Talk:Hagen–Poiseuille flow from the Navier–Stokes equations

Hagen-poiseuille equation;

$$ \Delta P= (32\mu v L)/ D^2$$

See equation 7 of Three approaches to calculating the velocity profile of a laminar incompressible fluid flow in a hollow tube

C. Neipp1, A. Hernández1, T. Beléndez2, J. J. Rodes1, and A. Beléndez1 1Departamento de Física, Ingeniería de Sistemas y Teoría de la Señal, Universidad de Alicante, E-03080 Alicante, Spain 2Departamento de Ciencia y Tecnología de los Materiales, Universidad Miguel Hernández de Elche, Avda. del Ferrocarril, s/n. E-03202 Elche Alicante, Spain

link http://scitation.aip.org/getpdf/servlet/GetPDFServlet?filetype=pdf&id=AJPIAS000071000001000046000001&idtype=cvips&doi=10.1119/1.1517596&prog=normal

the other steps are correct ( I think)

Jgreeter (talk) 02:35, 5 May 2011 (UTC)


 * I do not see anything wrong with the 'equation for pressure drop' provided in the article. Please explain if you found it incorrect. Salih  ( talk ) 03:51, 5 May 2011 (UTC)

Pressure gradient
"Assuming that the pressure decreases linearly across the length of the pipe, we have $$ - \frac{\partial p}{\partial z} = \frac{\Delta p}{L} $$ (constant)"

This sentence is actually imprecise. By the z-component of the equation of motion

$$ \frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial u_z}{\partial r}\right)= \frac{1}{\mu} \frac{\partial p}{\partial z}$$ ,

indeed, the pressure gradient must be constant, since – by incompressibility – the left-hand side is a function of $$r$$ only, and the right-hand side is a function of $$z$$ only. — Preceding unsigned comment added by Aspro89 (talk • contribs) 12:29, 4 December 2017 (UTC)