EKT103
ELECTRICAL ENGINEERING
Chapter 1
Three-Phase System
1
COURSE OUTCOME (CO)
CO1: Ability to define and explain the
concept of single-phase and three-phase
system.
2
Revision
• A sinusoid is a signal that has the form of the sine or cosine
function.
• A general expression for the sinusoid,
v(t )  Vm sin( t   )
where
Vm = the amplitude of the sinusoid
ω = the angular frequency in radians/s
Ф = the phase
3
Revision
A periodic function is one that satisfies v(t) = v(t + nT), for
all t and for all integers n.
T
2

f 
1
Hz
T
  2f
• Only two sinusoidal values with the same frequency can be
compared by their amplitude and phase difference.
• If phase difference is zero, they are in phase; if phase
difference is not zero, they are out of phase.
4
Revision
Example 1
Given a sinusoid, 5 sin( 4t  60 o ), calculate its
amplitude, phase, angular frequency, period, and
frequency.
5
Revision
Example 1
Given a sinusoid, 5 sin( 4t  60 o ), calculate its
amplitude, phase, angular frequency, period, and
frequency.
Solution:
Amplitude = 5, phase = –60o, angular frequency
= 4 rad/s, Period = 0.5 s, frequency = 2 Hz.
6
Revision
Example 2
Find the phase angle between i1  4 sin( 377t  25o )
and i2  5 cos(377t  40 o ), does i1 lead or lag i2?
7
Revision
Example 2
Find the phase angle between i1  4 sin( 377t  25o )
and i2  5 cos(377t  40 o ), does i1 lead or lag i2?
Solution:
Since sin(ωt+90o) = cos ωt
i2  5 sin( 377t  40o  90o )  5 sin( 377t  50o )
i1  4 sin( 377t  25o )  4 sin( 377t  180o  25o )  4 sin( 377t  205o )
therefore, i1 leads i2 155o.
8
Revision
Impedance transformation
Z  3ZY
1
ZY  Z
3
9
Single-Phase Circuit
A single phase circuit consists of a generator connected through a pair of wires
to a load
Two wire system
Three wired system
• same magnitude
• same phase
Two-Phase Circuit
a
A
Three wired
system
Second source
with 90° out
of phase
Three wired system
• same magnitude
• different phase
What is a Three-Phase Circuit?
•
It is a system produced by a generator consisting of three sources
having the same amplitude and frequency but out of phase with each
other by 120°.
Three sources
with 120° out
of phase
Four wired
system
12
Balance Three-Phase Voltages
•
A three-phase generator consists of a rotating magnet
(rotor) surrounded by a stationary winding (stator).
A three-phase generator
The generated voltages
13
Balance Three-Phase Voltages
•
Two possible configurations:
Three-phase voltage sources: (a) Y-connected ; (b) Δ-connected
14
Balance Three-Phase Voltages
Phase sequences
a) abc or positive sequence
sequence
b) acb or negative
15
Balance Three-Phase Voltages
If the voltage source have the same amplitude
and frequency ω and are out of phase with each
other by 120o, the voltage are said to be balanced.
Van  Vbn  Vcn  0
Van  Vbn  Vcn
Balanced phase voltages are equal in
magnitude and out of phase with each other by
120o
16
Balance Three-Phase Voltages
abc sequence or positive sequence:
Van  Vp 00
Vbn  Vp   1200
Vp is the
effective or
rms value
Vcn  Vp   2400  Vp   1200
acb sequence or negative sequence:
Van  Vp 00
Vcn  Vp   1200
Vbn  Vp   2400  Vp   1200
17
Balance Three-Phase Voltages
Example 1
Determine the phase sequence of the set of
voltages.
van  200 cos(t  10)
vbn  200 cos(t  230)
vcn  200 cos(t  110)
18
Balance Three-Phase Voltages
Solution:
The voltages can be expressed in phasor form as
Van  20010 V
Vbn  200  230 V
Vcn  200  110 V
We notice that Van leads Vcn by 120° and Vcn in turn leads Vbn
by 120°.
Hence, we have an acb sequence.
19
Balance Three-Phase Voltages
Two possible three-phase load configurations:
a) a wye-connected load
b) a delta-connected load
20
Balance Three-Phase Voltages
A balanced load is one in which the phase impedances
are equal in magnitude and in phase.
For a balanced wye connected load:
Z1  Z2  Z3  ZY
For a balanced delta connected load:
Za  Z b  Zc  Z 
Z  3ZY
1
ZY  Z
3
21
Balance Three-Phase Connection
•
Four possible connections
1. Y-Y connection (Y-connected source with a Yconnected load)
2. Y-Δ connection (Y-connected source with a Δconnected load)
3. Δ-Δ connection
4. Δ-Y connection
22
Balance Y-Y Connection
• A balanced Y-Y system is a three-phase system with a
balanced y-connected source and a balanced y-connected
load.
Balance Y-Y Connection
Z s  Source impedance
Z   Line impedance
Z L  Load impedance
Z Y  Total impedance per phase
Since all impedance are in series,
Thus
ZY  Z s  Z   Z L
ZY  ZL
24
Balance Y-Y Connection
Van
Ia 
ZY
Balance Y-Y Connection
Applying KVL to each phase:
Van
Ia 
ZY
Vbn Van   1200
Ib 

 I a   1200
ZY
ZY
Vcn Van   2400
Ic 

 I a   2400
ZY
ZY
VnN  Zn I n  0
I a  I b  I c  I n  0
26
Balance Y-Y Connection
Line to line voltages or line voltages:
Vab  3Vp 300
Vbc  3Vp   900
Vca  3Vp   2100
Magnitude of line voltages:
VL  3Vp
Vp  Van  Vbn  Vcn
VL  Vab  Vbc  Vca
Balance Y-Y Connection
Example 2
Calculate the line currents in the three-wire Y-Y system
shown below:
28
Balance Y-Y Connection
Example 2
Calculate the line currents in the three-wire Y-Y system
shown below:
Ans
I a  6.81  21.8 A
I b  6.81  141.8 A
I c  6.8198.2 A
29
Balance Y-Δ Connection
• A balanced Y-Δ system is a three-phase system with a
balanced y-connected source and a balanced Δ-connected
load.
30
Balance Y-Δ Connection
A single phase equivalent circuit
Van
Van
Ia 

ZY Z  / 3
Z
ZY 
3
Balance Y-Δ Connection
A single phase equivalent circuit
Line voltages:
Vab  3Vp 300  VAB
Vbc  3Vp   900  VBC
Vca  3Vp   2100  VCA
Balance Y-Δ Connection
A single-phase equivalent circuit of a balanced Y- circuit
Line currents:
I a  I AB  I CA  3I AB   30
I b  I BC  I AB  3I AB   150
I c  I CA  I BC  3I AB 90
Phase currents:
I AB
V AB

Z
I BC
V BC

Z
I CA
VCA

Z
Balance Y-Δ Connection
A single-phase equivalent circuit of a balanced Y- circuit
ICA  I AB  2400
Ia  I AB  ICA  I AB (1  1  2400 )
Ia  I AB 3  300
Magnitude line currents:
I L  3I p
I L  Ia  I b  Ic
I p  IAB  I BC  ICA
Balance Y-Δ Connection
Example 3
A balanced abc-sequence Y-connected source with ( Van  10010 ) is
connected to a Δ-connected load (8+j4) per phase. Calculate the
phase and line currents.
Solution
Using single-phase analysis,
Van
10010
Ia 

 33.54  16.57 A
Z  / 3 2.98126.57
Other line currents are obtained using the abc phase sequence
35
Balance Δ-Δ Connection
• A balanced Δ-Δ system is a three-phase system with a
balanced Δ -connected source and a balanced Δ -connected
load.
36
Balance Δ-Δ Connection
Line voltages:
Phase currents:
Line currents:
Vab  VAB
I a  I AB  I CA  3I AB   30
Vbc  VBC
I b  I BC  I AB  3I AB   150
Vca  VCA
I c  I CA  I BC  3I AB 90
Magnitude line currents:
IL  Ip 3
Total impedance:
Z
ZY 
3
I AB
V AB

Z
I BC
V BC

Z
I CA 
VCA
Z
37
Balance Δ-Δ Connection
Example 4
A balanced Δ-connected load having an impedance 20-j15  is connected
to a Δ-connected positive-sequence generator having ( Vab  3300 V).
Calculate the phase currents of the load and the line currents.
Ans:
The phase currents
I AB  13.236.87 A; I BC  13.2  81.13 A; I AB  13.2156.87 A
The line currents
I a  22.866.87 A; Ib  22.86  113.13 A; Ic  22.86126.87 A
38
Balance Δ-Y Connection
• A balanced Δ-Y system is a three-phase system with a
balanced y-connected source and a balanced y-connected
load.
39
Balance Δ-Y Connection
Applying KVL to loop aANBba:
Ia  Ib 
V p 0 0
ZY
From:
I b  I a   120 0
Ia  I b  Ia 3300
Line currents:
Vp
Ia 
3
  30 0
ZY
40
Balance Δ-Y Connection
Replace Δ connected source to equivalent Y
connected source.
Phase voltages:
Van 
Vbn 
Vcn 
Vp
3
Vp
3
Vp
3
  30 0
  150 0
  90 0
41
Balance Δ-Y Connection
A single phase equivalent circuit
Vp
Van
Ia 

ZY
3
  30 0
ZY
42
Balance Δ-Y Connection
Example 5
A balanced Y-connected load with a phase impedance 40+j25  is
supplied by a balanced, positive-sequence Δ-connected source with a
line voltage of 210V. Calculate the phase currents. Use Vab as reference.
Answer
The phase currents
I AN  2.57  62 A;
I BN  2.57  178 A;
I CN  2.5758 A;
43
Power in a Balanced System
• Comparing the power loss in (a) a single-phase system,
and (b) a three-phase system
PL2
P'loss  2 R 2 , single - phase
VL
Ploss  2I L 
2
PL2
R  2 R 2 , single - phase
VL
P'loss  3I 'L 
2
PL2
PL2
R'  3R' 2  R' 2 , three - phase
3VL
VL
• If same power loss is tolerated in both system, three-phase
system use only 75% of materials of a single-phase system
44
Power in a Balanced System
For Y connected load, the phase
voltage:
v AN 
2V p cos t
vBN 
2V p cos(t  120 0 )
vCN 
2V p cos(t  120 0 )
45
Power in a Balanced System
Phase current lag phase voltage by θ.
If
ZY  Z
The phase current:
ia 
2 I p cos(t   )
ib 
2 I p cos(t    1200 )
ic 
2 I p cos(t    1200 )
46
Power in a Balanced System
Total instantaneous power:
p  p a  p b  p c  v ANi a  v BNi b  v CN i c
p  3Vp I p cos
Average power per phase:
Reactive power per phase:
Pp  Vp I p cos
Qp  Vp I p sin 
Apparent power per phase:
Complex power per phase:
Sp  Vp I p
Sp  Pp  jQ p  Vp I*p
47
Power in a Balanced System
Total average power:
P  3Pp  3Vp I p cos  3VL I L cos
Total reactive power:
Q  3Qp  3Vp I p sin   3VL I L sin 
Total complex power:
2
3
V
S  3Sp  3Vp I*p  3I 2p Zp  *p
Zp
S  P  jQ 
3VL I L 
48
Power in a Balanced System
Power loss in two wires:
2
P
Ploss  2I 2L R  2R L2
VL
Power loss in three wires:
Ploss
2
2
P
P
 3I L2 R  3R L 2  R L2
3VL
VL
PL : power absorbed by the load
IL : magnitude of line current
VL : line voltage
R : line resistance
49
Example 6
A three-phase motor can be regarded as a balanced
Y-load. A three-phase motor draws 5.6 kW when the
line voltage is 220 V and the line current is 18.2 A.
Determine the power factor of the motor.
50
Example 6
A three-phase motor can be regarded as a balanced
Y-load. A three-phase motor draws 5.6 kW when the
line voltage is 220 V and the line current is 18.2 A.
Determine the power factor of the motor.
The apparent power is
S  3VL I L  3 (220)(18.2)  6935.13VA
P  S cos  5600W
The real power is
The power factor is
P
pf  cos    0.8075
S
51
Exercise 6
Calculate the line current required for a 30-kW
three-phase motor having a power factor of 0.85
lagging if it is connected to a balanced source
with a line voltage of 440 V.
52
Exercise 6
Calculate the line current required for a 30-kW
three-phase motor having a power factor of 0.85
lagging if it is connected to a balanced source
with a line voltage of 440 V.
Answer :
I L  46.31
53
Exercise 7
For the Y-Y circuit in Exercise 2, calculate the
complex power at the source and at the load.
54