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Help, i need to get all solutions for a system of equations.

JS_10858783
2-Explorer

Help, i need to get all solutions for a system of equations.

Hi, i have the follow system of 3 equations 

JS_10858783_0-1711393454984.png

i need to find the operating region for the system where vrd, vrqRe, Le, Nc, Cdc and w are constant parameters, Md and Mq are parameters that i can change and i want to introduce in the system to get all the Isq, Isq and Vdc solutions. In resume, i want to vary Md and Mq and get the corresponding Isd, Isq and Vdc for each par of Md and Mq input, anyone can help me with this? i will be very gratefull.

 

PD: Sorry if my english is not so good.

ACCEPTED SOLUTION

Accepted Solutions
Werner_E
25-Diamond I
(To:JS_10858783)

 

 to get all the Isq, Isq and Vdc solutions.

 

Guess you meant to solve for Isd, not twice for Isq 😉

 

Here is the parametrized solve block Terry was talking about.

You may provide better guess value but as we can see, even my bad guesses (all 1) gives a solution, so the system seems to be well determined.

Werner_E_0-1711399511141.png

You may even evaluate the solve block function symbolically to see the functional solution

Werner_E_1-1711399574076.png

And you now may define functions for each of the three variables you are looking for dependent on the two input values

Werner_E_2-1711399653692.png

 

You may also consider using units ... that's something Mathcad is pretty good at

 

BTW, If you define an symbolically evaluate the solve block before you define your variables, you may see the solution dependent on all constants and input variables

Werner_E_0-1711400088043.png

 

View solution in original post

7 REPLIES 7

Hi,

What you need is a parametrized Solve Block but first.

Mathcad 15?

lsd,,lsq, Vdc?

 

Can you provide your sheet with the values for Re, Le, Nc, Cdc, w

What range is Md and Mq varied over?

 

Cheers

Terry

 

Hi Terry, thankss

 

yep, the version is Mathcad 15, i will give you more context, im modeling an electrical device (an STATCOM), these three equations are the equations of the electrical LVK and LCK that modeling the system, so i want to know what are all the values of current of q-axys that i can get (Isd and Isq are currents). Vdc is the DC voltage of each sub-module.

 

I need to get every Md and Mq values that meet the following condition:

 

JS_10858783_0-1711396758092.png

and store those values in pairs in a vector or something similar and then, reeplace these values in the system of equations. So that could be the range.

 

thanks again Terry

Here is the sheet, sorry i forgot upload it.

Werner_E
25-Diamond I
(To:JS_10858783)

 

 to get all the Isq, Isq and Vdc solutions.

 

Guess you meant to solve for Isd, not twice for Isq 😉

 

Here is the parametrized solve block Terry was talking about.

You may provide better guess value but as we can see, even my bad guesses (all 1) gives a solution, so the system seems to be well determined.

Werner_E_0-1711399511141.png

You may even evaluate the solve block function symbolically to see the functional solution

Werner_E_1-1711399574076.png

And you now may define functions for each of the three variables you are looking for dependent on the two input values

Werner_E_2-1711399653692.png

 

You may also consider using units ... that's something Mathcad is pretty good at

 

BTW, If you define an symbolically evaluate the solve block before you define your variables, you may see the solution dependent on all constants and input variables

Werner_E_0-1711400088043.png

 

Hi,

My attempt

Graphs results.

But the condition is

Werner_E_0-1711412257144.png

and not

Werner_E_4-1711412633490.png

 

which means we have to deal with all points in the Md-Mq-plane inside and on the circle with origin as center and radius sqrt (3/2).

But I doubt that the resulting 3D plots are of any value as the values seem to be much too large to handle...

I scaled them by 10^-20 to get a better 3D plot

Werner_E_5-1711412713297.png
Werner_E_6-1711412762190.png

 

Thank you very much to both, I am truly very grateful. I will stick with this answer, but both have been very helpful to me, thank you.

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