Past exam of the flows, fluctuations and complexity and symmetry and relativity paper of Part B of the physics course of the University of Oxford / 2011 / 2 / 4

Table of contents
- Solution 4

Solution

Put

A = 2 μ a F_{0} k / (c f)

. In the specified near-wall limit the outer temperature is

T - T_{0} ≃ A / (b z)

, so

q_{C} ≃ \frac{2 κ A}{b z ^{3}}, \frac{∣ q _{C} ∣}{q _{R}} ≃ \frac{4 κ μ a}{ρ c f b z ^{3}} = \frac{4 κ ν a}{c f b z ^{3}}, χ = \frac{κ}{ρ c} .

Thus the near-wall radiative heating layer has scale

0 < z ≲ δ_{C}, δ_{C} = (\frac{4 κ μ a}{ρ c f b})^{1/3} = (\frac{4 κ ν a}{c f b})^{1/3} .

The comparison is an estimate of the breakdown of the outer solution, not a new inner temperature solution. It requires

δ_{C} ≪ b

and

k ρ δ_{C} ≪ 1

; neglecting longitudinal relative to transverse thermal conduction additionally requires

δ_{C} ≪ a

in the region to which the estimate is applied.

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