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Three Phase System

3.0 INTRODUCTION

This chapter is explaining about the three phase system. The learning outcome is student

should be able to apply the principles of three phase systems, balanced load, star and

delta types of connection to solve electrical circuit correctly.

3.1 THREE PHASE SYSTEM (3 )

Three-phase system is a common method of alternating current electric power generation,

transmission, and distribution. It is a type of poly phase system which has two or more

voltage source with fixed phase angle difference. Three phase system is a system that has

three source voltage of the same magnitude produced by the three separate coil conductor

which space 120

o

apart as shown in Figure 3.1. Normally a three phase supply has three(3) live wire or phase conductor of Red phase (R), Yellow phase (Y) and Blue phase (B).

North Pole

Y120

oB

R Oo

R

240

o

B Y

South Pole

Figure 3.1 : Three Phase Conductor Coil

The emf produced by each coils;

RR coil, Em sin 0 = zero, YY coil, Em sin 1200

= 0.866Em volt and BB coil , Em sin

2400

equal = -0.866 Em volt.

3

CHAPTERTHREE PHASE SYSTEM

(3 )

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The coils were rotating at a constant speed to generate three emf waveform with same

peak and frequency as in Figure 3.2.

Emf

Em ER EY EB

360O

0O

120o

240o

Figure 3.2 : Waveform of 3 System

Equations for each waveforms ;

Red Line Voltage, sinmR EE = .

Yellow Line Voltage, )120sin(o

mY EE = (3.1)

Blue Line Voltage, )240sin(o

mB EE =

The above equations can be represent in a form phase as below ;

omB

omY

omR

EE

EE

EE

2402

1202

02

=

=

=

Where the value of ER's, EYs and EBs is a value of rmsE ( root mean square).

(3.2)

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Three Phase System

3.2 DIFFERENT BETWEEN SINGLE PHASE AND THREE PHASE SYSTEM

Table 3.1: Difference Between Single Phase and Three Phase Systems

SYSTEM 1 SYSTEM 3

Definition:

System with 2 cable connecting from source

to load: Life cable (L) and neutral cable (N).

Definition:

System with 3 live cabels from supply to load :

Red phase(R), Yellow phase (Y) and Blue

phase (B).

Connection :

Live

240V beban

Neutral

Connection:

R

Y

240V 415V

B

N

Waveform :

V

240V

360o

0 180

Waveform:

V

R Y B

415V

360

0

120o 2400

Three phase system basically used in starting the 3phase machines, supply power to large

factories and for transmission and distribution 3 phase voltage. The advantages of 3

phase over single phase systems are:

i. More economical due to less construction and maintenance costs. The conductorused is smaller, thus the needs of copper can be reduced

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Three Phase System

ii. Easy to install due to smaller and lighter cable used. Therefore the smaller structuressupporters can be used and the distance between the supporters can be relatively

long.

iii. The characteristics of 3phase equipment have a better operating starting due to itsstable and fixed power.

iv. Electrical energy supplied is higher than single phase.v. Most large motors used for induction motor 3phase can stand alone without the

additional equipment such as capacitors, centrifugal switches or other additional

circuits

vi. The output Production and equipment quality of work for 3phase system is betterthan single phase because of the higher efficiency and power factor.

3.3 THREE PHASE LOAD CONNECTION

The load for three phase may be joined together either in star or delta connection. The

load in each phase is usually in balanced which means that the load/impedance at eachphase are the same.

There are 2 methods of load connection in three phase system:

1. Star or Wai connection2. Delta connection

3.3.1 Star ConnectionStar Connection, also known as Wai (Y) connection. This type of connection has four-

conductor wires where three for life line (Red(R), Yellow(Y) and Blue(B)) and one for

neutral(N) line as shown in Figure 3.3.

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Three Phase System

R (Red Line)

Load

N (Neutral Line)

Load Load

Y (Yellow Line)

B (Blue Line)

Figure 3.3: STAR Connection

i. VoltageLine voltage ( LV ) is the voltage measured between two life lines. While, the phase

voltage ( PV ) is the voltage measured between the life line with the neutral line as shown

in Figure 3.4.

R A

RNPEV =

LV

N

E

LV YNP EV =

BNP EV =

Y

B C

LV

Figure 3.4: Voltage in STAR connection

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Three Phase System

ERN , EYN and EBN are called phase voltage ( PV ). Phase difference between phase

voltages is equal to the 120. The mathematical relationship between the line voltage to

phase voltage is as equation 3.3 below;

( 3.3 )

ii. CurrentLine current ( LI ) is the current that flow in the line, while the phase currents ( PI ) are

defined as current which flow in the phase.

R LR II = (Red line current)

PPRII = (Red phase current)

LNII = (Neutral line current)

N

PPBII = (Blue phase current)

LYII = (Yellow line current)

YPPY

II = (Yellow phase current)

B

LBII = (Blue line current)

Figure 3.5 : Current in STAR connection

Refer to Figure 3.5 above, phase current ( PI ) which are PRI , PYI and PBI is equal to the

line current )( LI which are RI , YI and BI . In the form of mathematical equations, the

relationship between line currents and phase currents can be written as equation 3.4.

Line Voltage = 3 Phase Voltage

3=LV PV

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(3.4)

Example 3.1

Three uniform impedance each resistance 10 and inductance 0.019H, supply 415V,50Hz. Calculate the line current, phase current, phase voltage and line voltage if the three

Impedance connected in star.

Solution 3.1

Line voltage , =LV 415 V (Supply voltage 3 = Line voltage)

Phase voltage , 6.2393

415

3===

LP

VV V (star connection)

Impedance ,22

LP XRZ += ............................... equation. 1

Where, fLXL 2= = )019.0)(50(2 = 5.97 .

=+= 65.1197.51022

P

Z

Phase current,65.11

6.239==

p

PP

Z

VI = 20.57 A

Line current, PL II = = 20.57 A (star connection)

3.3.2 Delta ConnectionIn the delta ( ) connection, there are only three (3) conductor wire-line, ie red red

line(R), yellow line (Y) and the blue line (B). Point of AB, BC and AC are the points in

the phase.

Line current= Phase current

PL II =

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Three Phase System

Red line A

R

Load Load

Yellow line Load

Y

B C

Blue line

B

Figure 3.6: DELTA Connection

i. Current

RI = LI

R

1I = PI 2I = PI

YI = LI

Y 3I = PI BI = LI

B

Figure 3.7: Current in DELTA Connection

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Three Phase System

From Figure 3.7 above, current 1I , 2I and 3I are the currents of each phase and known as

phase current ( PI ). Meanwhile, RI , YI and BI are the current per line and known as

line current ( LI ). Since the loads are in balance, the current flow through each phase and

each line is the same as equation 3.5 and by 120 out of phase with each other.

(3.5)

In delta connection, the relationship between line currents and phase currents can be

written in the form of mathematical expressions such as equation 3.6.

ii. Voltage

R

PL VV =

Y

PL VV =

B

Figure 3.8: Voltage in DELTA Connection

LBYR

P

IIII

IIII

===

=== 321

Line Current = 3 Phase Current

PL II 3= (3.6)

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In the delta connection, the line voltage and phase voltage is the same for both measured

between two (2) live wires and it can be expressed in equations 3.7 below.

Example 3.2

Three uniform impedance each resistance 10 and inductance 0.019H, the supply 415V,50Hz. Calculate the line current, phase current, phase voltage and line voltage if the three

impedances connected in delta.

Solution 3.2

Line voltage , =LV 415 V (Supply voltage 3 = Line voltage)

Phase Voltage, VVP 415= (delta connection)

Impedance ,22

LP XRZ +=

where, fLXL 2= = )019.0)(50(2 = 5.97 .

=+= 65.1197.51022

PZ

Phase current,65.11

415==

p

PP

Z

VI = 35.9 A

Line current,PL II 3= = )9.35(3 = 62.18 A (delta connection)

3.4 THREE PHASE POWER

Three-phase power system is equal to the power that lies within the power of one-phase

is the apparent power, the active or real power and reactive power .

Line Voltage = Phase Voltage

=LV PV

(3.7)

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3.4.1 Apperent power

Apparent power is a measure of alternating current power that is computed by

multiplying the root-mean-square current by the root-mean-square voltage. The symbol

for apparent power is S and the unit is volt-ampere (VA)

Apperent power in every phase :

PPP IVS = (3.8)

Total apparent power in the system 3 :

PxSS 33 =

= PPIV3

= LLIV3 (3.9)

3.4.2 Real PowerReal power also called the average power or active power. The symbol for real power is

P and the unit is watt (W).

Real power per phase:

cosPPP IVP = (3.10)

Total active power in three-phase system:

PxPP 33=

= cos3 PPIV

From equation 3.7 for DELTA connection PL VV = and3

LP

II =

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Three Phase System

PPIVP 33 = = cos3

3 xI

xV LL

= cos3 LLIV

(3.11)

This is the same formula for the STAR connection.

3.4.3

Reactive Power

Reactive power is also called the power of imagination. The symbol for reactive power is

R and the unit is volt-ampere reactive (VAR)

Reactive power per phase:

sinPPP IVQ = (3.12)

Total reactive power in three-phase system:

PxQQ 33 =

= PPIV3 sin (3.13)

From equation 3.7 for DELTA connection PL VV = and3

LP

II =

sin33 PPIVQ = = cos33 xI

xV LL

= sin3 LLIV

=3P cos3 LLIV

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(3.14)

This is also the same formula for the STAR connection.

3.4.4 Power Triangle diagram

All of three power described above can be illustrated using a triangle diagram as shown

in Figure 3.9. This triangle is called power triangle.

S = VI

Q

P

Figure 3.9: Power Triangle diagram

Example 3.3

Based on Example 3.2, calculate the total apparent power, active power and reactive

power

Solution 3.3

i). Total apparent power, LLIVS 33 = = )18.62)(415(3 = 44.7 kVA

ii). Total active power, cos33 LLIVP = = ))(18.62)(415(3Z

R

= 44695.1(65.11

10) = 38.4 kW

=3Q sin3 LLIV

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iii). Total reactive power , sin33 LLIVQ = = )18.62)(415(3 sin 30.9o

= 22.9 kVAR

REFERENCE

Bakshi, U.A and Godse, A.P, (2007) Basic Electrical and Electronics Engineering 1st

Edition,

Technical Publication Pune, India

PROBLEMS

1. Three balanced load with 10 resistance and 0.5H inductor, stars connected supplied

with three phase 500v, 50Hz. Calculate:i. Line current

ii. Apparent Power

iii. Real Power

(1.83 < -95.9 A, 1 589 VA, 102.2 W)

2. Three impedances of 30 resistance and 50 inductance reactance supply with 415V,

50Hz. The impedances is connected in star connection. Calculate:

i. Line voltage

ii. Phase voltage

iii. Phase current

iv. Line current

v. Power factorvi. Apparent Power

(415< 0 V, 239.6< 0 V, 4.1 < -65.6 A, 4.1 < -65.6 A, 0.514, 2 947 VA)

3. By referring to question 2, if the three impedances connected in delta. Calculate (i), (ii),(iii), (iv), (v) and (vi).

(415< 0 V, 415< 0 V, 7.11 < -65.6 A, 12.32 < -65.6 A, 0.514, 8 855 VA)

4. Three balanced load, 10 inductance reactance and 12 resistance of impedance with

star connected was supplied with 400V, 50Hz. Calculate:

i. Line current

ii. Power Factor

iii. Apparent power in kVA

iv. Real Power in kW

(14.78 < -44.23 A, 0.768, 10.23 kVA, 8.05 kW)

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5. Calculate the values of (i), (ii), (iii) and (iv) in question 4 for delta connection.

(44.35 < -44.23 A, 0.768, 30.73 kVA, 23.59 kW)

6. Three balanced load connected in delta with three-phase supply 415V, 50Hz. The line

current is 15A and 0.8 lagging power factor, calculate:

i. Phase current, phase voltage and line voltage

ii. Power in watt

(8.66 < 40.96 A, 415< 0 V, 415< 0 V, 8 625W)

7. Three balanced load with 0.15H inductance and 15 resistance connected in delta,

supply with three-phase 415V, 50Hz. Calculate:

i. Line Voltage

ii. Phase Current

iii. Line Current

iv. Power Factor

v. Reactive power

(415 < 0 V, 8.39 < 80.37 A, 14.54 < 80.37 A, 0.303, 10 451 VAR)

8. Three balanced load resistance 15, 0.8H inductor and 10F capacitors for each phase. Ifthe supply voltage three-phase 415V, 50Hz is connected in a star. Calculate:

i. Phase Current

ii. Line Current

iii. Power Factor

iv. Real power

(3.49 < -85.97 A, 3.49 < -85.97 A, 0.218, 546.8 W)

9. Three balanced load connected in delta to a three-phase supply, using a total power of 5kW 415V and take a line current of 12A, specify:

i. power factorii. Impedance of each load

iii. Apparent power and reactive power

iv. Real power if the loads are star connected.

(0.579, 59.88 < 53.34 , 8 625 VA, 6 919 VAR, 5 000W)

10. Three balanced load connected in delta to a three-phase 415V power supply using thepower of 5750w and the line current of 10A.

i. Draw the load connection

ii. Calculate power factor

iii. Calculate Impedance of each load

(0.667, 71.92 < 53.54 )

11. Three balanced load are connected in delta to a three-phase power supply 400V, 50Hz.The line current is 13A and 0.8 power factor leading. Calculate:

i. phase current

ii. Impedance

iii. Inductance

(7.5 < -40.96 A, 53.33 < 40.96 , 0.1 H)

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12. Three balance loads with 12 resistance and 9 inductance reactance, connected in deltato three phase 440v supply. Calculate:

i. Line current

ii. Phase voltage

iii. Power factor

iv. kVA used

v. kW used

(29.33 < -40.96 A, 440 < 0 V, 0.8, 12.445 kVA, 9.956 kW)

13. Calculate (i), (ii), (iii), (iv) and (v) in question 12 by using star connection.(16.93 < -40.96 A, 254 < 0 V, 0.8, 12.9 kVA, 10.32 kW)