1. An Alternative Derivation of the Hagen-Poiseuille Equation
Group 14
Jill Brown
Sabine Knedlik
Jonathan Lok

2. Alternative Derivation of the Hagen-Poiseuille Equationz
Control volume
r+r Po PL
Looking at the geometry
of this system, we choose
cylindrical coordinates. z

4. 0 = - r [P/z] + r [rz/r]
We can factor out r and simplify to get:
0 = - r [P/z] + r [rz/r]
Now we shrink the control volume element to infinitesimal size by taking the limit as r and z
The pressure and shear stress are functions of only z and r. When flow is fully developed, the P is constant = dP/dz.
0 = - r dP/dz + r drz/dz
This becomes a separable differential equation:
r dP/dz = r drz/dr
∫ r dP/dz dr = ∫ r drz
rz = (dP/dz) r/2 + C/r

5. The constant C can be evaluated if the value of rz is known at some radius r; we can take r=0 at the center (axis). Since rz must be a finite value at r=0, C must be equal to zero.
rz = (dP/dz) r/2
Shear stress varies linearly in pipe from r=0 to r=R. We assume a Newtonian viscosity relationship for laminar flow.
rz =  dvz/dr
 dvz/dr = (dP/dz) r/2

6. Rearranging, we get:
Integrate:
We have a no-slip boundary condition
, r = R
By plugging in C value and factoring, we get:

7. The velocity is at a maximum when r=o (center of pipe)
Shear stress is greatest on wall!
Velocity profile = parabolic
So if we rewrite,
The average

8. For a parabolic profile:
Rearranging we get: if
This only applies for a Newtonian
fluid, steady laminar flow, fully
developed and incompressible!!!!
Then:

9. THE HAGEN- POISEUILLE EQUATION!!!!
The velocity profile across the pipe is parabolic. We have to
determine the mass flow rate Q of fluid passing in each second
through any cross section of the pipe.
A mass of
passes each second through
element
of the cross-sectional area.
where
THE HAGEN-
POISEUILLE
EQUATION!!!!
Hence: