Hey,
I´m not sure if it´s possible for an OPF to do so, because I think the
nearer you get a value to its constraint, the more it becomes "penalized".
It gets weighted with a factor mu (see mpc.gen(n , 22:25 )?), so the
objective function does not obtain the best solution with the power of
one generator near its maximum.
Also you have to consider the other constraints, like nodal voltage. The
calculation trys to minimize the objective funtion and find the ideal
balance between all constraints.
At least that´s what I think it does. I´m not into the opf mechanisms
very deep yet.
Greetings
Am 3/21/2011 1:13 PM, schrieb Guilherme Lira:
Simon,
Thanks for the input.
Yes, Pmax for the generator at bus 1 is 35 MW, which is 15 MW short of
being able to supply the load alone, but (leaving both online) in this
substation emulation it is desirable to fully use its capacity as its
connection to the load has the lowest impedance. I'm thinking that,
even though they are being modeled as substation busbars, the
generators' operating points might be limited by their capability
curves, is that correct?
Regards,
Guilherme Lira
On Mon, Mar 21, 2011 at 6:36 AM, Simon Schneider<[email protected]> wrote:
Hi Guilherme,
if you turn off the second gen (mpc.gen(2,8) = -1) you see, that several
other constraints are violated and the opf doesn´t converge.
The Generator at 1 isn´t powerfull enough to satisfy the load alone.
Greetings
Simon
Am 3/20/2011 10:33 PM, schrieb Guilherme Lira:
Ray/others,
In order to test the power loss minimization, I have tried modeling
the generators with equal linear costs simulating substation busbars.
Please have a look at the attached m-file, it contains a simple test
case consisting of two generators connected to a load, with different
branch impedance values. Theoretically, minimizing losses would mean
that the generator (substation busbar) connected to the load through
the branch of lowest impedance would be serving the load at full power
(Pmax), right? Can you please explain why it is not?
Thanks,
Guilherme
