Hello Mirish,
*A branch can have negative reactive power losses*
The shunt charging susceptance is /positive/ so it /cannot/ /produce/
reactive power but it will always /consume/ reactive power:
0 < Q_branch_shunt_loss = b/2 * |Vf|² + b/2 * |Vt|² = b/2 * (
(V_magnitude_f)² + (V_magnitude_t)² )
so it can only be the inductance that produced reactive power. And
indeed: The series element susceptance is -1/X ( since 1/j = -j )
0 > Q_series_loss = -1/x * | Vf - Vt |²
so the series element actually /generates/ reactive power (/negative/
loss). The second Line has a slightly higher X in relation to B (0.91%
against 0.87%) and therefore has a net reactive power production which
is even positive. Impedances can produce reactive power if their
suscepance is negative. Why can't they produce real power? Because the
resistance R is usally positive (There is few exceptions to that as well
though: https://en.wikipedia.org/wiki/Negative_resistance). You must
free yourself from the idea that reactive power is something only a
generator produces.
I believe what also confuses you, is power flow and current flow.
Current flow and power flow must not have the same direction. If you for
example have the:
* complex voltages Uf = -Ut
* and the complex currents If = -It
(see Figure 3-1, http://www.pserc.cornell.edu//matpower/manual.pdf)
Then the currents If = -It point in the same direction but the power flows:
Pf = Uf * conj(If) = -Ut * conj(-It) = Ut * conj(It) = Pt
point in opposite directions.
*How to calculate branch losses
*There seems to be a little bit of a difference of opinions here. I a
agree with your approach: To me, the loss of a branch is:
P_loss = Pf+Pt
Q_loss = Qf+Qt.
The to equations a simple conservation of energy. And yes the
conservation of energy fortunately also extends to reactive powers.
You have to be careful though: In the pretty printed output of the
runpf() method, the branch losses are /only /the series element losses
(I^2 * Z). That might cause some confusion.
I hope this helps and i hope my statements are correct, since I only
work on power grids for 5 months now.
P.S.: Next time please also include the 'from' and 'to' bus entry. That
would make it easier for me to give you a practical example.
Best regards
Dirk
On 28.02.2016 02:36, Mirish Thakur wrote:
Hello Matpower friends,
I have a question over similar sign of flows at both nodes of a
branch. I have performed PF with PG=0, PD=0 BRANCH_R=0; means total
reactive power model (reactive power generation and reactive demand
exist) and I got successful convergence. When I observed reactive
power flows in branches I got some surprising results in branch field
which I would like to share-
[ 2 44 0 0.00103 0.1179
501.0822 0 0 0 0 1
-360 360 0 2.53088 0 -14.8590]
[ 2 44 0 0.001491 0.16406 501.0822
0 0 0 0 1 -360 360
0 -2.53088 0 -14.6211]
So in above two branches (2 to 44) I can understand first branch flow,
but in second one both flows (QF and QT) have same negative sign. I
didn’t understand why this happened? Is this due to more charging
susceptance of second line or what? Another question is how can I
estimate MVAr losses in these scenarios? By simply adding both flows
in first transmission line gives Q loss = -12.3281 MVAr or by using
average power formula
From = sqrt(result.branch(:, 14).^2 + result.branch(:, 15).^2);
To = sqrt(result.branch(:, 16).^2 + result.branch(:, 17).^2);
Losses = (To + from)./2 = 8.6949 MVAr; ?
From my point of view simple addition will give me MVAr loss value but
I’m confused how to calculate MVAr loss in second line? Thanks in advance.
Regards
Mirish.
