The Hagen–Poiseuille equation can be derived from the
Navier–Stokes equations. The
laminar flow through a pipe of uniform (circular) cross-section is known as Hagen–Poiseuille flow. The equations governing the Hagen–Poiseuille flow can be derived directly from the
Navier–Stokes momentum equations in 3D cylindrical coordinates by making the following set of assumptions: • The flow is steady ( ). • The radial and azimuthal components of the fluid velocity are zero ( ). • The flow is axisymmetric ( ). • The flow is fully developed ( ). Here however, this can be proved via mass conservation, and the above assumptions. Then the angular equation in the momentum equations and the
continuity equation are identically satisfied. The radial momentum equation reduces to , i.e., the
pressure is a function of the axial coordinate only. For brevity, use instead of u_x. The axial momentum equation reduces to : \frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)= \frac{1}{\mu} \frac{\mathrm{d} p}{\mathrm{d} x} where is the dynamic viscosity of the fluid. In the above equation, the left-hand side is only a function of and the right-hand side term is only a function of , implying that both terms must be the same constant. Evaluating this constant is straightforward. If we take the length of the pipe to be and denote the pressure difference between the two ends of the pipe by (high pressure minus low pressure), then the constant is simply :-\frac{\mathrm{d}p}{\mathrm{d}x} = \frac{\Delta p}{L} = G defined such that is positive. The solution is : u = -\frac{Gr^2}{4\mu} + c_1 \ln r + c_2 Since needs to be finite at , . The no slip
boundary condition at the pipe wall requires that at (radius of the pipe), which yields . Thus we have finally the following
parabolic velocity profile: : u = \frac{G}{4\mu} \left(R^2 - r^2\right). The maximum velocity occurs at the pipe centerline (), . The average velocity can be obtained by integrating over the pipe
cross section, : {u}_\mathrm{avg}=\frac{1}{\pi R^2} \int_0^R 2\pi ru \mathrm{d}r = \tfrac{1}{2} {u}_\mathrm{max}. The easily measurable quantity in experiments is the volumetric flow rate . Rearrangement of this gives the Hagen–Poiseuille equation : \Delta p = \frac{8\mu Q L}{\pi R^4}. Although more lengthy than directly using the
Navier–Stokes equations, an alternative method of deriving the Hagen–Poiseuille equation is as follows.
Liquid flow through a pipe Assume the liquid exhibits
laminar flow. Laminar flow in a round pipe prescribes that there are a bunch of circular layers (lamina) of liquid, each having a velocity determined only by their radial distance from the center of the tube. Also assume the center is moving fastest while the liquid touching the walls of the tube is stationary (due to the
no-slip condition). To figure out the motion of the liquid, all forces acting on each lamina must be known: • The
pressure force pushing the liquid through the tube is the change in pressure multiplied by the area: . This force is in the direction of the motion of the liquid. The negative sign comes from the conventional way we define . •
Viscosity effects will pull from the faster lamina immediately closer to the center of the tube. •
Viscosity effects will drag from the slower lamina immediately closer to the walls of the tube.
Viscosity When two layers of liquid in contact with each other move at different speeds, there will be a
shear force between them. This force is
proportional to the
area of contact , the velocity gradient perpendicular to the direction of flow , and a proportionality constant (viscosity) and is given by : F_\text{viscosity, top} = - \mu A \frac{\Delta v_x}{\Delta y}. The negative sign is in there because we are concerned with the faster moving liquid (top in figure), which is being slowed by the slower liquid (bottom in figure). By
Newton's third law of motion, the force on the slower liquid is equal and opposite (no negative sign) to the force on the faster liquid. This equation assumes that the area of contact is so large that we can ignore any effects from the edges and that the fluids behave as
Newtonian fluids.
Faster lamina Assume that we are figuring out the force on the lamina with
radius . From the equation above, we need to know the
area of contact and the velocity
gradient. Think of the lamina as a ring of radius , thickness , and length . The area of contact between the lamina and the faster one is simply the surface area of the cylinder: . We don't know the exact form for the velocity of the liquid within the tube yet, but we do know (from our assumption above) that it is dependent on the radius. Therefore, the velocity gradient is the
change of the velocity with respect to the change in the radius at the intersection of these two laminae. That intersection is at a radius of . So, considering that this force will be positive with respect to the movement of the liquid (but the derivative of the velocity is negative), the final form of the equation becomes : F_\text{viscosity, fast} = - 2 \pi r \mu \, \Delta x \, \left. \frac{\mathrm{d}v}{\mathrm{d}r} \right|_r where the vertical bar and subscript following the
derivative indicates that it should be taken at a radius of .
Slower lamina Next let's find the force of drag from the slower lamina. We need to calculate the same values that we did for the force from the faster lamina. In this case, the area of contact is at instead of . Also, we need to remember that this force opposes the direction of movement of the liquid and will therefore be negative (and that the derivative of the velocity is negative). : F_\text{viscosity, slow} = 2 \pi (r + \mathrm{d}r)\mu \, \Delta x \left. \frac{\mathrm{d}v}{\mathrm{d}r} \right|_{r+\mathrm{d}r}
Putting it all together To find the solution for the flow of a laminar layer through a tube, we need to make one last assumption. There is no
acceleration of liquid in the pipe, and by
Newton's first law, there is no net force. If there is no net force then we can add all of the forces together to get zero : 0 = F_\text{pressure} + F_\text{viscosity, fast} + F_\text{viscosity, slow} or : 0 = - \Delta p 2 \pi r\,\mathrm{d}r - 2 \pi r \mu \,\Delta x \left. \frac{\mathrm{d}v}{\mathrm{d}r} \right|_r + 2 \pi (r + \mathrm{d}r) \mu \,\Delta x \,\left. \frac{\mathrm{d}v}{\mathrm{d}r} \right \vert_{r+\mathrm{d}r}. First, to get everything happening at the same point, use the first two terms of a
Taylor series expansion of the velocity gradient: : \left. \frac{\mathrm{d}v}{\mathrm{d}r} \right|_{r+\mathrm{d}r} = \left. \frac{\mathrm{d}v}{\mathrm{d}r} \right|_r + \left. \frac{\mathrm{d}^2 v}{\mathrm{d}r^2} \right|_r \,\mathrm{d}r . The expression is valid for all laminae. Grouping like terms and dropping the vertical bar since all derivatives are assumed to be at radius , : 0 = - \Delta p 2 \pi r \, \mathrm{d}r + 2 \pi \mu \, \mathrm{d}r \, \Delta x \frac{\mathrm{d}v}{\mathrm{d}r} + 2 \pi r \mu \, \mathrm{d}r \, \Delta x \frac{\mathrm{d}^2 v}{\mathrm{d}r^2} + 2 \pi \mu (\mathrm{d}r)^2 \, \Delta x \frac{\mathrm{d}^2 v}{\mathrm{d}r^2}. Finally, put this expression in the form of a
differential equation, dropping the term quadratic in . : \frac{1}{\mu} \frac{\Delta p}{\Delta x} = \frac{\mathrm{d}^2 v}{\mathrm{d}r^2} + \frac{1}{r} \frac{\mathrm{d}v}{\mathrm{d}r} The above equation is the same as the one obtained from the Navier–Stokes equations and the derivation from here on follows as before.
Startup of Poiseuille flow in a pipe When a
constant pressure gradient is applied between two ends of a long pipe, the flow will not immediately obtain Poiseuille profile, rather it develops through time and reaches the Poiseuille profile at steady state. The
Navier–Stokes equations reduce to :\frac{\partial u}{\partial t} = \frac{G}{\rho} + \nu \left(\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r}\right) with initial and boundary conditions, :u(r,0) = 0, \quad u(R,t) = 0. The velocity distribution is given by :u(r,t) = \frac{G}{4\mu}\left(R^2-r^2\right) - \frac{2GR^2}{\mu} \sum_{n=1}^\infty \frac{1}{\lambda_n^3} \frac{J_0(\lambda_n r/R)}{J_1(\lambda_n)} e^{-\lambda_n^2 \nu t/R^2}, \quad J_0\left(\lambda_n\right)=0 where is the
Bessel function of the first kind of order zero and are the positive roots of this function and is the
Bessel function of the first kind of order one. As , Poiseuille solution is recovered. ==Poiseuille flow in an annular section==