ieee bus-reg

[keerthanaa.11 Nallasamy]

Hi,
how to create a case format for ieee 37 node test feeder ? or otherwise
if u having it already kindly send it to my mail id.
-- 
Keerthanaa.VN

Re: PV curve using CPF

[nilesh patel]

Sir,If i want to increase load in continuation power flow by step of 1 MW, What 
should be the step size of Lamda. My system base case load is 5000 MW. As CPF 
accuracy depends on step-size.
Thanks.


From: Jose Luis Marin 
Sent: Mon, 10 Aug 2015 18:53:11 
To: MATPOWER discussion forum 
Subject: Re: PV curve using CPF

Shruti is right, the value you obtain for lambda is valid for all the network, 
since voltage collapse is a global phenomenon (in other words, you';ll see a 
nose point at the same value of lambda regardless of which bus you choose to 
plot).  Remember that lambda represents a fraction along the vector of 
injections linearly iterpolating [P_base, Q_base]  to  [P_target, 
Q_target].  The value of Lambda at the nose point is NOT the maximum 
loading point for that bus; rather, it is the maximum loading value along the 
path to the particular load/gen profile chosen as a target.

Of course, one may wonder about this other problem: for a given profile 
[P_base, Q_base], what is the target direction [P_target, Q_target] for which 
one would obtain the shortest value of critical lambda?  If this is what 
you';re thinking about, then it is in general a hard problem.  I suggest 
these references by Ian Dobson, on the concept of "shortest distance" to 
voltage collapse:
http://www.ece.wisc.edu/~dobson/PAPERS/publications.html#loading

-- 
Jose L. Marin
Gridquant España SL
Grupo AIA



On Mon, Aug 10, 2015 at 6:23 AM, nilesh patel  
wrote:

Sir,When we run continuation power flow for particular system, we get p-v curve 
for selected bus. using this p-v curve, we can find Voltage stability Margin 
(in MW) on that bus by difference of operating point to nose point lamda.  
        I agree lambda at nose point provides maximum 
loading value but that is for that bus only for which p-v curve is 
plotted. 
My question is How to find Voltage Stability Margin for whole Network using P-V 
curve ? I mean how to find maximum lamda for whole network using  p-v 
curve?
Thanks.
From: "Abhyankar, Shrirang G." 
Sent: Fri, 07 Aug 2015 22:31:31 
To: MATPOWER discussion forum 
Subject: Re: PV curve using CPF
 


 

I donⴠquite understand your question, can you please elaborate.



The maximum value of loading scaling parameter ᬡmbda⠧ives a measure of how much 
power can be transferred for a given transfer direction. So, lambda is also a 
measure of the nose point for the whole network. 



Shri





From: nilesh patel 

Reply-To: MATPOWER discussion forum 

Date: Friday, August 7, 2015 at 8:46 AM

To: matpower-l , MATPOWER-L 


Subject: PV curve using CPF







Dear Sir,
P-V curve solution using continuation power flow gives nose point (maximum 
loading point) for individual bus.



My question is - How to get nose point for whole network (all buses) using PV 
curve ?  I want to find network voltage stability margin rather than 
individual bus margin using CPF.



Thanks.






Nilesh Patel


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Re: convergence problem in runpf.

[Jose Luis Marin]

Mirish,

I couldn't help notice that you're building this model from scratch (well,
from a database) and you mentioned *"**To make the problem simple I used
all buses as PQ buses except one slack bus"*.   This actually makes it
harder to converge, unless you have *very* accurate data on what the
reactive injections Q (on generator buses) should be.

May I suggest a different, incremental approach:

   1. Start by keeping all generator buses you can as PV, instead of PQ.
   They will help holding up the voltage profile.  After all, a PV node is a
   slack bus in what regards the reactive power injection.
   2. For the loads, start by zeroing out PD (real power demand), but
   keeping QD (reactive demand)
   3. For generators, set the scheduled PG to zero
   4. For lines & transformers, zero out the resistance R
   5. The resulting network will be a "purely reactive power" model. Now
   run a powerflow.  If this doesn't have a feasible powerflow solution, it is
   because some branches have an X parameter that is too large (or
   equivalently, some load QD is too large).  Ramp down the profile of QD
   until you see convergence.
   6. Look at the resulting Q flows across branches, and try to detect
   anomalously large values (i.e. clear outliers). They will help you uncover
   values of X that may be wrong (too large).  Also, keep an eye on negative X
   coming from equivalents such as 3-winding transformers; they may also be
   wrong.
   7. Once you get that working, ramp up the values of PD on loads and PG
   on generators (keeping an eye on the swing's resulting PG, in order to
   redistribute big excesses).
   8. Finally ramp up the resistance on lines.

The whole idea is based on the fact that, for transmission networks (lines
with R<
wrote:

> Dear Mr.Shree,
>
> Thank you very much for your help. As per your suggestion and FAQ I tried
> to find out the problems.
> The results I got-
> 1) Fast-decoupled power flow did not converge in 30 iterations.
> 2) By following   http://www.pserc.cornell.edu/matpower/#pfconvergence  I
> tried to runcpf to get good  initial guess and i got results like
> step   1 : lambda =  0.084, corrector did not converge in 10 iterations.
> Where lambda is < 1 and for reducing steady state loading limitation I
> reduced demand less than 60 % which also failed to converge the power flow.
> 3) Also I tried to run an optimal power flow according to Dr. Ray's
> explanation  given in following link-
>
>
> *https://www.mail-archive.com/search?l=matpower-l@cornell.edu&q=subject:%22Re%5C%3A+%5C%5BMatpower%5C%5D+3500+bus+simulation%22&o=newest
> *
>
> but got the results like-
>
> MATPOWER Version 5.1, 20-Mar-2015 -- AC Optimal Power Flow
> MATLAB Interior Point Solver -- MIPS, Version 1.2, 20-Mar-2015
>  (using built-in linear solver)
>  itobjective   step size   feascond gradcond compcond
> costcond
>    -   
> 
>   0 1200199.7 2.41677 0.71  536.762
>  0
>   1 946197.39 15.531   1.3682  1.75871  525.914
> 0.209885
>   2 954529.91 15.405 0.766107 0.203773  297.341
> 0.00871422
>   3  954849.8 12.849 0.7277120.0545952  258.471
> 0.00033166
>   4 954629.03  13035  0.69114 0.107402  258.048
>  0.000228815
>   5 954614.88  33406 0.692682 0.255673  257.828
>  1.46744e-05
>   6 954525.69  14111 0.579613 0.143897  256.765
>  9.24569e-05
>   7 954539.42  61648 0.581139 0.501345  255.994
>  1.42362e-05
>   8 954518.93  22452 0.573652 0.478609  255.465
>  2.12443e-05
>   9 954494.92 8540.4 0.556318 0.403754  254.653
>  2.48944e-05
>  10 954523.58  20366 0.556265 0.570707  254.104
>  2.97206e-05
>  11 954522.07 6142.4 0.554989 0.647881  256.561
>  1.57288e-06
>  12 954573.42 6192.9 0.513972 0.716706  253.604
>  5.32434e-05
>  13 954575.97 5912.1 0.509457 0.699751  252.612
>  2.64406e-06
>  14 954576.23  16534 0.509454 0.674865  253.278
>  2.64555e-07
>  15 954579.65  12324 0.509394 0.812237  252.966
>  3.54362e-06
>  16 954579.86 7650.3 0.509391  0.80973  252.948
>  2.18359e-07
>  17 954579.87 8185.1 0.509391 0.809591  252.947
>  1.48635e-08
>  18 954579.88 8696.2 0.509391 0.809411  252.945
>  1.31087e-08
>  19  954579.9 9392.5  0.50939  0.80927  252.943
> 1.3818e-08
> Numerically Failed
>
> Did not converge in 19 iterations.
>
> >  Did NOT converge (3.71 seconds)  <
>
> 4) But when I used spy(J) , to look jacobian matrix it gives me some
> strange distribution. H