Transcript Chapter 3 - Three Phase System

BASIC ELECTRICAL TECHNOLOGY
DET 211/3
Chapter 3 - Three
Phase System
INTRODUCTION TO THREE
PHASE SYSTEM
In general, three phase systems are preferred over
single phase systems for the transmission of the power
system for many reasons, including the following:
• Thinner conductors can be used to transmit the same kVA at the
same voltage, which reduces the amount of copper required
(typically about 25% less) and turn reduces construction and
maintenance costs.
• The lighter lines are easier to install, and the supporting
structures can be less massive and farther apart.
• In general, most larger motors are three phase because they are
essentially self starting and do not require a special design or
additional starting circuitry.
Three phase voltages
A 3-phase generator basically consists of a
rotating magnet (called the rotor) surrounded by a
stationary winding (called the stator). Three
separate windings or coils with terminals a-a’, b-b’
and c-c’ are physically placed 120o apart around
the stator.
Generated Voltages
The three phase generator can supply
power to both single phase and three phase
loads
The sinusoidal expression for each of the phase
voltages
v AN  Vm( AN ) sin t
v BN  Vm ( BN ) sin(t  120 o )
vCN  Vm(CN ) sin(t  240 o )  Vm (CN ) sin(t  120 o )
The phasor diagram of the phase voltages
The effective value of each is determined by
V AN 
VBN 
Vm( AN )
2
Vm ( BN )
VCN 
2
Vm (CN )
2
 0.707 Vm( AN )
V AN  V AN ( m )0 o
 0.707 Vm ( BN )
VBN  VBN ( m )  120 o
 0.707 Vm (CN )
VCN  VCN ( m )  120 o
If the voltage sources have the same amplitude and frequency
ω and are out of the phase with each other by 120o, the
voltages are said to be balanced. By rearranging the phasors as
shown in figure below, so
V AN  VBN  VCN  V AN ( m) 0 o  VBN ( m)   120 o  VCN ( m)   120 o
 Vm (1.0  0.5  j 0.866  0.5  j 0.866 )  0
Where
| VAN || VBN || VCN | Vm
Generator and Load Connections
Each generator in a 3-phase system maybe either Yor D-connected and loads may be mixed on a power
system.
Z
Z
Z
Z
Z
Z
Wye Connected Generator
Applying KVL around the indicated loop in figure above, we obtain
I L  Ig
VAB  VAN  VBN  VAN  VNB
VBC  VBN  VCN  VBN  VNC
VCA  VCN  VAN  VCN  VNA
Wye Connected Generator
For line-to-line voltage VAB is given by:
VAB  VA  VB
 V 0 0  V   120 0
 1
3 

 V   V  j
V
 2

2


3
3
 V  j
V
2
2
 3
1 

 3V
j
 2
2 

 3V 30 0
Wye Connected Generator
VAB  V AB30 0  3V AN 30 0
VCA  3VCN 150 0
Phasor Diagram
VBC  3VBN 270 0
Wye Connected Generator
The relationship between the magnitude of the line-to-line and
line-to-neutral (phase) voltage is:
VLL  3V
The line voltages are shifted 300 with respect to the phase
voltages. Phasor diagram of the line and phase voltage for the Y
connection is shown below.
VCN
VAB
VCA
Rearrange
VAN
VBN
VBC
Line-to-line voltages
Phase voltages
Delta Connected Generator
For line-to-line voltage VAB is given
by:
I A  I AB  I CA
 I 0 0  I   240 0
 1
3 

 I   I  j
I
 2

2



V LL  V
3
3
I  j
I
2
2
 3
1 

 3I
j
 2

2


 3 I    30 0
Delta Connected Generator
The relationship between the magnitude of the line
and phase current is:
I L  3I
The line currents are shifted 300 relative to the
corresponding phase current. Phasor diagram of the
line and phase current for the Y connection is shown
IC below.
ICA
IAB
IB
IA
IBC
Line-to-line currents
Phase currents
Phase sequence
The phase sequence is the order in which the
voltages in the individual phases peak.
VC
VB
VA
VB
abc phase sequence
VA
VC
acb phase sequence
Power relationshipphase quantities
The power equations applied to Y-or D load in a
balanced 3-phase system are:
P  3V I  cos
P  3I2 Z cos
Real power
Unit=Watts(W)
Q  3V I  sin 
Q  3I2 Z sin
Reactive power
Unit=Volt-Amps-Reactive (VAR)
S  3V I 
S  3I 2 Z
Apparent power
Unit=Volt-Amps (VA)
 - angle between voltage and current in any phase of the load
Power relationshipLine quantities
The power equations applied to Y-or D load in a
balanced 3-phase system are:
P  3VLL I L cos
Real power
Q  3VLL I L sin 
Reactive power
S  3VLL I L
Apparent power
 - angle between phase voltage and phase current
in any phase of the load
Since both the three-phase source and the threephase load can be either Y- or D- connected, we
have 4 possible connections:
i) Y-Y connections (i.e: Y-connected source with a Yconnected load).
ii) Y-D connection.
iii) D-D connection
iv) D-Y connection
(i) Y connected generator/source with Y
connected load
Z1  Z 2  Z 3
I g  I L  I L
V  E
EL  3V
(ii) Y-D Connection
A balanced Y-D system consists of a balanced Yconnected source feeding a balanced D-connected
load
Z/3
Z/3
Z
Z
Z
Z/3
ZD
ZY 
3
D must consists of three equal impedances
(iii) ∆-∆ Connection
A balanced ∆-D system consists of a balanced ∆connected source feeding a balanced D-connected
load
Z
Z
Z
Z
Z
Z
(iv) DY Connection
A balanced D-Y system consists of a balanced Dconnected source feeding a balanced Y-connected load
Z
Z
Z
Z/3
Z/3
Z/3
Example 1
A 208V three-phase power system is shown in Figure 1. It
consists of an ideal 208V Y-connected three-phase generator
connected to a three-phase transmission line to a Y-connected
load. The transmission line has an impedance of 0.06+j0.12W
per phase, and the load has an impedance of 12+j9W per
phase. For this simple system, find
(a) The magnitude of the line current IL
(b) The magnitude of the load’s line and phase voltages VLL and VL
(c) The real, reactive and apparent powers consumed by the load
(d) The power factor of the load
(e) The real, reactive and apparent powers consumed by the
transmission line
(f) The real, reactive and apparent powers supplied by the generator
(g) The generator’s power factor
Example 1
0.06W
+
0.06W
i0.12W
i0.12W
V
Z
Vcn=120-2400
Z
Van=12000
208V
Z=12+ i9W
Z
+
Vbn=120-1200
_
0.06W
Figure 1
i0.12W
Solution Example 1
(a) The magnitude of the line current IL
I line
Vline

Z line  Z load
1200V

(0.06  j 0.12)W  (12  j 9)W
1200
1200


12.06  j 9.12 15.1237.1
 7.94  37.1 A
So, the magnitude of the line current is thus 7.94 A
Solution Example 1
(b) The magnitude of the load’s line and phase voltages VLL and VL
The phase voltage on the load is the voltage across one phase of
the load. This voltage is the product of the phase impedance and
the phase current of the load:
VL  I L Z L
 (7.94  37.1 A)(12  j 9W)
 (7.94  37.1 A)(1536.9W)
 119.1  0.2V
Therefore, the magnitude of the load’s phase voltage is
VL  119.1V
and the magnitude of the load’s line voltage is
VLL 
3VL  206.3V
Solution Example 1
(c) The real power consumed by the load is
PLoad  3V I  cos 
 3(119.1V )(7.94 A) cos 36.9
 2270W
The reactive power consumed by the load is
Q Load  3V I  sin 
 3(119.1V )(7.94 A) sin 36.9
 1702 var
The apparent power consumed by the load is
S Load  3V I 
 3(119.1V )(7.94 A)
 2839VA
Solution Example 1
PFLoad  cos 
(d) The load power factor is
 cos 36.9
 0.8lagging
(e) The current in the transmission line is 7.94  37.1 A
and the impedance of the line is (0.06  j 0.12)W or 0.13463.4W
per phase. Therefore, the real, reactive and apparent powers
consumed in the line are:
PLine  3I  Z cos 
QLine  3I  Z sin 
 3(7.94 A) (0.134W) cos 63.4
 11.3W
 3(7.94 A) 2 (0.134W) sin 63.4
 22.7 var
2
2
2
S Line  3I  Z
2
 3(7.94 A) 2 (0.134W)
 25.3VA
Solution Example 1
(f) The real and reactive powers supplied by the generator are the
sum of the powers consumed by the line and the load:
Pgen  Pline  Pload
 11.3W  2270W
 2281W
Q gen  Qline  Qload
 22.7 var  1702 var
 1725 var
The apparent power of the generator is the square root of the sum
of the squares of the real and reactive powers:
S gen 
Pgen  Q gen
 2860VA
2
2
Solution Example 1
(g) From the power triangle, the power factor angle  is
 gen  tan
1
Q gen
Pgen
1725VAR
 tan
 37.1
2281W
1
Therefore, the generator’s power factor is
PFgen  cos 37.1  0.798lagging
Assignment 2
A 208V three-phase power system is shown in Figure 2. It
consists of an ideal 208V Y-connected three-phase generator
connected to a three-phase transmission line to a D-connected
load. The transmission line has an impedance of 0.06+j0.12W
per phase, and the load has an impedance of 12+j9W per
phase. For this simple system, find
(a) The magnitude of the line current IL
(b) The magnitude of the load’s line and phase voltages VLL and VL
(c) The real, reactive and apparent powers consumed by the load
(d) The power factor of the load
(e) The real, reactive and apparent powers consumed by the
transmission line
(f) The real, reactive and apparent powers supplied by the generator
(g) The generator’s power factor
Assignment 2
Figure 2