ilto:matpowe...@list.cornell.edu>>
Date: Wednesday, April 19, 2017 at 1:32 AM
To: MATPOWER discussion forum
mailto:matpowe...@list.cornell.edu>>
Subject: Re: Line loading limit
and what in case of 'rundcopf'
On Wed, Apr 19, 2017 at 11:39 AM, Abhyankar, Shrirang G.
mailto:abhy...@anl.g
and what in case of 'rundcopf'
On Wed, Apr 19, 2017 at 11:39 AM, Abhyankar, Shrirang G.
wrote:
> runpf does not enforce any line limits. You need to run optimal power flow
> (runopf).
>
> Shri
>
> From: Akash Tyagi
> Reply-To: MATPOWER discussion forum
> Date: Tuesday, April 18, 2017 at 9:57 P
runpf does not enforce any line limits. You need to run optimal power flow
(runopf).
Shri
From: Akash Tyagi mailto:akashtyagi0...@gmail.com>>
Reply-To: MATPOWER discussion forum
mailto:matpowe...@list.cornell.edu>>
Date: Tuesday, April 18, 2017 at 9:57 PM
To: MATPOWER discussion forum
mailto:m
When using AC modeling, the power flow is complex and it is not that same at
the two ends of a branch, so you may want to use the average MVA as the flow.
define_constants;
r = runpf(…);
Af = sqrt(r.branch(:, PF)^2 + r.branch(:, QF)^2);
At = sqrt(r.branch(:, PT)^2 + r.branch(:, QT)^2);
flow = (Af
Hallo Basem,
I guess, except if it's already somewhere else calculated, you can just grab
the abs(mpc.branch(:,14)) (power injected at one end)and divide by the
mpc.branch(:,6) (rating A). I
Fot the losses just calculate the current and multiply it by the losses and
save it at some vector
chee
Thank you
--- On Mon, 3/21/11, Ray Zimmerman wrote:
From: Ray Zimmerman
Subject: Re: Line Loading
To: "MATPOWER discussion forum"
Date: Monday, March 21, 2011, 9:21 PM
The apparent power flow (MVA flow) at the "from" and "to" ends of each line can
be computed a
The apparent power flow (MVA flow) at the "from" and "to" ends of each line can
be computed as ...
define_constants;
results = runopf(...);
MVA_flow_from = sqrt(results.branch(:, PF).^2 + results.branch(:, QF).^2);
MVA_flow_to = sqrt(results.branch(:, PT).^2 + results.branch(:, QT).^2);
--
Ra
