

P

VAVIS

39065.0600cos

cos

6005120

=×==

=

=×==

θ

θ

var4563906002222

22

=−=−=

+=

PSQ

QPS

var45646.45sin600

46.45)65.0(cos)(cos

cos

sin

11

==

°===∴

=

=

−−

Q

PF

PF

SQ

θ

θ

θ


Example 7.9

A 50µF paper capacitor is placed across the motor terminals. Calculate:-

a) The reactive power generated by the capacitor

b) The active power absorbed by the motor

c) The reactive power absorbed by the line

d) The new line current


a) XC = 1/ jωC = 1/ j2πfC = 53Ω

so the current in the capacitor

I = V/ XC = 120 / 53 = 2.26 A

Therefore Qc = VI = 120(2.26) = 271 Var


b) The active power absorbed by the motor remains

the same.

c) Qnet = 456 – 271 =185 var

d)

903.0120

432cos

6.3120

432

432185390 2222

===

===

=+=+=

S

P

AV

SI

VAQPS

L

θ


S= 600VA

432VA

390 W

456 Var

185 var

271 var



PF correction

• It can be seen that apparent power is larger than P if PF is less than 1. So, the current I must be supplied is larger for PF<1, eventhough P supplied is the same.

• This larger current add cost to suppliers.

• Power companies charge industrial customers higher who operate at low power factor. (residential operate at unity PF)

• Must install capacitors to improve power factor.

Documents

EE207 Electrical Power - Lecture 2