thank you doctor.

بتاريخ ٢٠١٨/٠٤/٢٦ ١١:٢٢ م، كتب "Ray Zimmerman" <[email protected]>:

> The DC OPF in MATPOWER neglects losses. If you want to include losses, you
> will have to use the AC OPF. Or you could estimate the losses from the DC
> OPF solution by computing the average current in each branch and
> multiplying by the corresponding square of resistance in the line.
>
>     Ray
>
>
>
> On Apr 23, 2018, at 11:47 AM, Mohammed Alhajri <[email protected]> wrote:
>
> Hello all
>
> how i can include the losses in my code?
>
> my code is
>
> ++++++++++++++++++++++++++++++++++++++++++++++
> clear all;
> clc;
> define_constants;
> factor = [0.827315297246503; 0.870524914020136; 0.89225313150181;
> 0.874238893371013; 0.841447907563717; 0.84108393381471; 0.79110106954204;
> 0.797588753768103; 0.801421029392904; 0.822018440318329;
> 0.864051234184459; 0.876248439711015; 0.888800014120353;
> 0.922571691519356; 0.978924730305068; 1; 0.975195764982014;
> 0.903702013699042; 0.8424313185939; 0.786194728460529; 0.846555864529582;
> 0.851944597303249; 0.834213349185084; 0.84694392020701;];
> mpc0 = loadcase('OETC_OPF');
> nt = length(factor);
> diary R;
> for t = 1:nt
>     mpc(t) = mpc0;
>     [mpc(t).bus] = ...
>         scale_load(factor(t), mpc(t).bus);
>     results(t) = rundcopf(mpc(t));
> end
>
> ++++++++++++++++++++++++++++++++++++++++++++++
>
>  but it does not calculate the loses!
>
> On 23 April 2018 at 18:09, Mohammed Alhajri <[email protected]> wrote:
>
>> thank you doctor
>>
>> بتاريخ ٢٠١٨/٠٤/٢٣ ٦:٠٧ م، كتب "Ray Zimmerman" <[email protected]>:
>>
>>> The function most_summary() is only for summarizing the output of a
>>> call to most(). It does not apply to runopf().
>>>
>>>    Ray
>>>
>>> On Apr 21, 2018, at 6:27 AM, Mohammed Alhajri <[email protected]> wrote:
>>>
>>> thank you, i did it and it works
>>>
>>> ++++++++++++++++++++++++++++++++++++++++++++
>>>
>>> clear all;
>>> clc;
>>> define_constants;
>>> factor = [0.827315297246503; 0.870524914020136; 0.89225313150181;
>>> 0.874238893371013; 0.841447907563717; 0.84108393381471;
>>> 0.79110106954204; 0.797588753768103; 0.801421029392904;
>>> 0.822018440318329; 0.864051234184459; 0.876248439711015;
>>> 0.888800014120353; 0.922571691519356; 0.978924730305068; 1;
>>> 0.975195764982014; 0.903702013699042; 0.8424313185939;
>>> 0.786194728460529; 0.846555864529582; 0.851944597303249;
>>> 0.834213349185084; 0.84694392020701;];
>>> mpc0 = loadcase('OETC_OPF');
>>> nt = length(factor);
>>> for t = 1:nt
>>>     mpc(t) = mpc0;
>>>     [mpc(t).bus] = ...
>>>         scale_load(factor(t), mpc(t).bus);
>>>     results(t) = runopf(mpc(t));
>>> end
>>>
>>> ++++++++++++++++++++++++++++++++++++++++++++
>>>
>>> put how i can do the same job but using the function
>>>
>>>  most_summary()
>>>
>>> can you write the code, because i tried a lot but it did not work using
>>> the function  most_summary()
>>>
>>> On 19 April 2018 at 23:56, Ray Zimmerman <[email protected]> wrote:
>>>
>>>> Use the scale_load()
>>>> <http://www.pserc.cornell.edu/matpower/docs/ref/matpower6.0/scale_load.html>
>>>>  function
>>>> in a loop, something like shown here: https://www.mail-archive
>>>> .com/[email protected]/msg06477.html
>>>>
>>>>    Ray
>>>>
>>>>
>>>> On Apr 17, 2018, at 12:27 PM, Mohammed Alhajri <[email protected]>
>>>> wrote:
>>>>
>>>> Hello all,
>>>>
>>>> i want to run dc opf for 24 hr
>>>>
>>>> i have the load scale for 24hr, and i want to multiply each load bus
>>>> with this load scale to get the load for 24hr in MW
>>>>
>>>> how can i do that?
>>>>
>>>> i have attached:
>>>>
>>>> - case file
>>>> -load scale
>>>> -dcopf code
>>>>
>>>>
>>>>
>>>> <load scale.xlsx><OETC_DCOPF_2.m><OETC_OPF.m>
>>>>
>>>>
>>>>
>>>
>>>
>
>

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