Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 321 / 3 / b

Table of contents
- Solution b

Solution

Dot the momentum equation with

u

. The Coriolis acceleration does no work, while the buoyancy-variable equation gives

\frac{D}{D t} \frac{1}{2} N^{2} θ^{2} = N^{2} θ u_{z},

which cancels the buoyancy work

- N^{2} θ u_{z}

. Thus

\frac{D E}{D t} = - \frac{1}{ρ _{0}} u \cdot \nabla P + 3 Ω^{2} x u_{x} .

Using incompressibility,

u \cdot \nabla P = \nabla \cdot (P u), 3 Ω^{2} x u_{x} = \nabla \cdot (\frac{3}{2} Ω^{2} x^{2} u) .

Therefore

\partial_{t} E + \nabla \cdot F = 0, F = u (E + \frac{P}{ρ _{0}} - \frac{3}{2} Ω^{2} x^{2}) .

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version