Hi Ray,
In the branch data for the #2 transformer (the one that is on 161:13.14kV tap) what should TAP be? Bank #1 ratio matches the ratio of bus voltage bases (161/13.8) so its ratio is set to 1.0 So, I think the #2 transformer TAP (ratio ) should be (13.8/13.14 = 1.0502). Is that correct? Best regards, Russ From: [email protected] [mailto:[email protected]] On Behalf Of Ray Daniel Zimmerman Sent: Monday, December 21, 2020 11:44 AM To: MATPOWER-L Subject: Re: circulating current (MVAR loss) I suggest double-checking your calculations against the code in makeYbus.m, which is pretty straightforward, and the model described in the User’s Manual <https://matpower.org/docs/MATPOWER-manual-7.1.pdf> see Figure 3-1 and equation (3.2). Be sure to keep in mind the orientation of the taps in the model. Ray On Dec 16, 2020, at 3:52 PM, Russ Patterson <[email protected]> wrote: Carlos – thank you. Very helpful. The YBus I get for my case is below. I expected Y(1,1) to equal the of this sum: (1/j0.1) + (1/j0.09522) + (1/-j1.991) = j 19.9997 (negative sign is per coder preference). Is attached (page 1) not how MATPOWER would modify the bank #2 impedances before creating YBUS? Yb = Compressed Column Sparse (rows = 2, cols = 2, nnz = 4 [100%]) (1, 1) -> 0 - 19.0663i (2, 1) -> 0 + 19.5217i (1, 2) -> 0 + 19.5217i (2, 2) -> 0 - 20i Best regards, russ From: [email protected] [mailto:[email protected]] On Behalf Of Carlos E Murillo-Sanchez Sent: Wednesday, December 16, 2020 4:12 PM To: MATPOWER discussion forum Subject: Re: circulating current (MVAR loss) Russ Patterson wrote: Hi - I am still trying to hand calculate the flow into branch 2 from bus 1 to bus 2. I can’t get my results to match MATPOWER. I get Q into the banks from bus 1 of, Bank #1: 24.00 MVAR Bank #2: -25.02 MVAr Attached is my short calculation and the .m file. Is there a way to have MATPOWER barf out the YBUS matrix? >> help makeYbus If buses are numbered consecutively starting from 1 in the bus table (see ext2int if not), simply type: >> mpc = loadcase('mycase'); >> [Yb, Yf, Yt] = makeYbus(mpc) To get all the relevant current injections in the solved case, simply do >> mpc = runpf(mpc); >> define_constants; >> V = mpc.bus(:, VM) .* exp(1i * mpc.bus(:, VA)*pi/180); >> Ibus = Yb * V >> Ifrom = Yf * V; >> Ito = Yt * V; >From there, compute power injections as >> Sbusinj = V .* conj(Yb * V); >> Sfrominj = V(mpc.branch(:, F_BUS)) .* conj(Yf * V); >> Stoinj = V(mpc.branch(:, T_BUS)) .* conj(Yt * V); carlos. <power.pdf>
function mpc = circulating
% This example is to show circulating MVAR losses in 2 parallel banks
% with different turns ratios (no load connected)
%
% 161kV Bus 1 is infinite and is 1 pu and zero degrees
% 13.8kV Bus 2 has no load connected
%
% Transformer 1 between bus 1 & 2 is j0.1 pu (100MVA, 161kV base) and turns
ratio matches ratio of voltage bases (1.0)
% 161kV:13.8kV
% Transformer 2 between bus 1 & 2 is j0.1 pu (100MVA, 161kV base) and turns
ratio is 1.05 of voltage base ratio
% 161kV:13.14kV (bank 1)
%
%
% The voltage difference is (13.8kV � 13.14kV)/13.8kV = 0.0478 pu which is
the driving voltage for the circulating current.
% The circulating current will be 0.0478/0.2pu = 0.239pu A. This circulating
current will produce a var drop of
% Q = (0.239)(0.239) x (0.2) = 0.01144 pu. This is 1.14MVAr.
% - - - - - - - - - - -
%
\%% MATPOWER Case Format : Version 2
mpc.version = '2';
\%%----- Power Flow Data -----\%%
\%% system MVA base
mpc.baseMVA = 100;
\%% Vm under mpc.gen sets the slack bus (1) voltage magnitude
\%% Va under mpc.bus sets the slack bus (1) voltage angle
\%%
% bus_i type Pd Qd Gs Bs area Vm Va
baseKV zone Vmax Vmin
mpc.bus = [
1 3 0 0 0 0 1 1 0 161 1 1
1;
2 1 0 0 0 0 1 1 0 13.8 1 0
0;
];
\%% generator data
% bus Pg Qg Qmax Qmin Vg mBase status
Pmax Pmin Pc1 Pc2 Qc1min Qc1max Qc2min Qc2max ramp_agc
ramp_10 ramp_30 ramp_q apf ?
mpc.gen = [
1 0 0 0 0 1.0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0
0;
];
\%% branch data
% fbus tbus r x b rateA rateB rateC
ratio angle status angmin angmax
mpc.branch = [
1 2 0 .1 0 0 0 0 0 0 1
0 0;
% the second branch is via transformer that has 1.05 higher turns ratio as
compared to 161/13.8
1 2 0 .1 0 0 0 0 1.0502 0 1
0 0;
];
per_gross.pdf
Description: Adobe PDF document
