Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 3 / 16C (Fluid Dynamics)

Table of contents
- a 16C
  - Solution a
- b 16C
  - Solution b
- c 16C
  - Solution c
- d 16C
  - Solution d
- e 16C
  - Solution e

a

Solution

Axisymmetry and incompressibility give

0 = \nabla \cdot u = \frac{1}{r} \frac{d}{d r} (r u_{r}) .

Thus

r u_{r} = C

. Finiteness at the origin forces

C = 0

, so

u_{r} = 0 .

b

Solution

Using the polar-coordinate curl with

u_{θ} = r Ω (r)

gives

ω = \nabla \times u = (2Ω + r \frac{d Ω}{d r}) e_{z} .

c

Solution

Take the curl of the Navier-Stokes equation. The pressure term disappears, curl commutes with the Laplacian, and the incompressible vector identity

\nabla \times ((u \cdot \nabla) u) = (u \cdot \nabla) ω - (ω \cdot \nabla) u

gives

\frac{\partial ω}{\partial t} + (u \cdot \nabla) ω = (ω \cdot \nabla) u + ν \nabla^{2} ω .

Therefore the material derivative form is

\frac{D ω}{D t} = (ω \cdot \nabla) u + ν \nabla^{2} ω .

d

Solution

The vorticity is normal to the plane, so the stretching term vanishes. Axisymmetry also makes angular advection vanish, and part (a) gives no radial advection. Hence the scalar vorticity equation is the radial diffusion equation

\frac{\partial ω}{\partial t} = ν (\frac{\partial ^{2} ω}{\partial r ^{2}} + \frac{1}{r} \frac{\partial ω}{\partial r}) .

e

Solution

Set the similarity variable

ξ = \frac{r}{ν t}, ω (r, t) = W (ξ) .

Then

ω_{t} = - \frac{ξ}{2 t} W^{'}, ν (ω_{rr} + \frac{1}{r} ω_{r}) = \frac{1}{t} (W^{''} + \frac{1}{ξ} W^{'}) .

The factors of

t

cancel, leaving the self-similar ordinary differential equation

W^{''} + (\frac{1}{ξ} + \frac{ξ}{2}) W^{'} = 0 .

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