+ All Categories
Home > Documents > 3phase Sys n Power Concept -lect5

3phase Sys n Power Concept -lect5

Date post: 15-Jun-2015
Category:
Upload: hafiz-azman
View: 355 times
Download: 0 times
Share this document with a friend
Popular Tags:
52
EEE3233 POWER SYSTEM CHAPTER 3: 3-Phase System and Power Concept nur diyana kamarudin
Transcript
Page 1: 3phase Sys n Power Concept -lect5

EEE3233POWER SYSTEM

CHAPTER 3:3-Phase System and

Power Concept

nur diyana kamarudin

Page 2: 3phase Sys n Power Concept -lect5

Introduction The generation, transmission

and distribution of electric power is accomplished by means of three-phase circuit.

At the generating station, three sinusoidal voltage have equal magnitude and an equal 1200-phase difference between any two phase.

This is called a balanced source.

Page 3: 3phase Sys n Power Concept -lect5

Positive phase sequence when the voltages reaches their peak value in the sequential ABC

Negative phase sequence when the phase is order is ACB.

ECn

EBn

EAn

EAn

ECn

EBn

Positive or ABC phase sequence

Negative or ACB phase sequence

Page 4: 3phase Sys n Power Concept -lect5

Y-Connected Load

A Y-connecter generator supplying a Y-connected load.

Page 5: 3phase Sys n Power Concept -lect5

Y-Connected Load (cont.)

Assume a positive sequence to find the relationship between the line voltage (line-to-line voltage) and phase voltage (line-to-neutral voltage).

Line –to-neutral voltage of the phase is chosen as a reference.

|Vp|= magnitude of phase voltage

Page 6: 3phase Sys n Power Concept -lect5

Y-Connected Load (cont.)

The line voltages at the load terminals in terms of the phase voltage are found by application of Kirchhoff’s voltage law.

Page 7: 3phase Sys n Power Concept -lect5

Comparison of these line-to-line voltages with the line-to-neutral voltages of leads to the following conclusion:

In balanced three-phase Y-connected system with positive-sequence source, the line-to-line voltage are √3 times the line-to-neutral voltages and lead by 30°. That is;

Vab = √3Van +30°

Y-Connected Load (cont.)

Vbc = √3Vbn +30°

Vca = √3Vcn +30°

Page 8: 3phase Sys n Power Concept -lect5

The relationship between the line voltage and phase voltage is shown as below:

Y-Connected Load (cont.)

Phasor diagram showing phase and line voltages

Vab

Vbc

Vca

Van

Vbn

Vcn

Voltages Triangle

Page 9: 3phase Sys n Power Concept -lect5

RMS value of any of the line voltages is denoted by VL, then one of the important characteristic of the Y-connected tree-phase load may be expressed as

Y-Connected Load (cont.)

Three-phase current also possess three-phase symmetry and are given by

IL = IP

Page 10: 3phase Sys n Power Concept -lect5

Example

Calculate the line voltage and line current of a Y-Y connection

Page 11: 3phase Sys n Power Concept -lect5

A balanced ∆-connected load (with equal phase impedances) is shown in below:

∆-Connected Loads

Line voltage are the same as phase voltages.

Page 12: 3phase Sys n Power Concept -lect5

The phase current Iab is chosen as reference

∆-Connected Loads (cont.)

The relationship between phase and line currents can be obtained by applying Kirchhoff’s current law at the corner of ∆.

Ip = magnitude of phase current

Page 13: 3phase Sys n Power Concept -lect5

The RMS of any of the line currents and phase is denoted by IL

∆-Connected Loads (cont.)

The magnitude of line current is √3 times the magnitude of the phase current, and the line currents lags of phase currents by 30°.

Page 14: 3phase Sys n Power Concept -lect5

The relationship between the line currents and phase currents is showing in phase diagram.

∆-Connected Loads (cont.)

Page 15: 3phase Sys n Power Concept -lect5

Three Phase example

Page 16: 3phase Sys n Power Concept -lect5

Three Phase example(cont.)

Page 17: 3phase Sys n Power Concept -lect5

∆-Y Transformation For analyzing network problems, it is

convenient to replace the ∆-connected circuit with an equivalent Y-connected circuit.

Consider the Y-connected circuit of ZY Ω/phase which is equivalent to a balanced ∆-connected circuit of Z ∆ Ω/phase.

∆-connection Y-connection

Page 18: 3phase Sys n Power Concept -lect5

∆-Y Transformation(cont.) For ∆-connected circuit

(1)

From the phasor diagram, the relationship between balanced phase and line-to-line voltage; (2)

Page 19: 3phase Sys n Power Concept -lect5

∆-Y Transformation(cont.)

Substituting eq.(2) into eq.(1), we get

(3)or

(4)

For Y-connected circuit, we have

(5)

Substituting eq.(4) into eq.(5), we find that

(6)

Page 20: 3phase Sys n Power Concept -lect5

A balanced, positive-sequence, Y-connectedvoltage source with Eab = 480 0° volts is

applied to a balanced-∆ load with Z∆ = 30 0° Ω. The

line impedance between the source and load is ZL= 1 85°Ω for each phase. Calculate the

line current, the ∆-load currents, and the

voltages at the load terminals.

example

Page 21: 3phase Sys n Power Concept -lect5

Power in single-phase AC circuits

Figure show a single-phase sinusoidal voltage supplying a load

Instantaneous voltage be,

and Instantaneous current be,

The instantaneous power p(t) delivered to the load is the product of voltage v(t) and i(t) given by

By using the trigonometric identity

Page 22: 3phase Sys n Power Concept -lect5

Which results in

RMS value of v(t) is |V|=Vm/√2, RMS value of i(t) is |I|=Im/√2 . Let θ=(θv – θi). The above equation, in terms of the rms values, is reduce to

Page 23: 3phase Sys n Power Concept -lect5

Instantaneous power has been decomposed into two components. The first component is

Since the average value of this sinusoidal function is zero, the average power delivered to the load is given by

Also called Active power or Real powerThe second component is

The amplitude of this pulsing power is called reactive power

Q = |V|I|sin θ

Θ = θv– θi > 0 ;Q=+ve

Θ = θv– θi < 0; Q=-ve

Page 24: 3phase Sys n Power Concept -lect5

The Power Factor (pf) is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of load impedance.

The power factor may also be regarded as the ratio of the real power dissipated to the apparent power of the load.

Apparent Power

cos

Power Factor

( )v i

Ppf

S pS

P f

Page 25: 3phase Sys n Power Concept -lect5

Example

Calculate the power factor seen by the source and the average power supplied by the load

Page 26: 3phase Sys n Power Concept -lect5

Purely resistive load (R)

θv– θi = 0, Pf = 1 P/S = 1, all power are consumed

Purely reactive load (L or C)

θv– θi = ±90o, pf = 0

P = 0, no real power consumption

Resistive and reactive load (R and L/C)

θv– θi > 0 ;Q=+veθv– θi < 0; Q=-ve

• Lagging - inductive load• Leading - capacitive loadP/S < 1, Part of the apparent

power is consumed

Characteristics of the instantaneous power

Page 27: 3phase Sys n Power Concept -lect5

For circuits operating in sinusoidal-steady state, real and reactive power are conveniently calculated from complex power, defined below.

Complex power S is product of voltage and the conjugate of the current:

Complex Power

IVIVS iv

sincos IVjIV

Page 28: 3phase Sys n Power Concept -lect5

Thus, the complex power designated by S is given by

Complex Power(cont.)

jQPVIS

The magnitude of S, 22 QPS

* its unit is volt-amperes and the larger units are kVA or MVA The reactive power ,Q :

positive : θ(impedance angle) between V and I is +ve . In this case, load impedances is inductive (I lag V).

negative : θ(impedance angle) between V and I is -ve . In this case, load impedances is capacitive (I leadsV).

The impedances of the complex power S is given by

S

VZ

2

Page 29: 3phase Sys n Power Concept -lect5

Power Triangle

a) Power Triangle b) Impedance Triangle

Power Triangle

The COMPLEX Power is represented by the POWER TRIANGLE similar to IMPEDANCE TRIANGLE. Power triangle has four items: P, Q, S and θ.

0 Resistive Loads (Unity )

0 Capacitive Loads (Leading )

0 Inductive Loads (Lagging )

Q Pf

Q Pf

Q Pf

Page 30: 3phase Sys n Power Concept -lect5

Complex Power(cont.) The Complex Power contains all the information pertaining to the power absorbed by a given load.

2 2

Complex Power

Apparent Power

Real Pow

1

er

Reactive

= ( )2

=

= Re cos( )

=Q Im s Power

Power Fa

in( )

= =cctor os( )

Rms Rms v i

Rms Rms

v i

v i

v i

P jQ V I

S V I P Q

P S

S

P

S

S VI

S

S

S

• Real Power is the actual power dissipated by the load.

• Reactive Power is a measure of the energy exchange between source and reactive part of the load.

Page 31: 3phase Sys n Power Concept -lect5

Complex Power(cont.) Examples

Page 32: 3phase Sys n Power Concept -lect5

Complex Power(cont.)Solution

First method

Page 33: 3phase Sys n Power Concept -lect5

Second method

Solution complex power (cont.)

Third method

Page 34: 3phase Sys n Power Concept -lect5

Solution complex power (cont.)

Current phasor diagram and the complex power vector representation.

Page 35: 3phase Sys n Power Concept -lect5
Page 36: 3phase Sys n Power Concept -lect5
Page 37: 3phase Sys n Power Concept -lect5
Page 38: 3phase Sys n Power Concept -lect5

Power Factor Correction It is necessary to get PF is close to the unity. If

the PF < 1, the larger current I is supplied and additional cost to the utility company.

In order to maintain the power factor close to the unity, power companies install banks of capacitor throughout the network as needed.

A method of connecting a capacitor in parallel with an inductive load is known as power factor correction.

The effect of the capacitor is to increase the power factor of the source that delivers power to the load. The source apparent power also decreases.

Page 39: 3phase Sys n Power Concept -lect5

Power Factor Correction(cont.)

Original Inductive Load Inductive Load with improved power factor correction

Effect of capacitor on total current

Power triangle of power factor correction

Increasing the power factor without altering the voltage or current to the load is called Power Factor Correction

Page 40: 3phase Sys n Power Concept -lect5

Power Factor Correction(cont.)

Qc = Q1 – Q2

= P (tan θ1 - tan θ2)

= ωCVrms2

P = S1 cos θ1 Q2 = P tan θ2

c 1 22 2rms rms

Q P (tan θ tan θ )C

ωV ω V

Q1 = S1 sin θ1 = P tan θ1

Increasing the power factor without altering the voltage or current to the load is called Power Factor Correction.

Page 41: 3phase Sys n Power Concept -lect5

Power Factor Correction(Cont.)

Page 42: 3phase Sys n Power Concept -lect5

Power Factor Correction(Cont.)

Page 43: 3phase Sys n Power Concept -lect5

Power Factor Correction(cont.)

Page 44: 3phase Sys n Power Concept -lect5

2 cos( ) 2 cos( 120 )

2 cos( 120 )

2 cos( ) 2 cos( 120 )

2 cos( 120 )

cos( ) cos( ) cos( 120 )cos( 120 )2

cos( 120 )co

AN p BN p

CN p

a p b p

c p

a b c AN a BN b CN c

p p

v V t v V t

v V t

i I t i I t

i I t

p p p p v i v i v i

t t t tp V I

t

Using the above identity and simplifying, =2 t- we obtain

s( 120 )

1cos cos [cos( ) cos( )]

2 that:

13cos

2

cos 2 oos 3c c sp pp p

t

A B A B A B

VV I Ip

Consider a balanced three-phase source supplying a balanced Y-or ∆-connected load with the following instantaneous voltage and current .

Balanced Three-Phase Power(cont.)

Page 45: 3phase Sys n Power Concept -lect5

Balanced Three-Phase Power(cont.)

Three-phase reactive power is

The complex three-phase power is

So the real power of the three-phase is

ivLLpp whereIVIVP ;cos3cos33

sin3sin33 LLpp IVIVQ

333 jQPs or

ppIVs 33

Page 46: 3phase Sys n Power Concept -lect5

Balanced Three-Phase Power(cont.)ExampleA three-phase line has an impedance of 2 + j4Ω as shown in figure below

The line feeds two balanced three-phase loads that are connected in parallel. The first load is Y-connected and has an impedance of 30 + j40Ω per phase. The second load is ∆-connected and has impedance of 60-j45Ω. The lines is energized at the sending end from a three-phase balanced supply of line voltage 207.85V. Taking the phase voltage Va as reference, determine:

Page 47: 3phase Sys n Power Concept -lect5

Balanced Three-Phase Power(cont.)

a) The current, real power, and reactive power drawn from the supply.

b) The line voltage at the combined loads.

c) The current per phase in each load.

d) The total real and reactive powers in each load and the line.

Page 48: 3phase Sys n Power Concept -lect5

Solution

Page 49: 3phase Sys n Power Concept -lect5

Solution

(cont.)

Page 50: 3phase Sys n Power Concept -lect5

Unbalanced Three Phase Systems

An unbalanced system is due to unbalanced voltage sources or unbalanced load. In a unbalanced system the neutral current is NOT zero.

Unbalanced three phase Y connected load.

Line currents DO NOT add up to zero.

In= -(Ia+ Ib+ Ic) ≠ 0

Page 51: 3phase Sys n Power Concept -lect5
Page 52: 3phase Sys n Power Concept -lect5

THANK YOU FOR YOUR ATTENTION!!


Recommended