Hi Mirish and Dirk,
I believe Dirk is close, but the signs are incorrect. It is in fact the line
charging capacitance (like a any other shunt capacitor used for voltage
support) that produces reactive power (negative reactive losses) rather than
consuming it. And the series component always consumes reactive power rather
than generating it. If you look at Fig 3-1 and do the math carefully, the
series reactive losses are:
series_Q_losses = |Vf – Vt|^2 * 1/x
shunt_Q_losses = – (|Vf|^2 + |Vt|^2) * b/2
The total reactive losses in the line including both components can always be
found by adding Qf + Qt.
So, yes, the reason for the overall negative reactive losses in the second line
is that the shunt components are generating more reactive power than the series
element is consuming. As Dirk said, it’s important to realize that reactive
power is generated by network elements such as transmission lines and shunt
capacitors, not just generators.
Ray
> On Feb 28, 2016, at 1:03 PM, Dirk Toewe <[email protected]> wrote:
>
> 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
> <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
> <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.
>
