1. Passive Elements and Phasor Diagrams
Resistor
Inductor
Capacitor v i
I V v i I V v i I V
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Salvador Acevedo

2. Transformer feeding load:
Ideal Transformer
Transformer feeding load:
V2 = V1/a
I2 = V2/Z
I1= I2/a
V V1 I1 I2
Assuming a RL load
connected to secondary
and ideal source to primary
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Salvador Acevedo

3. Two Winding Transformer Model
The linear equivalent model of a real transformer consists of an ideal transformer and some passive elements
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Salvador Acevedo

4. AC Generators and Motors
AC synchronous generator
Single-phase equivalent
AC synchronous motor
AC induction motor (rarely used as generator)
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Salvador Acevedo

5. Steady-state Solution
In sinusoidal steady-state a circuit may be solved using phasors q vs i
p p
Iy=I sinq
Ix=I cosq q I VS I q
Rectangular form Polar form
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Salvador Acevedo

6. Single-phase Power Definitions
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7. Real Power P = S cos q = V I cos q watts
Power Triangle
Ssin q=Q S
P=Scosq q
Real Power P = S cos q = V I cos q watts
Reactive Power Q = S sin q = V I sin q vars
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Salvador Acevedo

8. Power Consumption by Passive Elements
Resistive Load
A resistor absorbs P
Purely Inductive Load
An inductor absorbs Q
Purely Capacitive Load
A capacitor absorbs negative Q. It supplies Q.
Industrial Power Systems
Salvador Acevedo

9. Advantages of Three-phase Systems
Creation of the three-phase induction motor
Efficient transmission of electric power
3 times the power than a single-phase circuit by adding an extra cable
Savings in magnetic core when constructing
Transformers
Generators
Single-phase
Load + v - i
Three-phase
Load va vb vc ia ib ic
p = vi p = va ia + vb ib + vc ic
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Salvador Acevedo

10. vb(t) = Vm sin (wt - 2p/3) = Vm sin (wt - 120°) volts
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°) volts or
vb(t) = Vm sin (wt + 2p/3) = Vm sin (wt + 120°) volts
w = 2 p f w: angular frequency in rad/sec f : frequency in Hertz Va Vb Vc
120°
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11. Y-connected Voltage Source
Star Connection (Y)
Y-connected Voltage Source
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12. D-connected Voltage Source
Delta Connection (D )
D-connected Voltage Source
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13. Y-connected Load
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14. D-connected Load
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15. Y-D Equivalence
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16. Power in Three-phase Circuits
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17. Three-phase Power
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18. Three-phase Transformers
Use of one three-phase transformer unit
Use of 3 single phase transformers to form a “Transformer Bank
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19. Physical Overview
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20. Three-phase Transformers Connections Y-Y; D-D; Y- D; D -Y
A single-phase transformer ®
Example:10 KVA, 38.1KV/3.81KV
Three-phase Transformers Connections Y-Y; D-D; Y- D; D -Y
Bank of
3 single-phase
transformers
Primary
terminals
Secondary
terminals
The Bank Transformation Ratio is defined as:
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Salvador Acevedo

21. using 3 single-phase transformers:
Y-Y connection
Ratings for Y-Y bank
using 3 single-phase transformers:
3x10KVA = 30 KVA
66 KV / 6.6 KV
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22. D-D connection Ratings for D-D bank: 30 KVA; 38.1 KV / 3.81 KV
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23. Y- D connection Ratings for Y-D bank: 30 KVA; 66 KV / 3.81 KV
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Salvador Acevedo

24. D -Y connection Ratings for D-Y bank: 30 KVA; 38.1 KV / 6.6 KV
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25. Power lines operate at kilovolts (KV)
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
** The percent value and the per unit value help the analyzer visualize how close the operating conditions are to their nominal values.
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26. V: voltage VBASE I: current IBASE S: power SBASE Z: impedance ZBASE
Defining bases
4 quantities are needed to model a network in per unit system:
V: voltage VBASE
I: current IBASE
S: power SBASE
Z: impedance ZBASE
Given two bases, the other two quantities are easily determined.
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Salvador Acevedo

27. Three phase bases
In three-phase systems it is common to have data for the three-phase power and the line-to-line voltage.
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 Ö3 incorrect !!!
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28. Sbase=300 MVA (three-phase power) Vbase=100 KV (line-to-line voltage)
Example
The following data apply to a three-phase case:
Sbase=300 MVA (three-phase power)
Vbase=100 KV (line-to-line voltage)
Three-phase load
270 MW
100 KV
pf=0.8 a b c +
V=1 p.u. -
Single-phase equivalent:
I=1.125 p.u.
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29. Transformers in per unit calculations
With an ideal transformer
High Voltage Bases Low Voltage Bases
Sbase1 = 5 KVA Sbase2 = 5KVA
Vbase1= 2400 V Vbase2 = 120 V
Ibase1 = 5000/2400=2.083 A I base2 = 5000/120= A
Zbase1= 2400/2.083=1152 W Z base2 = 120/41.667=2.88 W
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.
The load in per unit is:
Z=(5Ð 30°)/Zbase2 = Ð 30° p.u.
The current in the circuit is:
I=(1.0 Ð0°)/ ( Ð 30°) = Ð-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.
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30. One line diagrams
A one line diagram is a simplified representation of a multiphase-phase circuit.
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31. Nodal Analysis
Suppose the following diagram represents the single-phase equivalent of a three-phase system
Finding Norton equivalents and representing impedances as admittances:
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
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32. General form of the nodal analysis
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.
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Salvador Acevedo