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Lesson 1: Phasors and Complex Arithmetic ET 332b Ac Motors, Generators and Power Systems

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Learning Objectives

After this presentation you will be able to:

 Write time equations that represent sinusoidal voltages and currents found in power systems.  Explain the difference between peak and RMS electrical quantities.  Write phasor representations of sinusoidal time equations.  Perform calculations using both polar and rectangular forms of complex numbers.

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Ac Analysis Techniques

Time function representation of ac signals

Time functions give representation of sign instantaneous values

v(t)  Vmax sin(t  v ) Voltage drops

e(t)  Emax sin(t  e ) Source voltages

Currents i(t)  Imax sin(t  i )

Where Vmax = maximum ( peak) value of voltage Emax = maximum (peak) value of source voltage Imax = maximum (peak) value of current v, e, i = phase shift of voltage or current  = frequency in rad/sec Note: =2pf

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Ac Signal Representations

Ac power system calculations use effective values of time waveforms (RMS values) Therefore: V E I V  max E  max I  max RMS 2 RMS 2 RMS 2 1 Where  0.707 2

So RMS quantities can be expressed as:

VRMS  0.707Vmax

ERMS  0.707Emax

IRMS  0.707Imax

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Ac Signal Representations

Ac power systems calculations use phasors to represent time functions

Phasor use complex numbers to represent the important information from the time functions (magnitude and phase angle) in vector form.

Phasor Notation  I  I  V  VRMS RMS or or  I  I  V  VRMS RMS

Where: VRMS, IRMS = RMS magnitude of voltages and currents = phase shift in degrees for voltages and currents

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Ac Signal Representations

Time to phasor conversion examples, Note all signal must be the same frequency

Time function-voltage Find RMS magnitude  v(t) 170sin(377t 30 ) VRMS  0.707170 120.2 V

Phasor V 120.230 V Time function-current Find RMS magnitude

 i(t)  25sin(377t  20 ) IRMS  0.70725 17.7 A

Phasor I 17.7 20 A

Phase shift can be given in either radians or degrees. To convert, this conversion: use 1 degree = p/180 radians.

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Complex Number Representations

Polar to rectangular conversion Rectangular form of complex number

+j a  rcos() z  a  jb

r  b  rsin() -real +real

To convert between polar form and rectangular form, use calculator P-R and R-P keys or remember concepts from -j trigonometry

R-P conversion P-R conversion Magnitude: z  a 2  b2 z  r(cos()  jsin())

 b  Phase Angle:   tan1   a  lesson1_et332b.pptx 7

Complex Number Representations

Example 1-1 Rectangular-to-polar (R-P) conversion using trigonometry. Find the polar equivalent of the complex number z.

z  30  j20

Solution a  30 b  20 Magnitude z  a 2  b2  302  202  36.1

 b  Phase angle   tan1   a   20    tan1   33.7  30 

Watch for the location of the phasor when making P-R or R-P conversions. Some calculators return tan-1 into 1st and 4th quadrants only. lesson1_et332b.pptx 8

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Rectangular-to-Polar Conversion

Check the angle given from R-P conversion by geometric interpretation.

If real and imaginary parts both negative,

angle is in Quadrant III

Example 1-2: Find the polar form of z=-4+j5 Find r z  (4)2  52  6.4

This tan-1 function computes the Compute 1 5   tan    51.3 angle in Quadrant IV. The actual phase angle   4  angle is 180 degrees from this value ( - real, + imaginary is Quadrant II) lesson1_et332b.pptx 9

Polar-to-Rectangular Using Trig Functions

Conversion Equation z  a  jb  r[cos()  jsin()]

Example 1-3: Convert I  5053.1 to rectangular form

a  jb  r [cos()  jsin()] a  jb  50[cos(53.1 )  jsin(53.1 )]

a  jb  30  j40 Ans

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Complex Number Arithmetic

Properties of the imaginary operator j=√-1

+j The operator j translates physically into a 90o phase shift

j = 90o ... a 90 degree phase lead -j = -90o ... a 90 degree phase lag

90˚ lead

-j Also 1/j = -j and 1/-j = j 90˚ lag j (-j) = 1

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Complex Number Arithmetic

Complex Conjugate-reflection about the real axis

+j Conjugate (rectangular form) * z (a  jb)  a  jb Change sign on imaginary part b  Conjugate (polar form)  (z  )*  z  -b Change sign on angle z* -j

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Complex Number Arithmetic Addition and Subtraction of Complex Numbers For calculators without complex number arithmetic 1) convert both numbers to rectangular form 2) add/subtract real parts of both numbers and imaginary parts of both numbers Multiplication and Division of Complex Numbers For calculators without complex number arithmetic 1) convert both numbers to polar form 2) multiply/divide magnitudes 3) add angles for multiplication, subtract angles for division Inverting a Complex Number

1) convert number to polar form z   2) perform division (10o) / (z ) = 1/z  - lesson1_et332b.pptx 13

Complex Number Arithmetic

Example 1-4: Given the sinusoidal time functions and complex numbers below:  v1(t)  340sin(377 t 10 )  v2 (t)  277sin(377 t  30 ) Z  70  j20 I  3 j2

Find V2+V1, V2-V1, V1/Z, I(Z) give the results in polar form for all calculations

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Example 1-4 Solution (1)

Convert v1(t) and v2(t) into phasors

340 277 V   240.4 V  195.9 Find magnitudes 1 2 2 2

  V1  240.4 10 V2 195.9 -30

Find V1+V2 Convert phasors to rectangular form

  V1  240.4[cos( 10 )  jsin( 10 )]

V1  236.75  j41.745

  V2 195.9[cos( -30 )  jsin( -30 )]

V2 169.65  j97.95

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Example 1-4 Solution (2)

Add real and imaginary parts Vs  V1  V2

Vs  (236.75 169.65)  j(41.745  (97.95))

Vs  406.4  j56.2

Convert to polar form

2 2 Vs  406.4  (56.2)  410.3

1  56.2   s  tan    7.87  406.4 

 Vs  410.3 7.87 Ans

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Example 1-4 Solution (3)

Find V1-V2 subtract real and imaginary parts Vd  V1  V2

Vd  (236.75 169.65)  j(41.745  (97.95))

Vd  67.1 j139.7

Convert to polar form

Vd  67.1 j139.7 2 2 Vd  67.1 139.7 154.9

1139.7   d  tan    64.3  67.1   Vd 154.9 64.3 Ans

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Example 1-4 Solution (4)

Compute the quantity V1/Z and give the results in polar form Z  70  j20 Convert Z to polar form

Z  702  202  72.8

1 20   Z  tan   15.95  70 

To compute the quotient, divide magnitudes and subtract phase angles

V 240.410 1  Z 72.8 15.95

V1  240.4      10 15.95  3.3  5.95 Ans Z  72.8 

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Example 1-4 Solution (5)

Compute the quantity I(Z) and give the results in polar form

Convert I to polar form I  3 j2

I  32  22  3.61

1  2   I  tan    33.7  3 

Multiply magnitudes and add phase angles to get result

I Z  (3.61 33.7 )(72.815.95 ) I Z  3.6172.8 33.7 15.95 I Z  262.8 17.75 Ans

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ET 332b Ac Motors, Generators and Power Systems

END LESSON 1: PHASORS AND COMPLEX ARITHMETIC

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