1. 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:

2. 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.

3. 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.

4. ( ) [ ] [ ] 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