# Maximum velocity in a viscous fluid

Situation: A viscous fluid is flowing through a tube, and the rate of ejection is x $m^3s^{-1}$

Finding the velocity of the ejected fluid would be simple enough, by dividing the rate of ejection by the area from which the fluid is being ejected. However, what about for finding the maximum velocity of the fluid? (which is obviously at the center)

My logic: The velocity of a viscous fluid decreases as you move towards the walls. It is maximum at the center and zero at the point in contact with the wall. The velocity we calculated using the rate of ejection should be the average of the two. Is this correct?

• For laminar flows, yes: en.wikipedia.org/wiki/… . For turbulent flows the problem is much, much more complex. – Bernhard Nov 26 '14 at 15:23
• For turbulent flows, calculating this from theory would require computational fluid dynamics (potentially a lot of computation time). In practice, there's a lot of empirical data oriented towards doing this kind of calculation.. Friction factors can be found in a Moody Chart. – user3823992 Nov 26 '14 at 16:18
• Ummm @Bernhard would you mind explaining the difference between turbulent and laminar flows? – Gummy bears Nov 27 '14 at 13:54
• @Gummybears I recommend you to consult a textbook on fluid mechanics. Otherwise, start here: en.wikipedia.org/wiki/Reynolds_number (and follow the link to the pages "Laminar flow" and "Turbulence") – Bernhard Nov 27 '14 at 13:58
• Bernhard: Er, no? The velocity is distributed parabolically $$u=u_{max)(1-r/^2R^2)$$ – Philip Roe Apr 19 '17 at 2:00