Hagen-Poiseuille Equation(Fluid flow in a tube) The Hagen-Poiseuille equation relates volumetric flowrate in a tube to various parameters. z r z=0 z=L c L R Pressure: Po PL Arrows indicate flowing fluid Fluid is flowing in a tube of radius R. Performing a simple force balance on the liquid contained between z=0 and z=L will lead to the Hagen-Poiseuille equation and help us understand the factors involved. At equilibrium,  F = 0 The pressure on the liquid at z=0 acts to push the fluid down the tube; the force will simply be the area times the pressure. Force due to pressure at z=0: R2Po { plus z-direction

The pressure at z=L acts to push against the flow. Force due to pressure at z=L: R2PL { minus z-direction At the wall of the tube everywhere between z=0 and z=L, the viscous or frictional force resists the flow through the tube. This force per area is w, the shear stress at the wall. Force due to viscosity: 2RL w {Assume w acts in the plus z-direction (We have not included any gravitational force, but these can be added easily.) Substituting into the force balance: R2Po - R2PL + 2RL w = 0 Thus, w = - R(Po - PL) = -RP/2L 2L When a Newtonian fluid flows slowly in a tube, it flows in concentric cylindrical surfaces. Frictional forces are exerted as one layer slides over the next.

viscous or frictional force is that between the adjacent layers. The same equation or force balance as shown above can be applied at any radius. Now, the viscous or frictional force is that between the adjacent layers. z r R c L z=0 z=L arbitrary radius r Pressure: Po PL With the force balance, we now get:  = -r(Po - PL) (*) 2L Recall that  is related to the velocity gradient. For this coordinate system, that is dvz/dr. In words, how fast is one layer traveling relative to the next. (How would you expect this to affect the shear stress?)  = (dvz/dr) Substituting into (*), we obtain: (dvz/dr) = -r(Po - PL) This is a separable differential equation. dvz = -(Po - PL) r dr  vz = -(Po - PL) r2/2 + C1 2L 2L A boundary condition is used to find C1. We use what is called the “no-slip” boundary condition; that is, we require the velocity of the liquid to be equal to that of the tube wall at r = R (i.e. vz = 0 at r = R).

vz(R) = 0 = -(Po - PL) R2/2 + C1  C1 = (Po - PL)R2 2L 4L and vz = (Po - PL)R2 [ 1 - r2/ R2] < This is the velocity profile or velocity as a 4L function of r. The velocity profile for laminar flow of a fluid in a tube is parabolic! z r The fluid in this cap represents the volume of fluid flowing across plane z = 0 during some time t. Z = 0 To get the Hagen-Poiseuille Equation, we must calculate the volumetric flowrate Q in the tube. Let’s look at the cross-section at z = 0. The volume of fluid that has gone through dA in time t is: dV = vz(r, ) dA. t or dQ = vz(r, ) dA where dA = r d dr {Review polar or cylindrical coordinates--See MAPs} dA  r

Q = R4 (Po - PL) Hagen-Poiseuille Equation 8L Integrating, R 2 R 2 Q =  dQ =   vz(r, ) r d dr =   (Po - PL) R2( 1 - r2/ R2) r d dr      4L 0 0 0 0 R R Q =  R2 (Po - PL)  ( r - r3/ R2) dr =  R2 (Po - PL) [r2/2 - r4/(4R2)] 2L  2L 0 Q = R4 (Po - PL) Hagen-Poiseuille Equation 8L Note the presence of parameters for each of the primary factors. A tube can be a simple viscometer. Cost of viscometers: $100 to $105