This is a pipe flow equation.

Tobyjug

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Hi, the equation I have is for gas flow in a pipe 'Q' being the flow rate in m³/s. I am having great difficulty in transposing to get D as the subject (D being the internal pipe diameter), can anyone help. The Number pd is the pressure drop Pa/m

q= -2.695*sqrt(pd)*(D)^2.5*log(((4.05*10^-7)/D)+(2.3*10^-5)/sqrt(pd)/D^1.5))

This equation is correct for flow 'Q' I would just like to make D the subject.
I am just rubbish at transposing. Can anyone help.

Thank you
Tobyjug
 
q= -2.695*sqrt(pd)*(D)^2.5*log(((4.05*10^-7)/D)+(2.3*10^-5)/sqrt(pd)/D^1.5))

This equation is correct for flow 'Q' I would just like to make D the subject.
If I'm reading the equation correctly (guessing at the meaning of the triple "fraction"), and assuming that "q" is meant to be "Q", then the equation is as follows:

. . . . .\(\displaystyle \displaystyle{Q\, =\, -2.695\sqrt{pd\, }D^{2.5}\, \log\left(\frac{4.05\times 10^{-7}}{D}\, +\, \frac{2.3\times 10^{-5}}{\sqrt{pd\, }D^{1.5}}\right)}\)

I'm not at all certain that this can be solved for "D=". Is that what the assignment requires? ;)
 
Since D appears both "in" the logarithm and out of it, this equation clearly cannot be solved for D in terms of elementary functions. It is possible that you might be able to re-write it as an exponential and so solve in terms of "Lambert's W function", defined as the inverse function to \(\displaystyle f(x)= xe^x\).
 
If I'm reading the equation correctly (guessing at the meaning of the triple "fraction"), and assuming that "q" is meant to be "Q", then the equation is as follows:

. . . . .\(\displaystyle \displaystyle{Q\, =\, -2.695\sqrt{pd\, }D^{2.5}\, \log\left(\frac{4.05\times 10^{-7}}{D}\, +\, \frac{2.3\times 10^{-5}}{\sqrt{pd\, }D^{1.5}}\right)}\)

I'm not at all certain that this can be solved for "D=". Is that what the assignment requires? ;)

Thank you for your reply. I had a correct answer for q (being flow rate in this equation, D being internal bore of tubing and pd being pressure drop in Pa/m)
I had to cheat, I put the equation in excel and made a spinner for Pa with a percentage error count. which gave me the right answer by changing pa using the spinner. Although I would have preferred an equation.
Thanks again
Tobyjug
 
Thank you

Thank you I will try some of this math. Regards Tobyjug

I'm changing your Q to q, your d to a, your D to d.

u = 4.05 * 10^(-7)
v = 2.3 * 10^(-5)
w = p*a
x = -2.695 * w
y = u/d + v/(w*d^1.5)

Your equation can now be arranged this way:
q/x = d^2.5 * LOG(y)

a, p and q are givens; no idea what they can be, but assuming:
a=2, p=4.5, q=1, then:
d = .17862

I calculated that as a "hit and miss" looper.
Lambert method would be closer...I think:confused:

Tried q=36 (instead of 1): d = .69187
 
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