Hagen-Poiseuille flow in a circular tube

Assumptions

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid in a horizontal circular pipe of radius $R$ and length $L$. Velocity is axial only and depends on $r$ only.

        r
        ^   _____________________________
   R    |  /      ->  ->   ->
        | /   ->  ->  ->   ->   -> u(r)
   0 ---+--------------------------------->  x
        | \   ->  ->  ->   ->
        |  \_______________________________
           p1                       p2

Force balance on a cylinder of radius r

Pressure force $= (p_1 - p_2)\pi r^2 = \Delta p\,\pi r^2$. Shear force on the cylindrical surface $= \tau\,2\pi r L$. For equilibrium:

$$\tau = \frac{\Delta p\, r}{2L}$$

This is linear in $r$: zero at the axis, maximum at the wall.

Velocity distribution

With $\tau = -\mu\,du/dr$ (velocity decreases as $r$ increases):

$$\frac{du}{dr} = -\frac{\Delta p}{2\mu L}r$$

Integrating, with $u = 0$ at $r = R$:

$$u = \frac{\Delta p}{4\mu L}\left(R^2 - r^2\right)$$

The profile is a paraboloid.

Maximum velocity (at $r = 0$)

$$u_{max} = \frac{\Delta p R^2}{4\mu L}$$

Discharge

$$Q = \int_0^R u\,2\pi r\,dr = \frac{\pi \Delta p}{2\mu L}\left[\frac{R^4}{2} - \frac{R^4}{4}\right] = \frac{\pi \Delta p R^4}{8\mu L} = \frac{\pi \Delta p D^4}{128\mu L}$$

This is the Hagen-Poiseuille equation.

Average velocity

$$\bar V = \frac{Q}{\pi R^2} = \frac{\Delta p R^2}{8\mu L} = \frac{u_{max}}{2}$$

Wall shear stress

$$\tau_w = \frac{\Delta p R}{2L} = \frac{\Delta p D}{4L} = \frac{8\mu\bar V}{D}$$

Head loss

$$h_f = \frac{\Delta p}{\rho g} = \frac{32\mu \bar V L}{\rho g D^2}$$

Friction factor

Equating with the Darcy-Weisbach equation $h_f = \dfrac{f L \bar V^2}{2 g D}$:

$$f = \frac{64\mu}{\rho \bar V D} = \frac{64}{Re}$$

The result is valid only for laminar flow, $Re < 2000$. The head loss is directly proportional to the mean velocity and independent of the pipe roughness.