Re: Help on radial LV distribution network

[Ray Zimmerman]

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  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  > 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  >:
> 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
> 1.3. 0 0 0 0
> 2.1.0.00120.0004 0 0
> 3.1.0.00190.0006 0 0
> 4.1.0.00060.0002 0 0
> 5.1.0.00240.0008 0 0
> 6.1.0.00120.0004 0 0
> 7.1.0.00100.0003 0 0
> 8.1.0.00230.0008 0 0
> 9.1.0.00050.0002 0 0
>10.1.0.00060.0002 0 0
>11.1.0.00120.0004 0 0
>12.1. 0 0 0 0
>   Columns 7 through 12
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
>   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  > 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  > 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  
>  |
> 
> 

Re: Help on radial LV distribution network

[Nazurah Nasir]

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  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 :
>
>> 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
>> 1.3. 0 0 0 0
>> 2.1.0.00120.0004 0 0
>> 3.1.0.00190.0006 0 0
>> 4.1.0.00060.0002 0 0
>> 5.1.0.00240.0008 0 0
>> 6.1.0.00120.0004 0 0
>> 7.1.0.00100.0003 0 0
>> 8.1.0.00230.0008 0 0
>> 9.1.0.00050.0002 0 0
>>10.1.0.00060.0002 0 0
>>11.1.0.00120.0004 0 0
>>12.1. 0 0 0 0
>>   Columns 7 through 12
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>> 1.1. 00.41501.1.1000
>>   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 
>> 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  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
 Generators11 

Re: Help on radial LV distribution network

[Jose Luis Marín]

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 :

> 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
> 1.3. 0 0 0 0
> 2.1.0.00120.0004 0 0
> 3.1.0.00190.0006 0 0
> 4.1.0.00060.0002 0 0
> 5.1.0.00240.0008 0 0
> 6.1.0.00120.0004 0 0
> 7.1.0.00100.0003 0 0
> 8.1.0.00230.0008 0 0
> 9.1.0.00050.0002 0 0
>10.1.0.00060.0002 0 0
>11.1.0.00120.0004 0 0
>12.1. 0 0 0 0
>   Columns 7 through 12
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
> 1.1. 00.41501.1.1000
>   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 
> 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  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
>>> Generators11 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

Re: Help on radial LV distribution network

[Nazurah Nasir]

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
1.3. 0 0 0 0
2.1.0.00120.0004 0 0
3.1.0.00190.0006 0 0
4.1.0.00060.0002 0 0
5.1.0.00240.0008 0 0
6.1.0.00120.0004 0 0
7.1.0.00100.0003 0 0
8.1.0.00230.0008 0 0
9.1.0.00050.0002 0 0
   10.1.0.00060.0002 0 0
   11.1.0.00120.0004 0 0
   12.1. 0 0 0 0
  Columns 7 through 12
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
1.1. 00.41501.1.1000
  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 
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  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
>> Generators11 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  1.0000.000* 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  

Re: Help on radial LV distribution network

[Nazurah Nasir]

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  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
> Generators11 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  1.0000.000* 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  1.0000.000   - - - -
>       
>Total:  1.23  0.46  1.00  0.30
>
> 
> 
> | Branch Data
>  |
> 
> 
> Brnch   From   ToFrom Bus Injection   To Bus Injection Loss (I^2 *
> Z)
>   # BusBusP (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 

Re: Help on radial LV distribution network

[marwan ahmed]

what name type matlab program   can run this  my matlab R2013a  dont's run

On 26 October 2016 at 07:26, Saranya A  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
> Generators11 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  1.0000.000* 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  1.0000.000   - - - -
>       
>Total:  1.23  0.46  1.00  0.30
>
> 
> 
> | Branch Data
>  |
> 
> 
> Brnch   From   ToFrom Bus Injection   To Bus Injection Loss (I^2 *
> Z)
>   # BusBusP (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 
> wrote:
>
>>
>>
>> Hi all MatPower community,
>>
>> I am trying to develop a simple LV network in radial network
>> distribution. However, my 

Re: Help on radial LV distribution network

[Saranya A]

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
Generators11 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  1.0000.000* 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  1.0000.000   - - - -
      
   Total:  1.23  0.46  1.00  0.30


| Branch Data
   |

Brnch   From   ToFrom Bus Injection   To Bus Injection Loss (I^2 *
Z)
  # BusBusP (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 
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 

Help on radial LV distribution network

[Nazurah Nasir]

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


LV10.m
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