Solving Orifice Plate Calculation Equation - Help Needed

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  • Thread starter Thread starter mattaddis
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Discussion Overview

The discussion revolves around solving an equation related to orifice plate calculations, specifically focusing on determining the variable d2. Participants are engaged in mathematical reasoning and providing assistance for a mechanical engineering application.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Matt seeks help to solve the equation for d2: α = π(d2/2)²√(1/(1-(d2/d1)⁴)).
  • One participant suggests squaring the equation to eliminate the square root and provides a series of algebraic manipulations leading to a polynomial form.
  • Another participant proposes a potential solution for d2: d2 = 1/√[4]{(π/(4α))² + 1/d1⁴}.
  • Matt tests both proposed formulas and finds that one yields the desired result while the other does not correctly solve for d2.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct formula for d2, as Matt indicates that one equation provides the correct result while the other does not.

Contextual Notes

There are unresolved aspects regarding the manipulation of the original equation and the conditions under which the proposed solutions are valid.

mattaddis
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Hello there,

I am a mechanical engineer and am attempting to build an Excel document for calculations purposes.

Currently I am stuck on orifice plate calculations.

I was wondering if someone could help me solve the following equation for d2?

[tex]\alpha[/tex] = [tex]\pi[/tex](d2/2)2 [tex]\sqrt{}[/tex]1/1-(d2/d1)4

Any help would be appreciated.

Thanks in advance.

(Apologies if I have posted this in the wrong place)

Matt
 
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Sure thing,

[tex]\alpha = \pi\left(\frac{d_2}{2}\right)^2\sqrt{\frac{1}{1-\left(\frac{d_2}{d_1}\right)^4}}[/tex]

square the entire equation to rid yourself of the square root:

[tex]\alpha^2 = \pi^2\left(\frac{d_2}{2}\right)^4\left(\frac{1}{1-\left(\frac{d_2}{d_1}\right)^4}\right)[/tex]

Multiply through by that denominator:

[tex]\alpha^2\left(1-\left(\frac{d_2}{d_1}\right)^4}\right) = \pi^2\left(\frac{d_2}{2}\right)^4[/tex]

Expand the left side, and move the right to the left side:

[tex]\alpha^2-\alpha^2\left(\frac{d_2}{d_1}\right)^4} -\pi^2\left(\frac{d_2}{2}\right)^4=0[/tex]

This can be more easily visualized as:

[tex]\alpha^2-d_2^4\frac{\alpha^2}{d_1^4} -d_2^4\frac{\pi^2}{16}=0[/tex]

Factorize out the required variable:

[tex]\alpha^2-d_2^4\left(\frac{\alpha^2}{d_1^4} +\frac{\pi^2}{16}\right)=0[/tex]

Well you can probably finish it from here, and you might want to manipulate some things so you don't have fractions in fractions.
 
Check if it's
[tex]d_2=\frac{1}{\sqrt[4]{\left(\frac{\pi}{4\alpha}\right)^2+\frac{1}{d_1^4}}}[/tex]
 
Thankyou both for yor prompt reply.

I have tested the two formulas based on existing figures and Mentallic's formula/equation gives the desired result.

Gerenuk, you equation gives d1, not d2.

Thank you very much for your assistance.

Matt
 
You're welcome :smile:
 

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