Three-Phase Systems
3
i3
2 v23
·
v12
1
·
i2
·
G
v31
·
U
·
i1
·
The genera)on and the distribu)on of electrical energy is usually done by three-­‐
phase systems. There are three wire system connected to a generator consis)ng of three AC sources having the same amplitude and frequency (mostly 50 Hz, ω = 314 rad/s, Ve-­‐ S230/400 in Europe, most of Asia and Australia, 60 Hz, ω = 377 rad/s, Ve-­‐ S120/210, D120/240 in Nord America, Canada, and others) but out of phase with each other by 120°. Three are the main mo)va)ons: 1.  Normally the electric power is generated by three phase electrical machines (alternators – synchronous generators). 2.  In balanced three phase systems the total instantaneous power is constant (not pulsa)ng). This results in an uniform transmission and less vibra)ons. 3.  For the same amount of transported power, a three phase system is more economical than a single phase. Three-Phase Systems
The Set of voltages and currents in three phase systems are the following. The line currents i1(t), i2 (t), and i3 (t) are the currents flowing in each of the three lines. From the KCL it follows:
i1(t) + i2 (t) + i3 (t) = 0 
I1 + I 2 + I 3 ! = !0
The line tensions (or line voltages, line to line tensions) v12 (t), v23 (t), and v31 (t) are also the poten)al differences between the terminals 1 and 2, 2 and 3, and 3 and 1. From the KTL it follows: I 1
1
•
E 1
•O
E 3
E 2

V
31

V
12
I 2
2
•·
3

V
23
•·
I 3
v12 (t) + v23 (t) + v31 (t) = 0  +V
 +V
 ! = !0
V
12
23
31
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
1
Three-Phase Systems
The#three#phase#system#is#isofrequential.#The#three#phase#currents#and#
the#three#phase#voltages#must#fulfill#the#Kirchhoff's#laws.#In#the#phasor#
plane#it#follows:
###################################I 1 + I 2 + I 3 # = #0
 +V
 +V
 # = #0
##################################V
12
23
31
Therefore#the#three#phase#line#currents#are#represented#by#the#tringle(of(
the(line(currents#and#the#the#three#phase#line#tensions#are#represented#by#
the#tringle(of(the(line(tensions.

V
23
I 1
I 2

V
12

V
31
I 3
Triangle of the line tensions
Triangle of the line currents
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
Three-Phase Systems
Balanced line tension system
2π
3
α = e

V
12

V
31
2π
3
2π 3
2π
3
j(2π/3)

V
23
2π
3

V
31

V
12

V
23
2π
3
Positive sequence
 " to "V
 " and
(Clockwise rotation from V
12
23
 " to V
 )
from "V
23
31
Negative sequence
 " to
(Counterclockwise rotation from V
12
 " and from "V
 " to V
 )
"V
23
23
31
 ;""V
 " = "V
 e&j(2π /3) ;""V
 " = "V
 "e&j(4π /3)
V
12
23
12
31
12
 ;""V
 " = "V
 e j(2π /3) ;""V
 " = "V
 "e j(4π /3)
V
12
23
12
31
12
 ;""V
 " = " α 2 V
 ;""V
 " = ""α V
 "
V
12
23
12
31
12
 ;""V
 " = " α V
 ;""V
 " = " α 2 V
 "
V
12
23
12
31
12
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2
Three-Phase Systems
Balanced line current system
2π
3
α = e
2π
3
j(2π/3)
I 1
I 1
I 3
2π
3
2π
3
2π 3
I 3
I 2
I 2
2π
3
Positive sequence
(Clockwise rotation from I 1 " to "I 2 " and
from "I 2 " to I 3 )
Negative sequence
(Counterlockwise rotation from I 1 "
" to "I 2 " and from "I 2 " to I 3 )
I 1 ;""I 2 " = "I 1 e&j(2π /3) ;""I 3 " = "I 1 "e&j(4π /3)
I 1 ;""I 2 " = "I 1 e j(2π /3) ;""I 3 " = "I 1 "e j(4π /3)
I 1 ;""I 2 " = " α 2I 1 ;""I 3 " = ""α I 1 "
I 1 ;""I 2 " = " α I 1 ;""I 3 " = " α 2I 1 "
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
Three Phase System Supply
1
The phase tensions E 1 ,"E 2 ,"E 3 (or
phase voltages) are the potential
difference between the center O of
the Y connected tension system and
the terminals 1, 2, and 3.
E 1
V 31
 ! = !E
 !"!E

!!!!!!!!!!!!V
12
1
2
 ! = !E
 !"!E

!!!!!!!!!!!!V
23
2
3
 = !E
 -E

!!!!!!!!!!!!V
31
3
1
23
V 12
O
• E 2
2
E 3
3
 ,"E
 ,"E

Phase tensions:"""E
1
2
3
 ,V
 ,"V

Line tensions:""""""V
12
31

V
31
•
E 3
•
V 23
•
Triangle of the line tensions
and the phase tensions

E
1
O
•

V
12

E
2
O center of the
tension system

V
23
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3
Three Phase System Supply
Balanced Phase Tension
Balanced line tension system
(positive sequence):"
 ;""V
 " = " α 2 V
 ;""V
 " = ""α V

"""""V
12
23
12
31
12
"""
Balanced phase tension system
(positive sequence):"

α V
12

E
1

V
12
2
·
• α E 1
O

α E
1
 ;""E
 " = " α E
 ;""E
 " = ""α E

"""""E
1
2
1
3
1
"""""""""""
2

α 2 V
12
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
Three Phase System Supply
Balanced line tension system
(positive sequence):"
 " = "V
"""""""""V
12
(
(
)
)
V
1 + j" 3 "
2
 " = ""V
 "e%j4π /3 " = "V"e%j4π /3 " = " V %1 + j" 3
"""""""""V
31
12
2
"""
 "="V
 "e%j2π /3 " = "V"e%j2π /3 " = "%
"""""""""V
23
12
Balanced phase tension system
(positive sequence):"
"""""""" E 1 " = "E"e%j"π /6 """" where ""E" = " V 3
"""""""""E 2 " = " E""e%j"π /6 "e%j2π /3 " = "E"e%j5π /6
"""""""""E 3 " = " E""e%j"π /6 "e%j4π /3 " = "E"e%j9π /6
"""""""""""

α V
12


α 2 V
α E
12
1
•
O

α E
1
2

E
1

V
12
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
4
Three Phase System Loading
Y-Connected load
1
2
3
•
•
•

I 2
I3
1
2
3
I 3
I 23 Z 23
•
There are ∞2 set
of three
impedances which
deliver the same
set of currents
when connected
to the same set of
line tensions
 =Z
 I - Z
 I
⎧V
12
1 1
2 2
⎪⎪
 =Z
 I - Z
 I
⎨V
23
2 2
3 3
⎪I + I + I = 0
1
2
3
⎪⎩
→
 =Z
 I
⎧E
⎪  1  1 1
E
=
Z
I
⎨ 2
2 2
 =Z
 I
⎪⎩E
3
3 3
•
I 2
•
I 1
Z 12 I 12
Z 31
I 31
O
•
•
•·
•
Z 1
Z 2
Z 3
I 1
Δ-Connected Load
 =Z
 I
⎧V
12
12 12
⎪⎪


V
=
Z
I
⎨ 23
23 23
⎪
 
⎩⎪ V31 = Z 31 I 31
→
⎧I 1 = I 12 - I 31
⎪  
⎨I 2 = I 23 - I12
⎪⎩I 3 = I 31 - I 23
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
Three Phase System Loading
Y-Connected load
1
2
3
•
•
•

I 2
I3
Z 2
Z 3
I 1
D-Connected Load
1
2
3
•·
I 23 Z 23

Z
1
•
•
I 3
·
•
Z 2
Z 3
•

V
12
I 2
•
·
•

V
23
I 1
•
I 3
I 23 Z 23
•
I 31
I 1
Z 12 I 12
Z 31
I 31

Z
1
O
I 2
I 3
O

V
23
•
•

V
12
·
•
•
I 2
•
I 1
Z 12 I 12
Z 31
5
Wye and Delta Load Connections
•
•
·
·
ZY2
•
·
ZY1
ZΔ3
ZY3
3
ZΔ2
3
•
·
•
ZΔ1
2
•
2
1•
·
1
·
For impedances as for
resistors each impedance of
the wye connection is the
product of the two
impedances of the delta
connection connected to the
same node, divided by the
sum of the three impedances
of the delta connection.
Each impedance of the delta connection is the sum of all the products of the impedances of
the wye connection two by two, divided by the impedance in the opposite branch of the wye
connection.
!!!!!!!!!!!!!!!!!!!!!!!!!Z Y1 =
Z Δ1Z Δ3
Z Δ2Z Δ3
Z Δ1Z Δ2
;""Z Y2 =
;!!Z Y3 =
Z Δ1 + Z Δ2 + Z Δ3
Z Δ1 + Z Δ2 + Z Δ3
Z Δ1 + Z Δ2 + Z Δ3
Z Z + Z Z + Z Z
Z Z + Z Z + Z Z
Z Z + Z Z + Z Z
Z Δ1 = Y1 Y2 Y2 Y3 Y3 Y1 ;##Z Δ2 = Y1 Y2 Y2 Y3 Y3 Y1 ;##Z Δ3 = Y1 Y2 Y2 Y3 Y3 Y1
Z
Z
Z
Y3
Y1
When Ża = Żb = Żc = ŻΔ and Ż1 = Ż2 = Ż3= ŻY the
load is said a balanced load. In this case It is:
Y2
Z
Z Y = Δ
3
!;!!!Z Δ = 3 Z Y
Three Phase System with Neutral
•
1

E
1
I n
n
O •
E 3
I 1
·

E
2
2
•
3
•
I 2
I 3
A three phase generator is equivalent to
three y-connected mono-phase generators
with a symmetric tension system. The node
of center of the y-connection is linked to a
fourth line. The relation between the
phase tensions E and the line tensions V
in a balanced system is:
E=
V
3
The tension with an effective (rms) voltage
Ee = 230 V is obtained by means of a three
phase symmetrical system with an effective
line tension Ve = 400 V. In this case it is:
I 1 + I 2 + I 3 + I n ! = !0
ü  The line tension system with neutral can afford the same utilization of the
system without neutral.
ü  Both the line tension and the phase system are available for the use.
6
Power in Three Phase Systems
In a balanced three phase system with a posi)ve sequence the phase voltages are v p1 = 2 E cos ω t;"""v p2 = 2 E cos ω t −120° ;"""v p3 = 2 E cos ω t + 120°
(
)
(
)
where E is the rms effec)ve value of the phase voltage: E = V 3 where V is jϕ
the rms value of the line tension. In Y-­‐connected balanced load with Z
 Y = Ze
the currents lag behind their corresponding phase voltages by ϕ. Thus
(
)
(
)
i p1 = 2I cos ω t − ϕ ;"""i p2 = 2I cos ω t − ϕ − 120° ;"
""""""""i p3 = 2I cos ω t − ϕ + 120°
(
)
where I is the rms value of the current. The total instantaneous power
absorbed by Y-connected load is
()
p t = p1 + p 2 + p3 = v p1i p1 + v p2i p2 + v p3i p3
(
)
) (
) (
!!!!!!!!!!!!!!!!!!! = 2EI ⎡⎣ cos ω t cos ω t − ϕ +
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!+ cos ω t − 120° cos ω t − ϕ − 120° +
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! + cos ω t + 120° cos ω t − ϕ + 120° ⎤⎦
(
(
)
)
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
Power in Three Phase Systems
By applying the trigonometric iden)ty 1
cos A cos B = ⎡⎣cos ( A + B ) + cos ( A − B )⎤⎦
2
it results ()
(
)
)
(
)
p t = EI ⎡⎣3cos ϕ + cos 2ω t − ϕ + cos 2ω t − ϕ − 240° +
!!!!!!!!!!!!!!!!! + cos 2ω t − ϕ + 240° ⎤⎦ =
!!!!!!!!!=EI ⎡⎣3cos ϕ + cos α + cos α cos 240°+ sin α sin 240° +
!!!!!!!!!!!!!!!!! + cos α cos 240°! sin α sin 240° ⎤⎦ =
!!!!!!!!!=EI ⎡⎣3cos ϕ + cos α + 2 cos α cos 240° ⎤⎦ =
⎡
⎤
⎛ 1⎞
!!!!!!!!!=EI ⎢3cos ϕ + cos α +2 ⎜ − ⎟ cos α ⎥ = 3EI cos ϕ
⎝ 2⎠
⎣
⎦
(
where α =2ωt-ϕ.
Ø  Thus the total instantaneous power in a balanced three phase system does not change with )me as the instantaneous power of each phase does. This result is true for a balanced system with balanced loads weather the loads are Y-­‐connected or Δ-­‐connected. This is one of the mo,va,ons to use three phase systems to generate and distribute power.
7
Power in Three Phase Systems
The total average power and the reactive power in a three phase balanced
system are (E and V are the rms effective values of the phase and line tensions
respectively):
P = 3EI cos ϕ = 3 VI cos ϕ
and
Q = 3EI sin ϕ = 3VI sin ϕ
The power factor is
⎛
Q⎞
cos ϕ = cos ⎜ tan -1 ⎟
(1)
P⎠
⎝
The power factor is defined by eq. 1 for every three phase system independently
from the connection.
For unbalanced system the power factor is given by eq. 1 where the average and
the reactive power can be calculated from:
l
l
k=1
k=1
P!! = !! ∑ R k !I 2k !!!and!!!!Q!! = !! ∑ X k !I 2k
where Rk and Xk are the resistances and the reactances of the circuit.
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
Power in Three Phase Systems
In a unbalanced system the phase shift angle between phase voltage and
line current usually is different for each phase. In this case the shift angle
correction has to be done for each phase.
For unbalanced system the power factor is given by
⎛
Q⎞
cos ϕ = cos ⎜ tan -1 ⎟ =
P⎠
⎝
P
=
P
N
(

N=P+jQ
and N= P 2 + Q 2
P +Q
where the average and the reactive power can be calculated from the
resistances and the reactances of the circuit.
In this case φ is the the angle corresponding
to the rotation of the the current system with
Ė2
reference to the voltage system in order to
İ2
obtain the maximal value of the average
power.
İ3
Ė1
This corresponds to the shift angle φ of a
Ė3
single phase system for which P = IV cos φ.
İ1
When φ is equal to zero P is maximal.
2
2
)
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
8
Power in Three Phase Systems
In a unbalanced system the phase shift angle between phase voltage and
line current usually is different for each phase. In this case the shift angle
correction has to be done for each phase.
For unbalanced system the power factor is given by
⎛
Q⎞
cos ϕ = cos ⎜ tan -1 ⎟ =
P⎠
⎝
P
=
P
N
(

N=P+jQ
and N= P 2 + Q 2
P +Q
where the average and the reactive power can be calculated from the
resistances and the reactances of the circuit.
In this case φ is the the angle corresponding
to the rotation of the the current system with
φ
Ė2
reference to the voltage system in order to
İ2
obtain the maximal value of the average
power.
İ3
Ė1
This corresponds to the shift angle φ of a
Ė3
single phase system for which P = IV cos φ.
İ1
When φ is equal to zero P is maximal.
2
2
)
Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna
Power in Three Phase Systems
In a unbalanced system the phase shift angles between the three phase
voltages and the corresponding line currents are φ1, φ2, and φ3. After the
rotation they are: φ1- φ, φ2- φ, and φ3- φ. The average power for a rotation of
φ is
3
(
3
)
(
P = ∑ E k I k cos ϕ k − ϕ ; Q = ∑ E k I k sen ϕ k − ϕ
k=1
k=1
)
The maximum of P corresponds to ϕ for which:
3
dP
= − ∑ E k I k sen ϕ k − ϕ = 0
dϕ
k=1
(
→
3
- cos ϕ ∑ E k I k sen ϕ k + sen ϕ ∑ E k I k cos ϕ k = 0
k=1
→
)
3
sen ϕ
tan ϕ =
=
cos ϕ
k=1
3
∑ E I sen ϕ
k k
k
∑ E I cosϕ
k
k=1
3
k=1
k k
φ
İ2
Q
=
P
The power factor is thus defined as
⎛
Q⎞
P
P
→
cos ϕ = cos ⎜ tan -1 ⎟ =
=
2
2
P⎠
N
⎝
P +Q
Ė2
İ3
Ė1
İ1
Ė3
9
Terminology
Balanced current system Sistema di correnti equilibrato Balanced load Carico equilibrato Balanced tension system Sistema simmetrico Line tension Tensioni concatenate Negative sequence Sequenza negativa Phase tension Tensioni d fase Positive sequence Sequenza positiva Three-­‐phase electric Sistema elettrico system trifase 19
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10