# poiseuille's law derivation

\begin{align*} 6 \pi \eta rv & = \frac 43 \pi r^3 \rho g - \frac 43 \pi r^3 \sigma g \\ &= \frac 43 \pi r^3 (\rho - \sigma) g \\ \text {or,} \: \eta &= \frac {2r^2 (\rho - \sigma ) g}{9v} \dots (ii) \end{align*}\begin{align*} 6 \pi \eta rv & = \frac 43 \pi r^3 \rho g - \frac 43 \pi r^3 \sigma g \\ &= \frac 43 \pi r^3 (\rho - \sigma) g \\ \text {or,} \: \eta &= \frac {2r^2 (\rho - \sigma ) g}{9v} \dots (ii) \end{align*}=43πr3ρg−43πr3σg=43πr3(ρ−σ)g=2r2(ρ−σ)g. Stay connected with Kullabs. The viscous force, F in a direction opposite to the direction of motion of the body. i.e; V α r, Inversely proportional to the coefficient of viscosity, η of the liquid. CBSE class 11 Physics notes with derivations are best notes by our expert team. directly proportional to the fourth power of the radius of a tube. So, the net force acting on the body is zero and the ball starts to fall with a constant velocity, i.e. Community smaller than society. Consider a solid cylinder of fluid, of radius r inside a hollow cylindrical pipe of radius R. The driving force on the cylinder due to the pressure difference is: The viscous drag force opposing motion depends on the surface area of the cylinder (length L and radius r): Hagenbach was the first who called this law the Poiseuille's law. Derivation of Poiseuille’s formula by dimensional analysis. When a small spherical body is dropped in a viscous medium, the layer in contact with it starts moving with the same velocity as that of the body whereas the layer at a considerable far distance will be at rest. Now the equation of continuity giving the volume flux for a variable speed is: Substituting the velocity profile equation and the surface area of the moving cylinder: Peter's Index  The coefficient of viscosity of the medium. i.e;V α 1/l, The pressure gradient (P/l) (i.e. Our notes has covered all topics which are in NCERT syllabus plus other topics which are required for Board Exams. 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It is a network of social relationships which cannot see or touched. Now by substituting vales of a, b and c, in equation (iv) we get, \begin{align*} V &= K\left (\frac {P}{ l}\right ) r^4 \eta ^{-1} \\ &= K \frac {Pr^4}{nl} \\ V &= \frac {\pi}{8} \frac {Pr^4}{nl} \dots (v) \\ \end{align*}. The value of proportionality is found to be $$\frac {\pi }{8}$$ from experiment. As time passes the velocity of falling body increases and viscous force acting on it also increases in that proportion. We know empirically that the velocity gradient should look like this: At the centre Figure $$\PageIndex{4}$$: (a) If fluid flow in a tube has negligible resistance, the speed is the same all across the tube. The pressure gradient of a tube. The theoretical derivation of a slightly different form of the law was made independently by Wiedman in 1856 and Neumann and E. Hagenbach in 1858 (1859, 1860). \: V \propto \eta ^c \dots (iii) , \begin{align*} V &\propto \left (\frac {P}{ l}\right )^a r^b \eta ^c \\ V &= K\left (\frac {P}{ l}\right )^a r^b \eta ^c \dots (iv) \\ \end{align*}, Writing dimensional formula on both sides we have,[M^0 L^3 T^{-3}] = \left [ \frac {ML^{-1}T^{-2}}{L} \right ]^a . Derivation of Poiseuille’s Formula by Dimensional Analysis.