📂Fluid Mechanics

Proof of Hagen–Poiseuille Law

Law ¹

Definition

An incompressible Newtonian fluid flowing through a pipe of constant diameter $D = 2R$ is considered. Let the distance from the pipe center in the direction perpendicular to the pipe axis be $r$. At the pipe wall ($r = R$) the velocity is $\mathbf{0}$, and at the pipe center ($r = 0$) the velocity is maximal. When the velocity distribution as a function of $r$ forms a parabola, the flow is called Hagen–Poiseuille flow.

Result

In this flow, the velocity $u(r)$ at a distance $r$ from the pipe center is given by: $$ u = - {\frac{ 1 }{ 4 \mu }} {\frac{ d p }{ d x }} \left( R^{2} - r^{2} \right) $$ Here, $\mu$ is the kinematic viscosity and $p$ is the pressure. The volumetric flow rate $Q$ through the pipe is $$ Q = - {\frac{ \pi R^{4} }{ 8 \mu }} {\frac{ d p }{ d x }} $$ which is known as the Hagen–Poiseuille law.

Explanation

In simple terms, for the rightward flow shown above: if the velocity distribution is uniform whether near the wall or not (left figure), the fluid behaves as an ideal (inviscid) fluid; if the velocity decreases as one approaches the wall (right figure), the flow is Hagen–Poiseuille.

Proof

Signs can be confusing throughout the derivation; a simple rule is to pay attention only to whether the force acts inward on the infinitesimal cylinder and take the sign accordingly. I will not explain every sign case-by-case.

Consider an infinitesimal cylinder of length $l$ and radius $r$. Assume the flow is in the $x$ direction. If the red pressure acting on the cylinder from the left is $p_{1}$, then the blue pressure acting from the right, having advanced by $l$, is $- \left( p_{1} + {\frac{ d p }{ d x }} l \right)$. On the lateral surface of the cylinder there is a yellow shear stress; this contributes the product of the stress $\tau$ and the lateral area $2 \pi r l$. The top and bottom face areas are both $\pi r^{2}$, and the force balance reads: $$ {\color{red} p_{1} \pi r^{2}} - {\color{blue} \left( p_{1} + {\frac{ d p }{ d x }} l \right) \pi r^{2}} - {\color{orange} \tau 2\pi r l} = 0 $$ Canceling several terms and dividing both sides by $\pi r l$ yields: $$ \ - {\frac{ d p }{ d x }} r - 2 \tau = 0 $$

Newton’s law of viscosity: $$ \tau = \mu \left( \nabla \mathbf{u} + \left( \nabla \mathbf{u} \right)^{T} \right) $$

If Newton’s law of viscosity is taken in one-dimensional form considering only the $x$ direction, it becomes $\tau = - \mu du / dr$, and we obtain the differential equation: $$ {\frac{ d u }{ d r }} = {\frac{ 1 }{ 2 \mu }} {\frac{ d p }{ d x }} r $$ Integrating both sides with respect to $r$ gives, up to the integration constant $C$, the following: $$ u = {\frac{ 1 }{ 4 \mu }} {\frac{ d p }{ d x }} r^{2} + C $$ By the definition of Hagen–Poiseuille flow, the velocity at the pipe wall ($r = R$) must be $0$; substituting $r = R$ yields: $$ C = - {\frac{ 1 }{ 4 \mu }} {\frac{ d p }{ d x }} R^{2} $$ With the integration constant determined and substituted back into the velocity expression, we obtain: $$ u = - {\frac{ 1 }{ 4 \mu }} {\frac{ d p }{ d x }} \left( R^{2} - r^{2} \right) $$

Consider a thin annulus at radial distance $r$ from the center, of width $dr$. If the area of the infinitesimal annulus $S$ is $dA$ and the velocity there is constant $u$, then the infinitesimal volumetric flow is $d Q = u d A$, and the total flow rate is obtained by the area integral: $$ Q = \int_{S} d Q = \int_{S} u d A $$ Since the annulus area is $dA = 2 \pi r dr$, the flow-rate expression becomes: $$ \begin{align*} Q =& \int_{0}^{R} - {\frac{ 1 }{ 4 \mu }} {\frac{ d p }{ d x }} \left( R^{2} - r^{2} \right) 2 \pi r dr \\ =& - {\frac{ \pi }{ 2 \mu }} {\frac{ d p }{ d x }} \int_{0}^{R} \left( R^{2} r - r^{3} \right) dr \\ =& - {\frac{ \pi }{ 2 \mu }} {\frac{ d p }{ d x }} \left[ {\frac{ R^{4} }{ 2 }} - {\frac{ R^{4} }{ 4 }} \right] \\ =& - {\frac{ \pi R^{4} }{ 8 \mu }} {\frac{ d p }{ d x }} \end{align*} $$

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