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Optimization and Computational Fluid Dynamics - Department of ...

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2 A Few Illustrative Examples <strong>of</strong> CFD-based <strong>Optimization</strong> 39<br />

a hydrodynamic fluctuation term ˜p [33, 53]:<br />

p (x,t)=pu (t)+˜p(x,t) . (2.5)<br />

For the presented application (a domestic gas burner), the numerical domain<br />

is considered open, hence pu equals the atmospheric pressure <strong>and</strong> is<br />

constant. If acoustic waves may propagate in the gas mixture, then an additional<br />

acoustic pressure term has to be considered. Since acoustic-flame<br />

interactions are not addressed here, it is assumed from now on that acoustic<br />

waves are either nonexistent or <strong>of</strong> negligible effect on the flame structure<br />

<strong>and</strong> on the flow. Readers particularly interested in acoustic-flame interactions<br />

may refer to [35, 47] for further specific details.<br />

Within the low-Mach number approximation, the density appears as a<br />

function <strong>of</strong> temperature T <strong>and</strong> mean molar weight W. When the numerical<br />

domain is opened to the atmosphere, the influence <strong>of</strong> the hydrodynamic pressure<br />

pu on the density must be neglected. The full problem is then described<br />

by the following set <strong>of</strong> balance equations, written in conservative form for<br />

mass, momentum, mass fractions <strong>and</strong> enthalpy, solved in the present case:<br />

∂ρ<br />

+ ∇ · (ρv) = 0 (2.6)<br />

∂t<br />

∂(ρv)<br />

∂t + ∇ · (ρvv)=−∇˜p + ∇ · � μ � ∇v +(∇v) T��<br />

(2.7)<br />

∂(ρYk)<br />

∂t + ∇ · (ρYkv)=−∇ · (ρYkV k)+Wk ˙ωk , 1 ≤ k ≤ K − 1 (2.8)<br />

�<br />

K�<br />

�<br />

∂(ρh)<br />

+ ∇ · (ρhv)=∇ · λ∇T − ρ hkYkV k . (2.9)<br />

∂t<br />

The notations used here are st<strong>and</strong>ard in the combustion community <strong>and</strong><br />

are explained for example in [39]. Using a unity Lewis number assumption<br />

Le = λ/(ρcpD) = 1 for the molecular diffusion model, Eqs. (2.8) <strong>and</strong> (2.9)<br />

become much simpler:<br />

∂(ρYk)<br />

∂t + ∇ · (ρYkv)=∇ ·<br />

∂(ρh)<br />

∂t<br />

� λ<br />

k=1<br />

∇Yk<br />

cp<br />

� �<br />

λ<br />

+ ∇ · (ρhv)=∇ · ∇h<br />

cp<br />

�<br />

+ Wk ˙ωk<br />

(2.10)<br />

. (2.11)<br />

Equation (2.10) is solved for N − 1 species, because the last species (the<br />

nitrogen) is a non-reacting species (dilutant) simply determined using:<br />

K−1 �<br />

YN2 =1− Yk . (2.12)<br />

k=1

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