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8-Gravel

## Symbolic Equation

I am trying to set up a solution for a symbolic equation, but somehow the solution units are not working out correctly.  I have attached the worksheet and would appreciate any comments on what is wrong.  The result should be in feet, but as you can see, there are three variables that are not being handled correctly.

Also for some reason I get an error message that says "The contents of the attachment doesn't match its file type."

Do I need to save back to an earlier version?   I am running MathCAD 15.0.

ACCEPTED SOLUTION

Accepted Solutions
25-Diamond I
(To:wbell)

You had used the wrong equal sign a couple of times!

You used the boolean equal (the fat equal) instead of the assignment equal (the := ).

As a couple of your expressions are dependent on an unknown D, its more natural to turn them into functions of 😧

The symbolic result is too large to display because the different units in play are all seen as unknown variabled by the symbolic engine. You will get one real and two non-real results.

You may consider using numerical methods to get the value of D (either the function "root" or a solve block:

6 REPLIES 6
21-Topaz I
(To:wbell)

Hi,

You may have to zip it first. A serious bug in this forum often prevents from attaching a mathcad file. In Windows Explorer click with the right-mouse button on the mathcad file and select 'Send to' -> 'Compressed (zipped) folder'. The resulting .zip file can be attached

Zip file attached, thank you.

21-Topaz I
(To:wbell)

Hi,

I have redefined some variables as functions of D so they vary when you change D.  This resolves the first problem

Second is the equation you are trying to solve for 0 has units so they must be declared.

You are an engineer after a numerical solution use a solve block not the symbolic equation solver.

25-Diamond I
(To:wbell)

You had used the wrong equal sign a couple of times!

You used the boolean equal (the fat equal) instead of the assignment equal (the := ).

As a couple of your expressions are dependent on an unknown D, its more natural to turn them into functions of 😧

The symbolic result is too large to display because the different units in play are all seen as unknown variabled by the symbolic engine. You will get one real and two non-real results.

You may consider using numerical methods to get the value of D (either the function "root" or a solve block:

8-Gravel
(To:Werner_E)

Thank you Werner.  I will make the changes and press on with the analysis.

23-Emerald III
(To:wbell)

It's no problem to do it symbolically, as long as you do it properly:

Now we can go ahead and solve that equation for D, then substitute in the equation for K.p:

Of course the solution is much bigger, and you don't want to see all, but you should be able to work with its results:

Success!
Luc

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