LOADFLOW ANALYSIS LOG

>> 2003/10/8 <<
Opened log.

In this log, I intend to document my attempt to get the hang of power 
system loadflow analysis.

Starting tomorrow.....


>> 2003/10/9 <<
The Story So Far ....

Loadflow analysis was the only part of my postgraduate diploma which I 
actualy failed.  This has bugged me for a long time, and I decided to 
tackle it again, in the hopes of getting a job off the back of it, as 
much as anything else.

Given a power grid of n nodes, we may apply Kirtchoff's current law to 
give us equations of the form:

 [1]  Y12 ( V2 - V1 ) + Y13 ( V3 - V1 ) + Y14 ( V4 - V1 ) = I1
 [2]  Y21 ( V1 - V2 ) + Y23 ( V3 - V2 ) + Y24 ( V4 - V2 ) = I2
 [3]  Y32 ( V2 - V3 ) + Y31 ( V1 - V3 ) + Y34 ( V4 - V3 ) = I3
 [4]  Y42 ( V4 - V2 ) + Y43 ( V4 - V3 ) + Y41 ( V1 - V4 ) = I4
  .
  .
  .

Each of these is a string of three terms for currents coming into the 
node being equated to an extra-network current (say from a power station 
or to a load).  Note that any node with a shunt will have an extra term 
of the form Ys * Vn on the left hand side.  This is identical to the 
terms for between nodes, but obviously the far end voltage is zero.

If we multiply these out and then take out the voltages as common 
factors (and multiplying both sides by -1 for some reason), we obtain:

 [1] V1(Y12+Y13+Y14)-V2*Y12         -V3*Y13         -V4*Y14 = -I1
 [2] -V1*Y12        +V2(Y12+Y23+Y24)-V3*Y23         -V4*Y24 = -I2
 [3] -V1*Y13        -V2*Y23         +V3(Y13+Y23+Y34)-V4*Y34 = -I3
 [4] -V1*Y14        -V2*Y24         -V3*Y34         +V4(Y14+Y24+Y34)= -I4
  .
  .

(Note that Y12=Y21)

This can be put in matrix form:

    V [Ybus] = I

where:

     [ V1 ]
 V = [ V2 ]
     [ V3 ]
     [ V4 ]

        [ Y12+Y13+Y14    -Y12        -Y13        -Y14     ]
 Ybus = [    -Y12     Y12+Y23+Y34    -Y23        -Y24     ]
        [    -Y13        -Y23     Y13+Y23+Y34    -Y34     ]
        [    -Y14        -Y24        -Y34     Y14+Y24+Y34 ]

     [ -I1 ]
 I = [ -I2 ]
     [ -I3 ]
     [ -I4 ]

In real life, there is no need for this much rigmarole.  V & I are 
fairly easy to remember.  Ybus is actualy not much more difficult.  The 
terms on the leading diagonal in (column n, row n) are merely the sum of all 
admittances connected to node n (including any shunts).  The off 
diagonal terms for (column x, row y) are merely the negative of the 
admittance between nodes x & y.  The sharp eyed will have already 
spotted that Ybus is symetrical about the leading diagonal.

Loads could be represented by shunts across the appropreate nodes.  
Unfortunatly, loads on power networks are generaly given as so-many 
volt-ampares at such-and-such a power factor or in watts and 
volt-ampares reactive or whatever.

Given that P=IVcos(phi) and Q=IVsin(phi), where cos(phi) is the power 
factor, we can say that:
I=P/cos(phi)=Q/sin(phi)

Alternators should just be loads which inject current rather than 
removing it.


>> 2003/10/13 <<
It should be borne in mind that these calculations would be made in /per 
unit/ - that is to say that they would be normalised to a system voltage 
of 1 and a system aparent power of 1.  Thus, we need not worry 
excessivly about what size of values to put in.  Sensible voltages are 
0.9 to 1.1 per unit, which is nominal voltage +- 10% (common practice on 
the British national grid).

QUESTIONS:
  Is Ybus a Jacobian of anything?
  What (if any) is the significance of Ybus' eigenvectors and 
    eigenvalues?
  What significance can be attached to grad(), div() and curl()?


CALCULATING WHAT NODE VOLTAGES GIVE SENSIBLE LINE CURRENTS
Gauss-Jordan elimination
Newton-Raphson itteration
Gauss-Seidel itteration
Fast Decoupled itteration



GAUSSIAN ELIMINATION
We may apply Gaussian elimination to Ybus augmented with I to solve for V.  
This is the same Gaussian elimination as described by Lin-Huan(?)'s "Chiu 
ChangSuan Shu" in about 200BC and taught in secondary school.  We merely 
have the complication of a lot more lines and complex numbers in the matrix.

It is of note that, since power is dependant on voltage as well as 
current, nodes bearing loads will be constrained to have certain voltages.  
Generating nodes are not so constrained, as their currents are variables.

Thus, although theoreticaly we are solving for V, we actualy already know 
some of the elements of V and are unsure of some of the elements of I!  
The use of the augmented matrix should avoid confusion.

Once this is done, the line currents can be checked by direct calculation.  
Quite what to do if a line current is too high, I'm not sure.


NEWTON-RAPHSON ITTERATION
For a simple y=f(x), where f'(x) is the derivative of f(x):

a is a guess at where a root of the equation lies.
a' is a better approximation

  a' = a - f(a)/f'(a)