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ppal_25568718-Opal
18-Opal
August 21, 2016
Solved

HELP WITH GIVEN and FIND

  • August 21, 2016
  • 5 replies
  • 10308 views

I am trying to solve a set of equations using Given and Find.  I wish to define some boundary conditions. Eg. I may wish to limit some answers to a range between 0 and  1 or limit some values to positive numbers numbers. I don't know how. Your help will be appreciated. File is attached.

Basically I am using current and voltage data from relays at both ends of a line to find a fault location.

If we limit all values to real and  M is between  0 and 1

The expected answer for U  is about  0.75 - 0.76.

Best answer by -MFra-

I think you have not addressed the problem properly. You should consider the impedances of which the resistance is the real part and the reactance the imaginary part. The results must adhere to these definitions.

Z1&Zm.jpg

5 replies

StephenW
StephenW23-Emerald III
23-Emerald III
August 22, 2016

Toby Metcalf‌ Can you please move this question to the Mathcad section.

The PTC Community Help‌ section of the forum is for question about the forum itself. Toby will move the post to the right place and you'll get the help you need.

T
TmetcalfEmeritus
Emeritus
August 22, 2016

Thank you Stephen - moving to PTC Mathcad

W
Werner_E25-Diamond I
25-Diamond I
August 22, 2016

Why do you name the variable M when it should be U?

You can name it U in the solve block and in all your equations and can albeit name it U in the solve vector, too.

As for the conditions, you may simply write 0<M<1 somewhere in the solve block. You get U=0.534

If the guess values are real, Mathcads solver will stay with real numbers.

If you want non-reals, then you would have to provide a non-real guess value.

But it seems that as your formulas use the complex unit j, you get a complex value for M rsp. U

To force a variable R1 to be real you could add something like Im(R1)=0.

As you have just 3 equations but 6 variables to solve for, you have a great degree of freedom concerning the possible solutions.

Werner

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-MFra-21-Topaz IIAnswer
21-Topaz II
August 22, 2016

I think you have not addressed the problem properly. You should consider the impedances of which the resistance is the real part and the reactance the imaginary part. The results must adhere to these definitions.

Z1&amp;Zm.jpg

W
Werner_E25-Diamond I
25-Diamond I
August 22, 2016

No need to demand Im(M)=0 as M>0 implies M is real.

But of course it does not harm.

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