Industrial%20Power%20Systems

1
Passive Elements and Phasor Diagrams

• Resistor
• Inductor
• Capacitor

2
Ideal Transformer

Transformer feeding load
V2 V1/a I2 V2/Z I1 I2/a
Assuming a RL load connected to secondary and
ideal source to primary
3
Two Winding Transformer Model

• The linear equivalent model of a real transformer
  consists of an ideal transformer and some passive
  elements

4
AC Generators and Motors

• AC synchronous generator
• Single-phase equivalent
• AC synchronous motor
• Single-phase equivalent
• AC induction motor (rarely used as generator)

5
Steady-state Solution

• In sinusoidal steady-state a circuit may be
  solved using phasors

Rectangular form Polar form
6
Single-phase Power Definitions

7
Power Triangle

8
Power Consumption by Passive Elements

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

p vi p va ia vb ib vc ic
10
Three-phase Voltages

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
11
Star Connection (Y)

• Y-connected Voltage Source

12
Delta Connection (D )

• D-connected Voltage Source

13
Y-connected Load

14
D-connected Load

15
Y-D Equivalence

16
Power in Three-phase Circuits

17
Three-phase Power

18
Three-phase Transformers

• Use of one three-phase transformer unit
• Use of 3 single phase transformers to form a
  Transformer Bank

19
Physical Overview
20
Three-phase Transformers Connections Y-Y D-D
Y- D D -Y
Bank of 3 single-phase transformers
Primary terminals
Secondary terminals
21
Y-Y connection
Ratings for Y-Y bank using 3 single-phase
transformers 3x10KVA 30 KVA 66 KV / 6.6 KV
22
D-D connection
Ratings for D-D bank 30 KVA 38.1 KV / 3.81 KV
23
Y- D connection
Ratings for Y-D bank 30 KVA 66 KV / 3.81 KV
24
D -Y connection
Ratings for D-Y bank 30 KVA 38.1 KV / 6.6 KV
25
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.

26
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.

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.

28
Example

• The following data apply to a three-phase case
• Sbase300 MVA (three-phase power)
• Vbase100 KV (line-to-line voltage)

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 5000/24002.083 A I base2
5000/12041.667 A Zbase1 2400/2.0831152 W Z
base2 120/41.6672.88 W
From the circuit V12400 V. V2V1/aV1/20120
V. In per unit V11.0 p.u. V21.0 p.u.
The load in per unit is Z(5Ð 30)/Zbase2
1.7361 Ð 30 p.u. The current in the circuit
is I(1.0 Ð0)/ (1.7361 Ð 30) 0.576 Ð-30
p.u. The current in amperes is Primary
I10.576 x Ibase1 1.2 A. Secondary I20.576 x
Ibase2 24 A.
30
One line diagrams

• A one line diagram is a simplified representation
  of a multiphase-phase circuit.

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
I1y1 V1 y12(V1-V2) y13(V1-V3) 0 y12
(V2-V1) y2 V2 y23(V2-V3) I3y13(V3-V1)
y23(V3-V2) y3 V3
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.