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 Incoming links: Falling film flow

Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 66 / 2 / Solution 2026-10-03

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Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 1 / 5C / Solution Created 2026-09-24 Updated 2026-09-29

Take

x

down the plane and

y

normally away from it, with the solid at

y = 0

and the free surface at

y = h

. A steady unidirectional falling film flow has velocity

u = u (y) e_{x}

. The streamwise Navier-Stokes equation reduces to

ν u^{''} = - g sin α,

where

ν

is the kinematic viscosity. The lower no-slip boundary condition and upper stress-free boundary condition are

u (0) = 0, u^{'} (h) = 0.

Integrating gives the velocity profile

u (y) = \frac{g sin α}{ν} (h y - \frac{y ^{2}}{2}) .

The volume flux per unit width is

Q = \int_{0}^{h} u (y) d y = \frac{g h ^{3} sin α}{3 ν} .

If the upper surface is replaced by a stationary solid plane, the second boundary condition becomes

u (h) = 0

. The resulting plane-channel profile and flux are

u (y) = \frac{g s i n α}{2 ν} y (h - y), Q_{closed} = \frac{g h ^{3} sin α}{12 ν} = \frac{1}{4} Q .

The free surface exerts no tangential shear stress, whereas the stationary upper wall enforces no slip and exerts a retarding shear. This additional drag accounts for the reduced flux.

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