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 <mari...@aia.es> 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 <nurnazu...@gmail.com>:
>
>> 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 <nurnazu...@gmail.com>
>> 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 <asar...@gmail.com> 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 <nurnazu...@gmail.com>
>>>> 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
>>>>>
>>>>>
>>>>
>>>
>>
>
