ee207 electrical power - lecture 3

EE207 Electrical Power

Lecture 3

3 phase circuits

Rajparthiban Kumar EE207 Electrical Power 2

Introduction• Three phase power is preferred to single phase

power for the following reasons:-

a) Three phase motor, generators and transformers are simpler, cheaper and more efficient.

b) Three phase transmission line can deliver more power for a given weight and cost

c) The voltage regulation of 3 phase transmission line is inherently better.


Three Phase System

• As the magnet turns it sweeps across the conductors, inducing in them a voltage given as:

• Induced voltage will be maximum when the poles are in the position shown by the Fig. because the flux density is greatest at the centre of the pole.

• Induced voltage will be zero when the poles have rotated 90o.

S

N

ωm

Single Phase

Generator

BlvEa ====1

1

a

R


Three Phase System

• The voltage and current are

in phase as load is resistive.

• The Power supplied is

always positive and at twice

the voltage and current

frequency and also has a

peak:

Pp=VpIp

• The average power supplied:

90 180 270 360 450 θθθθ

0

Vp

0

-Vp

Ip

90 180 270 360 450 θθθθ

Pp

rmsrmsppavg IVIVP ========2

11φφφφ


• In a two phase generator

another set of windings

(b,2) is placed 90o from

the first set (a,1). Thus at

90o voltage will also be

induced and power will be

generated.

• In this case power will be

supplied almost in

continuous manner and

thus vibration no longer

exists.

Two Phase

Generator

SN

ωm

1

a

R

b2

R

Three Phase System


• The voltage in phase a and

b are given respectively

as:

• The total power Ptotal has

average value equal to the

peak power

)cos(Vv

cosVv

pb

pa

2

ππππθθθθ

θθθθ

++++====

====0 1.57 3.14 4.71 6.28 7.85 9.42 10.99-1

0

1

0 1.57 3.14 4.71 6.28 7.85 9.42 10.990

0.5

1

0 1.57 3.14 4.71 6.28 7.85 9.42 10.990

0.5

1

va vbIa Ib

Pa Pb

Pa+Pb=Ptotal

avgppavg PIVP φφφφφφφφ 12 2========

Three Phase System


Three Phase System

• Three phase generator is

similar to two phase generator

except that it has three coils

placed 120o from each other.

• The three phase voltages are

given as:

SN

ωm

1

a

R

b

2

R

c

3

R

Three Phase

Generator

)cos(Vv

)cos(Vv

cosVv

pc

pb

pa

3

4

3

2

ππππθθθθ

ππππθθθθ

θθθθ

−−−−====

−−−−====

====


Three Phase System

• Note that the voltage signals are phase shifted by 120o

• The total power supplied to a three phase load is:

• Note also that the sum of three phase voltages is Zeroat any instant.

• Thus if these voltages are applied across three equalimpedances, the sum of the resultant three phase currents is also Zero

0 60 120 180 240 300 360 420 480 540 600 660-1

-0.5

0

0.5

1

va vb vc

ppavgavg IV.PP 513 13 ======== φφφφφφφφ

0====++++++++ cba vvv


Three Phase System

• The windings of a three phase system are connected to three resistive loads using 6 wires.

• However the return wires form all the three phases can be joined in one conductor called the Neutral.

• Incase the loads in three phase system are equal (balanced) , the sum of the three currents at any instant will be Zero and thus the Neutral wire can be removed.

a

c b

1

2

3

R

R

R

va1

vb2

vc3

Ia

Ib

Ic

a

c b

12

3

R

R

R

va1

vb2

vc3

Ia

Ib

Ic

Ia +Ib+Ic=0


Wye (Y) Connection

• The voltages (van, vbn and

vcn) between line and neutral

are known as phase voltages

(vφφφφ)))).

• The voltage difference

between two phase voltages

(vab, vbc, vca) is known as

line voltage (vL). Line

voltages are given as:

a

cb

R

R

R

van

v bn

vcn

Ia

Ib

Ic

n vab

vca

vbc

ancnca

cnbnbc

bnanab

vvv

vvv

vvv

−−−−====

−−−−====

−−−−====

• Currents in the lines (Ia, Ib and Ic

are known as line currents.

• In aY connected system line

currents are equal to those currents

flow in the load phases.

Y System


van

-vnb

vbn

vcn

-vna

-vnc

bnbc vv 3====

γ=30γ=30γ=30γ=30οοοο

120o

120o

120o

)cos(Vv

)sincos(V

)sincos(V

)sincos(cosV

)cos(VcosVv

opab

p

p

p

ppab

303

2

1

2

33

2

3

2

3

2

3

2

1

3

2

++++====

−−−−====

−−−−====

−−−−++++====

−−−−−−−−====

θθθθ

θθθθθθθθ

θθθθθθθθ

θθθθθθθθθθθθ

ππππθθθθθθθθ

In a Y connected system The line

voltage is √√√√3 times the phase voltage

and it leads by 30o.

Wye (Y) Connection

Y system Phasor Diagram


• Example

• A 3-phase, 60 Hz, wye connected generator has a line voltage of 23900 V

Calculate

• the line-to-neutral voltage

• voltage induced in the individual windings

• the time interval between the positive peak voltage of phases A and B

• the peak value of the line voltage


VV

V LLLN 13800

3

23900

3===

5.55ms s 1/180 s) (1/60 1/3 T

120 is waveformsltagebetween vo angle The

s 1/60 takescyle One

===

°

VVV rmsLp 338002, ==


Delta Connection

• Three phase load can also be

connected to form what is

known as Delta System.

• In a ∆ the line voltages

appear across the load

phases. However, the

currents in the load phases

are not the same as the line

currents.

• The line current in ∆∆∆∆system is √√√√3 times the

phase current and it

lags by 30o.

R

R

R

Ia

Ib

Ic

vab

vbc

vca

a

b

c

Iφ1φ1φ1φ1

Iφ2φ2φ2φ2

Iφ3φ3φ3φ3

∆∆∆∆ System

23

12

31

φφφφφφφφ

φφφφφφφφ

φφφφφφφφ

III

III

III

c

b

a

−−−−====

−−−−====

−−−−====


• Note from the phasor

diagram that the phase

currents are in phase with the

line voltages because the load

is resistive.

• The line currents Ia, Ib and Ic

are phase shifted from each

other by 120o.

Iφ1φ1φ1φ1

γ=30γ=30γ=30γ=30οοοο

120o

120o

120o

Iφ2φ2φ2φ2

Iφ3φ3φ3φ3

-Iφ3φ3φ3φ3

vab

vca

vbc

-Iφ1φ1φ1φ1

Ib

Ic

θθθθIIa 3====

∆∆∆∆ system Phasor Diagram

Delta Connection


Example

Three identical impedances are connected in

delta across a 550 V line. The current drawn

is 10 A. Calculate

• the current in each impedance

• the impedance values


Ω===

==

9577.5

550

I

VZ

550V isresistor across Voltage

77.53

AI

I aφ


Power In Balanced Three Phase System

• If the magnitude of the phase voltages for a Y connected

system is :

• And if the magnitude of the phase currents for a delta

system is:

• Then the total three phase power is:

Where θ is the power factor angle (angle between the voltage and the current)

cnbnan vvvv ============φφφφ

cnbnan IIII ============φφφφ

θθθθφφφφφφφφφφφφ cosIvP 33 ====


• For a Y system is the magnitudes of the line voltage and

line current are |VL| and |IL| respectively. Then:

• The three phase power can be expressed in terms of the

line quantities as:

• Also the total Reactive power (Q) in a three phase system

is given as


φφφφφφφφ IIandV

v LL ========3

θθθθφφφφ cosIVP LL33 ====

θθθθ

θθθθφφφφφφφφφφφφ

sinIV

sinIVQ

LL3

33

====

====



• The total apparent Power (S) in a three phase system is

also given as:

• In a Delta system all the above equations are valid.

• For ∆ system, the total three phase power (P) is:

• However for ∆, the phase voltages and currents can be

express as line quantities as:

LL IVS 33 ====φφφφ

θθθθφφφφφφφφφφφφ cosIvP 33 ====

3

φφφφφφφφ

IIandVv LL ========



• Then the total power in a ∆ system is given as:

LL

LL

LL

IVS

and

sinIVQ

,similarly

cosIVP

3

3

3

3

3

3

====

====

====

φφφφ

φφφφ

φφφφ

θθθθ

θθθθ


Examples

• Example 1

For the circuit in the figure below calculate:

– the current in each line.

– The voltage across the inductor terminals

– The Power absorbed by the load

– The Reactive power

– The apparent Power

– The power factorR

R

R

j4Ω

j4Ω

j4Ω

3Ω

3Ω 3Ω

a

b

c

440V

3 phase line


Ω=+= 534 22Z

VV

V LLLN 254

3==

VIXV

AZ

VI

L

LN

2.203)4(8.50

inductor theacross Voltage

8.50

===

==

254V

3+4j


LL

LL

LL

L

IVS

IVIVQ

IVIVP

AI

3

sin3sin3

cos3cos3

87.368.5087.365

0254

=

==

==

−∠=∠

∠=

θθ

θθ

φφ

φφ


Examples

• Example 2

A manufacturer plant draws a total of 415kVA from 2400V 3

phase line. If the plant power factor is 87.5% lagging,

calculate:

• The impedance of the plant

• The phase angle between the line to neutral voltage and the

line current.

• The complete phasor diagram of the plant.


AI

IVS

VVV

L

LL

LN

100)2400(3

415000

3

13863

2400

==

=

=== φ

Ω===∴ 9.13100

1386branch)per (

I

VZ

φ


°=

=

29

875.0_

θ

factorpower


Example

Motor

3594 HP

93% efficient

Pf = 0.9

Capacitor

Bank

1800 kvar

4000V

3phase

• A 5000 HP wye connected motor is connected to a 4000V,

3phase, 60 Hz line.

• A delta connected capacitor rated at 1800 kvar is also

connected to the line.

• If the motor produces an output of 3594 hp at 93%

efficiency and power factor of 90% lagging, calculate:-


a) The active power absorbed by the motor

kWP

P

kWP

m 288393.0

2681

2681)746(3594

===

==

η


kVarPSQ

kVAPS

S

P

1395

32039.0/2883cos/

cos

22 =−=

===

=

θ

θ

b) The reactive power absorbed by the motor


• c) The reactive power supplied by

transmission line

var40513951800

var1395

var1800

kQ

kQ

kQ

net

m

c

−=+−=

=

−=


d) The apparent power supplied by the line

kVAQPS

kQ

kWP

2911

var405

2883

22 =+=

−=

=


e) The line current f) Motor line current

Ak

I

IVS

L

LL

420)4000(3

2911

33

==

=φA

kIm 462

)4000(3

3203==


• Phasor diagram

°=

=

8.25

9.0cos

mθθ

°=

===

8

99.02911

2883cos

θ

θLS

P

25.8°

8°2309V

IL = 420A

Im = 462 A

Ic = 260A


• Capacitor current

• Q =1800kvar

• V = 4000V

• Q = 3 VI sin θ

• Ic =260 A


Phase Sequence

3-phase systems have a property called phase sequence

– it determines the direction of

• rotation of 3-phase motors

– it means the order in which the three line

• voltages become successively positive

– two sequences: ABC & ACB

• positive sequence: ABC

• •negative sequence: ACB


• Phase sequence is determined by a simple

• circuit

• – one lamp always burns brighter

• – the sequence order is bright lamp, dim

• lamp, capacitor


• Wattmeters measure active power

• single-phase wattmeters have two pairs of terminals: voltage & current

• each terminal pair has polarity markings polarity markings determine the direction of power flow

• 3-phase, 3-wire circuits

• – power is measured by two single-phase wattmeters

• – total power is the sum of the two meters

• 3-phase, 4-wire circuits

• – power is measured by three single-phase wattmeters

• – total power is the sum of the three meters


• Questions

• 8.5

• 8.11

• 8.15

• 8.26

ee207 electrical power - lecture 3

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