PX285 - H6c - example in a cylinder

assuming a laminar, steady and viscous flow:

\vec{u} = u_{z} (r) \hat{z}

considering a fluid element in cylindrical geometry

area of the outer surface:

A_{O} = Δ z (r + Δ r) Δ ϕ

area of the inner surface:

A_{I} = Δ z r Δ ϕ

F_{v} = (r + Δ r) Δ z Δ ϕ μ \frac{\partial u_{z}}{\partial r} |_{r + Δ r} - r Δ ϕ Δ z μ \frac{\partial u_{z}}{\partial r} |_{r}

note: areas of the inner and outer surfaces of the chosen fluid element are different (in contrast with the plane case)
using tayor expansion:

\frac{\partial u_{z}}{\partial r} |_{r + Δ r} \approx \frac{\partial u_{z}}{\partial r} |_{r} + \frac{\partial^{2} u_{z}}{\partial r^{2}} |_{r} Δ r + \dots

ignoring all terms in $(Δ r)^{2}$ or smaller:

F_{v} = μ Δ z Δ ϕ Δ r (\frac{\partial u_{z}}{\partial r} |_{r} + r \frac{\partial^{2} u_{z}}{\partial r^{2}} |_{r}) = μ Δ z Δ ϕ Δ r \frac{\partial}{\partial r} (r \frac{\partial u_{z}}{\partial r})

the pressure force:

F_{P} = r Δ ϕ Δ r (P (z) - P (z + Δ z)) = - r Δ ϕ Δ r Δ z \frac{\partial P}{\partial z}

since it is a steady motion, there is no acceleration, so the net force must be zero:

\begin{matrix} μ \frac{\partial}{\partial r} (r \frac{\partial u_{z}}{\partial r}) = r \frac{\partial P}{\partial z} \\ μ \frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial u_{z}}{\partial r}) = \frac{\partial P}{\partial z} = - Q \end{matrix}

integrating twice and applying boundary conditions: $u_{z} (r = a) = 0$ and $\frac{\partial u_{z}}{\partial r} (r = 0) = 0$

u_{z} (r) = \frac{Q}{4 μ} (a^{2} - r^{2}) = \frac{P_{1} - P_{2}}{4 μ L} (a^{2} - r^{2})

this is again,the poisseuille flow