Chapter 1

The Power System: An Overview

Power System Analysis

Fundamentals of Power Systems
 basic models of power apparatus,
 transformers, synchronous machines, transmission lines
 simple systems
 one feeder radial to single load

What more is there?
 large interconnected systems
 multiple loads; multiple generators
 why have large interconnected systems?
 reliability; economics
 analysis of the large system
 flow of power and currents; control and stability of the system
 proper handling of fault conditions; economic operation
Power Systems I
Modern Power Systems

Power Producer
 generation station
 prime mover & generator
 step-up transformer

Transmission Company
 HV transmission lines
 switching stations
 circuit breakers
 transformers

Distribution Utility
 distribution substations
 step-down transformers
 MV distribution feeders
 distribution transformers
Power Systems I
Network Layout

HV Networks
 Large quantities of power
shipped over great distances
 Sharing of resources
 Improved reliability
 Economics of large scale

MV Networks
 Local distribution of power
 Numerous systems
 Economics of simplicity
 Autonomous operation

Loads
 Industrial & Commercial
 Residential
Power Systems I
System Control

Network Protection
 Switchgear
 instrumentation transformers
 circuit breakers
 disconnect switches
 fuses
 lightning arrestors
 protective relays

Energy Management
Systems
 Energy Control Center
 computer control
 SCADA - Supervisory Control
And Data Acquisition

Power Systems I
Computer Analysis

Practical power systems
 must be safe
 reliable
 economical

System Analysis
 for system planning
 for system operations
 requires component modeling
 types of analysis
 transmission line performance
 power flow analysis
 economic generation
scheduling
 fault and stability studies

Power Systems I
Chapter 2

AC Power
Single-Phase Power Consumption
i(t)
v (t ) = Vm cos(ω t + θ v )
i (t ) = I m cos(ω t + θ i ) v(t)
p (t ) = v (t ) i (t ) = Vm I m cos(ω t + θ v ) cos(ω t + θ i )
cos A cos B = 12 cos( A − B ) + 12 cos( A + B )
p (t ) = 12 Vm I m {cos(θ v − θ i ) + cos(2ω t + θ v + θ i )}
θ = θ v − θ i Vm = 2 V Im = 2 I
p (t ) = V I cosθ {1 + cos 2(ωt + θ v )}+ V I sin θ sin 2(ωt + θ v )

energy flow into energy borrowed and

the circuit returned by the circuit
Power Systems I
Average Active (Real) Power
p(t ) = V I {1 + cos 2(ω t + θ v )}cosθ + V I sin 2(ω t + θ v )sin θ
1 2π
P=
2π ∫0
p(t ) dt
2π
=V I ∫ {1 + cos 2(ω t + θ )}cosθ + sin 2(ω t + θ )sin θ dt
0
v v

2π 2π
∫ cos(ω t ) dt = 0 ∫ sin (ω t ) dt = 0
0 0

P = V I cosθ
P
pf = cosθ =
V I

Power Systems I
Apparent Power

P = V I cosθ
S=V I

p(t ) = V I {1 + cos 2(ω t + θ v )}cosθ + V I sin 2(ω t + θ v )sin θ

pR (t ) = V I {1 + cos 2(ω t + θ v )}cosθ = P {1 + cos 2(ω t + θ v )}
p X (t ) = V I sin 2(ω t + θ v )sin θ = S sin θ sin 2(ω t + θ v )

Power Systems I
Reactive Power
p X (t ) = V I sin θ sin 2(ωt + θ v ) = S sin θ sin 2(ωt + θ v )
Q ≡ S sin θ = V I sin θ
p X (t ) = Q sin 2(ωt + θ v )

for a pure resistor
 the impedance angle is zero, power factor is unity
 apparent power and real power are equal

for a purely inductive circuit
 the current lags the voltage by 90°, average power is zero
 no transformation of energy

for a purely capacitive circuit
 the current leads the voltage by 90°, average power is zero
Power Systems I
AC Power

Example
 the supply voltage is given by v(t) = 480 cos t

 the load is inductive with impedance Z = 1.20∠60°
 determine the expression for the instantaneous current i(t) and
instantaneous power p(t)
 plot v(t), i(t), p(t), pR(t), pX(t) over an interval of 0 to 2

π
Power Systems I
Complex Power

Real Power, P
 RMS based - thermally equivalent to DC power

Reactive Power, Q
 Oscillating power into and out of the load because of its reactive
element (L or C).
 Positive value for inductive load (lagging pf)

Complex Power, S

V I * = V I ∠(θ v − θ i ) = V I ∠θ = S
S = V I cosθ + j V I sin θ = P + jQ
S = P2 + Q2
Power Systems I
Complex Power
V

θv S
I
θ Q
θi θ
Leading Power Factor
P
I
Lagging Power Factor θi V
θ
θv P
θ
Q
S
Power Systems I
The Complex Power Balance

From the conservation of energy
 Real power supplied by the source is equal to the sum of the real
powers absorbed by the load and the real losses in the system
 Reactive power must also be balanced
 The balance is between the sum of leading and the sum of lagging
reactive power producing elements
 The total complex power delivered to the loads in parallel is the
sum of the complex powers delivered to each
0 = ∑ Pgen − ∑ Ploads − ∑ Plosses
0 = ∑ Qleading + ∑ Qcaps − ∑ Qlagging − ∑ Qind
0 = ∑ S gen − ∑ Sloads − ∑ Slosses
Power Systems I
Complex Power

Example
 in the circuit below, find the power absorbed by each load and
the total complex power
 find the capacitance of the capacitor to be connected across the
loads to improve the overall power factor to 0.9 lagging

V I1 I2 I3
1200 V
Z1=60+j0 Z2=6+j12 Z3=30-j30

Power Systems I
Complex Power Flow
Z = R+j X =|Z|∠γ

Consider the following circuit

 For the assumed direction

of current V1 I12 V2
V1 = V1 ∠δ 1 V2 = V2 ∠δ 2
V1 ∠δ 1 − V2 ∠δ 2 V1 V2
I12 = = ∠(δ 1 − γ ) − ∠(δ 2 − γ )
Z ∠γ Z Z
 The complex power
 V1 V2 
S12 = V1 I = V1 ∠δ 1  ∠(γ − δ 1 ) −
*
12 ∠(γ − δ 2 )
Z Z 
2
V1 V1 V2
= ∠γ − ∠(γ + δ 1 − δ 2 )
Power Systems I Z Z
Complex Power Flow
 The real and reactive power at the sending end
2
V1 V1 V2
P12 = cos γ − cos(γ + δ 1 − δ 2 )
Z Z
2
V1 V1 V2
Q12 = sin γ − sin (γ + δ 1 − δ 2 )
Z Z
 Transmission lines have small resistance compared to the
reactance. Often, it is assumed R = 0 (Z = X∠90°)

P12 =
V1 V2
sin (δ 1 − δ 2 ) Q12 =
V1
[V 1 − V2 cos(δ 1 − δ 2 ) ]
X X

Power Systems I
Complex Power Flow

For a typical power system with small R / X ratio, the
follow observations are made
 Small changes in δ1 or δ2 will have significant effect on the real
power flow
 Small changes in voltage magnitude will not have appreciable
effect on the real power flow
 Assuming no resistance, the theoretical
maximum power (static transmission V1 V2
capacity) occurs when the angular Pmax =
difference, δ, is 90° and is given by: X
 For maintaining stability, the power system operates with small
load angle δ
 The reactive power flow is determined by the magnitude
difference of the terminal voltages
Power Systems I
Three-Phase Power

Balanced three-phase power
 Assumes balanced loads
 Assumes voltage and currents with phases that have 120°
separation

P3φ = 3 V p I p cosθ = 3 VLL I L cosθ

Q3φ = 3 V p I p sin θ = 3 VLL I L sin θ
S3φ = 3 V p I ∗p = 3 VLL I L∗

Power Systems I
 Chapter 3

Power Apparatus Modeling

System Modeling

Systems are represented on a per-phase basis
 A single-phase representation is used for a balanced system
 the system is modeled as one phase of a wye-connected network
 Symmetrical components are used for unbalanced systems
 unbalance systems may be caused by: generation, network
components, loads, or unusual operating conditions such as faults
 The per-unit system of measurements is used

Review of basic network component models
 generators
 transformers
 loads
 transmission lines

Power Systems I
Generator Models

Generator may be modeled in three different ways
 Power Injection Model - the real, P, and reactive, Q, power of the
generator is specified at the node that the generator is connected
 either the voltage or injected current is specified at the connected
node, allowing the other quantity to be determined
 Thevenin Model - induced AC voltage, E, behind the synchronous
reactance, Xd
E Xd Node

 Norton Model - injected AC current, IG, in parallel with the

synchronous reactance
IG
Node
Xd
Power Systems I
Transformer Model

Equivalent circuit of a two winding transformer

R1 X1 N1 : N2 R2 X2

E1 E2
V1 Rc Xm V2

Power Systems I
Transformer Model

Approximate circuit referred to the primary

REQ1 XEQ1

N1
V1 Rc Xm V2′ = V2
N2

Power Systems I
Load Models

Models are selected based on both the type of analysis
and the load characteristics

Constant impedance, Zload
 Load is made up of R, L, and C elements connected to a network
node and the ground (or neutral point of the system)

Constant current, Iload
 The load has a constant current magnitude I, and a constant
power factor, independent of the nodal voltage
 Also considered as a current injection into the network

Constant power, Sload
 The load has a constant real, P, and reactive, Q, power
component independent of nodal voltage or current injection
 Also considered as a negative power injection into the network
Power Systems I
Per Unit System

Almost all power system analyses are performed in per-
units x ( engr. unit ) x%
x per unit ( pu ) = engineering ( actual )
=
xbase ( engr. unit ) 100

Per unit system for power systems
 Based on a per-phase, wye-connect, three-phase system
 3-phase power base, S3φ
 common power base is 100 MVA
 Line-to-line voltage base, VLL S3φ −base
 voltage base is usually selected I L−base =
from the equipment rated voltage 3 VLL−base
(VLL−base ) (VLN −base )
 Phase current base, IL 2 2
 Phase impedance base, Z Z base = =
S3φ −base S1φ −base
Power Systems I
Per Unit System

Equipment impedances are frequently given in per units
or percentages of the impedance base
 The impedance base for equipment is derived from the rated
power and the rated voltage
 When modeling equipment in a system, the per unit impedance
must be converted so that the equipment and the system are on
a common base
old new
ZΩ Sbase ZΩ Sbase
Z old = old = Z Ω old 2 new
Z pu = new = Z Ω new 2
pu
Z base (Vbase ) Z base (Vbase )
Z new
=
S new
base
⋅
(
V )
old 2
base
⋅Z old
=Z old S new
base
V
⋅ 
old
base


2

pu
(V )
new 2
base
S old
base
pu pu
S old
base V
new
base 
new
Sbase
 It is normal for the voltage bases to be the same: Z
new
pu =Z old
pu old
Sbase
Power Systems I
Per Unit System

The advantages of the per unit system for analysis
 Gives a clear idea of relative magnitudes of various quantities
 The per-unit impedance of equipment of the same general type
based upon their own ratings fall in a narrow range regardless of
the rating of equipment.
 Whereas their impedances in ohms vary greatly with the ratings.
 The per-unit impedance, voltages, and currents of transformers
are the same regardless of whether they are referred to the
primary or the secondary side.
 Different voltage levels disappear across the entire system.
 The system reduces to a system of simple impedances
 The circuit laws are valid in per-unit systems, and the power and
voltages equations are simplified since the factors of √3 and 3
are eliminated in the per-unit system
Power Systems I
Per Unit System

Example
 the one-line diagram of a three-phase power system is shown
 use a common base of 100 MVA and 22 kV at the generator
 draw an impedance diagram with all impedances marked in per-unit
 the manufacturer’s data for each apparatus is given as follows
 G: 90 MVA 22 kV 18% G
 T1: 50 MVA 22/220 kV 10%
 L1: 48.4 ohms
 T2: 40 MVA 220/11 kV 6% T1 T3
 T3: 40 MVA 22/110 kV 6.4%
 L2: 65.43 ohms L1 L2
 T4: 40 MVA 110/11 kV 8%
 M: 66.5 MVA 10.45 kV 18.5% T2 T4
 Ld: 57 MVA 10.45 kV 0.6 pf lag

Power Systems I M Ld