refresher course on electrical fundamentals 090310

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  • 7/29/2019 Refresher Course on Electrical Fundamentals 090310

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    Refresher course on Electrical fundamentals

    (Basics of A.C. Circuits)

    by B.M.Vyas

    A specifically designed programme for

    Da Afghanistan Breshna Sherkat (DABS)Afghanistan

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    2

    Areas Covered Under this Module

    1. Sinusoidal AC waveform

    2. Instantaneous current and Voltage

    3. Series RL circuit

    4. Types of power and energy in ac circuits, powertriangle

    5. Vector diagrams as a analysis tool

    6. Power flow principles

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    Components of A.C. Circuits

    Generation, transmission and distribution, all deal withsinusoidal voltages and currents

    All loads may be represented as series/ parallel combination

    ofresistance, inductance and capacitance. We want to obtain the steady state performance of the circuit.

    Simplest method of calculation is to use notations ofPhasorAlgebra.

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    VOLTAGE / CURRENT SINE WAVE

    The equation of instantaneous value isv = Vm sinv =Vm sint

    v = instantaneous voltageVm = maximum /peak voltage

    = angle in degrees or radians= 2 f

    What about ac sine current waveform?

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    Phase Shift

    Red is Leading

    Red is lagging

    This gap givesthe amount ofphase shift

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    How much is Phase Shift of B with respect to A ?

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    The RMS value of an alternating current is given by the steady (DC)Current which when flowing through a given circuit for a given timeProduce the same heat as produced by the alternating current whenFlowing through same circuit for the same duration.

    RMS current

    I = Im / 2

    PEAK current

    Im

    RMS CURRENT

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    AVERAGE CURRENT

    The average value of an AC is expressed by that steady current (DC)

    Which transfers across any circuit, the same charge as is transferred byThat of AC

    Avg. currentIav= 2Im /

    PEAK currentIm

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    Scalar

    A scalar is a quantity with magnitude only. Ithas no direction. Example of scalar quantities are

    Length Area

    Volume Speed Mass Energy

    Work Power Temperature Pressure

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    Vector or phasorA vector

    quantity hasboth

    magnitude

    and direction.

    E.g.

    Displacement

    , Velocity,

    force,acceleration

    etc.

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    Addition & Subtraction of two Vectors

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    A

    B

    A-B

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    PURE RESISTIVE CIRCUIT

    AC Power Waveform

    v = Vm sint

    i = Im sint

    Power P= V I

    Both V and I areRMS values

    Power is positive inboth half cycles

    VI

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    INDUCTIVE CIRCUIT

    AC Power Waveform

    v = Vm sint

    i = Im sin (t /2)

    Power P= vi

    Vm Im= - ---------- sin 2t2

    The shaded areas in this diagramshow how poweris absorbed and

    returned to the circuit.

    The shaded areas above the baseline( + levels) represent power that isabsorbed by the inductor.

    The shaded areas below the baseline(- levels) representpower that isreturned to the circuit.

    P (over the cycle) = 0

    V

    I

    900

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    CAPACITIVE CIRCUIT

    AC Power Waveform

    v = Vm sint

    i = Im sin (t +/2)

    Power P= vi

    Vm Im= ---------- sin 2t2

    The shaded areas in this diagram showhow poweris absorbed and returned to

    the circuit.

    The shaded areas above the baseline( + levels) represent power that isabsorbed by the inductor.

    The shaded areas below the baseline(- levels) representpower that isreturned to the circuit.

    P = 0

    V

    I

    900

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    Expression for a sinusoidal voltagesource:

    For a linearcircuit, in thesteady state, the current is alsosinusoidal and can beexpressed as:

    R-L circuit

    i = Im sin (t - )

    v = Vm sint

    V

    I

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    Is the phase difference.

    += sincos IjII

    Impedance Triangle

    LjRZ +=

    222 LRZ +=

    R

    L 1tan=

    R

    Z

    XL

    RIZIVIPav22

    coscos ===

    Thus, all the power consumedin the R-L circuit is actually thepower dissipated in theresistance

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    Phase Relationship:

    pure resistive element: Impedance= 0R

    The voltage and currentphasors are in phase.

    pure Inductive element:Impedance=

    90L

    the current lags the voltage by 90o

    pure Capacitive element:Impedance=

    901

    C

    the current leads the voltage by 90o

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    Power Factor

    The power factorin the circuit is a factor by which V.I(Apparent Power) should be multiplied to get the Active(or real) power

    PowerApparent

    PowerActive

    S

    PCos ==

    By convention, it is assumed, that if the circuit is inductive,the reactive power is positive,and for capacitive circuit, the reactive power is negative.

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    Three Types of Power

    V

    I

    I cos

    ISin

    P - Active

    QR

    eactive

    Power Triangle

    Type Formula UnitApparent VI volt-ampere or VA

    Active VI Cos watt or W

    Reactive VI Sin reactive volt-ampereor VAR

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    The reactive power is the power thattravels back and forth between thesupply and the circuit due to thepresence of reactive elements(inductance and capacitance).

    Over a cycle, its average is zero.By convention, it is assumed, that ifthe circuit is

    inductive, the reactive power ispositive, and for

    capacitive circuit, the reactivepower is negative.

    Reactive Power

    AC Power Waveform for

    pure capacitance

    v = Vm sint

    i = Im sin (t +/2)

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    Power Flow Quadrants: IEC62053-23

    Export Reactive Power

    V

    Import Active Power

    Import Reactive Power

    Export Active Power

    Q1PF: lagging /

    Inductive

    Q2PF: Leading /

    capacitive

    Q3PF: lagging

    /Inductive

    Q4PF: Leading

    /capacitive

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    Power Flow

    Power System Power System

    Electricity

    Interconnected system of generation and load centers

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    Measurement Reference

    Power flow to Voltage bus is taken positive (import) Current transformer connection polarity (p2 to bus side)

    Utility A

    A B

    Utility B

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    EHV BUS

    P1 P2

    HV BUS

    P2 P1

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    Power Components

    Quadrature component

    In phase component

    V ( reference)

    I

    00

    900

    1800

    2700

    I Sin

    I Cos

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    Three phase system

    1200

    1200

    1200

    1

    23

    12

    23

    31

    1,2,3 phase voltages

    12,23,31 Line voltages

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    Three phase system

    1200

    1200

    1200

    1

    23

    23

    31

    1,2,3 phase voltages

    12,23,31 Line voltages

    How to getphase voltages byPhasor additions?

    V12= V1-V2=V1+(-V2) 12

    23

    31

    30

    30

    30

    12

    -V2

    Using Law ofParallelogram

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    Three-phase Y configurations

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    Three-phase Y and configurations

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    Phasor representation of a 3 phase load

    Er,Ey, Eb voltages.

    Ir,Iy,Ib currents.

    1,2,3 phaseangles (lagging)

    er

    eyeb

    ir

    iy

    ib

    1

    23

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    Thank you