Basics on Electric Power System Analysispaginaspersonales.deusto.es/enrique.zuazua/MTM2017... · • Definition of active power, apparent power, reactive power • Use of complex

Basics on

Electric Power

System Analysis

Jon Andoni Barrena

Eneko Unamuno

01/05/2018

Mondragon Unibertsitatea

04.06.18 Mondragon Unibertsitatea 2

Research Project

This presentation is an activity funded by the Spanish ministry of economy and

competitiveness:


“Control para la Mejora de la Respuesta Inercial y

Estabilidad de Redes Hibridas AC/DC”

“Control y Estabilidad de Redes Eléctricas Hibridas

AC/DC: Ecuaciones Diferenciales y Ecuaciones en

Derivadas Parciales para el Análisis de Estabilidad de

Redes”

MTM2017-82996-C2-1-R

MTM2017-82996-C2-2-R

Contents

• Basic concepts:– Voltage, current, power, Kirchhoff's circuit laws, voltage source, current source.

– Passive elements: Instantaneous voltage, current, power, energy equations (ohm's law)

• Direct Current and Alternating Current: definitions– Analysis of single-phase ac systems:

• RMS values: Definition of impedance and admittance of passive elements

• Definition of active power, apparent power, reactive power

• Use of complex values for calculations in ac systems: Phasor representation of electrical quantities

– Analysis of three phase ac systems:

• Voltages and currents in balanced three-phase circuits

• Power in balanced three-phase circuits

• Electric Power systems:– Single-line or one-line representation of power systems

– Impedance and reactance diagrams

– Per-unit quantities

– Node equations: Admittance and impedance model and network calculations.

– Power-Flow solutions: The Newton-Raphson method

• Operation and control of electric power systems– Description of the hierarchical control in power systems

– Inertial response and primary frequency regulation


Basic Concepts

1

Basic Concepts

• Voltage: The difference in electric potential between two points. The voltage

between two points is equal to the work done per unit of charge against a

static electric field to move a test charge between two points. This is

measured in units of volts (a joule per coulomb); moving 1 coulomb of charge

across 1 volt of electric potential requires 1 joule of work.

• Current: An The SI unit for measuring an electric current is the ampere, which

is the flow of electric charge across a surface at the rate of one coulomb per

second. electric current is a flow of electric charge.

• Electric power is the rate, per unit time, at which electrical energy is

transferred by an electric circuit. The SI unit of power is the watt, one joule per

second.


𝑝 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒 =𝑣 ∙ 𝑞

𝑡= 𝑣 ∙ 𝑖

Basic Concepts

• Resistance: The electrical resistance of an electrical conductor is a measure

of the difficulty to pass an electric current through that conductor. The inverse

quantity is electrical conductance, and is the ease with which an electric

current passes. The SI unit of electrical resistance is the ohm (Ω), while

electrical conductance is measured in siemens (S).

• Ohm's law states that the current through a conductor between two points is

directly proportional to the voltage across the two points. Introducing the

constant of proportionality, the resistance.


𝑖(𝑡) =𝑣(𝑡)

𝑅

Basic Concepts

• Joule heating, also known as Ohmic heating and resistive heating, is the

process by which the passage of an electric current through a conductor

produces heat. Joule's first law, also known as the Joule–Lenz law, states that

the power of heating generated by an electrical conductor is proportional to

the product of its resistance and the square of the current.


𝑝(𝑡) = 𝑖(𝑡)2 ∙ 𝑅

Basic Concepts

• Kirchhoff's circuit laws are two equalities that deal with the current and

potential difference:

– Kirchhoff's current law: At any node (junction) in an electrical circuit, the sum of

currents flowing into that node is equal to the sum of currents flowing out of that

node.

– Kirchhoff's voltage law: The directed sum of the electrical potential differences

(voltage) around any closed network is zero.


1

1 4 2 3

1 2 3 4

0

0

n

k

k

I

I I I I

I I I I

=

=

+ = +

+ + + =

1

1 2 3

1 2 3

0

0

n

k

k

s

s

V

V V V V

V V V V

=

=

= + +

− − − =

Basic Concepts

• Passive elements:

– Inductor: An inductor, also called a coil, choke or reactor, is a passive two-terminal

electrical component that stores energy in a magnetic field when electric current

flows through It. An electric current flowing through a conductor generates a

magnetic field surrounding it. The total amount of magnetic field through a circuit,

the magnetic flux Φ , generated by a given current I depends on the geometric

shape of the circuit. The ratio of these quantities is the inductance. In the

International System of Units (SI), the unit of inductance is the henry (H).


2 2N N SL

l

l

S

= =

=

L: Inductance

N: Number of turns

l: Length of the magnetic circuit

S: Area of the magnetic field cross-section

μ: Permeability

ℜ: Reluctance

Basic Concepts


– Inductor: Any change in the current through an inductor creates a changing flux,

inducing a voltage across the inductor. By Faraday's law of induction, the voltage

induced by any change in magnetic flux through the circuit is:

– The energy stored in the magnetic field is:


𝑣𝐿 =𝑑𝜙

𝑑𝑡= 𝐿 ∙

𝑑𝑖𝐿𝑑𝑡

𝐸𝐿 =1

2𝐿 ∙ 𝑖𝐿

2

Basic Concepts


– Capacitor: A capacitor, is a passive two-terminal electrical component that stores

potential energy in an electric field. Most capacitors contain at least two electrical

conductors often in the form of metallic plates or surfaces separated by a dielectric

medium. When these two conductors experience a potential difference, an electric

field develops across the dielectric, causing a net positive charge to collect on one

plate and net negative charge to collect on the other plate. The ratio of the electric

charge on each conductor to the potential difference between them is the

capacitance. The unit of capacitance in the International System of Units (SI) is the

farad (F), defined as one coulomb per volt (1 C/V). The capacitance of a capacitor

is proportional to the surface area of the plates (conductors) and inversely related

to the gap between them.


0 r

SC

l =

Basic Concepts


– Capacitor: The capacitance describes how much charge can be stored on one

plate of a capacitor for a given voltage drop.

– The equation that relates the current, voltage and capacitance of a capacitor can

be derived from the previous equation:

– The energy stored in the electric field is:


𝐶 =𝑑𝑄

𝑑𝑉𝑐

𝑖𝐶 = 𝐶 ∙𝑑𝑣𝐶𝑑𝑡

𝐸𝐶 =1

2𝐶 ∙ 𝑣𝐶

2

Basic Concepts

• Active elements:

– A voltage source is a two-terminal device which can maintain a fixed voltage independent of

the load resistance or the output current.

– The internal resistance of an ideal voltage source is zero; it is able to supply or absorb any

amount of current. The current through an ideal voltage source is completely determined by

the external circuit.

– A current source is a two terminal device that delivers or absorbs an electric current which is

independent of the voltage across it.

– The internal resistance of an ideal current source is infinite. An independent current source

with zero current is identical to an ideal open circuit. The voltage across an ideal current

source is completely determined by the circuit it is connected to.


Basic Concepts

• Direct Current system (DC):

– Direct current (DC) is the unidirectional flow of electric charge. The electric current

flows in a constant direction, distinguishing it from alternating current (AC). The

term DC is used to refer to power systems that use only one polarity of voltage or

current, and to refer to the constant, zero-frequency, or slowly varying local mean

value of a voltage or current. For example, the voltage across a DC voltage source

is constant as is the current through a DC current source.


Basic Concepts

• Direct Current system (DC):

– A direct current circuit is an electrical circuit that consists of any combination of

constant voltage sources, constant current sources, and resistors. In this case, the

circuit voltages and currents are independent of time. A particular circuit voltage

or current does not depend on the past value of any circuit voltage or current. This

implies that the system of equations that represent a DC circuit do not involve

integrals or derivatives with respect to time.


Basic Concepts

• Alternating Current system (AC):

– Alternating current (AC) is an electric current which periodically reverses direction,

in contrast to direct current (DC) which flows only in one direction. The usual

waveform of alternating current in most electric power circuits is a sine wave.

– Electrical energy is distributed as alternating current because AC voltage may be

increased or decreased with a transformer. This allows the power to be transmitted

through power lines efficiently at high voltage, which reduces the energy lost as

heat due to resistance of the wire, and transformed to a lower, safer, voltage for

use. Use of a higher voltage leads to significantly more efficient transmission of

power.


Single-Phase

AC Systems

2

Single-phase ac system

• RMS or effective values of current and voltage:

• For a cyclically alternating electric current, the effective value of the current is

equal to the value of the direct current that would produce the same average

power dissipation in a resistive load.

– Let’s consider a cyclical current i(t) of period T. The instantaneous power

dissipated in a resistance R:

– The average value of the power, known also as active power:

– Therefore:


𝑝(𝑡) = 𝑖(𝑡)2 ∙ 𝑅

𝑃 =1

𝑇න0

𝑇

𝑝 𝑡 𝑑𝑡 =1

𝑇න0

𝑇

𝑖(𝑡)2𝑅 𝑑𝑡 = 𝐼𝑒𝑓2 ∙ 𝑅

𝐼𝑒𝑓 =1

𝑇න0

𝑇

𝑖(𝑡)2 𝑑𝑡 = 𝐼𝑅𝑀𝑆


• RMS or effective values of current and voltage:

– For a sinusoidal current waveform:

• The instantaneous power dissipated in a resistance is:

• The RMS value of a sinusoidal waveform is:


𝐼𝑅𝑀𝑆 =1

𝑇න0

𝑇

𝑖(𝑡)2 𝑑𝑡 =𝐼𝑀

2

𝑖 𝑡 = 𝐼𝑀 ∙ sin(𝜔𝑡)

𝑝 𝑡 = (𝐼𝑀∙ sin 𝜔𝑡 ) 2 ∙ 𝑅 =𝐼𝑀

2 ∙ 𝑅

2− 𝐼𝑀

2 ∙ 𝑅 ∙ cos(2𝜔𝑡)


• Impedance: Electrical impedance is the measure of the opposition that a

circuit presents to a current when a voltage is applied. Definition by A. E.

Kennelly in 1893: “The impedance of a conductor is its apparent resistance,

and is expressible in ohms. More strictly, it has been defined as the ratio of the

effective E. M. F. between the terminals of the conductor, to the effective

current strength it carries.”

• Impedance is a complex number, with the same units as resistance, for which

the SI unit is the ohm (Ω). Its symbol is usually Z, and it may be represented

by writing its magnitude and phase in the form |Z|∠θ. However, cartesian

complex number representation is often more powerful for circuit analysis

purposes.

• The reciprocal of impedance is admittance, whose SI unit is the siemens,



• Impedance: Electrical impedance is the measure of the opposition that a

circuit presents to a current when a voltage is applied. Definition by A. E.

Kennelly in 1893: “The impedance of a conductor is its apparent resistance,

and is expressible in ohms. More strictly, it has been defined as the ratio of the

effective E. M. F. between the terminals of the conductor, to the effective

current strength it carries.”

• Impedance is a complex number, with the same units as resistance, for which

the SI unit is the ohm (Ω). Its symbol is usually Z, and it may be represented

by writing its magnitude and phase in the form |Z|∠θ. However, cartesian

complex number representation is often more powerful for circuit analysis

purposes.

• The reciprocal of impedance is admittance, whose SI unit is the siemens,



• Resistive impedance:

– If a sinusoidal current is applied to a resistance:

– The voltage drop at the terminals of the resistance is:

– So the resistive impedance (the ratio between the effective voltage and effective

current) is:

– In this context, a resistive admittance is the inverse of the impedance:


𝑖𝑅 𝑡 = 𝐼𝑀 ∙ sin(𝜔𝑡)

𝑣𝑅 𝑡 = 𝑖𝑅 𝑡 ∙ 𝑅 = 𝐼𝑀 ∙ 𝑅 ∙ sin 𝜔𝑡 = 𝑉𝑀 ∙ sin 𝜔𝑡

𝑍𝑅 = 𝑅

𝑌𝑅 =1

𝑍𝑅


• Inductive impedance:

– If a sinusoidal current is applied to an inductor, the voltage drop at the terminals of

the inductor is:

– So the inductive impedance (the ratio between the effective voltage and effective

current in an inductor) is:

– The voltage is a sinusoidal waveform that leads by 90º the current. We can

express it in polar form:


𝑣𝐿 = 𝐿 ∙𝑑𝑖𝐿𝑑𝑡

= 𝐿 ∙𝑑(𝐼𝑀∙ sin(𝜔𝑡))

𝑑𝑡= 𝐼𝑀 ∙ 𝐿 ∙ 𝜔 ∙ cos 𝜔 ∙ 𝑡

𝑍𝐿 =𝑉𝐿𝐼𝐿= 𝐿 ∙ 𝜔 Ω

𝒁𝑳 =𝑉𝐿∠90º

𝐼𝐿∠0º= 𝐿𝜔∠90º


• Capacitive impedance:

– If a sinusoidal voltage is applied to a capacitor, the voltage drop at the terminals of

the capacitor is:

– So the resistive impedance (the ratio between the effective voltage and effective

current) is:

– The current is a sinusoidal waveform that leads by 90º the voltage. We can

express it in polar form:


𝑖𝑐 = 𝐶 ∙𝑑𝑣𝐶𝑑𝑡

= 𝐶 ∙𝑑(𝑉𝑀∙ sin(𝜔𝑡))

𝑑𝑡= 𝑉𝑀 ∙ 𝐶 ∙ 𝜔 ∙ cos 𝜔𝑡

𝑍𝐶 =𝑉𝐶𝐼𝐶

=1

𝐶 ∙ 𝜔Ω

𝒁𝑪 =𝑉𝐶∠0º

𝐼𝐶∠90º=

1

𝐶𝜔∠ − 90º


• Complex voltage and current, phasor representation

– To simplify calculations, sinusoidal voltage and current waves are commonly

represented as complex-valued functions of time, as rotating vectors in the

complex plane, usually called phasors. Their arithmetic operations (addition,

multiplication, differentiation, integration) can be more conveniently carried out.

• For a sinusoidal current and voltages:

• The complex current is defined as:

• Aplying Euler’s formula, it can be deduced that:


𝑖 𝑡 = 𝐼𝑀 ∙ cos 𝜔𝑡 + 𝜑 and 𝑣 𝑡 = 𝑉𝑀 ∙ cos(𝜔𝑡 + 𝜙)

𝑰 = 𝐼𝑀 ∙ 𝑒𝑗 𝜔𝑡+𝜑 and 𝑽 = 𝑉𝑀 ∙ 𝑒𝑗(𝜔𝑡+𝜙)

𝑖 𝑡 = 𝑅𝑒 𝑰 = 𝐼𝑀∙ cos 𝜔𝑡 + 𝜑

𝑣 𝑡 = 𝑅𝑒 𝑽 = 𝐼𝑀∙ cos(𝜔𝑡 + 𝜙)



– Aligned voltage and current (e.g. through a resistor):

– The power is not constant and has an average value and an oscillatory component

(only the average or “active” component provides real power)


https://www.electronics-tutorials.ws/accircuits/power-in-ac-circuits.html

( ) ( ) ( ) ( )cos cos 2 = = − −

average oscillatory

p t v t i t VI VI t



– Current lagging the voltage by 90º (e.g. through an inductor):

– The average power is zero and has an oscillatory component



( ) ( ) ( ) ( ) ( )cos 2 90 sin 2p t v t i t VI t VI t = = − + = −


– Current leading the voltage by 90º (e.g. through a capacitor):

– The average power is zero and has an oscillatory component




( ) ( ) ( ) ( ) ( )cos 2 90 sin 2p t v t i t VI t VI t = = − − =

• Complex voltage and current, power definitions

– We can rearrange the previous instantaneous power equation as follows

where P and Q are the average active and reactive power

– In this context, the apparent power is defined as:



( ) ( ) ( ) ( )

( )( ) ( )

( )( ) ( )

cos cos 2

cos 1 cos 2 sin sin 2

1 cos 2 sin 2

average oscillatory

QP

oscillatoryaverage

p t v t i t VI VI t

VI t VI t

P t Q t

= = − − =

= − − =

= − −

*S P jQ= = +VI

Three-Phase

AC Systems

3

Three-phase systems

Why three-phase?

• At classical power plants energy is generated by generators driven by

turbines, and these generators need more than one phase to rotate.

• For the same conductor size more power can be transferred, or if the same

power is transferred the losses can be reduced.

• The transferred instantaneous power (for balanced three-phase systems) is

constant, compared to the pulsating power of single-phase systems.


Three-phase systems

• Single-phase system

• Three-phase system


At a balanced three-phase power system, the sum of voltages with respect to the

neutral point is zero:

where the phase voltages are represented as as

In this context, the neutral point is defined as the point to which the sum of

referenced voltages is zero.

( ) ( ) ( ) 0+ + =an bn cnv t v t v t

Three-phase systems


( ) ( )

( ) ( )

( ) ( )

sin

sin 120

sin 120

=

= −

= +

an M

bn M

cn M

v t V t

v t V t

v t V t

Three-phase systems

Phase voltages of a balanced three-phase power system can be represented in a

phasor diagram


Three-phase systems

Line voltages of a balanced three-phase power system can be represented as

follows in the phasor diagram


( ) ( ) ( ) 0+ + =ab bc cav t v t v t

Three-phase systems

Line and phase voltages of a balanced three-phase power system can be related

as follows


3 3= = = =ab ca bc an Mv v v v V

Three-phase systems

The power of a three-phase system is the sum of powers of each line

It can be demonstrated that the instantaneous power equals the average power over one

period, meaning that it is constant

On the other hand, the instantaneous reactive power is oscillatory, and therefore the use

of the average reactive power is preferred, which is defined as


( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

= + +

= + +

a b c

an a bn b cn c

p t p t p t p t

v t i t v t i t v t i t

( ) 3 cos = =p t P V I

3 sinQ V I =

Three-phase systems

The power triangle illustrates the relation between the active, reactive and

apparent power

where the apparent power, similar to single-phase systems, is defined as

From these relations, the concept of power factor is defined as the ratio between the

active power and the apparent power:


cos =P

S

2 2 *3 = + =S P Q V I

Single-Line

Representation

of Power

Systems

4

Single-line representation

• The basic components of a power system are generators,transformers, transmission lines, and loads. The interconnectionsamong these components in the power system may be shown ina so-called single-line or one-line diagram.

• The advantage of such a one-line representation is its simplicity:

– One phase represents all three phases of the balanced system;

– the equivalent circuits of the components are replaced by theirstandard symbols; and the completion of the circuit through theneutral is omitted.

• For analysis, the equivalent circuits of the components are shownin a reactance diagram or impedance diagram.



• Symbols used to represent the typical components of a power

system.




system.


NODE or BUS

(substation)BRANCHES

(lines)

NETWORK

(but unloaded and

unsupplied)



system.


LOAD: Extracts MW out of the

node (injects negative MW into

the node)

GENERATOR: Injects MW

into the node

NETWORK

(loaded and

supplied)


• Example of a one-line diagram for a power system consisting of

two generating stations connected by a transmission line














Impedance and

Reactance

Diagrams,

Per-Unit Quantities

5

Impedance and reactance diagrams

• The impedance diagram of a power system is a single phase

diagram representing all the equivalent circuits of the

components of the power system. This is called reactance

diagram if resistances are neglected.


Per-unit quantities

• Computations for a power system having two or more voltage

levels become very cumbersome when it is necessary to convert

currents to a different voltage level wherever they flow through a

transformer (the change in current being inversely proportional

to the transformer turns ratio).

• In an alternative and simpler system, a set of base values, or

base quantities, is assumed for each voltage class, and each

parameter is expressed as a decimal fraction of its respective

base.


Per-unit quantities

• Computations for a power system having two or more voltage

levels become very cumbersome when it is necessary to convert

currents to a different voltage level wherever they flow through a

transformer (the change in current being inversely proportional

to the transformer turns ratio).

• In an alternative and simpler system, a set of base values, or

base quantities, is assumed for each voltage class, and each

parameter is expressed as a decimal fraction of its respective

base.


Per-unit quantities

• In the per-unit system, the voltages, currents, powers,impedances, and other electrical quantities are expressed on aper-unit basis by the equation:

• It is customary to select two base quantities to define a givenper-unit system. The ones usually selected are voltage andpower.


Quantity per unit =Actual value

Base value of quantity

Per-unit quantities

• A minimum of four base quantities is required to completely define a per=unit

system; these are voltage, current, power, and impedance (or admittance). If

two of them are set arbitrarily, then the other two become fixed. The following

relationships hold on a per-phase basis:


Per-unit quantities

• For representing a three-phase power system in per-unit

quantities, the base power should be defined:

where

• The voltage base is different on each side of the transformer:


IVSSSb 33 13 === −−

3/)(lineLneutraltoline VVV == −−

Lcurrentline III == −

LLb IVS 3=

LVLbLV VV ,=HVLbHV VV ,=

Per-unit quantities

• So the base current and base impedance are already fixed:


b

b

b

bbb

S

V

S

VV

I

VZ

2)(3

3===

b

bb

V

SI

3=

Per-unit quantities

• Example:


Generador

100 MVA

22 kV

X=90%

Transformador

100 MVA

22:110 kV

X=10%

Línea de transmisión

Z = j0.8403 pu @ 120

kV y 50 MVA

Carga

datos de operación:

V=110 kV

S=10 MVA

fp = 1

Transformador

100 MVA

120:24 kV

X=12.6%

Generador

80 MVA

22 kV

X=1.48 pu


Z = j60.5 ohms


X = 60.5 ohms

Per-unit quantities

• Example:1.- Choose an appropriate power base, for example 100MVA


Generador

100 MVA

22 kV

X=90%

Transformador

100 MVA

22:110 kV

X=10%


Z = j0.8403 pu @ 120

kV y 50 MVA

Carga


V=110 kV

S=10 MVA

fp = 1

Transformador

100 MVA

120:24 kV

X=12.6%

Generador

80 MVA

22 kV

X=1.48 pu


Z = j60.5 ohms


X = 60.5 ohms

Per-unit quantities

• Example:

2.- Define the voltage base for every area of the system (delimited by

transformers)


Per-unit quantities

• Example:

2.- In this example a base voltage of 110kV is selected for the high-

voltage area.


Generador

100 MVA

22 kV

X=90%

Transformador

100 MVA

22:110 kV

X=10%


Z = j0.8403 pu @ 120

kV y 50 MVA

Carga


V=110 kV

S=10 MVA

fp = 1

Transformador

100 MVA

120:24 kV

X=12.6%

Generador

80 MVA

22 kV

X=1.48 pu


Z = j60.5 ohms


X = 60.5 ohms

Sbase = 100 MVA

Vbase = 110 kV

Per-unit quantities

• Example:

2.- The base power and voltage of the whole system is already defined

applying the transformation ratio of the transformers.


22:110 kV 120:24 kV

Sbase = 100 MVA

Vbase = 110 kV

Sbase = 100 MVA

Vbase = 110 x (24/120) =22 kV

Sbase = 100 MVA

Vbase = 22 kV

Per-unit quantities

• Example:

3.- Convert all the impedances to p.u.


Generador 1

100 MVA

22 kV

X=90%

Transformador

100 MVA

22:110 kV

X=10%

Per-unit quantities

• Example:




Z = j0.8403 pu @ 120

kV y 50 MVA

Carga


V=110 kV

S=10 MVA

fp = 1


Z = j60.5 ohms


X = 60.5 ohms

( )sistema

placapu

sistemabase

linea

sistemabase

lineabaseplacapu

LL puj

MVA

kV

j

MVA

kV

MVA

kVj

Z

Z

Z

ZjjXZ 2

100

)110(

242

100

)110(

50

)120(8403.0

8403.0

:lineUpper

22

2

=

=

==

==

−

−

−

−

−−

Per-unit quantities

• Example:




Z = j0.8403 pu @ 120

kV y 50 MVA

Carga


V=110 kV

S=10 MVA

fp = 1


Z = j60.5 ohms


X = 60.5 ohms

sistema

sistemabase

lineaLL puj

MVA

kV

j

Z

ZjXZ 5.0

100

)110(

5.60

:lines Bottom

2=

===

−

−

Per-unit quantities

• Example:



Transformador

100 MVA

120:24 kV

X=12.6%

Generador 2

80 MVA

22 kV

X=148%

( )

sistema

placapu

g

sistemabase

generador

sistemabase

generadorbaseplacapu

g

pu

MVA

kV

MVA

kV

X

Z

Z

Z

ZX

Generator

85.1

100

)22(

80

)22(48.1

48.1

:

2

2

2

2

=

=

=

=

−

−

−

−

−−

Per-unit quantities

• Example:



Transformador

100 MVA

120:24 kV

X=12.6%

Generador 2

80 MVA

22 kV

X=148%

( )

sistema

placapu

t

sistemabase

transf

sistemabase

transfbaseplacapu

t

pu

MVA

kV

MVA

kV

X

Z

Z

Z

ZX

rTransforme

15.0

100

)22(

100

)24(126.0

126.0

:

2

2

2

2

=

=

=

=

−

−

−

−

−−

Per-unit quantities

• Example:

3.- Impedance diagram in p.u.


+

V1= 1 p.u.

-

zg1=j0.9

z13=j2 p.u.

z12=j0.5 p.u. z23=j0.5 p.u.

z2=10 p.u.

zt2=j0.15

+

V3= -j1 p.u.

-

1 3

2

zg2=j1.85zt1=j0.1

45

Electric Power

Systems

6

Node equations, admittance

matrix.

Node equations

• Let’s consider the four bus system shown in the figure, being 𝐼𝑖the current injected to each node by the generators and loads.


I1 y12

y14

y34

y23

y13

I2

I4

I3

Node equations

• Kirchoff’s current law: sum of the currents at any node must be zero.


I1 y12

y14

y34

y23

y13

I2

I4

I3

V1

V4

V2

V3

I14

I13

I12

y13

1413121 IIII ++=

Node equations

• Applying Ohm’s law, line currents are:


I1 y12

y14

y34

y23

y13

I2

I4

I3

V1

V4

V2

V3

I14

I13

I12

y13

)( jiijij VVyI −=

Node equations

• Therefore:


I1 y12

y14

y34

y23

y13

I2

I4

I3

V1

V4

V2

V3

I14

I13

I12

y13

)()()( 4114311321121 VVyVVyVVyI −+−+−=

Node equations

• Therefore:


I1 y12

y14

y34

y23

y13

I2

I4

I3

V1

V4

V2

V3

I14

I13

I12

y13

)()()()( 14413312214131211 yVyVyVyyyVI −+−+−+++=

Node equations

• Repeating for the other four buses:


)()()()( 14413312214131211 yVyVyVyyyVI −+−+−+++=

)()()()( 24423324232122112 yVyVyyyVyVI −+−++++−=

)()()()( 34434323133223113 yVyyyVyVyVI −++++−+−=

)()()()( 43443424134224114 yVyyyVyVyVI −++++−+−=

Node equations

• Writing the previous equations in matrix form:

• Being the admittance matrix of the power system or Y-bus is defined as:


++−−−

−++−−

−−++−

−−−++

=

4

3

2

1

434241434241

343432313231

242324232121

141312141312

4

3

2

1

V

V

V

V

yyyyyy

yyyyyy

yyyyyy

yyyyyy

I

I

I

I

++−−−

−++−−

−−++−

−−−++

=

434241434241

343432313231

242324232121

141312141312

yyyyyy

yyyyyy

yyyyyy

yyyyyy

Y

Node equations

• We’ll define the elements of Y-bus as:

• So:


=

44434241

34333231

24232221

14131211

YYYY

YYYY

YYYY

YYYY

Y

=

4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

V

V

V

V

YYYY

YYYY

YYYY

YYYY

I

I

I

I

Electric Power

Systems

7

Power-flow solutions

Power-flow Calculation

• For a power system of n nodes, the apparent power injected to the i-th node

is defined by

• We define


+

=

−

i

ik ik ik

ji i i i

ik i k

Y G jB

V V e V

*

* * *

1 1= =

= = =

n n

i i i i ik k i ik kk k

S V I V Y V V Y V


• And we substitute

• Solving the real and imaginary parts we have


* *

1 1

1

( )

(cos sin )( )

= =

=

= + = = −

= + −

ikn n

ji i i i ik k i k ik ik

k k

n

i k ik ik ik ikk

S P jQ V Y V V V e G jB

V V j G jB

1

( cos sin ) =

= + = −n

i i k ik ik ik ik Gi Dik

P V V G B P P

1

( sin cos ) =

= − = −n


Q V V G B Q Q

cos sin = +je j


• In the method based on Newton-Raphson for solving power-flows, the

Newton iterative method is employed to determine the magnitude and phase of

each node voltage.

• We need to solve the power balance equations:


1

1

( cos sin )

( sin cos )

=

=

= + = −

= − = −

n


n


P V V G B P P

Q V V G B Q Q


• We assume that the compensation node (slack) is the first node (the

magnitude and phase are fixed). We need to determine the voltage

magnitude/angle of the rest of nodes:


2 2 2 2

n

2 2 2 2

( )

( )( )

( )

( )

− +

− + = = − +

− +

G D

n Gn Dn

G D

n n Gn Dn

P P P

P P Pf

V Q Q Q

V Q Q Q

x

xx x

x

x


• The power-flow is solved using the process described previously.

• For v=0, an initial guess is made for X, X(v)


( )

( 1) ( ) ( ) 1 ( )

While ( ) Do

( ) ( )

1

End While

+ −

= −

= +

v

v v v v

v v

f x

x x J x f x


• The most challenging stage of the process is determining and inverting the

Jacobian matrix of n by n, J(x)


1 1 1

1 2

2 2 2

1 2

1 2

( ) ( ) ( )

( ) ( ) ( )

( )

( ) ( ) ( )

=

n

n

n n n

n

f f f

x x x

f f f

x x x

f f f

x x x

x x x

x x x

J x

x x x


• The elements of the Jacobian matrix are derived differentiating each function,

fi(x), by each variable.

• For instance, if fi(x) is the equation of the active power:


1

1

( ) ( cos sin )

( )( sin cos )

( )( sin cos ) ( )

=

=

= + − +

= − +

= −

n


ni

i k ik ik ik iki k

k i

ii j ik ik ik ik

j

f x V V G B P P

f xV V G B

f xV V G B j i


• Example:

– For the two node system shown in the figure, solve the power-flow equations via

the Newton-Raphson method to determine the voltage magnitude and angle from

the second node.

– Assume that the first node is the slack node and that SBase = 100 MVA.


Line Z = 0.1j

One Two 1.000 pu 1.000 pu

200 MW

100 MVR

0 MW

0 MVR

2

2

10 10

10 10

− = = −

bus

j j

V j jx Y


• General power balance equations:

• Power balance equations from the second node:


1

1

( cos sin )

( sin cos )

=

=

= + = −

= − = −

n


n


P V V G B P P

Q V V G B Q Q

2 1 2

22 1 2 2

(10sin ) 2.0 0

( 10cos ) (10) 1.0 0

+ =

− + + =

V V

V V V


• Now we calculate the Jacobian matrix:


2 2 2

22 2 2 2

( ) (10sin ) 2.0 0

( ) ( 10cos ) (10) 1.0 0

= + =

= − + + =

P V

Q V V

x

x

2 2

2 2

2 2

2 2

2 2 2

2 2 2 2

( ) ( )

( )( ) ( )

10 cos 10sin

10 sin 10cos 20

=

= − +

P P

V

Q Q

V

V

V V

x x

J xx x


• For v=0, we assume

• We calculate

• Solving


(0) 0

1

=

x

2 2(0)

22 2 2

2 2 2(0)

2 2 2 2

(10sin ) 2.0 2.0( )

1.0( 10cos ) (10) 1.0

10 cos 10sin 10 0( )

10 sin 10cos 20 0 10

+ = =

− + +

= = − +

Vf

V V

V

V V

x

J x

1(1) 0 10 0 2.0 0.2

1 0 10 1.0 0.9

−−

= − =

x


Done!


(1)

2

(1)

1(2)

0.9(10sin( 0.2)) 2.0 0.212( )

0.2790.9( 10cos( 0.2)) 0.9 10 1.0

8.82 1.986( )

1.788 8.199

0.2 8.82 1.986 0.212 0.233

0.9 1.788 8.199 0.279 0.8586

(

−

− + = =

− − + +

− = −

− − − = − = −

f

f

x

J x

x

(2) (3)

(3)

0.0145 0.236)

0.0190 0.8554

0.0000906( )

0.0001175

− = =

=

f

x x

x 2 0.8554 13.52= − V


• Once the voltage in the second node has been calculated, the rest of values

of the power system can be also calculated, e.g. the line currents or the

reactive power of the generator:


Line Z = 0.1j


200 MW

100 MVR

200.0 MW

168.3 MVR

-13.522 Deg

200.0 MW 168.3 MVR

-200.0 MW-100.0 MVR


• This example has two solutions. The second “low voltage” solution can be

found from an initial low voltage value.

• For v=0, we assume

• We calculate


(0) 0

0.25

=

x

2 2(0)

22 2 2

2 2 2(0)

2 2 2 2

(10sin ) 2.0 2( )

0.875( 10cos ) (10) 1.0

10 cos 10sin 2.5 0( )

10 sin 10cos 20 0 5

+ = = −− + +

= = − + −

Vf

V V

V

V V

x

J x


• Iterating…


1(1)

(2) (2) (3)

0 2.5 0 2 0.8

0.25 0 5 0.875 0.075

1.462 1.42 0.921( )

0.534 0.2336 0.220

−−

= − = − −

− − = = =

f

x

x x x

Line Z = 0.1j


200 MW

100 MVR

200.0 MW

831.7 MVR

-49.914 Deg

200.0 MW

831.7 MVR

-200.0 MW

-100.0 MVR


• The figure illustrates the convergence region for the two solutions for different

initial guesses in the proposed example. The red region converges to a “high

voltage” solution and the yellow converges to the “low voltage” solution.



• For the following example we would have:


2 2 2 2

3 3 3 3

2 2 2

( )

( ) ( ) 0

V ( )

− + = = − + =

+

G D

G D

D

P P P

f P P P

Q Q

x

x x x

x

Line Z = 0.1j

Line Z = 0.1j Line Z = 0.1j

One Two 1.000 pu

0.941 pu

200 MW

100 MVR170.0 MW

68.2 MVR

-7.469 Deg

Three 1.000 pu

30 MW

63 MVR

Operation and

control of power

systems

8

Operation and control of power systems

• Conventional power system



• Hierarchical control of conventional ac power systems



• In the electric power system, the consumed active power is “instantaneously”

equal to the generated active power.


losseslinePPP LoadGenE +==


• Example: Hydro generator


σΓ = 𝐽 ∙ 𝛼 → Γ𝑚𝑒𝑐 − Γ𝑒𝑙𝑒𝑐 = 𝐽 ∙ 𝛼

𝑃𝑚𝑒𝑐−𝑃𝑒𝑙𝑒𝑐

Ω= 𝐽 ∙ 𝛼




𝛼 ≈ −𝐾 ∙Δ𝑃𝑒𝑙𝑒𝑐𝐽

< 0

• If the mechanical input power is not changed, the rotating speed (and system frequency) willdecrease indefinitely.

• The rate of change of the frequency (angular acceleration) is directly proportional to thepower variation and inversely proportional to the system inertia.




𝛼 ≈ −𝐾 ∙Δ𝑃𝑒𝑙𝑒𝑐𝐽

< 0

• The frequency primary regulation acts on the prime mover of the generator (in this particularcase, increases the water flow into the turbine) increasing the mechanical input power. Usuallyit is a proportional controller, also known as “droop control”.

• Using this “droop control”, the mechanical input power is modified proportionally to thefrequency variation


• Hierarchical control of conventional ac power systems


Jon Andoni [email protected]

Eneko [email protected]

Goiru, 2. Apartado 23

20500 Arrasate – Mondragon

T. +(34) 678360038

[email protected]

Eskerrik asko

Muchas gracias

Thank you

Documents

Basics on Electric Power System Analysispaginaspersonales.deusto.es/enrique.zuazua/MTM2017... · • Definition of active power, apparent power, reactive power • Use of complex