Transcript Elements

Passive Elements and Phasor Diagrams
Resistor
v  Ri
V = RI
R
i
+
v
-
v
i
I
Inductor
i
di
dt
V = jwL I
vL
L
+
v
V
-
v
i
V
I
Capacitor
+
v
i
dv
dt
I = jwC V
C
i
v
iC
-
I
V
ELEC 371
Three-phase systems
1
Industrial Power Systems
Salvador Acevedo
Ideal Transformer
i1
+
v1
-
v1 i 2 N1
a
 
v 2 i1 N2
V1 I2 N1
a
 
V2 I1 N2
i2
+
v2
-
N1:N2
Transformer feeding load:
I1
+
V1
-
I2
+
V2
-
Z
V2 = V1/a
I2 = V2/Z
I1= I2/a
V2
Assuming a RL load
connected to secondary
and ideal source to primary
V1
I1
I2
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Three-phase systems
2
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Two Winding Transformer Model
 The linear equivalent model of a real
transformer consists of an ideal transformer and
some passive elements
i1
i2
+
+
v1
v2
-
ELEC 371
Three-phase systems
3
N1:N2
-
Industrial Power Systems
Salvador Acevedo
AC Generators and Motors
 AC synchronous generator
 Single-phase equivalent
 AC synchronous motor
 Single-phase equivalent
 AC induction motor (rarely used as generator)
ELEC 371
Three-phase systems
4
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Steady-state Solution
In sinusoidal steady-state a circuit may be solved using phasors
I
R
vs
+
VS
-

jwL
i
0
VS = R I
p
2p
Ix=I cos
+ jwL I
VS = (R + jwL) I
VS

VS = (R + jX) I

VS = Z I
VS
Z
Vmax 0
I 
Z 
I
I =
Iy=I sin
Rectangular form
I  I max   
I
Polar form
I = Ix + j Iy = Imax  
From rectangular form to polar form:
I =
Ix 2 + Iy 2
 Iy 

Ix 
  tan 1 

Magnitude
Angle or phase
From polar form to rectangular form:
Ix  Icos
Iy  Isin
ELEC 371
Three-phase systems
5
Real part
Reactive or imaginary part
Industrial Power Systems
Salvador Acevedo
Single-phase Power Definitions
i(t) = Im sin (wt+
i) amps
+
-
v(t) = Vm sin(wt+
v)
volts
Load: any R,L,C
combination
w : angular frequency in rad/sec
f: frequency in Hz
w=2pf
Instantaneous power

p( t )  v ( t ) i ( t )  Vm sin wt  v
p( t ) 
I
m

  )
sin( wt  i )
1
V I cos(v  i )  cos( 2 wt  v
2 m m
i
Average Power (or REAL POWER)
1
P
T
T

p( t ) dt 
0
1
V I cos  VrmsI rms cos
2 m m
Apparent Power
S = VrmsI rms
Power Factor
REAL POWER
P
pf 

APPARENT POWER
S
For this circuit, the power factor is
V
I
cos
pf = rms rms
 cos
Vrms I rms
ELEC 371
Three-phase systems
6
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Power Triangle
S
Ssin =Q

P=Scos
Real Power
P = S cos  = V I cos 
Reactive Power
Q = S sin   V I sin  vars
watts
Complex Power
S = S  
P + j Q
S = VI cos + jVIsin
If
 = v - i
v = 0
and assuming a reference
then
 = -i
therefore

S = V Icos( ) -
S = V Icos(-i ) + jIsin(-i )
i
jIsin(i )


S  V I*
The magnitude
is called Apparent Power:
S = VI
ELEC 371
Three-phase systems
7
volt - amperes (VA)
Industrial Power Systems
Salvador Acevedo
Power Consumption by Passive Elements
Impedance: Z = R + jX = Z 

Resistive Load
Z = R = R  0o
P = VI cos0o = VI = I 2 R = V 2 / R
Q = VI sin 0o = 0
watts
vars
A resistor absorbs P
Purely Inductive Load
Z = jwL = jX L  X L90o
P  VI cos(90o ) = 0
watts
Q = VI sin(90o ) = VI = I 2 X L = V2 / X L
var s
An inductor absorbs Q
Purely Capacitive Load
Z=
1
= -jX C  X C   90o
jwC
P  VI cos(-90o ) = 0
watts
Q = VI sin(-90o ) = -VI = -I 2 X L = -V 2 / X L
var s
A capacitor absorbs negative Q. It supplies Q.
ELEC 371
Three-phase systems
8
Industrial Power Systems
Salvador Acevedo
Advantages of Three-phase Systems
 Creation of the three-phase induction motor
Starting torque
Three-phase
induction motor
Single-phase
induction motor
yes
no
needs auxiliary
starting circuitry
Steady state torque Constant
Oscillating causing
vibration
 Efficient transmission of electric power
 3 times the power than a single-phase circuit by
adding an extra cable
i
va
+
v
-
Single-phase
Load
vb
vc
p = vi
ia
ib
Three-phase
Load
ic
p = va ia + vb ib + vc ic
 Savings in magnetic core when constructing
 Transformers
 Generators
ELEC 371
Three-phase systems
9
Industrial Power Systems
Salvador Acevedo
Three-phase Voltages
va
vb
vc
va(t) = Vm sin wt
volts
vb(t) = Vm sin (wt - 2p/3) = Vm sin (wt - 120)
volts
vc(t) = Vm sin (wt - 4p/3) = Vm sin (wt - 240)
or
volts
vb(t) = Vm sin (wt + 2p/3) = Vm sin (wt + 120)
w=2p f
w: angular frequency in rad/sec
volts
f : frequency in Hertz
Vc
120

120
Va
120
ELEC 371
Three-phase systems
10
Vb


Industrial Power Systems
Salvador Acevedo
Star Connection (Y)
 Y-connected Voltage Source
a
- n
c
+
+
Van
-
Vbn
+
Vcn
b
Line - to - neutral voltages Van, Vbn, Vcn.
(phase voltages for Y - connection)
same magnitude: V P
VP  Van  Vbn  Vcn
Line - to - line voltages Vab, Vbc, Vca
same magnitude: VLL
Vab = Van - Vbn
VLL =
ELEC 371
Three-phase systems
11
3 VP
Industrial Power Systems
Salvador Acevedo
Delta Connection (D )
 D-connected Voltage Source
a
Vca
+
Vab
-
+
c
Vbc
+
b
Line - to - line voltages Vab, Vbc, Vca.
(phase voltages for D - connection)
same magnitude: VLL  VP
Phase currents Iab, Ibc, Ica.
same magnitude: I P
Line currents Ia, Ib, Ic.
same magnitude: I L
IL = 3 IP
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Three-phase systems
12
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Y-connected Load
Ia
a
- n
c
+
ia
+
Van
-
Vcn
Za
Vbn
+
n'
b
Zc
Zb
Ib
Ic
ia
ia
Balanced case: Za = Zb = Zc = Z
Ia + Ib + Ic = 0
Ib = Ia -120
Ic = Ia - 240
ELEC 371
Three-phase systems
13
Industrial Power Systems
Salvador Acevedo
D-connected Load
Ia
a
- n
c
+
ia
+
Van
-
Vcn
Zca
Vbn
+
b
Zab
Zbc
Ib
Ic
ia
ia
ELEC 371
Three-phase systems
14
Industrial Power Systems
Salvador Acevedo
Y-D Equivalence
Za
Zca
Zab
n'
Zc
Zb
Zbc
Balanced case:
Za = Zb = Zc = Zy
Z D = 3Zy
Zab = Zbc = Zca = Z D = 3Zy
ELEC 371
Three-phase systems
15
Industrial Power Systems
Salvador Acevedo
Power in Three-phase Circuits
Three - phase voltages and currents:
va  Vm sin wt  v 
ia  I m sin wt  i 
vc  Vm sin wt  v  240
ic  I m sin wt  i  240
vb  Vm sin wt  v  120
ib  I m sin wt  v  120
The three - phase instantaneous power is:
p(t )  p3  va ia  vb ib  vc ic
sin wt  v  sin wt  i   sin wt  v  120 sin wt  v  120
p3  Vm I m 

 sin wt  v  240 sin wt  i  240


This expression can easily be reduced to:
p3  23 Vm I m cosv  i 
Since the instantaneous power does not change with the time,
its average value equals its intantaneous value:
P3  p3
P3  3VP I P cos
where: VP 
ELEC 371
Three-phase systems
16
Vm
2
IP 
Im
2
  v  i
Industrial Power Systems
Salvador Acevedo
Three-phase Power
In a Y - connection
VLL  3 VP
IL  IP
V 
P3  3VP I P cos  3 LL  I L cos  3 VLL I L cos
 3
In a D - connection
VLL  VP
IL  3 IP
I 
P3  3VP I P cos  3VLL  L  cos  3 VLL I L cos
 3
Regardless of the connection (for balanced systems),
the average power (real power) is :
P3  3 VLL I L cos
watts
Similarly, reactive power and apparent power expressions are:
Q 3  3 VLL I L sin 
vars
S3  3 VLL I L
VA
ELEC 371
Three-phase systems
17
Industrial Power Systems
Salvador Acevedo
Three-phase Transformers
 Use of one three-phase transformer unit
 Use of 3 single phase transformers to form a
“Transformer Bank
ELEC 371
Three-phase systems
18
Industrial Power Systems
Salvador Acevedo
Physical Overview
ELEC 371
Three-phase systems
19
Industrial Power Systems
Salvador Acevedo
N1 : N 2
A single-phase transformer 
Example:10 KVA, 38.1KV/3.81KV
+
v2
-
+
v1
-
a=N1/N2=V1/V2
Three-phase Transformers Connections
Y-Y; DD; Y- D; D -Y
Primary
terminals
Bank of
3 single-phase
transformers
Secondary
terminals
The Bank Transformation Ratio is defined as:
primary voltage
a' line  to  line
line  to  line secondary voltage
ELEC 371
Three-phase systems
20
Industrial Power Systems
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Y-Y connection
A
B
C
N
N1
N2
+ v1 -
+ v2 -
N1
N1
N2
N2
n
a
b
c
Ratings for Y-Y bank
using 3 single-phase transformers:
3x10KVA = 30 KVA
66 KV / 6.6 KV
ELEC 371
Three-phase systems
21
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DD connection
A
N1
N2
+ v1 -
+ v2 -
a
B
C
N1
N1
N2
N2
b
c
Ratings for DD bank:
30 KVA; 38.1 KV / 3.81 KV
ELEC 371
Three-phase systems
22
Industrial Power Systems
Salvador Acevedo
Y- D connection
A
B
C
N
N1
N2
+ v1 -
+ v2 -
a
N1
N1
N2
N2
b
c
Ratings for Y-D bank:
30 KVA; 66 KV / 3.81 KV
ELEC 371
Three-phase systems
23
Industrial Power Systems
Salvador Acevedo
D -Y connection
A
N1
N2
+ v1 -
+ v2 -
B
C
N1
N1
N2
N2
n
a
b
c
Ratings for D-Y bank:
30 KVA; 38.1 KV / 6.6 KV
ELEC 371
Three-phase systems
24
Industrial Power Systems
Salvador Acevedo
Per unit modelling
Power lines operate at kilovolts (KV)
and kilowatts (KW) or megawatts (MW)
To represent a voltage as a percent of a reference
value, we first define this BASE VALUE.
Example:
Base voltage: Vbase = 120 KV
Circuit voltage
Percent of
base value
Per unit value
108 KV
90%
0.9
120 KV
100%
1.0
126 KV
105%
1.05
60 KV
50%
0.5
per unit quantity =
Voltage_1=
actual quantity
base quantity
108
 0.9 p.u.
120
** The percent value and the per unit value help
the analyzer visualize how close the operating
conditions are to their nominal values.
ELEC 371
Three-phase systems
25
Industrial Power Systems
Salvador Acevedo
Defining bases
4 quantities are needed to model a network in per unit system:
V:
I:
S:
Z:
Vpu 
S pu 
voltage
current
power
impedance
VBASE
IBASE
SBASE
ZBASE
Vactual
Vbase
I pu 
Sactual
S base
Z pu 
I actual
I base
Z actual
Z base
Given two bases, the other two quantities are easily determined.
If base voltage and base power are known:
Vbase  100 KV, S base  100 MVA
then, base current and base impedance are:
S base
100,000,000
I base =
I base 
 1000 A.
Vbase
100,000
Vbase
100,000
Z base =
Z base 
 100 
I base
1000
Another way to express base impedance is:
Z base
Vbase
Vbase
=


I base
 S base 


 Vbase 
V
base

2
S base
Real power base and reactive power base are:
Pbase = S base = 100 MW
Q base = S base = 100 MVAR
ELEC 371
Three-phase systems
26
Industrial Power Systems
Salvador Acevedo
Three phase bases
In three-phase systems it is common to have data for the
three-phase power and the line-to-line voltage.
Sbase -3   3S base 1
Vbase  LL 
3Vbase  LN
The bas e c urrent and impedanc e
f or the three - phas e c as e are :
I base
 S base 3

3
 
 Vbase  LL

3







S base 3
3Vbase  LL
2
Z base
 Vbase  LL 


2

Vbase  LL 
3




S base 3
 S base 3 


3


In per unit,
line - to - neutral v oltage = line - to - line v oltage
VLN(pu) = VLL(pu)
w hy ?
With p.u. calculations, three-phase values of voltage,
current and power can be used without undue anxiety
about the result being a factor of 3 incorrect !!!
ELEC 371
Three-phase systems
27
Industrial Power Systems
Salvador Acevedo
Example
The following data apply to a three-phase case:
Sbase=300 MVA
(three-phase power)
Vbase=100 KV
(line-to-line voltage)
a
b
c
Three-phase load
270 MW
100 KV
pf=0.8
Using the per unit method:
270
P=
 0.9 p.u.
300
V = 1.0 p.u.
Normally, we' d say:
P = 3 VL I L cos = 3 VL I L pf
I=
P

3 VL pf
270x106
3 (100x103 ) (0.8)
 1948.5 A.
Single-phase equivalent:
I=1.125 p.u.
P = V I pf
then
P
0.9
I=

 1125
.
p. u.
V pf (10
. )(0.8)
+
V=1 p.u.
-
This current is 12.5% higher than its base value!
 300,000 
To check: 1.125xIbase = (1.125)
. x 1732  1948.5 A.
  1125
 3 100 
ELEC 371
Three-phase systems
28
Industrial Power Systems
Salvador Acevedo
Transformers in per unit calculations
 With an ideal transformer
+
V1
-
2400 V.
+
V2
-
4.33 + j 2.5 ohms
2400:120 V
5 KVA
High Voltage Bases
Sbase1 = 5 KVA
Vbase1= 2400 V
Ibase1 = 5000/2400=2.083 A
Zbase1= 2400/2.083=1152 
Low Voltage Bases
Sbase2 = 5KVA
Vbase2 = 120 V
I base2 = 5000/120=41.667 A
Z base2 = 120/41.667=2.88 
From the circuit:
V1=2400 V.
V2=V1/a=V1/20=120 V.
In per unit:
V1=1.0 p.u.
V2=1.0 p.u.
+
1.0
-
+
1.0
-
The load in per unit is:
Z=(5 30)/Zbase2 =1.7361  30 p.u.
The current in the circuit is:
I=(1.0 0/ (1.7361  30 =0.576 -30 p.u.
The current in amperes is:
Primary:
I1=0.576 x Ibase1= 1.2 A.
Secondary:
I2=0.576 x Ibase2= 24 A.
ELEC 371
Three-phase systems
29
Industrial Power Systems
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One line diagrams
 A one line diagram is a simplified
representation of a multiphase-phase circuit.
TRANSFORMER
Transmission line
TRANSFORMER
GENERATOR
GENERATOR
Transmission line
Transmission line
LOAD
ELEC 371
Three-phase systems
30
Industrial Power Systems
Salvador Acevedo
Nodal Analysis
Suppose the following diagram represents the single-phase equivalent of
a three-phase system
z13=j2 p.u.
z3=j2 p.u.
z1=j1 p.u.
z12=j0.5 p.u.
z23=j0.5 p.u.
V1= 1 p.u.
V3= -j1 p.u.
z2=10 p.u.
Finding Norton equivalents and representing impedances as admittances:
y13=-j0.5 p.u.
1
y12=-j2 p.u.
2
y1=-j1 p.u.
3
y23=-j2 p.u.
y2=0.1 p.u.
y3=-j0.5 p.u.
I1= -j1 p.u.
I3= -0.5 p.u.
I1=y1 V1 + y12(V1-V2) + y13(V1-V3)
0 = y12 (V2-V1) + y2 V2 + y23(V2-V3)
I3=y13(V3-V1) + y23(V3-V2) + y3 V3
In matrix form:
 y1 + y12 + y13

- y12


- y13
- y12
y12 + y 2 + y 23
 j 3.5
j2

0.1  j 4
 j2
 j 0.5
j2
ELEC 371
Three-phase systems
31
- y 23
j 0.5  V1
 
j 2   V2  
 j 3  V3
  V1
 
- y 23
  V2  
y13 + y 23 + y 3   V3
- y13
 - j1 


 0 
- 0.5
I1
 
0
I3
 V1
 
solving  V2  
 V3
0.77  24


 0.73  35  p. u.
 0.71  44 
Industrial Power Systems
Salvador Acevedo
General form of the nodal analysis
The system of equations is repeated here to find a general solution technique:
y1 + y12 + y13
  V1 I1
- y12
- y13

   
y
y
+
y
+
y
y
12
12
2
23
23

 V2   0 

- y13
- y 23
y13 + y 23 + y 3   V3 I3
or
Y11 Y12 Y13   V1  J1

   
Y
Y
Y
21
22
23

 V2  J2
Y31 Y32 Y33   V3 J3
In general:
N
Yii =  y ij
i  1,2... N
j=1
Yij = -y ij
i  1,2... N ;
j  1,2... N ;
i j
J i =  I i (from current sources flowing into the node) i  1,2... N
Once the voltages are found, currents and powers are
easily evaluated from the circuit. We have solved one
of the phases of the three-phase system (e.g. phase
‘a’). Quantities for the other two phases are shifted
120 and 240 degrees under balanced conditions.
Actual quantities can be found by multiplying the per
unit values by their corresponding bases.
ELEC 371
Three-phase systems
32
Industrial Power Systems
Salvador Acevedo