Basic Concepts Presentation

7/31/2019 Basic Concepts Presentation

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Ontoseno Penangsang1)

1) Electrical Department, Sepuluh Nopember of Institute TechnologySurabaya, 60111, Indonesia


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Single Phase Circuits

Review of Phasors

Real and Reactive PowerReal and Reactive Loads

Power Triangle

Real and Reactive Power

FlowOntoseno [email protected]


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Review of Phasors

Goal of phasor analysis is to simplify theanalysis of constant frequency ac systems

v(t) = Vmax cos(wt + qv)i(t) = Imax cos(wt + qI)

Root Mean Square (RMS) voltage of sinusoid

2 max

0

1( )

2

TV

v t dt T

Ontoseno [email protected]


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Phasor Representationj

( )

Euler's Identity: e cos sin

Phasor notation is developed by rewriting

using Euler's identity

( ) 2 cos( )

( ) 2 Re VV

j t

j

v t V t

v t V e

q

w q

q q

w q



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PhasorRepresentation, contd

Then drop the constant terms

( ) Re 2

V cos sin

I cos sin

VjV

j t

V V

I I

V V e V

v t Ve

V j V

I j I

q

w

q

q q

q q



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Advantages of Phasor Analysis

0

2 2

Resistor ( ) ( )

( )Inductor ( )

1 1Capacitor ( ) (0)

C

Z = Impedance

R = Resistance

X = Reactance

XZ = =arctan( )

t

v t Ri t V RI

di tv t L V j LI

dt

i t dt v V I j C

R jX Z

R X

R

w

w

Device Time Analysis Phasor



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Sine Wave Basics

RMS a method for computing the effective valueof a time-varying e-m wave.



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Real, Reactive and ApparentPower in AC Circuits

in DC circuits: P=VI but= in AC circuits: averagepower supplied to the load will be affected by thephase angle q between the voltage and the current.

If load is inductive the phase angle (also calledimpedance angle) is positive; (i.e, phase angle ofcurrent will lag the phase angle of the voltage) andthe load will consume both real and positive reactivepower

If the load is capacitive the impedance angle will benegative (the phase angle of the current will lead thephase angle of the voltage) and the load willconsume real power and supply reactive power.



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Resistive and Reactive Loads



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Impedance Angle, CurrentAngle & Power

Inductive loads positive impedanceangle, current angle lags voltage angle

Capacitive loads negative impedanceangle, current angle leads voltage angle

Both types of loads consume real powerOne (inductive) consumes reactive as

well while the other (capacitive) suppliesreactive power



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Real and Reactive PowerEquations

First term is energy flow into the circuit Average =

Real power (P)

Second term is power transferred back and forthbetween source and load (Reactive power Q)

v(t) = 2 V cos t

i(t) = 2 I cos (t-)

p(t) =v

(t)i(t) = 2VIcos t cos (t-

)

p(t) = VI cos(1 + cos 2t) + VI sinsin (2t)



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Real term averages to P = VI cos (+)

Reactive term (amplitude) to Q = VI sin

(+/-)

Reactive power is the power that is first stored and thenreleased in the magnetic field of an inductor or in the

electric field of a capacitor

Apparent Power (S) is just = VI

Apparent PowerEquation



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Voltage, Current and Poweras a function of time

Voltage

Current(lagging)

Power

Real Power



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Voltage

Current(in phase)

Power

Real Power



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Voltage

Current(lags 90%)

Reactive Power



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Loads withConstant Impedance

V = IZ

Substituting into real and reactive powerequations

P = I2Z cos q Q = I2Z sin q

S= I2Z

Since Z = R +jX = Z cos q + jZ sin q P = I2R and Q = I2X real and

reactive power consumed by load



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Complex Power andKey Relationship ofPhase Angle to V&I

S= P + jQ

S= VI(complex conjugate operator)

If V= V30o and I= I15oTHEN.. COMPLEX POWER SUPPLIED TO

LOAD :

S= (V30o)(I-15o) = VI (30o-15o )

= VI cos(15o ) + jVI sin(15o )

NOTE: Since Phase Angle q = qv - qiS= VI cos(q) + jVI sin(q) = P + jQ



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S= P + jQ

S= VI cosq + j VI sinq

S= VI(cosq + j sinq)

S= VIej OR S=VIq

Since V= V00 and I= I-q

S = VI*

V0

I-q

q

Complex Power


i


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Review V, I, Z

If load is inductive then the Phase Angle(Impedance Angle Zqo) is positive, Ifphase angle is positive, the phase angleof the current flowing through the loadwill lag the voltage phase angle acrossthe load by the impedance angle Zqo.

0

0

0

Z

V

Z

0V

Z

VI



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The Power Triangle



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Real Power Flow

0VV

q II

qcosIq

V

I

If the component of I along the axis of E is in phase with E, thepower is generated powerwhich is being deliveredto the system,for this component of current is always flowing away from the

positively marked terminal

P, the real part of VI* (Vicos), is positive.

The Source is a Generator

System



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q II

0VVqcosI

V

I

q

System

If the component of current along the axis of E is negative (1800 outof phase with E), power is being absorbed

P, the real part of VI* (Vicos), is negative.

The Source is a Motor

Real Power Flow



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Reactive Power Flow

0VV

90II

V

I90

LX

Positive reactive power equal to |I|2X is supplied to the inductancesince inductance draws positive Q.

Q, the imaginary part of VI* (Visin), is positive

I lags V by 900



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Reactive Power Flow

q II

0VV V

I90

Negative Q must be supplied to the capacitance of the circuit, orthe source with the emf E is receiving positive Q from the capacitor

Q, the imaginary part of VI* (Visin), is negative

I leads V by 900



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Example

V = 1200o V

Z = 20-30o

Calculate current I, Power Factor (is itleading or lagging), real, reactive,

apparent and complex power suppliedto the load



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Three-Phase (3- ) Circuits

What are they?

Generating 3- Voltages andCurrents

Balanced systems

Wye (Y) and delta () connectionsThree Phase Power



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What does Three-Phase mean?

A 3- circuit is a 3- AC-generation

system serving a 3- AC load

3 - 1- AC generators with equal voltagebut phase angle differing from the

others by 120o



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One Cycle (1/50th sec.)


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Common Neutral

A 3- circuit can have the negativeends of the 3- generators connected to

the negative ends of the 3- AC loadsand one common neutral wire cancomplete the system

If the three loads are equal (orbalanced) what will the return current bein the common neutral?



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If loads are equal.

the return current can be calculated tobe

ZERO!

Neutral is actually unnecessary in abalancedthree-phase system (but isprovided since circumstances maychange)



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BalancedThree Phase Systems

Balanced

Same amplitude

120phase diff.

Phase shift

ia lags ua angle



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Three-phase currents

Balance: Ia+ Ib+ Ic=0

No return current Losses reduced

No return conductor

a

b

c



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Wye (Y) connection



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Wye (Y) connection

Ic

Ia

Ib

a

LOADSOURCE

a

bb

c c

nn

n : neutral

Vab = Vbc = Vca = VL : line to line voltage

Van = Vbn = Vcn = Vp : line to neutral (phase) voltage

Ic

Ia

Ib

Ia = Ib = Ic = IL : line current

= Ip : phase current


V l C


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Voltage CurrentRelationships

IL = Ip

VL = 3 VpVL, Vp, IL , Ip : rms values of

voltage and current



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Phasor Diagram

SOURCE = LOAD

Vab

30o

opcn 120VV

opan 0VV

o

pbn VV 120



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Delta ( ) connection



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Delta ( ) connection

Ic

Ia

Ib

SOURCE

ab

c

Vab = Vbc = Vca = VL : line to line voltage

= Vp : phase voltage

Iab

Ibc

Ica

Ic

Ia

Ib

LOAD

ab

c

Iab

Ibc

Ica

Iab = Ibc = Ica = Ip : phase current

Ia = Ib = Ic = IL : line current


V lt C t


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Voltage CurrentRelationships

VL = Vp

IL = 3 IpVL, Vp, IL , Ip : rms values of

voltage and current



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o

pbc II 120

o

pca II 120 bI

opab II 0

o30

o

pbc II 120

o

pca II 120

o

pab II 0

aI

o30

Phasor Diagram

SOURCE LOAD



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=uiR

=uiL

Three Phase Power



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Three Phase Power

pa(t) , pb(t) , pc(t) : phase power

ptotal(t)= pa(t)+ pb(t)+ pc(t)

three phase power

pa(t) = pb(t) = pc(t)

If voltages and currents balanced

cosj need not be zero



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ppp

ppp

SinIVQ

CosIVP

j

j

j

j

3

3

3

3

FOR A BALANCED THREE PHASE SYSTEM :

p : phase angle between

phase voltage and phase

current (lagging)

Three Phase Power

THREE PHASE POWER : 3 X PHASE POWER

Watt : real power

Var : reactive power



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Rumus Daya 3 Fasa

pLL

pLL

SinIVQ

CosIVP

j

j

j

j

3

3

3

3

Three Phase Power

LpL

p IIV

V ;3

3; LpLp

IIVV

For Y connection

For connection

Watt : real power

Var : reactive power



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LLIV

QPS

3

22

VA

Three Phase Power

APPARENT POWER :

Basic Concepts Presentation

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