Re: Help on radial LV distribution network
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 On-line Capacity 250.0-300.0 to
300.0
Committed Gens
Re: Help on radial LV distribution network
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
>>> Inter-ties 0 Total Inter-tie Flow 0.0
Re: Help on radial LV distribution network
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 0.770 -4.116 - -0.
Re: Help on radial LV distribution network
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 Nasir
> wrote:
>
>>
>>
Re: Help on radial LV distribution network
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 model did not converge or if I scale the R and
