The problem is that case14 already includes a non-zero BS for one of the buses
in the original data, so you have to add your constant impedance load to what
is already there. Try this …
define_constants
opt = mpoption('out.all' , 0, 'verbose', 2);
solved = runpf(case14(), opt);
solved2 = solved;
solved2.bus(:,GS) = solved2.bus(:,GS) + solved.bus(:, PD)./ (solved.bus(:,
VM).^2);
solved2.bus(:,BS) = solved2.bus(:,BS) - solved.bus(:, QD)./ (solved.bus(:,
VM).^2);
solved2.bus(:,PD) = 0;
solved2.bus(:,QD) = 0;
solved2 = runpf(solved2, opt);
norm(solved.bus(:, VM)-solved2.bus(:, VM))
You may also want to look at the experimental options for ZIP load handling in
MATPOWER 6, namely exp.sys_wide_zip_loads.pw and exp.sys_wide_zip_loads.pw. For
example …
solved = runpf(case14(), opt);
total_load(solved, 'all', [], opt)
solved2 = solved;
solved2.bus(:,PD) = solved.bus(:, PD) ./ (solved.bus(:, VM).^2);
solved2.bus(:,QD) = solved.bus(:, QD) ./ (solved.bus(:, VM).^2);
opt = mpoption(opt, 'exp.sys_wide_zip_loads.pw', [0 0 1],
'exp.sys_wide_zip_loads.qw', [0 0 1]);
solved2 = runpf(solved2, opt);
total_load(solved2, 'all', [], opt)
One advantage to doing it this way is that it’s still listed as “load” in the
output.
Ray
> On Jul 6, 2016, at 10:49 PM, Georgiadis Dionysios
> <[email protected]> wrote:
>
> I see, subtle point but it is my fault for not noticing in the documentation.
> Thank you.
>
> Still however, the issue persists. Here is a minimal ingredients example:
>
> opt = mpoption('out.all' , 0);
>
> solved = runpf(case14(), opt);
>
> solved2 = solved;
>
> solved2.bus(:,GS) = solved.bus(:, PD)./ (solved.bus(:, VM).^2);
> solved2.bus(:,BS) = -solved.bus(:, QD)./ (solved.bus(:, VM).^2);
>
> solved2.bus(:,PD) = solved2.bus(:,PD).*0;
> solved2.bus(:,QD) = solved2.bus(:,QD).*0;
>
> solved2 = runpf(solved2, opt);
> norm(solved.bus(:, VM)-solved2.bus(:, VM))
>
> One may notice not only that the solver iterates but also that the norm in
> the final row is nonzero, proving that a new solution is obtained (which is
> slightly different).
>
> From: <[email protected]
> <mailto:[email protected]>> on behalf of Jose Luis
> Marín <[email protected] <mailto:[email protected]>>
> Reply-To: MATPOWER discussion forum <[email protected]
> <mailto:[email protected]>>
> Date: Wednesday, 6 July 2016 6:57 pm
> To: MATPOWER discussion forum <[email protected]
> <mailto:[email protected]>>
> Subject: Re: Replacing constant power with constant impedance loads
>
>
> 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
> <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] <mailto:[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
>
