The Hagen-Poiseuille law

$$ \frac{\Delta V}{\Delta t} \left(Q \right) = \Delta P \cdot \frac {\pi}{8} \cdot \frac {1}{\eta} \cdot \frac {R^4}{L}$$

The volume flow Q is directly proportional to the pressure difference at the beginning and end of the tube (&Delta;P) and to the fourth power of its radius (of practical importance when reducing the diameter of the arteriole).

Derivation
'' For the derivation you need to understand the basics of integral calculus. ''

The derivation is based on Newton's law of tangential stresses in a liquid. The Hagen-Poiseuille law therefore applies to ``Newtonian fluids''. Newton's law for a liquid flowing in laminar flow in a tube with velocity in the direction of the x axis, on which the wall of the tube is located, while the diameter of the tube is located on the y axis, has the form:
 * $$\tau = - \eta \frac{ \text{d}v }{ \text{d}y}$$.

The tangential stress $$\tau$$, created by the friction between the wall and the flowing liquid, is transferred to the other layers of the liquid, causing a pressure drop between the beginning and the end of the tube. The force of friction will therefore arise at the interface between the tube wall and the liquid and can be expressed as:
 * $$F_t = \text{circumference} \cdot \text{tube length} \cdot \tau = 2 \pi y \cdot L \cdot \tau = - 2 \pi y \cdot L \cdot \eta \frac { \text{d}v }{ \text{d}y} $$.

This force will cause a decrease in the compressive force:
 * $$F_p = \pi y^2 \Delta P$$.

From the equality of these forces we then get:
 * $$\text{d}v =- \frac{ \Delta P}{2 L \eta} \cdot y \text{d}y$$,

and after integration:
 * $$v = \int \text{d}v = - \frac{ \Delta P}{ 2 L \eta} \int y \text{d}y = - \frac{ \Delta P}{2 L \eta} \cdot \frac{y^2}{2} + C$$,

and substituting for the integration constant C according to the initial conditions (there is zero velocity at the wall, y = r):
 * $$v = \frac{ \Delta P}{ 4 L \eta} \cdot (r^2 - y^2)$$.

So the velocity has a parabolic profile.

For speed:
 * $$v = \frac{\text{d}Q}{\text{d}S}$$,

so for the volume flow we get:
 * $$Q = \int_0^r \text{d}Q = \int_0^r v \text{d}S = \int_0^r \frac{ \Delta P}{ 4 L \eta} \cdot (r^ 2 - y^2) 2\pi y \text{d} y = \frac{\Delta P \pi}{2 L \eta} \cdot (\int_0^r r^2 y \text{d}y - \ int_0^r y^3 \text{d}y) = \frac{\Delta P \pi}{2 L \eta} \cdot (\frac{r^4}{2} - \frac{ r^4 }{ 4})$$.

By which we obtained the resulting form of the Hagen-Poiseuille law:
 * $$Q = \frac{\Delta P \pi r^4}{8 L \eta}$$.

Related Articles

 * Stokes' Law
 * Viscosity
 * Newtonian fluid