I think that you have to invert the sign of QD. From
http://www.pserc.cornell.edu//matpower/docs/ref/matpower6.0b1/idx_bus.html:
columns 1-13 must be included in input matrix (in case file)
1 BUS_I bus number (positive integer)
2 BUS_TYPE bus type (1 = PQ, 2 = PV, 3 = ref, 4 = isolated)
3 PD Pd, real power *demand* (MW)
4 QD Qd, reactive power *demand* (MVAr)
5 GS Gs, shunt conductance (MW *demanded* at V = 1.0 p.u.)
6 BS Bs, shunt susceptance (MVAr *injected* at V = 1.0 p.u.)
7 BUS_AREA area number, (positive integer)
8 VM Vm, voltage magnitude (p.u.)
9 VA Va, voltage angle (degrees)
10 BASE_KV baseKV, base voltage (kV)
11 ZONE zone, loss zone (positive integer)
12 VMAX maxVm, maximum voltage magnitude (p.u.)
13 VMIN minVm, minimum voltage magnitude (p.u.)
--
Jose L. Marin
Grupo AIA
2016-07-06 12:07 GMT+02:00 Georgiadis Dionysios <
[email protected]>:
> Dear all,
>
>
> After some quick calculations one can see that it is possible to replace a
> constant power load with a constant impedance load of exactly equal power
> consumption.
>
> Here is an example, say bus 1 absorbs *S1* MVAs under *V1* volts. Bus 1
> is constant power.
> This would be equal to a constant impedance load, with impedance
> *conjugate(Z1)
> = (Vn/V1)^2 * S1,* where Vn is the nominal voltage.
>
> Thus, we can set the PD and QD entries of the case to zero, and replace
> them with the right values for GS and BS.
>
> Specifically, after solving a case with all constant power loads we will
> obtain the Vm vector (voltage amplitudes) and the Pd and Qd vectors. Then,
> we can set all PD and QD values to zero and place* (1/Vm).^2 * Pd *and
> *(1/Vm).^2
> * Qd* in the GS and BS columns respectively.
>
> From my understanding the output should be exactly the same. The solved
> case should remain a perfectly good solution to the PF equations. Yet, when
> I do so, the solver iterates.
>
> Can someone shed light on this? I apologise if I am asking the question in
> the wrong place.
>
>
> Best regards,
> Dionysios Georgiadis
>
