Poiseuille flow in tubes of circular cross-section

Consider now the steady, laminar flow of an incompressible Newtonian liquid through a uniform cylindrical tube, of internal radius a and length L. Let the axis of the tube be in the z direction. If is the liquid velocity component in the direction of the axis, then, assuming that there is no radial liquid flow, the equation of continuity gives du ldz = 0 and, further, the pressure is constant over any cross-section. Since the flow is steady bu /bt = 0 and this, with bu jbz = 0, ensures that Dw /Df is zero and the solution obtained is a creeping solution. 

Under steady flow conditions the net force acting on the Uquid contained in a cylindrical element of radius r and length L, coaxial with the tube, must be zero. 

Let there be a constant body force Z per unit mass, acting in the z direction. The forces acting on the liquid element in the z direction are the body force, the pressure force arising from the difference between the pressure pi at the inlet end of the tube and the pressure P2 at the outlet end, and the viscous force. The magnitude of the body force acting on the liquid element is wr LpZ, where p is the density of the liquid, and the pressure force is (pi —p2) r - Since the velocity gradient at the surface of the element is everywhere (dw /dr) ., the viscous drag exerted on the curved surface of the element by the remainder of the liquid is  

The pressure gradient (pi — P2)lL must be at most a constant since it cannot vary with r. Let it be denoted by +G. Then  

The derivation of Poiseuille s equation assumes that, for the whole length of the tube, the liquid is travelling with constant 

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Poiseuille flow in tubes of circular cross-section... 

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