Transcript Chapter 3 - Three Phase System

EET 103
Chapter 3
(Lecture 1)
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 )
Phase expression
In phase expression
EM
EA 
0
2
EB 
EM
2
  120
Where
EM : peak value
EA, EB and EC : rms value
7
EC 
EM
2
120
The phasor diagram of the phase voltages
The effective value of
each is determined
by
Vm( AN )
V AN 
VBN 
2
Vm ( BN )
2
 0.707 Vm( AN )
 0.707 Vm ( BN )
V AN  V AN ( m )0 o
VBN  VBN ( m )  120 o
VCN 
Vm (CN )
2
 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
VAN  VBN  VCN
 VAN ( m) 0o  VBN ( m)   120o  VCN ( m)   120o
 Vm (1.0  0.5  j 0.866  0.5  j 0.866 )  0
Where
| V AN || VBN || VCN | Vm
Connection in Three Phase System
A 3 phase system is equivalent to three single
phase circuit
Two possible configurations in three phase
system
1. Y - connection (star connection)
2. ∆ - connection (delta connection)
10
Three phase Voltages Source
Y-connected source
11
∆-connected source
Three phase Load
Y - connected load
12
∆ - connected load
Generator and Load Connections
Each generator in a 3 phase system maybe either
Y or 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 I
L
g
VAB  VAN  VBN  VAN  VNB
VBC  VBN  VCN  VBN  VNC
VCA  VCN  VAN  VCN  VNA
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
Phasor Diagram
VAB  V AB30 0  3V AN 30 0
VCA  3VCN 150 0
VBC  3VBN 270 0
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
VAN
VBN
VBC
Rearrange
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


3
3
 I  j
I
2
2
V LL  V
 3
1 

 3I
j
 2
2 

 3 I    30 0
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
I
I
shown below.
CA
C
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
EXAMPLE 3.1
Calculate the line currents in the three-wire Y - Y
system as shown below.
21
Solution 3.1
Single Phase Equivalent Circuit
Phase ‘a’ equivalent circuit
22
I Aa
VAN

; ZT  (5  j 2)  (10  j8)  16.15521.8
ZT
I Aa
1100

 6.81  21.8
16.15521.8
I Bb  I Aa   120
 6.81  141.8A
I Cc  I Aa   240
 6.81  261.8  6.8198.2A
23
EXAMPLE 3.2
A balanced delta connected load having an
impedance 20 - j15  is connected to a delta
connected, positive sequence generator
having VAB = 3300 V. Calculate the phase
currents of the load and the line currents.
24
Solution 3.2
 ZΔ  20  j15   25  36.87
 VAB  3300 V
25
Phase Currents
Vab
3300
I ab 

 13.236.87A
ZΔ 25  38.87
I bc  I ab   120  13.2 - 83.13A
I ca  I ab   120  13.2156.87A
26
Line Currents
I Aa  I ab 3  30

 13.236.87 3  30
 22.866.87 A

I Bb  I Aa   120  22.86 - 113.13 A
I Cc  I Aa   120  22.86126.87 A
27
∆ - Connected Generator with a Y Connected Load
28
EXAMPLE 3.3
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.
29
Solution 3.3
the load impedance, ZY and the source voltage, VAB are
 ZY  40  j25  47.1732 
 VAB  2100 V
30
When the ∆ - connected source is transformed
to a Y - connected source,
VAB
Van 
  30
3
2100

 1  30
3
 121.2 - 30 V
31
The line currents are
I Aa
Van 121.2  30


 2.57 - 62 A
ZY
47.1732
I Bb  I Aa   120  2.57 - 182 A
I Cc  I Aa   120  2.5758 A
32
Summary of Relationships in Y and
∆ - connections
Y-connection
Voltage
magnitudes
Current
magnitudes
Phase
sequence
∆-connection
VL  3Vφ
VL  Vφ
IL  Iφ
I L  3I φ
VL leads Vφ by IL lags Iφ by 30°
30°
33
EET 103
Chapter 3
(Lecture 2)
Three Phase System
Power
Y - Connected Balanced Load
Average Power
The average power delivered to each phase
The total power to the balanced load is
Reactive Power
The reactive power of each phase is
The total reactive power of the load is
Apparent Power
The apparent power of each phase is
The total apparent power of the load is
Power Factor
The power factor of the system is
∆ - Connected Balanced Load
Average Power
Reactive Power
Apparent Power
Power Factor
EXAMPLE 3.4
Determine the total power (P), reactive power
(Q) and complex power (S) at the source and at
the load.
Solution 3.4 Single Phase Equivalent Circuit
Phase ‘a’ equivalent circuit
Known quantities
Vg =VAN= 1100 V
ZY = 10 + j8 
Zline = 5 - j2 
Line / Phase Currents
VAN
IA 
Zline  Z Y
1100
IA 
16.15521.8
 6.81  21.8 A
Source & Load Power

φ φ
Ssource  3V I
 (2087  j834.6)VA
 Ps  2087 W, Qs  834.6 VAR
2
SLoad  3 I φ Z 
 (1392  j1113)VA
 PL  1392 W, Q L  1113 VAR
EXAMPLE 3.5
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
Known Quantities
• PLoad = 5600 W
• VL
= 220 V
• IL
= 18.2 A
Solution 3.5
Power factor = cos 
S  3Vφ I φ
 3 VL I L
 6935.13 VA
|S|
Q

P  S cos θ
P
P
5600
 cos θ  
 0.8
S 6935.13
Example 3.6
For the Y - connected load in Figure
a) find the average power to each phase and
the total load
b) determine the reactive power to each phase
and the total reactive power
c) find the apparent power to each phase and
the total apparent power
d) find the power factor of the load
Figure
Solution 3.6
a) The average power to each phase is
P  V I cos 
V
I
 100  20  cos 53.13
 1200 W
Total load
PT  3  P  3 1200W  3600 W
b) The reactive power to each phase is
Q  V I sin 
V
I
 100  20  sin 53.13
 1600 VAR
Total reactive power
QT  3  Q  3 1600  4800VAR
c) The apparent power to each phase is
S  V I
 100  20
 2000 VA
Total apparent power
ST  3  S  3  2000  6000 VA
d) The power factor
PT
FP 
ST
3600 W

6000 VA
 0.6 lagging
Power relationship - Phase 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
Watts (W)
Q  3V I  sin 
Q  3I2 Z sin
Reactive power
Volt-Amps-Reactive (VAR)
S  3V I 
S  3I 2 Z
Apparent power
Volt-Amps (VA)
 - angle between voltage and current in any phase of the load
Power relationship - Line 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 three
phase load can be either Y or D connected, we
have 4 possible connections
1. Y - Y connections (Y - connected source with
Y - connected load)
2. Y - D connection (Y - connected source with
D - connected load)
3. D - D connection (D - connected source with
D - connected load)
4. D - Y connection (D - connected source with
Y - connected load)
1. Y connected generator / source with Y
connected load
I g  I L  I L
V  E
EL  3V
Z1  Z 2  Z 3
2. Y - D Connection
A balanced Y - D system consists of a
balanced Y - connected 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
3. ∆ - ∆ Connection
A balanced ∆ - D system consists of a
balanced ∆ - connected source feeding a
balanced D - connected load
Z
Z
Z
Z
Z
Z
4. D  Y Connection
A balanced D - Y system consists of a
balanced D - connected source feeding a
balanced Y - connected load
Z
Z
Z
Z/3
Z/3
Z/3
Example 3.7
Each transmission line of the 3 wire, three phase
system in Figure has an impedance of 15 Ω + j
20 Ω. The system delivers a total power of 160
kW at 12,000 V to a balanced three-phase load
with a lagging power factor of 0.86.
a. Determine the magnitude of the line voltage
EAB of the generator.
b. Find the power factor of the total load
applied to the generator.
c. What is the efficiency of the system?
Figure
Solution 3.7
a.
Vø (load) =
VL 12000V

 6936.42 V
1.73
3
PT (load) = 3 Vø Iø cos θ
and
PT
160000W
I

 8.94 A
3V cos  36936.420.86
Since θ = cos-1 0.86 = 30.68o (lagging)
And assigning V  V 0 , a lagging power
factor results in I  8.94A  30.68
For each phase, the system will appear as
shown in figure below.
E AN  I  Z line  V  0
Or
E AN  I Zline  V
 8.94A - 30.682553.13  6936.42V0
 223.5V22.45  6936.42V0
 206.56V  j85.35V  6936.42V
 7142.98V  j85.35V
 7143.5V0.68
Then
E AB  3Eg  (1.73)(7143.5V)  12358.26 V
PT  Pload  Plines
b.
 160kW  3( I L ) 2 Rline
 160kW  38.94A  15
 160,000 W  3596.55W
 163,596.55W
2
PT  3VL I L cos  T
And
PT
163,596.55W
cos  T 

3VL I L 1.7312,358.26V 8.94A 
And
F p  0.856
< 0.86 of load
c.
Po
Po
160 kW
 

 0.978  97.8%
Pi Po  Plosses 160 kW  3596.55 W
Example 3.8
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 + j0.12 per phase and
the load has an impedance of 12 + j9 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
0.06
+
0.06
i0.12
i0.12
V
Z
Z
Vcn=120-2400
Van=12000
208V
Z=12+ i9
Z
+
Vbn=120-1200
_
0.06
Figure 1
i0.12
Solution 3.8
(a)The magnitude of the line current IL
I line
Vline

Z line  Z load
1200V

(0.06  j 0.12)  (12  j 9)
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
(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 9)
 (7.94  37.1 A)(1536.9)
 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
(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
(d) The load power factor is
(e)
The current in the
transmission line is
7.94  37.1 A
The impedance per phase of the line is
PFLoad  cos 
 cos 36.9
 0.8lagging
(0.06  j 0.12)
or
0.13463.4
Therefore, the real, reactive and apparent powers consumed in
the line are
2
PLine  3I  Z cos 
QLine  3I  Z sin 
 3(7.94 A) (0.134) cos 63.4
2
 11.3W
S
 3I Z
 3(7.94 A) 2 (0.134) sin 63.4
 22.7 var
2
2
Line

 3(7.94 A) 2 (0.134)
 25.3VA
(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
(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
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 + j0.12 per phase and
the load has an impedance of 12 + j9 per phase. For this simple
system, find
a. The magnitude of the line current IL
Assignment 3.1
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
Figure 2