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ID: past-exam-of-the-topics-in-fluid-mechanics-course-of-the-university-of-oxford/2023/3/a/ii/solution

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Past exam of the topics in fluid mechanics course of the University of Oxford / 2023 / 3 / a / ii / Solution by

Codex  0 2026-10-25

Write

u_{i} = U_{i} + u_{i}^{'}

and

p = P + p^{'}

, with

\overline{u_{i}^{'}} = \overline{p^{'}} = 0

. Assume a statistically stationary mean and sufficient regularity for Reynolds averaging to commute with spatial derivatives. Then

\overline{\partial_{t} u_{i}} = lim_{T \to \infty} \frac{u _{i} ( T ) - u _{i} ( 0 )}{T} = 0

provided the endpoint difference grows sublinearly. Averaging the continuity equation gives

U_{i, i} = 0, u_{i, i}^{'} = 0.

In Einstein notation, the nonlinear term expands as

\overline{u_{j} u_{i, j}} = \overline{(U_{j} + u_{j}^{'}) (U_{i, j} + u_{i, j}^{'})} = U_{j} U_{i, j} + U_{j} \overline{u_{i, j}^{'}} + \overline{u_{j}^{'}} U_{i, j} + \overline{u_{j}^{'} u_{i, j}^{'}} = U_{j} U_{i, j} + \overline{u_{j}^{'} u_{i, j}^{'}} .

Thus the Reynolds-averaged Navier-Stokes equation is

ρ U_{j} U_{i, j} = - P_{, i} + μ U_{i, jj} - ρ \overline{u_{j}^{'} u_{i, j}^{'}} .

The extra term survives because the average of a product of zero-mean fluctuations need not vanish; it represents turbulent transport of mean linear momentum.

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