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Transport Process

Uploaded: 5 years ago
Contributor: nyota
Category: Chemical Engineering
Type: Lecture Notes
Rating: N/A
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Filename:   06.pptx (691.78 kB)
Page Count: 28
Credit Cost: 2
Views: 111
Last Download: N/A
Transcript
Shell Momentum Balances Outline Convective Momentum Transport Shell Momentum Balance Boundary Conditions Flow of a Falling Film Flow Through a Circular Tube Convective Momentum Transport Recall: MOLECULAR MOMENTUM TRANSPORT Convective Momentum Transport: transport of momentum by bulk flow of a fluid. Outline Convective Momentum Transport Shell Momentum Balance Boundary Conditions Flow of a Falling Film Flow Through a Circular Tube Shell Momentum Balance Steady and fully-developed flow is assumed. Net convective flux in the direction of the flow is zero. Outline Convective Momentum Transport Shell Momentum Balance Boundary Conditions Flow of a Falling Film Flow Through a Circular Tube Boundary Conditions Recall: No-Slip Condition (for fluid-solid interfaces) Additional Boundary Conditions: For liquid-gas interfaces: “The momentum fluxes at the free liquid surface is zero.” For liquid-liquid interfaces: “The momentum fluxes and velocities at the interface are continuous.” Flow of a Falling Film Liquid is flowing down an inclined plane of length L and width W. ? – film thickness Vz will depend on x-direction only Why? z x y Assumptions: Steady-state flow Incompressible fluid Only Vz component is significant At the gas-liquid interface, shear rates are negligible At the solid-liquid interface, no-slip condition Significant gravity effects Flow of a Falling Film z x y ? W L ?xz ? x ?xz ? x + ? z x y ?ij ? flux of j-momentum in the positive i-direction Flow of a Falling Film z x y ? W L z x y ?ij ? flux of j-momentum in the positive i-direction ?yz ? y=0 ?yz ? y=W Flow of a Falling Film z x y ? W L z x y ?ij ? flux of j-momentum in the positive i-direction ?zz ? z=0 ?zz ? z=L ?g cos ? Flow of a Falling Film P(W??)|z=0 – P(W??)|z=L + (?xz? x )(W*L) – (?xz ? x +?x )(W?L) + (?yz? y=0 )(L*?) – (?yz ? y=W )(L??) + (?zz ? z=0)(W* ?) – (?zz ? z=L)(W??) + (W?L??)(?gcos ?) = 0 Dividing all the terms by W?L?? and noting that the direction of flow is along z:

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