Dear Shri,
Thank you for your email. I tried this method, however it says that it does not
converge.
I am also looking to plot the voltage against the reactive power load. Is it
possible to do this using the continuation power flow method?
Thank you
Hiranya
On 24 May 2015, at 23:28, Abhyankar, Shrirang G. <abhy...@anl.gov> wrote:
> Hi,
> You might want to try MATPOWER’s continuation power flow feature instead of
> rolling out your own load increment loop. The reason is that as you increase
> the load the Jacobian matrix conditioning worsens and running a simple power
> flow may yield a divergent solution. For your case, I would run the
> continuation power flow as follows:
>
> mpcbase = loadcase(‘case95’);
> mpctarget = mpcbase;
> mpctarget.bus(5,[PD,QD]) = mpctarget.bus(5,[PD,QD])*3.0;
>
> outresults = runcpf(mpcbase,mpctarget);
>
> The results of the continuation power flow can be found in the cpf field of
> the outresults struct.
>
> You can further customize continuation power flow based on your needs (for
> e.g. vary the step-size, add custom monitors, stop at a given loading, and
> others). Read the MATPOWER manual to know more.
>
> Shri
>
>
> From: Hiranya Seneviratne <hiranya.seneviratn...@gmail.com>
> Reply-To: MATPOWER discussion forum <matpowe...@list.cornell.edu>
> Date: Saturday, May 23, 2015 at 7:57 AM
> To: MATPOWER discussion forum <matpowe...@list.cornell.edu>
> Subject: Fwd: increasing Reactive power load by a load factor
>
>
>>
>> Dear Ray,
>>
>> I am trying to implement a code in matpower, that increases the reactive
>> power load by a factor and then see what happens to the voltage.
>> I have written this code, however the value of y and q don’t seem to change
>> appropriately. will you be please able to let me know how i can implement
>> this?
>>
>>
>>
>> clear all
>> close all
>> define_constants;
>> results = loadcase('case95');
>>
>> for i = 1:10
>> define_constants;
>>
>> y = results.bus(5,[PD QD]);
>> results.bus(5,[PD QD]) = y + 0.2*y;
>>
>> x = runpf(results);
>>
>> v(i) = x.bus(5,VM)
>> q(i) = x.bus(5,QD)
>> p(i) = x.bus(5,PD)
>>
>>
>>
>>
>> end
>>
>
