FYI, please start a new thread for a new question, as opposed to just replying
to an existing thread.
If they are considered as fixed P and Q they can be added to any existing load
in the bus matrix or they can be added as a negative generator. If you add it
as a shunt (in the bus matrix) it will be treated as a constant impedance
element.
Ray
On Nov 10, 2023, at 12:30 PM, Muhammad Junaid <[email protected]> wrote:
Hi Sir. I hope you are doing well. If EVs are to be added as loads on buses
that ave P and Q. Should the load be added in bus data or should I be added as
shunt element or can it be used as negative generator?
From: Ray Daniel Zimmerman
Sent: Wednesday, 8 November 2023 9:14 PM
To: MATPOWER-L
Reply To: MATPOWER discussion forum
Subject: Re: Finding the maximum positive (and negative) power injection at a
bus
See Section 6.4.2 on Dispatchable Loads in the MATPOWER User’s
Manual<https://nam12.safelinks.protection.outlook.com/?url=https%3A%2F%2Fmatpower.org%2Fdocs%2FMATPOWER-manual-7.1.pdf&data=05%7C01%7CMATPOWER-L%40list.cornell.edu%7C4dcf7d1317f04ea3925008dbe2527c97%7C5d7e43661b9b45cf8e79b14b27df46e1%7C0%7C0%7C638352616118627603%7CUnknown%7CTWFpbGZsb3d8eyJWIjoiMC4wLjAwMDAiLCJQIjoiV2luMzIiLCJBTiI6Ik1haWwiLCJXVCI6Mn0%3D%7C3000%7C%7C%7C&sdata=rciuQXBVdWWY3iOIYtl%2BkqZuDAboCd5etlOBASsrmHk%3D&reserved=0>.
A dispatchable load is exactly what you are looking for. It’s a negative
generator with a constant power factor constraint. Any generator with PMIN < 0
and PMAX == 0 will be treated as a dispatchable load and the constant power
factor constraint will be added automatically.
Ray
On Nov 5, 2023, at 6:12 PM, Ronald Cabaoig <[email protected]> wrote:
Dear MATPOWER team,
In connection with placing a dummy generator to maximize/minimize the power
injection at the bus of interest, is it also possible in MATPOWER to impose a
power factor constraint on the said dummy generator?
Examples:
- Scenario 1: minimize the negative power injection in bus X, with a power
factor constraint of 0.80 lagging pf.
- Scenario 2: minimize the negative power injection in bus X, with a power
factor constraint of 0.80 leading pf.
Cheers,
Ronald
On Sat, Oct 14, 2023 at 6:29 AM Ronald Cabaoig
<[email protected]<mailto:[email protected]>> wrote:
Thank you very much!
Cheers,
Ronald
On Sat, Oct 14, 2023 at 1:33 AM Ray Daniel Zimmerman
<[email protected]<mailto:[email protected]>> wrote:
That’s correct. Except, if you are zeroing out all other costs, there is no
need to use such a huge magnitude for the cost. Any reasonable non-zero number
would do (like 10 or 100). When the problem includes coefficients with huge
ranges it can potentially result in numerical issues.
Ray
On Oct 12, 2023, at 6:26 PM, Ronald Cabaoig
<[email protected]<mailto:[email protected]>> wrote:
Hi sir,
Thanks for the suggestion!
Let me confirm my understanding:
1. Create a dummy generator with the following limits: -99999 MW <= Pdummy <=
99999 MW
2. To minimize the injection, set a very large positive cost coefficient (C =
99999Pdummy) and zero out all other costs, turning the objective function to:
min (99999Pdummy + 0Pg1 + 0Pg2 + ...)
3. To maximize the injection, set a very large negative cost coefficient (C =
-99999Pdummy) and zero out all other costs, turning the objective function to:
min (-99999Pdummy + 0Pg1 + 0Pg2 + ...)
4. Retain all other information
Thanks in advance!
Cheers,
Ronald
On Tue, Oct 10, 2023 at 5:23 AM Ray Daniel Zimmerman
<[email protected]<mailto:[email protected]>> wrote:
You can do this by placing a dummy generator at the bus of interest and using a
very large negative (to maximize the injection) or positive (to minimize the
injection) generation cost. You can either zero out all other costs, or simply
make the magnitude of the cost large enough to dominate all other costs.
Ray
> On Oct 5, 2023, at 8:48 PM, Ronald Cabaoig
> <[email protected]<mailto:[email protected]>> wrote:
>
> Dear MATPOWER team,
>
> Greetings!
>
> In the current MATPOWER version, is it possible to reformulate the objective
> function to find the maximum positive (and negative) power injection in a
> certain bus:
>
> Maximize the positive power injection in bus X (or could also be: minimize
> the negative power injection in bus X)
> Subject to the AC power flow equations, branch flow limits, generator output
> limits, voltage limits.
>
> The main decision variable in this case would be the power injection in bus
> X. Other values that could vary are the generator outputs, voltages, and the
> like (all subject to the constraints).
>
> Cheers,
> Ronald
