Download Derive from first principles the Poiseuille equation for

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.
Consider steady flow of a Newtonian fluid through a straight tube of circular crosssection 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.
P1
r
L
P2
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 pressure difference. In an ideal fluid with no viscosity, the fluid
will move in bulk. However in viscous fluids, there will be a resistance to the fluid
motion as shear force is induced between adjacent layers of fluid. In the figure, the shear
force retarding the motion is acting on the surface of the fluid element and given by
FT = τ2πrL
In fully-developed steady flow, these two forces balance each other and by equating the
above two expressions, we obtain
πr2(p1 – p2) = τ2πrL
and this equation can be reduced to
τ = r(p1 – p2)/(2L)
The constitutive law for a Newtonian fluid is given by:
du
  
dr
(1)
(2)
where u and μ are the velocity and viscosity of the fluid, respectively. In flow through
tubes, the velocity magnitude is zero near the wall and increases towards the center of the
tube. Thus, the ratio du/dr is negative and a negative sign is arbitrarily introduced.
Combining (1) and (2) and rewriting, we obtain:
 P 
rdr
du  
 2L 
with ΔP= p1 – p2
Integrating the above relationship, we obtain:
r 2 P
u
C
4L
where C is a constant
(3)
Since the tube wall does not move, the fluid adjacent to the wall is stationary. Thus the
boundary condition can be specified as u = 0 at r = R. Applying the boundary condition
to (3), we can obtain C:
R 2 P
C
4L
and the expression for the velocity profile over the tube section will be given by:
R 2  r 2 P
(4)
u
4L


Now we need to relate (4) with the flow Q. The flow in an annulus of thickness r is
given by:
So the flow can be written:
R 4 P
Q
8L
(5)
Equation 5 is referred to as the Poiseuille Equation and relates the flow rate to the
pressure drop across the tube of radius R and a Newtonian fluid with a viscosity
coefficient μ.
If the flow is laminar, from equation (4), the velocity profile will be parabolic with the
maximum velocity umax occurs at the center of the tube, as shown in the below figure:
From equation (4) the peak velocity (r=0) is
u max 
R 2 P
4L
The mean velocity can be calculated using equation (5)
u mean 
Q R 4 P
1
R 2 P

 2 
A
8L
8L
R
Hence the ratio
u mean
u max
R 2 P
1
8L
 2

R P 2
4 L
References:
1. http://www.efm.leeds.ac.uk/CIVE/CIVE1400/Examples/eg7_ans.htm
2. http://hyperphysics.phy-astr.gsu.edu/Hbase/pfric.html
3. Transport Phenomena, 3rd ed. by Bird et.al.