If you specify all powers in kW (and kVAr) including baseMVA, then all of your outputs will be in the same units as well. They just will be labeled incorrectly as MW and MVAr.
Ray > On Oct 26, 2016, at 12:08 PM, Nazurah Nasir <[email protected]> wrote: > > Hi Jose, > > thank you for the reply. I want to keep voltage in kV, hence the 0.415. I am > looking for alternatives that could allow me to run analysis without having > to divide my load demand by 1000 each time. And I thought probably by > changing the R+jX value, it might help. If I don't want to divide my load > power by 1000 each time, what should I change in the casefile? > > On Thu, Oct 27, 2016 at 12:20 AM, Jose Luis Marín <[email protected] > <mailto:[email protected]>> wrote: > > If you want to do things properly, I guess you would have to do all of the > following: > First, MATPOWER expects baseMVA to be in MW. If you want to use a p.u. > system based around Volts and 1 kW (instead of the traditional kV and 100 > MW), then you need baseMVA=0.001 at the beginning of your case file. > Expressing voltages in pu should be no problem, just divide the quantity in > volts by your voltage levels (in your case I see it's only one, 415 Volts) > Now, MATPOWER expects all power quantities (P, Q) to be expressed in MW. So > if your values are in kW, divide them by 1000. > Finally, to convert resistances and reactances (R, X), use your baseMVA and > voltage base as follows: take the initial quantity in Ohms, and multiply it > by: baseMVA / Vbase^2 = 1000 Watts / (415 Volts)^2 = 5.806357961968356e-03 > You don't have Bshunt values in your case, but if you had, the conversion > factor would be just the inverse of the one used for resistance and > reactance. > I hope I'm not missing anything, I think that's all you need in your case. > > -- > Jose L. Marin > Grupo AIA > > > > > 2016-10-26 14:44 GMT+02:00 Nazurah Nasir <[email protected] > <mailto:[email protected]>>: > But does that means I should not divide my input power data by 1000 to make > it in MW? If I do that, it won't converge. For example, these are my Power > input for one time: > > Columns 1 through 6 > 1.0000 3.0000 0 0 0 0 > 2.0000 1.0000 0.0012 0.0004 0 0 > 3.0000 1.0000 0.0019 0.0006 0 0 > 4.0000 1.0000 0.0006 0.0002 0 0 > 5.0000 1.0000 0.0024 0.0008 0 0 > 6.0000 1.0000 0.0012 0.0004 0 0 > 7.0000 1.0000 0.0010 0.0003 0 0 > 8.0000 1.0000 0.0023 0.0008 0 0 > 9.0000 1.0000 0.0005 0.0002 0 0 > 10.0000 1.0000 0.0006 0.0002 0 0 > 11.0000 1.0000 0.0012 0.0004 0 0 > 12.0000 1.0000 0 0 0 0 > Columns 7 through 12 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > 1.0000 1.0000 0 0.4150 1.0000 1.1000 > Column 13 > 0.9400 > 0.9400 > 0.9400 > 0.9400 > 0.9400 > 0.9400 > 0.9400 > 0.9400 > 0.9400 > 0.9400 > 0.9400 > 0.9400 > > Thank you very much for the help > > Yours sincerely, > Nur > > On Wed, Oct 26, 2016 at 11:14 PM, Nazurah Nasir <[email protected] > <mailto:[email protected]>> wrote: > Aren't I supposed to make the R and X in p.u. if I want to use them in > MATPOWER? Regardless, your simulation seems to be more sensible. But, I just > curious, so we don't necessarily change the R and X into p.u. values? > > Thanks for the response. > > On Wed, Oct 26, 2016 at 3:26 PM, Saranya A <[email protected] > <mailto:[email protected]>> wrote: > Hi Nur, > Dont divide R and X with the voltage. I get the following power flow without > those two lines. > > ------------ > runpf('LV10') > > MATPOWER Version 6.0b1, 01-Jun-2016 -- AC Power Flow (Newton) > > Newton's method power flow converged in 5 iterations. > > Converged in 0.02 seconds > ================================================================================ > | System Summary > | > ================================================================================ > > How many? How much? P (MW) Q (MVAr) > --------------------- ------------------- ------------- ----------------- > Buses 12 Total Gen Capacity 261.0 -302.0 to 302.0 > Generators 11 On-line Capacity 250.0 -300.0 to 300.0 > Committed Gens 1 Generation (actual) 1.2 0.5 > Loads 10 Load 1.0 0.3 > Fixed 10 Fixed 1.0 0.3 > Dispatchable 0 Dispatchable -0.0 of -0.0 -0.0 > Shunts 0 Shunt (inj) -0.0 0.0 > Branches 11 Losses (I^2 * Z) 0.23 0.16 > Transformers 0 Branch Charging (inj) - 0.0 > Inter-ties 0 Total Inter-tie Flow 0.0 0.0 > Areas 1 > > Minimum Maximum > ------------------------- -------------------------------- > Voltage Magnitude 0.717 p.u. @ bus 11 1.000 p.u. @ bus 1 > Voltage Angle -5.40 deg @ bus 11 0.00 deg @ bus 1 > P Losses (I^2*R) - 0.06 MW @ line 1-2 > Q Losses (I^2*X) - 0.04 MVAr @ line 1-2 > > ================================================================================ > | Bus Data > | > ================================================================================ > Bus Voltage Generation Load > # Mag(pu) Ang(deg) P (MW) Q (MVAr) P (MW) Q (MVAr) > ----- ------- -------- -------- -------- -------- -------- > 1 1.000 0.000* 1.23 0.46 - - > 2 0.950 -0.744 - - 0.10 0.03 > 3 0.905 -1.483 - - 0.10 0.03 > 4 0.864 -2.206 - - 0.10 0.03 > 5 0.828 -2.898 - - 0.10 0.03 > 6 0.797 -3.541 - - 0.10 0.03 > 7 0.770 -4.116 - - 0.10 0.03 > 8 0.749 -4.607 - - 0.10 0.03 > 9 0.733 -4.993 - - 0.10 0.03 > 10 0.722 -5.260 - - 0.10 0.03 > 11 0.717 -5.397 - - 0.10 0.03 > 12 1.000 0.000 - - - - > -------- -------- -------- -------- > Total: 1.23 0.46 1.00 0.30 > > ================================================================================ > | 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 1.23 0.46 -1.17 -0.42 0.055 0.04 > 2 2 3 1.07 0.39 -1.03 -0.36 0.046 0.03 > 3 3 4 0.93 0.33 -0.89 -0.30 0.038 0.03 > 4 4 5 0.79 0.27 -0.76 -0.25 0.030 0.02 > 5 5 6 0.66 0.22 -0.64 -0.20 0.023 0.02 > 6 6 7 0.54 0.17 -0.52 -0.16 0.016 0.01 > 7 7 8 0.42 0.13 -0.41 -0.13 0.011 0.01 > 8 8 9 0.31 0.10 -0.30 -0.09 0.006 0.00 > 9 9 10 0.20 0.06 -0.20 -0.06 0.003 0.00 > 10 10 11 0.10 0.03 -0.10 -0.03 0.001 0.00 > 11 1 12 0.00 0.00 0.00 0.00 0.000 0.00 > -------- > -------- > Total: 0.228 0.16 > > On Tue, Oct 25, 2016 at 10:31 PM, Nazurah Nasir <[email protected] > <mailto:[email protected]>> wrote: > > > Hi all MatPower community, > > I am trying to develop a simple LV network in radial network distribution. > However, my model did not converge or if I scale the R and X, the results is > too big (which means the R,X) scaling is wrong. I tried for one month now but > still could get around why it is not converging. > > I need to work on this PowerFlow to work inside my bilevel programming loop. > but it seems my code won't work because the power flow is not converging. > > I need help on verifying my parameter. Attached is my code that I build. As > the MatPower is in three phase balanced, I lumped my loads that connected to > a bus as one load, hence the voltage at the bus is 0.415kV. My input power > are all in kW, hence I change the impedance values accordingly by multiplying > it by 1000. The input power is just a dummy value of 0.1MW because it will > update itself in a loop. But since input power is in kW, I should divide that > by 1000 right? > > Thank you so much for the help. > > > Best regards, > Nur > > > > > >
