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POISEUILLE FLOW
Poiseuille flow is the steady, axisymmetric flow in an infinitely long, circular pipe of radius, R, as sketched in Figure 1. The flow is caused by a pressure gradient, dp/dx, in the axial direction, x. The resulting
Figure 1: Poiseuille flow.
axisymmetric continuity equation for an incompressible fluid yields
∂ux ∂x
= 0 (Bic1)
so that the axial velocity, ux(r), is a function only of r, the radial coordinate. Using this the axisymmetric Navier-Stokes equations for an incompressible fluid of constant and uniform viscosity reduce to
∂p ∂x
= μ
∂^2 ux ∂r^2
r
∂ux ∂r
(Bic2)
∂p ∂r
= 0 (Bic3)
The second of these shows that the pressure, p(x), is a function only of x and hence the gradient, dp/dx, is well defined and a parameter of the problem. This allows the first of these equations (Bic2) to be integrated so that the velocity, ux, can be written as
ux(r) =
r^2 4 μ
dp dx
where C 1 and C 2 are integration constants to be determined by the application of the boundary conditions. On the axis the velocity cannot be infinite therefore C 1 must be zero. Moreover, the no-slip boundary condition on the pipe wall requires that ux(R) = 0 and so
C 2 = −
R^2
4 μ
dp dx
(Bic5)
and so the solution to Poiseuille flow in a circular pipe is
ux(r) =
4 μ
dp dx
(R^2 − r^2 ) (Bic6)
where the pressures, p 1 and p 2 , could be measured at two different x locations a distance apart in order to determine dp/dx = (p 1 − p 2 )/. Parenthetically, we should add that the general solution (Bic4) also allows
construction of the axial flow between two cylinders in which the outer radius of the inner cylinder is R 1 and the inner radius of the outer cylinder is R 2. The no-slip boundary conditions at these two surfaces then require that R^21 4 μ
dp dx
- C 1 ln R 1 + C 2 = 0 (Bic7)
and R^22 4 μ
dp dx
- C 1 ln R 2 + C 2 = 0 (Bic8)
from which C 1 and C 2 can be determined and the solution constructed. Beyond that it is also possible to stipulate that one of the cylinders has an axial velocity, U, and to proceed to construct yet another axisymmetric flow of this type.
However, we confine the present discussion to the simple Poiseuille flow while recognizing that parallel analyses can be carried out for the other variants. Note first that the velocity distribution in Poiseuille flow is parabolic according to equation (Bic6) and has a maximum velocity on the axis of
ux(0) =
R^2
4 μ
dp dx
(Bic9)
The volume flow rate, Q˙, is
Q˙ =
∫ R
0
2 πr ux dr =
πR^4 8 μ
dp dx
(Bic10)
so that the average velocity of the flow, u, is
u =
Q˙
πR^2
R^2
8 μ
dp dx
(Bic11)
Thus the average is 1/2 of the maximum. This last equation is often written as
Δp =
8 μu R^2
(Bic12)
to yield the pressure drop, Δp, over a length, , of the pipe.
The shear stress distribution in the flow is best examined by applying the momentum theorem to a cylindrical control volume of radius, r, centered on the axis of the pipe and with length, . Since the velocities in an out of the end of this cylinder are identical there is no net flux of momentum in or out of this control volume and so the theorem says the axial forces must balance. The forces due to the pressure on one end of this cylinder will be πr^2 p while the force on the other end a distance downstream will be πr^2 (p − Δp) where Δp is the pressure drop between the two locations. Consequently the net axial force in the positive x direction due to the pressure forces on the ends will be πr^2 Δp. The only other force acting in the axial direction is due to the shear stress acting on the outer surface area of the control volume. Denoting that shear stress by σrr which, in accord with the sign convention used in section (Bhd), is positive in the positive x direction, this shear force will be equal to 2πrσrr. Therefore the balance of forces acting on the control volume requires that
2 πrσrr + πr^2 Δp = 0 (Bic13)
so that
σrr = −
rΔp 2
(Bic14)
Figure 2: The Moody diagram for the friction factor, f, as a function of Reynolds number, 2ρuR/μ.
