Navier - Stokes Equation Earlier we solve for velocity profile in a tube using a shell balance approach. This was long and tedious. Now we will use the Navier-Stokes equation. For Flow in Cylindrical Conduit: fully developed steady incompressible laminar flow r R z From Appendix E (p. 709), the r-component of the Navier-Stokes equation in cylindrical coordinate is:

horizontal orientation Simplifying, symmetry symmetry Steady state symmetry Similar result obtained for the -component. z-direction: ( ) ú û ù ê ë é + - = ¶ 2 1 z v r P t g q m Simplifying, Fully developed Fully developed ( ) ú û ù ê ë é + - = ¶ 2 1 z v r P t g q m symmetry Steady flow Assume horizontal orientation symmetry Pressure change in the  and r direction are negligible. \ Notice that the z-component is more interesting since the flow (or momentum) is axial.

Assuming constant pressure drop: In order for vz to be finite at r = 0, must equal zero. vz = 0 at r = R, so, This is the same result as before.

( ) [ ] [ ] t = - ( ) ( ) ( ) ( ) ( ) t = -m R D P rz 2 L F = t × A t To find the shear stress at wall: To find the force on pipe of length L: Start from Stokes relation: F = t × A [ ] pipe rz wall surface t = -m ¶ v + ¶ v z r rz ¶ r ¶ z F = t × 2 p RL t = -m dv pipe rz wall z rz ( ) dr = = - R D P × 2 p r R RL 2 L [ ] ( ) t = - m r dP rz w 2 m dz r = R ( ) t = - R dP rz 2 dz t dz = - R dP ò ò rz 2 ( ) ( ) t z - z = - R P - P rz 2 1 2 2 1 ( ) t = - R D P rz 2 L