Download Resistor Inductor Capacitor IVIVIV i C dv dt = I = jwC V

Passive Elements and Phasor Diagrams
Resistor
v = Ri
R
i
+
v
V = RI
-
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
EECE 458/571
Three-phase systems
1
© Salvador Acevedo, 2000
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
EECE 458/571
Three-phase systems
2
© Salvador Acevedo, 2000
Two Winding Transformer Model
à The linear equivalent model of a real
transformer consists of an ideal transformer and
some passive elements
i1
i2
+
+
v1
v2
-
EECE 458/571
Three-phase systems
3
N1:N2
-
© Salvador Acevedo, 2000
AC Generators and Motors
à AC synchronous generator
« Single-phase equivalent
à AC synchronous motor
« Single-phase equivalent
à AC induction motor (rarely used as generator)
EECE 458/571
Three-phase systems
4
© Salvador Acevedo, 2000
Steady-state Solution
In sinusoidal steady-state a circuit may be solved using phasors
I
R
vs
+
VS
-
θ
jwL
i
0
V S = R I
+
jw L
(R + jw L ) I
V S =
(R + jX ) I
V S =
Z I
V
I =
Ix=I cosθ
I = I
VS
θ
V S
Z
=
θ
I
∠ 0
Z ∠ θ
I
Iy=I sinθ
m ax
m
2π
I
V S =
I
π
Rectangular form
ax ∠ − θ
Polar form
I = Ix + j Iy = Imax ∠ θ
F r o m
I =
θ
r e c ta n g u la r f o r m
I x
=
ta n
2
+
− 1
I y
2
 I y 


 I x 
M
I y
=
I s in θ
EECE 458/571
Three-phase systems
5
p o la r f o r m :
a g n itu d e
A n g le
F r o m p o la r f o r m
I x = I c o s θ
to
to
o r p h a s e
r e c ta n g u la r f o r m :
R e a l p a rt
R e a c tiv e
o r im a g in a r y
p a rt
© Salvador Acevedo, 2000
Single-phase Power Definitions
i(t) = Im sin (wt+θi) amps
+
-
v(t) = Vm sin(wt+θv)
volts
Load: any R,L,C
combination
w: angular frequency in rad/sec
f: frequency in Hz
w=2πf
Instantaneous pow er
p (t ) = v (t ) i(t ) =
p (t ) =
[V
m
sin w t + θv
][ I
m
]
)}
sin ( w t + θi )
1
V m I m {c o s ( θ v − θ i ) − c o s ( 2 w t + θ v + θ i
2
A v e rage P o w er (o r R E A L P O W E R )
1
P =
T
T
∫
p(t ) d t =
0
1
V m I m c o s θ = V rm s I rm s c o s θ
2
A p p a ren t P o w e r
S = V rm s I rm s
P o w e r F a c to r
REAL POW ER
P
=
pf =
APPARENT POWER
S
F o r th i s c ircuit, th e p o w e r f a c to r i s
V rm s I rm s c o s θ
= cosθ
pf =
V rm s I rm s
EECE 458/571
Three-phase systems
6
© Salvador Acevedo, 2000
Power Triangle
S
Ssin θ=Q
θ
P=Scosθ
Real Power
P = S cos θ = V I cos θ watts
Reactive Power
Q = S sin θ = V I sin θ vars
C o m p le x P o w e r
S =
S
S
∠ θ =
P
V I c o sθ
=
+
+
j Q
jV I s in θ
If θ = θ v - θ i
a n d a s s u m in g a re fe re n c e θ
t h e n θ = -θ i
th e re fo re
S =
V
S =
V
S =
V
v
= 0
[I c o s ( - θ ) + j I s i n ( - θ ) ]
[I c o s ( θ ) - j I s i n ( θ ) ]
i
i
i
i
I *
T h e m a g n itu d e
is c a lle d A p p a re n t P o w e r:
S = V I v o lt - a m p e re s (V A )
EECE 458/571
Three-phase systems
7
© Salvador Acevedo, 2000
Power Consumption by Passive Elements
Impedance: Z = R + jX = Z ∠θ
Ω
Resistive Load
Z = R
=
R ∠ 0o
P = V I cos0o = V I = I2 R = V
Q = V I sin 0 o = 0
2
/ R
w a tts
v a rs
A resistor absorbs P
Purely Inductive Load
Z = jw L = jX
P = V I co s(9 0
Q = V I sin (9 0
L
o
o
= X
L
∠ 90
) = 0
o
w a tts
) = V I = I2X
L
= V
2
/ X
var s
L
An inductor absorbs Q
Purely Capacitive Load
Z =
1
= -jX
jw C
C
= X
P = V I co s(-9 0 o ) = 0
C
∠ − 90o
w a tts
Q = V I sin (-9 0 o ) = -V I = -I 2 X
L
= -V
2
/ X
L
var s
A capacitor absorbs negative Q. It supplies Q.
EECE 458/571
Three-phase systems
8
© Salvador Acevedo, 2000
Advantages of Three-phase Systems
à 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
à Efficient transmission of electric power
ð 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
à Savings in magnetic core when constructing
ð Transformers
ð Generators
EECE 458/571
Three-phase systems
9
© Salvador Acevedo, 2000
Three-phase Voltages
va
vb
vc
va(t) = Vm sin wt
volts
vb(t) = Vm sin (wt - 2π/3) = Vm sin (wt - 120°)
volts
vc(t) = Vm sin (wt - 4π/3) = Vm sin (wt - 240°)
or
volts
vb(t) = Vm sin (wt + 2π/3) = Vm sin (wt + 120°)
w=2πf
w: angular frequency in rad/sec
volts
f : frequency in Hertz
Vc
°
120
Va
120
120
EECE 458/571
Three-phase systems
10
Vb
°
°
© Salvador Acevedo, 2000
Star Connection (Y)
à Y-connected Voltage Source
a
- n
c
+
Vcn
+
Van
-
Vbn
+
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 =
EECE 458/571
Three-phase systems
11
3 VP
© Salvador Acevedo, 2000
Delta Connection (∆
∆)
à ∆-connected Voltage Source
a
Vca
+
Vab
-
+
c
Vbc
+
b
Line - to - line voltages Vab, Vbc, Vca.
(phase voltages for ∆ - 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
EECE 458/571
Three-phase systems
12
© Salvador Acevedo, 2000
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°
EECE 458/571
Three-phase systems
13
© Salvador Acevedo, 2000
∆-connected Load
Ia
a
- n
c
+
ia
+
Van
-
Vcn
Zca
Vbn
+
b
Zab
Zbc
Ib
Ic
ia
ia
EECE 458/571
Three-phase systems
14
© Salvador Acevedo, 2000
Y-∆
∆ Equivalence
Za
Zca
Zab
n'
Zc
Zb
Zbc
Balanced case:
Za = Zb = Zc = Zy
Z ∆ = 3Zy
Zab = Zbc = Zca = Z ∆ = 3Zy
EECE 458/571
Three-phase systems
15
© Salvador Acevedo, 2000
Power in Three-phase Circuits
Three-phase voltages and currents:
va =Vm sin( wt +θv )
ia = Im sin( wt +θi )
vc =Vm sin( wt +θv − 240°)
ic = Im sin( wt +θi − 240°)
vb =Vm sin( wt +θv −120°)
ib = Im sin( wt +θv −120°)
The three-phase instantaneous power is:
p(t) = p3φ = va ia + vb ib + vc ic
sin( wt +θv ) sin( wt +θi ) + sin( wt +θv −120°) sin( wt +θv −120°)
p3φ = VmIm

+ sin( wt +θv − 240°) sin( wt +θi − 240°)


This expression can easily be reduced to:
p3φ = 23 VmIm cos(θv −θi )
Since the instantaneous power does not change with the time,
its average value equals its intantaneous value:
P3φ = p3φ
P3φ = 3VP I P cosθ
where: VP =
EECE 458/571
Three-phase systems
16
Vm
2
IP =
Im
2
θ = θv −θi
© Salvador Acevedo, 2000
Three-phase Power
In a Y-connection
VLL = 3 VP
IL = IP
V 
P3φ = 3VP I P cosθ = 3 LL  I L cosθ = 3 VLL I L cosθ
 3
In a ∆ -connection
VLL = VP
IL = 3 IP
I 
P3φ = 3VP I P cosθ = 3VLL  L  cosθ = 3 VLL I L cosθ
 3
Regardless of the connection (for balanced systems),
the average power (real power) is :
P3φ = 3 VLL I L cosθ
watts
Similarly, reactive power and apparent power expressions are:
Q3φ = 3 VLL I L sinθ
vars
S3φ = 3 VLL I L
VA
EECE 458/571
Three-phase systems
17
© Salvador Acevedo, 2000
Per unit modelling
àPower 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.
EECE 458/571
Three-phase systems
18
© Salvador Acevedo, 2000
Defining bases
4 quantities are needed to model a network in per unit system:
V:
I:
S:
Z:
V
S
voltage
current
power
impedance
V a c tu a l
V base
=
pu
=
pu
VBASE
IBASE
SBASE
ZBASE
I
S a c tu a l
S base
Z
pu
pu
=
=
I a c tu a l
I base
Z a c tu a l
Z base
Given two bases, the other two quantities are easily determined.
If b a s e v o lta g e a n d b a s e p o w e r a r e k n o w n :
V
base
= 100 K V ,
S
= 100 M V A
base
th e n , b a s e c u r r e n t a n d b a s e im p e d a n c e a r e :
I
base
S
V
=
base
I
=
base
base
1 0 0 ,0 0 0 ,0 0 0
= 1000 A .
1 0 0 ,0 0 0
V base
1 0 0 ,0 0 0
Z base =
Z base =
= 100 Ω
I base
1000
A n o th e r w a y to e x p re s s b a s e im p e d a n c e is:
Z
base
=
V base
V
=
I base
 S

 V
base
base
base



=
(V
S
base
)
2
base
R e a l p o w e r b a s e a n d re a c tiv e p o w e r b a s e a re:
P
Q
base
base
= S
= S
EECE 458/571
Three-phase systems
19
base
base
= 100 M W
= 100 M V A R
© Salvador Acevedo, 2000
Three phase bases
In three-phase systems it is common to have data for the
three-phase power and the line-to-line voltage.
S
base -3 Φ
V base −
LL
= 3 S base −1Φ
=
3 V base −
LN
T h e b a s e c u r r e n t a n d im p e d a n c e
f o r th e th r e e - p h a s e c a s e a r e :
I base
Z
base
 S base − 3Φ


3
=
 V base − LL


3






 V base − LL


3
=
 S base − 3Φ


3






=
=
S base − 3Φ
3 V base −
(V
LL
)
base − LL
2
S base − 3Φ
I n p e r u n i t,
lin e - to - n e u tra l v o lta g e
V L N (pu) = V L L (pu)
=
lin e - to - lin e v o lta g e
w hy?
With
With p.u.
p.u. calculations,
calculations, three-phase
three-phase values
values ofof voltage,
voltage,
current
current and
and power
power can
can be
be used
used without
without undue
undue anxiety
anxiety
about
aboutthe
theresult
resultbeing
beingaafactor
factorofof√3
√3incorrect
incorrect!!!
!!!
EECE 458/571
Three-phase systems
20
© Salvador Acevedo, 2000
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=
= 09
. p.u.
300
V =1.0 p.u.
Normally, we' d say:
P = 3 VL I L cos θ = 3 VL I L pf
I=
P
=
3 VL pf
270x106
3 (100x10 3 ) ( 0.8)
Single-phase equivalent:
I=1.125 p.u.
P = V I pf
+
V=1 p.u.
-
then
I=
= 1948.5 A.
P
09
.
=
= 1125
.
p.u.
V pf (10
. )(08
.)
This current is 12.5% higher than its base value!
 300,000
To check: 1.125xIbase = (1.125)
. x 1732 = 19485
. A.
 = 1125
 3 100 
EECE 458/571
Three-phase systems
21
© Salvador Acevedo, 2000
Transformers in per unit calculations
à 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 Ω
Low Voltage Bases
Sbase2 = 5KVA
Vbase2 = 120 V
I base2 = 5000/120=41.667 A
Z base2 = 120/41.667=2.88 Ω
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.
EECE 458/571
Three-phase systems
22
© Salvador Acevedo, 2000
One line diagrams
à A one line diagram is a simplified
representation of a multiphase-phase circuit.
TRANSFORMER
Transmission line
TRANSFORMER
GENERATOR
GENERATOR
Transmission line
Transmission line
LOAD
EECE 458/571
Three-phase systems
23
© Salvador Acevedo, 2000
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
2
y12=-j2 p.u.
y1=-j1 p.u.
y23=-j2 p.u.
y2=0.1 p.u.
3
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:
 y 1 + y 12 + y 13

- y 12


- y 13
 − j 3.5

 j2
 j 0.5
j2
0.1 − j 4
j2
EECE 458/571
Three-phase systems
24
- y 12
y 12 + y 2 + y 23
- y 23
  V1   I1

  
- y 23
  V2  =  0 
+ y 23 + y 3   V3   I3 
- y 13
y 13
j 0.5   V1   - j1 
  

j 2   V2  =  0 
− j 3   V3   - 0.5
 V1   0.77 ∠ − 24 ° 
  

solving  V2  =  0.73∠ − 35°  p . u .
 V3   0.71∠ − 44 ° 
© Salvador Acevedo, 2000
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
- y23
y13 +y23 +y3 V3 I3
or
Y11 Y12 Y13 V1 J1

   
Y21 Y22 Y23 V2 = J2
Y31 Y32 Y33 V3 J3
In general:
N
Yii = ∑yij
i = 12
, ... N
Yij = -yij
i = 12
, ... N;
j=1
j = 12
, ... N;
i≠ j
, ... N
J i = ∑Ii (from current sources flowing into the node) i = 12
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.
EECE 458/571
Three-phase systems
25
© Salvador Acevedo, 2000