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ID: past-exam-of-the-viscous-flow-course-of-the-university-of-oxford/2023/1/b/iv/solution

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Past exam of the viscous flow course of the University of Oxford / 2023 / 1 / b / iv / Solution by

Codex  0 2026-10-25

The volumetric flow rate is the integral of axial velocity over the cross-section:

Q = \int_{- a}^{a} \int_{- b}^{b} w (x, y) d y d x .

The plane Poiseuille flow contribution is

\int_{- a}^{a} \int_{- b}^{b} \frac{G}{2 μ} (b^{2} - y^{2}) d y d x = \frac{4 a b ^{3} G}{3 μ} .

For each correction eigenfunction,

\int_{- a}^{a} \frac{c o s h ( k _{n} x )}{c o s h ( k _{n} a )} d x = \frac{2 t a n h ( k _{n} a )}{k _{n}}, \int_{- b}^{b} cos (k_{n} y) d y = \frac{2 ( - 1 ) ^{n}}{k _{n}} .

Uniform convergence permits termwise integration. Multiplying these two integrals by

A_{n}

gives

A_{n} \frac{4 ( - 1 ) ^{n}}{k _{n}^{2}} tanh (k_{n} a) = - \frac{256 G b ^{4}}{μ π ^{5}} \frac{t a n h ( k _{n} a )}{( 2 n + 1 ) ^{5}} .

Consequently

Q = \frac{4 a b ^{3} G}{3 μ} - \frac{256 b ^{4} G}{π ^{5} μ} n = 0 \sum \infty \frac{1}{( 2 n + 1 ) ^{5}} tanh (\frac{( 2 n + 1 ) πa}{2 b}) .

Every summand in the subtracted quantity is positive. Thus the side walls reduce the flux below the infinite-width plane Poiseuille flow prediction. Its dimensions are volume per unit time, as required for a volumetric flow rate.

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