lec10 (2) - One-Dimensional (Vertical) Chemistry-Transport...

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One-Dimensional (Vertical) Chemistry-Transport Model Take horizontal average of equation (5) and denote horizontal average with overbar and deviation from horizontal average with a prime: [] () ii i dd PL iV iw XMw dz dz d X'Mw X Mw dz −= = = =+ G i i d X'M'w ' MX'w ' wX'M' X'Mw dz + + (net vertical flux of air [ ] Mw 0 = ) ( i d ' dz ) i ( i X' 0 = and w0 = ) ( ) i d MX'w ' dz ± [] [] ( ) M' M ² i dX d Mz w dz dz ⎛⎞ −δ ⎝⎠ ± ' ( eddy diffusion approximation* ) i z dX d MK dz dz =− (9) ( z K = eddy diffusion coefficient z w' ) *eddy diffusion approximation : Consider case when loss only: [ ] i i PL L τ ( τ= chemical lifetime of i) i M'X' MX + τ i ( i = and [ ] M' 0 = ) z z+| δ z| -|w ' | (air moving down from above) (air moving up from below) +|w ' | z-| δ z X i w dX i (+ | δ z | )(- | w ' | ) = = - ' ' dz i | δ z||w ' | i (-| δ z|)(+|w ' |) Figure by MIT OCW.
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[ ] i MX τ ± [] () ii M'X' ² For brevity drop subscripts i and overbars and assume z KK = is independent of altitude and temperature is constant: [ ] 2 2 XM dd X KM dz dz dM dX d X dz dz dz ⎛⎞ = ⎜⎟ τ ⎝⎠ =+ [ ] 2 2 M dX d X Hd z d z =− + (In hydrostatic equilibrium: [ ] [ ] M dz H for constant temperature) Rearranging: 2 2 dX 1dX X 0 dz H dz K −− = τ ( 1 0 ) General solution is: zz X Aexp Bexp hh +− 1 2 2 11 1 1 h2 H4 H K ± + τ 0 (Note that h and + > h0 < ) Determine A and B from boundary conditions. Say X Æ 0 as z Æ , then and A0 = ( ) XX 0 = at z = 0 is given so ( ) BX0 = .
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lec10 (2) - One-Dimensional (Vertical) Chemistry-Transport...

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