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Lesson 1: Representation of Ac Voltages and Currents1 1/22/2016 Lesson 1: Phasors and Complex Arithmetic ET 332b Ac Motors, Generators and Power Systems lesson1_et332b.pptx 1 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. lesson1_et332b.pptx 2 1 1/22/2016 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 lesson1_et332b.pptx 3 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 0.707Vmax ERMS 0.707Emax IRMS 0.707Imax lesson1_et332b.pptx 4 2 1/22/2016 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 lesson1_et332b.pptx 5 Ac Signal Representations Time to phasor conversion examples, Note all signal must be the same frequency Time function-voltage Find RMS magnitude v(t) 170sin(377t 30 ) VRMS 0.707170 120.2 V Phasor V 120.230 V Time function-current Find RMS magnitude i(t) 25sin(377t 20 ) IRMS 0.70725 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. lesson1_et332b.pptx 6 3 1/22/2016 Complex Number Representations Polar to rectangular conversion Rectangular form of complex number +j a rcos() z a jb r b rsin() -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: tan1 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 j20 Solution a 30 b 20 Magnitude z a 2 b2 302 202 36.1 b Phase angle tan1 a 20 tan1 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 4 1/22/2016 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 jb r[cos() jsin()] Example 1-3: Convert I 5053.1 to rectangular form a jb r [cos() jsin()] a jb 50[cos(53.1 ) jsin(53.1 )] a jb 30 j40 Ans lesson1_et332b.pptx 10 5 1/22/2016 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 lesson1_et332b.pptx 11 Complex Number Arithmetic Complex Conjugate-reflection about the real axis +j Conjugate (rectangular form) * z (a jb) a jb Change sign on imaginary part b Conjugate (polar form) (z )* z -b Change sign on angle z* -j lesson1_et332b.pptx 12 6 1/22/2016 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 (10o) / (z ) = 1/z - lesson1_et332b.pptx 13 Complex Number Arithmetic Example 1-4: Given the sinusoidal time functions and complex numbers below: v1(t) 340sin(377 t 10 ) v2 (t) 277sin(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 lesson1_et332b.pptx 14 7 1/22/2016 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 ) jsin( 10 )] V1 236.75 j41.745 V2 195.9[cos( -30 ) jsin( -30 )] V2 169.65 j97.95 lesson1_et332b.pptx 15 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 lesson1_et332b.pptx 16 8 1/22/2016 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 1139.7 d tan 64.3 67.1 Vd 154.9 64.3 Ans lesson1_et332b.pptx 17 Example 1-4 Solution (4) Compute the quantity V1/Z and give the results in polar form Z 70 j20 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 240.410 1 Z 72.8 15.95 V1 240.4 10 15.95 3.3 5.95 Ans Z 72.8 lesson1_et332b.pptx 18 9 1/22/2016 Example 1-4 Solution (5) Compute the quantity I(Z) and give the results in polar form Convert I to polar form I 3 j2 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 )(72.815.95 ) I Z 3.6172.8 33.7 15.95 I Z 262.8 17.75 Ans lesson1_et332b.pptx 19 ET 332b Ac Motors, Generators and Power Systems END LESSON 1: PHASORS AND COMPLEX ARITHMETIC lesson1_et332b.pptx 20 10 .
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