Transcript Elements
Passive Elements and Phasor Diagrams
Resistor
i R + v -
v
Ri
V = R I v i I V
Inductor
v i i L + v -
Capacitor
v
v
L di dt
V = jwL I V I i i C + v I
i
C dv dt
I = jwC V ELEC 371 Three-phase systems 1 V Industrial Power Systems Salvador Acevedo
Ideal Transformer
i1 + v1 N1:N2 i2 + v2 -
a
v
1
v
2
i
2
i
1 N1 N2
a
V1 V2 I2 I1 N1 N2 Transformer feeding load: I1 I2 + V1 + V2 V2 = V1/a I2 = V2/Z I1= I2/a Z V2 V1 Assuming a RL load connected to secondary and ideal source to primary I1 I2 ELEC 371 Three-phase systems 2 Industrial Power Systems Salvador Acevedo
Two Winding Transformer Model
+ i1 The linear equivalent model of a real transformer consists of an ideal transformer and some passive elements i2 + v1 v2 N1:N2 ELEC 371 Three-phase systems 3 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 Industrial Power Systems Salvador Acevedo
Steady-state Solution
In sinusoidal steady-state a circuit may be solved using phasors R I vs + VS jwL i VS = R I + jwL I VS = (R + jwL) I VS = (R + jX) I VS = Z I I = I VS Z V max
Z
0 I I max 0 p 2p Ix=I cos VS I Iy=I sin Rectangular form Polar form
I
I = Ix + j Iy = Imax From rectangular form to polar form: I = Ix 2 + Iy tan 1
Iy Ix
2 Magnitude Angle or phase From polar form to rectangular form: Ix Iy Icos Isin Real part Reactive or imaginary part ELEC 371 Three-phase systems 5 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=2 p f w: angular frequency in rad/sec f: frequency in Hz Instantaneous power V m 1 2 V I m m cos(
v
sin
wt
v
I m sin(
wt
i
) cos( 2
wt
v
i
)
i
) Average Power (or REAL POWER) P 1
T
0
T
dt
1 2 V I m m cos V rms I rms cos Apparent Power S = V rms I rms Power Factor pf REAL POWER APPARENT POWER
P S
For this circuit, the power factor is pf = V rms I rms cos cos V rms I rms ELEC 371 Three-phase systems 6 Industrial Power Systems Salvador Acevedo
Power Triangle
Real Power Reactive Power S P=Scos Ssin =Q P = S cos = V I cos watts Q = S sin V I sin vars Complex Power S = S P + j Q S = VI cos + jVIsin If = v i and assuming a reference v then = i therefore i i ) + jIsin( i ) - jIsin( i ) ) S V I * = 0 The magnitude is called Apparent Power: S = VI volt - amperes (VA) ELEC 371 Three-phase systems 7 Industrial Power Systems Salvador Acevedo
Power Consumption by Passive Elements
Impedance: Z = R + jX = Z Resistive Load Z = R = R 0 o P = VI cos0 o Q = VI sin 0 o = 0 vars 2 / R A resistor absorbs P watts Purely Inductive Load Z = jwL = jX L P VI cos(90 o ) X = 0 L 90 o watts Q = VI sin(90 o ) L = V 2 / X L An inductor absorbs Q var
s
Purely Capacitive Load Z = 1 jwC = -jX
C
X C P VI cos(-90 o ) = 0 90 o watts Q = VI sin(-90 o ) 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 Three-phase induction motor Single-phase induction motor Starting torque yes Steady state torque Constant no needs auxiliary starting circuitry Oscillating causing vibration + v Efficient transmission of electric power 3 times the power than a single-phase circuit by adding an extra cable i ia va Single-phase Load vb ib Three-phase Load ic vc p = vi 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 vb(t) = Vm sin (wt - 2 p /3) = Vm sin (wt - 120 ) vc(t) = Vm sin (wt - 4 p /3) = Vm sin (wt - 240 ) or vb(t) = Vm sin (wt + 2 p /3) = Vm sin (wt + 120 ) w = 2 p f w: angular frequency in rad/sec volts volts volts volts f : frequency in Hertz Vc 120 120 120 Va ELEC 371 Three-phase systems 10 Vb Industrial Power Systems Salvador Acevedo
Star Connection (Y)
Y-connected Voltage Source c + a Vcn n + Van Vbn + b Line - to - neutral voltages Van, Vbn, Vcn.
(phase voltages for Y - connection) same magnitude: V P V P Van Vbn Vcn ELEC 371 Three-phase systems 11 Line - to - line voltages Vab, Vbc, Vca same magnitude: V LL Vab = Van - Vbn V = 3 V P Industrial Power Systems Salvador Acevedo
Delta Connection (
D
)
D -connected Voltage Source Vca c + Vbc + a b + Vab Line - to - line voltages Vab, Vbc, Vca.
same magnitude: V LL V P Phase currents Iab, Ibc, Ica.
same magnitude: I P Line currents Ia, Ib, Ic.
same magnitude: I L P ELEC 371 Three-phase systems 12 Industrial Power Systems Salvador Acevedo
Y-connected Load
c + a Vcn n + Van Vbn + b Ia ia Ic ia Za Zc n' Zb Ib 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
c + a Vcn n + Van Vbn + b Ia ia Ic ia Zca Zbc Zab Ib ia ELEC 371 Three-phase systems 14 Industrial Power Systems Salvador Acevedo
Y-
D
Equivalence
Za Zc n' Zb
Balanced case: Za = Zb = Zc = Zy
Zca Zbc Zab ELEC 371 Three-phase systems 15 Industrial Power Systems Salvador Acevedo
Power in Three-phase Circuits
Three - phase voltages and currents:
v a v b v c
V m V m V m
sin sin sin
wt wt wt
v
v
v
120 240
i b i c i a
I m I m I m
sin sin
wt wt
sin
wt
i
v
i
120 240 The three - phase instantaneous power is:
p
3
p
3
V I m m
v i a a
sin
wt
v i b b
v
v i c c
wt
sin
wt
i
v wt
240
v wt
120
i wt
240
v
120 This expression can easily be reduced to:
p
3 3 2
V I m m
cos
v
i
Since the instantaneous power does not change with the time, its average value equals its intantaneous value:
P
3
P
3
p
3 3
V I P P
cos where:
V P
V m
2 ELEC 371 Three-phase systems 16
I P
I m
2
i
Industrial Power Systems Salvador Acevedo
Three-phase Power
In a Y - connection
V LL
3
V P I L
I P P
3 3 cos 3
V LL
3
I L
cos 3
V LL I L
cos
V LL
V P I L
3
I P P
3 3
V I P P
cos 3
V LL
I L
3 cos 3
V LL I L
cos Regardless of the connection (for balanced systems), the average power (real power) is :
P
3 3
V LL I L
cos
watts
Similarly, reactive power and apparent power expressions are: Q 3 3
V LL I L
sin vars S 3 3
V LL 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
A single-phase transformer Example:10 KVA, 38.1KV/3.81KV
N1 : N2 + v1 + v2 a=N1/N2=V1/V2
Three-phase Transformers Connections
Y-Y; DD ; Y D ; D -Y Primary terminals Bank of 3 single-phase transformers Secondary terminals The Bank Transformation Ratio is defined as:
a'
line to line primary voltage line to line secondary voltage
ELEC 371 Three-phase systems 20 Industrial Power Systems Salvador Acevedo
A N1 + v1 N2 + v2 a Y-Y connection B N N1 N2 n b C N1 N2 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 Industrial Power Systems Salvador Acevedo
A N1 + v1 N2 + v2 a DD connection B N1 N2 C N1 N2 b c Ratings for DD bank: 30 KVA; 38.1 KV / 3.81 KV ELEC 371 Three-phase systems 22 Industrial Power Systems Salvador Acevedo
A N1 + v1 N2 + v2 Y D connection B N N1 N2 C N1 N2 a 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 B C N1 + v1 N2 + v2 a N1 N2 N1 N2 n 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 108 KV 120 KV 126 KV 60 KV Percent of base value 90% 100% 105% 50% Per unit value 0.9
1.0
1.05
0.5
per unit quantity = actual quantity base quantity Voltage_1= 108 120 0 9 ** 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: voltage V BASE I: S: Z: current power impedance I S Z BASE BASE BASE V pu S pu V actual V base S actual S base I pu 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: V base 100 KV, S base 100 MVA then, base current and base impedance are: I base Z base = = S base V base V base I base I base Z base 1000 1000 A.
100 Another way to express base impedance is: 2 Z base = V base V base V base I base S base S base V base Real power base and reactive power base are: P base Q base = S base = S base = 100 MW = 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.
S base 3
V base
LL
3
S base
1 3
V base
LN
The base current and impedance for the three phase case are :
I base
S base
3 3
V base
3
LL
S base
3 3
V base
LL Z base
V S base
3
base
LL
3 2 3
V base
LL S base
3 2 In per unit, line to neutral voltage = line to line voltage V LN(pu) = V LL(pu) why?
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
a b c
Example
The following data apply to a three-phase case: S base =300 MVA V base =100 KV (three-phase power) (line-to-line voltage) Three-phase load 270 MW 100 KV pf=0.8
Normally, we' d say: P = I = P 270x10 6 3 (100x10 3 0 8 1948.5 A.
Using the per unit method: P = 270 300 V = 1.0 p.u.
p.u.
P = V I pf then P I = V pf Single-phase equivalent: I=1.125 p.u.
This current is 12.5% higher than its base value!
To check: 1.125xIbase = (1.125) 300,000 3 100 x 1732 .
A.
ELEC 371 Three-phase systems 28
+
V=1 p.u.
-
Industrial Power Systems Salvador Acevedo
Transformers in per unit calculations
With an ideal transformer 2400 V.
+ V1 + V2 4.33 + j 2.5 ohms High Voltage Bases S base1 = 5 KVA V base1 = 2400 V I base1 = 5000/2400=2.083 A Z base1 = 2400/2.083=1152 2400:120 V 5 KVA Low Voltage Bases S base2 = 5KVA V base2 = 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 )/Z base2 =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: Secondary: I1=0.576 x I base1 = 1.2 A.
I2=0.576 x I base2 = 24 A.
ELEC 371 Three-phase systems 29 Industrial Power Systems Salvador Acevedo
One line diagrams
A one line diagram is a simplified representation of a multiphase-phase circuit.
GENERATOR TRANSFORMER Transmission line Transmission line Transmission line TRANSFORMER GENERATOR 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 y23=-j2 p.u.
3 I1= -j1 p.u.
y1=-j1 p.u.
y2=0.1 p.u.
y3=-j0.5 p.u.
I3= -0.5 p.u.
I 1 =y 1 V 1 + y 12 (V 1 -V 2 ) + y 13 (V 1 -V 3 ) 0 = y 12 (V 2 -V 1 ) + y 2 V 2 + y 23 (V 2 -V 3 ) I 3 =y 13 (V 3 -V 1 ) + y 23 (V 3 -V 2 ) + y 3 V 3 In matrix form: y + y 12 - y - y 12 13 + y 13 y 12 - y 12 - y 23 23 y 13 - y 13 - y 23 + y 23 + y 3 V1 V2 V3 I1 0 I3
j j j
2
j
2
j
4
j
2
j
j
2
j
3 V1 V2 V3 - j1 0 - 0.5
solving V1 V2 V3 .
.
24 35 44 ELEC 371 Three-phase systems 31 Industrial Power Systems Salvador Acevedo
General form of the nodal analysis
The system of equations is repeated here to find a general solution technique: y + y + y 13 - y - y 12 13 - y - y 12 23 23 - y - y 13 23 y + y + y 13 23 3 V1 V2 V3 I1 0 I3 or Y 11 Y 21 Y 31 Y Y Y 12 22 32 Y Y Y 13 23 33 V1 V2 V3 J1 J2 J3 In general: N j=1 y ij Y = -y ij ij
i
i
1 2
N j
1 2
i
j
I i (from current sources flowing into the node)
i
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