Hi friends,
I have a simple power flow case with an infinite bus and 2 transformers (no
load). The transformers have different turns ratios so I’ll get circulating
current and VAr drop in the banks.
#1 Transformer (100MVA)
161:13.8kV, X=0.1pu
#2 Transformer (100MVA)
161:13.14kV, X=0.1pu
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.
I ran the attached power flow case and got 1.13MVAR (nice). But, I don’t
understand the big MVAr drops being reported in the branch report (see below).
What is causing those big values?
================================================================================
| Branch Data |
================================================================================
Brnch From To From Bus Injection To Bus Injection Loss (I^2 * Z)
# Bus Bus P (MW) Q (MVAr) P (MW) Q (MVAr) P (MW) Q (MVAr)
----- ----- ----- -------- -------- -------- -------- -------- --------
1 1 2 0.00 23.81 0.00 -23.24 -0.000 0.57
2 1 2 0.00 -22.68 0.00 23.24 -0.000 0.57
-------- --------
Total: 0.000 1.13
I have a second minor question, how do you specify “branch 1” or “branch 2”
when you have 2 branches between 2 buses? Does MATPOWER just take them in
order they show up in the mpc.branch structure (e.g. the 1st occurrence as
“branch 1”)? I did “help idx_brch” but didn’t see anything specifying.
Thank you,
Russ
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 1.05 0 1
0 0;
];
