Circuit Analysis for Power Transmission Network Question

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  • Thread starter Constantinos
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  • #1
Hey!

Below you will find a piece of text I found on a paper. There are some things I don't get, perhaps you could help me.

What it means for the network to be lossless?
Also, why is the power flow equal to Re[Vi(t) Ii,k(t)*] and not outright Vi(t)*Ii,k(t) ? Why the complex conjugate?

I get the math, but not the definitions really!

attachment.php?attachmentid=46400&stc=1&d=1334832268.jpg

Homework Statement





Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
gneill
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A lossless network contains no resistances, so no real energy can be lost by that network (remember that reactances only store and release energy without loss).

The complex conjugate of the current is used in order to get the phosor angle correct for the power. It has to do with how phasors are multiplied.

Suppose:
##V = |V| \angle \phi_v##
##I = |I| \angle \phi_i ##

Multiplying:
##V I = |V||I| \angle \phi_v + \phi_i##

However, for the power we should have the angle corresponding to the power factor:

##\phi_v - \phi_i = \phi~~~~##

So we need to invert the sign of the current angle, hence the complex conjugate is taken.
 
  • #3
A lossless network contains no resistances, so no real energy can be lost by that network (remember that reactances only store and release energy without loss).
Ah yes I see, having susceptance means it also has reactance (I forgot about that).

The complex conjugate of the current is used in order to get the phosor angle correct for the power. It has to do with how phasors are multiplied.

Suppose:
##V = |V| \angle \phi_v##
##I = |I| \angle \phi_i ##

Multiplying:
##V I = |V||I| \angle \phi_v + \phi_i##

However, for the power we should have the angle corresponding to the power factor:

##\phi_v - \phi_i = \phi~~~~##

So we need to invert the sign of the current angle, hence the complex conjugate is taken.
I'm not sure I understand. Why would we need the angle of the power factor and not the angle of power itself? By power factor I assume you mean this correct?
 
  • #4
gneill
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20,874
2,837
I'm not sure I understand. Why would we need the angle of the power factor and not the angle of power itself? By power factor I assume you mean this correct?
If you expand VI* then:

##V I^* = |V||I| \angle \phi = |V||I| cos(\phi) + j|V||I| sin(\phi)##

But this is just the sum of the real and imaginary power P + jQ.

##VI^* = P + jQ##
 
  • #5
Yes I think I get it now, thanks for the clarifications!
 

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