Fluid Dynamics

# Poiseuille’s Formula

Poiseuille concluded that the volume of liquid flowing per second depends upon the following pressure gradient and gave a formula known as Poiseuille’s Formula.

​Through a horizontal tube, a liquid moves with cylindrical layers coaxial with the axis of the tube. Then, the velocity of the cylindrical layer is maximum along the axis of the tube. Its value goes on decreasing towards the wall of the tube and the velocity is minimum for the layer which is in contact with the wall of the tube.

​Poiseuille studied the rate of flow of liquid through a horizontal capillary tube and concluded that the volume of the liquid flowing per second $(V)$ depends upon;

1. Pressure gradient [the rate of change of pressure with length] $\text{i.e. } V ∝\left(\frac{P}{l}\right)^x$ 2. Radius of the capillary tube $\text{i.e. } V ∝r^y$ 3. Coefficient of viscosity of the liquid $\text{i.e. } V ∝η^z$

Thus, $V ∝\left(\frac{P}{l}\right)^x r^y η^z$ $V=k\left(\frac{P}{l}\right)^x r^y η^z \text{ ____(1)}$ where, $k =$ proportionality constant and $x$, $y$ and $z$ are the dimensions of $V$ in terms of $(P/l)$, $r$ and $η$.

Writing dimensions of each terms of $(1)$, $\left[L^3T^{-1}\right]=\left[ML^{-2}T^{-2}\right]^x \left[L\right]^y \left[ML^{-1}T^{-1}\right]^z$ $\left[L^3T^{-1}\right]=\left[M^{x+z}L^{-2x+y-z}T^{-2x-z}\right]$ equating, $∴x+z=0$ $x=-z$ $∴-2x-z=-1$ $-2x+x=-1$ $x=1$ $z=-1$ $∴-2x+y-z=3$ $-2+y+1=3$ $y=4$ putting the values of $x$, $y$ and $z$ in $(1)$, $V=k\left(\frac{P}{l}\right)^1 r^4 η^{-1}$ $V=k\frac{Pr^4}{lη}$ Experimentally, the value of $k$ is found to be $π/8$. $V=\frac{P π r^4}{8 η l}$ This is known as Poiseuille’s formula.

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