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klax9-Granite
9-Granite
November 2, 2016
Solved

AC induced voltages on buried pipelines.

  • November 2, 2016
  • 3 replies
  • 3240 views

I am struggling to show the result of the calculation in terms of the voltage and the phase angle (in degrees).  Vp calculation is correct but it does not show the phase angle.  The phase angle should be -121.4 degrees as shown on the sample calculation at the end of the works sheet.  I can produce the correct result in Matlab using the rad2deg function ( [abs(Vp) rad2deg(angle(Vp))] ) but I cannot find an equivalent function in Mathcad.  Any guidance gratefully received.

Thank you.

Best answer by RichardJ

If you take the absolute value it calculates the magnitude and you lose all the phase information. Get rid of the absolute operator and just format the result (on the math formatting tab) to magnitude and angle:

3 replies

L
LucMeekes23-Emerald IV
23-Emerald IV
November 2, 2016

X simply divide the expression you want to plot by deg should do the job.

Luc

K
klax9-GraniteAuthor
9-Granite
November 2, 2016

Thank you for the prompt reply.  I don't understand your response because If I divide Vp by 1 degree I do not get 16.9  and a phase angle of -121, which is the correct answer.  I get -505.147 and -832.498.  At this stage I am not interested in plotting it, just calculating it.

The real element (Re) is -8.816 and the imaginary element (Im) is -14.53

If I use the < symbol for the phase angle then the answer I wish to achieve is 16.9 < -121.4

Kind Regards

R
RichardJ19-TanzaniteAnswer
19-Tanzanite
November 2, 2016

If you take the absolute value it calculates the magnitude and you lose all the phase information. Get rid of the absolute operator and just format the result (on the math formatting tab) to magnitude and angle:

K
klax9-GraniteAuthor
9-Granite
November 2, 2016

Thank you Richard.  I did know about that feature at all!

Kind regards

N
nas0k1-Visitor
1-Visitor
November 2, 2016

I noticed that you didn't use units, which work quite well in Prime; as an example:

You may want to try it in your worksheet to clearly communicate your application's attributes.

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    -MFra-21-Topaz II
    21-Topaz II
    November 2, 2016

    Sorry if I dare to intervene in the discussion now ended. What I want to point out is that using the units, the results are the same, although it is not very clear what unit has the constant Dep  (but if the argument of the logarithm must be dimensionless, it follows that, having the constants Dab, Dbp, Dcp, units in meters, also Dep must be in meters).

    low frequency current induction.jpg

      K
      klax9-GraniteAuthor
      9-Granite
      November 2, 2016

      Spot on!  Thank you.

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