Derive from first principles the Poiseuille equation for pressure drop generated by the steady flow of a Newtonian fluid through a straight tube of circular cross-section. If the flow is laminar, what is the form of the velocity profile with in the tube? Show that the mean velocity is half the peak in such circumstances.
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Consider steady flow of a Newtonian fluid through a straight tube of circular cross-section with a radius R. Consider an element of fluid with a radius r and axial length L. Let the pressure at the end be p1 and p2 as shown in the figure.
The net force pushing the fluid is
FP =p1πr2 - p2πr2 = πr2(p1 - p2)
Note that the force inducing the motion of the fluid is the difference or gradient in pressure and does not depend upon the absolute magnitude of the pressure itself. In other words, even if the pressure in the tube is very large, there will be no motion of the fluid if there is no difference in pressure between the two ends and the motion will be in the direction of the positive ...
The solution provides step-by-step explanations and derivations for the Poiseuille equation in a 3-page Word document.