2009-08-26 Introduction to Power System Analysis Part II

# 2009-08-26 Introduction to Power System Analysis Part II -...

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Introduction to Power System Analysis Part II Jovan Ilić ECE 18-418 Fall 2009

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Phasors       ( ) 2 | | cos 2 | v t V t i t V t     cos sin j ej   Using Euler form:       Re 2 | | 2 | | | | 0 j t j t v t V e Ve where V V V        2 | | 2 | | | | j t j t j i t I e Ie where I I e I 
Rotating phasors idea: x axis =   Re 2 | | jt Ve y axis =   Im 2 | | Mag = 22 xy arctan y x    Angle =

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Lumped Parameter Elements
Lumped Parameter Time Characteristics [1]

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        ' ( ) 2 | | cos Re 2 | | jt v t V t V e v t     ' ' 22 j t j t vt V i t e Ie RR     V I R Resistor current in phasor domain Capacitor current in phasor domain         ' || 2 cos V i t i t t R    ' ' 2 dv i t C j CVe so I j CV dt          ' Re 2 | | cos 2 i t i t C V t    Inductor current in phasor domain   '' 1 j t j t VV i t v dt e Ie I L j L j L       ' Re 2 cos 2 V i t i t t L
1 1 1 jC Z R j L U I Z 1 Z R j L U I Z    ( ) 2 | | cos( ) v t V t   || V ( ) 2 | | cos( ) i t I t I  

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Calculations in Phasor Domain     | | Re Im V V V j V       22 | | Im V V V      1 Im tan V V        | | cos VV     | |sin       12 * 1 2 1 * | | | || | | | V V a c j b d V a j V a jb V c jd V V I b VI V I I II    
Phasors in Complex Plane V1 V2 V1+V2 -V1 V2-V1 conj(V2)

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3-Phase system in phasors domain A balanced 3-phase system can always be solved as a single-phase equivalent in phasors domain.
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## 2009-08-26 Introduction to Power System Analysis Part II -...

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