The Power of Peripheral Blood Smears: Apparent Diagnostic Clues
Basics on Electric Power System Analysispaginaspersonales.deusto.es/enrique.zuazua/MTM2017... ·...
Transcript of Basics on Electric Power System Analysispaginaspersonales.deusto.es/enrique.zuazua/MTM2017... ·...
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:
04.06.18 Mondragon Unibertsitatea 3
“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
04.06.18 Mondragon Unibertsitatea 4
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.
04.06.18 Mondragon Unibertsitatea 6
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.
04.06.18 Mondragon Unibertsitatea 7
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.
04.06.18 Mondragon Unibertsitatea 8
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.
04.06.18 Mondragon Unibertsitatea 9
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).
04.06.18 Mondragon Unibertsitatea 10
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
• Passive elements:
– 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:
04.06.18 Mondragon Unibertsitatea 11
𝑣𝐿 =𝑑𝜙
𝑑𝑡= 𝐿 ∙
𝑑𝑖𝐿𝑑𝑡
𝐸𝐿 =1
2𝐿 ∙ 𝑖𝐿
2
Basic Concepts
• Passive elements:
– 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.
04.06.18 Mondragon Unibertsitatea 12
0 r
SC
l =
Basic Concepts
• Passive elements:
– 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:
04.06.18 Mondragon Unibertsitatea 13
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.
04.06.18 Mondragon Unibertsitatea 14
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.
04.06.18 Mondragon Unibertsitatea 15
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.
04.06.18 Mondragon Unibertsitatea 16
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.
04.06.18 Mondragon Unibertsitatea 17
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:
04.06.18 Mondragon Unibertsitatea 19
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
Single-phase ac system
• 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:
04.06.18 Mondragon Unibertsitatea 20
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)
Single-phase ac system
• 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,
04.06.18 Mondragon Unibertsitatea 21
Single-phase ac system
• 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,
04.06.18 Mondragon Unibertsitatea 22
Single-phase ac system
• 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:
04.06.18 Mondragon Unibertsitatea 23
𝑖𝑅 𝑡 = 𝐼𝑀 ∙ sin(𝜔𝑡)
𝑣𝑅 𝑡 = 𝑖𝑅 𝑡 ∙ 𝑅 = 𝐼𝑀 ∙ 𝑅 ∙ sin 𝜔𝑡 = 𝑉𝑀 ∙ sin 𝜔𝑡
𝑍𝑅 = 𝑅
𝑌𝑅 =1
𝑍𝑅
Single-phase ac system
• 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:
04.06.18 Mondragon Unibertsitatea 24
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
Single-phase ac system
• 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:
04.06.18 Mondragon Unibertsitatea 25
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
Single-phase ac system
• 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:
04.06.18 Mondragon Unibertsitatea 26
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)
Single-phase ac system
• Complex voltage and current, phasor representation
– 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)
04.06.18 Mondragon Unibertsitatea 27
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
Single-phase ac system
• Complex voltage and current, phasor representation
– Current lagging the voltage by 90º (e.g. through an inductor):
– The average power is zero and has an oscillatory component
04.06.18 Mondragon Unibertsitatea 28
https://www.electronics-tutorials.ws/accircuits/power-in-ac-circuits.html
( ) ( ) ( ) ( ) ( )cos 2 90 sin 2p t v t i t VI t VI t = = − + = −
• Complex voltage and current, phasor representation
– Current leading the voltage by 90º (e.g. through a capacitor):
– The average power is zero and has an oscillatory component
Single-phase ac system
04.06.18 Mondragon Unibertsitatea 29
https://www.electronics-tutorials.ws/accircuits/power-in-ac-circuits.html
( ) ( ) ( ) ( ) ( )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:
Single-phase ac system
04.06.18 Mondragon Unibertsitatea 30
( ) ( ) ( ) ( )
( )( ) ( )
( )( ) ( )
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.
04.06.18 Mondragon Unibertsitatea 32
Three-phase systems
• Single-phase system
• Three-phase system
04.06.18 Mondragon Unibertsitatea 33
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
04.06.18 Mondragon Unibertsitatea 34
( ) ( )
( ) ( )
( ) ( )
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
04.06.18 Mondragon Unibertsitatea 35
Three-phase systems
Line voltages of a balanced three-phase power system can be represented as
follows in the phasor diagram
04.06.18 Mondragon Unibertsitatea 36
( ) ( ) ( ) 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
04.06.18 Mondragon Unibertsitatea 37
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
04.06.18 Mondragon Unibertsitatea 38
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
= + +
= + +
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:
04.06.18 Mondragon Unibertsitatea 39
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.
04.06.18 Mondragon Unibertsitatea 41
Single-line representation
• Symbols used to represent the typical components of a power
system.
04.06.18 Mondragon Unibertsitatea 42
Single-line representation
• Symbols used to represent the typical components of a power
system.
04.06.18 Mondragon Unibertsitatea 43
NODE or BUS
(substation)BRANCHES
(lines)
NETWORK
(but unloaded and
unsupplied)
Single-line representation
• Symbols used to represent the typical components of a power
system.
04.06.18 Mondragon Unibertsitatea 44
LOAD: Extracts MW out of the
node (injects negative MW into
the node)
GENERATOR: Injects MW
into the node
NETWORK
(loaded and
supplied)
Single-line representation
• Example of a one-line diagram for a power system consisting of
two generating stations connected by a transmission line
04.06.18 Mondragon Unibertsitatea 45
Single-line representation
• Example of a one-line diagram for a power system consisting of
two generating stations connected by a transmission line
04.06.18 Mondragon Unibertsitatea 46
Single-line representation
• Example of a one-line diagram for a power system consisting of
two generating stations connected by a transmission line
04.06.18 Mondragon Unibertsitatea 47
Single-line representation
• Example of a one-line diagram for a power system consisting of
two generating stations connected by a transmission line
04.06.18 Mondragon Unibertsitatea 48
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.
04.06.18 Mondragon Unibertsitatea 50
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.
04.06.18 Mondragon Unibertsitatea 51
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.
04.06.18 Mondragon Unibertsitatea 52
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.
04.06.18 Mondragon Unibertsitatea 53
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:
04.06.18 Mondragon Unibertsitatea 54
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:
04.06.18 Mondragon Unibertsitatea 55
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:
04.06.18 Mondragon Unibertsitatea 56
b
b
b
bbb
S
V
S
VV
I
VZ
2)(3
3===
b
bb
V
SI
3=
Per-unit quantities
• Example:
04.06.18 Mondragon Unibertsitatea 57
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
Línea de transmisión
Z = j60.5 ohms
Línea de transmisión
X = 60.5 ohms
Per-unit quantities
• Example:1.- Choose an appropriate power base, for example 100MVA
04.06.18 Mondragon Unibertsitatea 58
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
Línea de transmisión
Z = j60.5 ohms
Línea de transmisión
X = 60.5 ohms
Per-unit quantities
• Example:
2.- Define the voltage base for every area of the system (delimited by
transformers)
04.06.18 Mondragon Unibertsitatea 59
Per-unit quantities
• Example:
2.- In this example a base voltage of 110kV is selected for the high-
voltage area.
04.06.18 Mondragon Unibertsitatea 60
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
Línea de transmisión
Z = j60.5 ohms
Línea de transmisión
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.
04.06.18 Mondragon Unibertsitatea 61
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.
04.06.18 Mondragon Unibertsitatea 62
Generador 1
100 MVA
22 kV
X=90%
Transformador
100 MVA
22:110 kV
X=10%
Per-unit quantities
• Example:
3.- Convert all the impedances to p.u.
04.06.18 Mondragon Unibertsitatea 63
Línea de transmisión
Z = j0.8403 pu @ 120
kV y 50 MVA
Carga
datos de operación:
V=110 kV
S=10 MVA
fp = 1
Línea de transmisión
Z = j60.5 ohms
Línea de transmisión
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:
3.- Convert all the impedances to p.u.
04.06.18 Mondragon Unibertsitatea 64
Línea de transmisión
Z = j0.8403 pu @ 120
kV y 50 MVA
Carga
datos de operación:
V=110 kV
S=10 MVA
fp = 1
Línea de transmisión
Z = j60.5 ohms
Línea de transmisión
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:
3.- Convert all the impedances to p.u.
04.06.18 Mondragon Unibertsitatea 65
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:
3.- Convert all the impedances to p.u.
04.06.18 Mondragon Unibertsitatea 66
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.
04.06.18 Mondragon Unibertsitatea 67
+
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.
04.06.18 Mondragon Unibertsitatea 69
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.
04.06.18 Mondragon Unibertsitatea 70
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:
04.06.18 Mondragon Unibertsitatea 71
I1 y12
y14
y34
y23
y13
I2
I4
I3
V1
V4
V2
V3
I14
I13
I12
y13
)( jiijij VVyI −=
Node equations
• Therefore:
04.06.18 Mondragon Unibertsitatea 72
I1 y12
y14
y34
y23
y13
I2
I4
I3
V1
V4
V2
V3
I14
I13
I12
y13
)()()( 4114311321121 VVyVVyVVyI −+−+−=
Node equations
• Therefore:
04.06.18 Mondragon Unibertsitatea 73
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:
04.06.18 Mondragon Unibertsitatea 74
)()()()( 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:
04.06.18 Mondragon Unibertsitatea 75
++−−−
−++−−
−−++−
−−−++
=
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:
04.06.18 Mondragon Unibertsitatea 76
=
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
04.06.18 Mondragon Unibertsitatea 78
+
=
−
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
Power-flow Calculation
• And we substitute
• Solving the real and imaginary parts we have
04.06.18 Mondragon Unibertsitatea 79
* *
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
i i k ik ik ik ik Gi Dik
Q V V G B Q Q
cos sin = +je j
Power-flow Calculation
• 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:
04.06.18 Mondragon Unibertsitatea 80
1
1
( cos sin )
( sin cos )
=
=
= + = −
= − = −
n
i i k ik ik ik ik Gi Dik
n
i i k ik ik ik ik Gi Dik
P V V G B P P
Q V V G B Q Q
Power-flow Calculation
• 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:
04.06.18 Mondragon Unibertsitatea 81
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
Power-flow Calculation
• The power-flow is solved using the process described previously.
• For v=0, an initial guess is made for X, X(v)
04.06.18 Mondragon Unibertsitatea 82
( )
( 1) ( ) ( ) 1 ( )
While ( ) Do
( ) ( )
1
End While
+ −
= −
= +
v
v v v v
v v
f x
x x J x f x
Power-flow Calculation
• The most challenging stage of the process is determining and inverting the
Jacobian matrix of n by n, J(x)
04.06.18 Mondragon Unibertsitatea 83
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
Power-flow Calculation
• 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:
04.06.18 Mondragon Unibertsitatea 84
1
1
( ) ( cos sin )
( )( sin cos )
( )( sin cos ) ( )
=
=
= + − +
= − +
= −
n
i i k ik ik ik ik Gi Dik
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
Power-flow Calculation
• 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.
04.06.18 Mondragon Unibertsitatea 85
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
Power-flow Calculation
• General power balance equations:
• Power balance equations from the second node:
04.06.18 Mondragon Unibertsitatea 86
1
1
( cos sin )
( sin cos )
=
=
= + = −
= − = −
n
i i k ik ik ik ik Gi Dik
n
i i k ik ik ik ik Gi Dik
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
Power-flow Calculation
• Now we calculate the Jacobian matrix:
04.06.18 Mondragon Unibertsitatea 87
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
Power-flow Calculation
• For v=0, we assume
• We calculate
• Solving
04.06.18 Mondragon Unibertsitatea 88
(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
Power-flow Calculation
Done!
04.06.18 Mondragon Unibertsitatea 89
(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
Power-flow Calculation
• 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:
04.06.18 Mondragon Unibertsitatea 90
Line Z = 0.1j
One Two 1.000 pu 0.855 pu
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
Power-flow Calculation
• 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
04.06.18 Mondragon Unibertsitatea 91
(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
Power-flow Calculation
• Iterating…
04.06.18 Mondragon Unibertsitatea 92
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
One Two 1.000 pu 0.261 pu
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
Power-flow Calculation
• 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.
04.06.18 Mondragon Unibertsitatea 93
Power-flow Calculation
• For the following example we would have:
04.06.18 Mondragon Unibertsitatea 94
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
04.06.18 Mondragon Unibertsitatea 96
Operation and control of power systems
• Hierarchical control of conventional ac power systems
04.06.18 Mondragon Unibertsitatea 97
Operation and control of power systems
• In the electric power system, the consumed active power is “instantaneously”
equal to the generated active power.
04.06.18 Mondragon Unibertsitatea 98
losseslinePPP LoadGenE +==
Operation and control of power systems
• Example: Hydro generator
04.06.18 Mondragon Unibertsitatea 99
σΓ = 𝐽 ∙ 𝛼 → Γ𝑚𝑒𝑐 − Γ𝑒𝑙𝑒𝑐 = 𝐽 ∙ 𝛼
𝑃𝑚𝑒𝑐−𝑃𝑒𝑙𝑒𝑐
Ω= 𝐽 ∙ 𝛼
Operation and control of power systems
• Example: Hydro generator
04.06.18 Mondragon Unibertsitatea 100
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.
Operation and control of power systems
• Example: Hydro generator
04.06.18 Mondragon Unibertsitatea 101
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
Operation and control of power systems
• Hierarchical control of conventional ac power systems
04.06.18 Mondragon Unibertsitatea 102
Jon Andoni [email protected]
Eneko [email protected]
Goiru, 2. Apartado 23
20500 Arrasate – Mondragon
T. +(34) 678360038
Eskerrik asko
Muchas gracias
Thank you