Re: Replacing constant power with constant impedance loads

[Georgiadis Dionysios]

I see, I think I should had checked that before accessing the mailing list.

You example work perfectly.

From: 
<bounce-120612361-76097...@list.cornell.edu<mailto:bounce-120612361-76097...@list.cornell.edu>>
 on behalf of Ray Zimmerman <r...@cornell.edu<mailto:r...@cornell.edu>>
Reply-To: MATPOWER discussion forum 
<matpowe...@list.cornell.edu<mailto:matpowe...@list.cornell.edu>>
Date: Friday, 8 July 2016 3:55 am
To: MATPOWER discussion forum 
<matpowe...@list.cornell.edu<mailto:matpowe...@list.cornell.edu>>
Subject: Re: Replacing constant power with constant impedance loads

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 
<dionysios.georgia...@frs.ethz.ch<mailto:dionysios.georgia...@frs.ethz.ch>> 
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: 
<bounce-120608294-76097...@list.cornell.edu<mailto:bounce-120608294-76097...@list.cornell.edu>>
 on behalf of Jose Luis Marín <mari...@aia.es<mailto:mari...@aia.es>>
Reply-To: MATPOWER discussion forum 
<matpowe...@list.cornell.edu<mailto:matpowe...@list.cornell.edu>>
Date: Wednesday, 6 July 2016 6:57 pm
To: MATPOWER discussion forum 
<matpowe...@list.cornell.edu<mailto:matpowe...@list.cornell.edu>>
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:

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 
<dionysios.georgia...@frs.ethz.ch<mailto:dionysios.georgia...@frs.ethz.ch>>:
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