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Laminar Boundary Layer Order of Magnitude Analysis δ (x) u∞ µ∞ ρ∞ x y c Assumptions (in addition to incompressible, steady & 2-D) ∗ Changes in x − direction occur over a distance c ∂ 1 1 ∂ ⇒ ~ or, we write = O( ) ∂x C ∂x C ∗ Changes in y − direction occur over a distance δ 1 1 ∂ ∂ ~ or, = O( ) ⇒ ∂y ∂y δ δ ∗ δ << C (boundary layer is thin) ρ u c ∗ Re = ∞ ∞ >> 1 (Reynolds number is large) µ∞ ∗ Changes in x − velocity are proportional to u ∞ ∆u = O(u ∞ ) ∗ Changes in pressure are proportional to ρ ∞ u ∞2 ∆p = O( ρ ∞ u ∞2 ) Now, let’s do the order of magnitude analysis: Conservation of Mass ∂u ∂v + =0 ∂x ∂y Now, we introduce our assumed orders. For example: u ∂u = O( ∞ ) ∂x c Laminar Boundary Layer Order of Magnitude Analysis Since we have not assumed an order for v variations, just leave this as simply ∆v and say, ∂v ∆v = Ο( ) ∂y δ ∂u ∂v + = 0 , they obviously have equal (though opposite) magnitudes and ∂x ∂y therefore orders. Thus, the changes in v scale as: Since ∆v = Ο(u ∞ δ C ) Since v = 0 at a wall, ∆v = Ο(v) itself and we can say v δ = Ο( ) u∞ c x − momentum ρu ∂u ∂u ∂p ∂ 2u ∂ 2u + ρv =− +µ 2 +µ 2 ∂x ∂x ∂y ∂x ∂y Ο( ρ ∞ µ u µ u u ∞2 u2 u2 ) Ο ( ρ ∞ ∞ ) Ο ( ρ ∞ ∞ ) Ο( ∞ 2 ∞ ) Ο ( ∞ 2 ∞ ) c c c δ c If we compare the last two terms, clearly µ∞u∞ c 2 << µ∞u∞ δ2 Since we have assumed δ << C . Thus, we can assume that µ 16.100 2002 ∂ 2u is small. ∂y 2 2 Laminar Boundary Layer Order of Magnitude Analysis In order for any viscous terms to remain, this requires that µ ∂ 2u be equal in ∂x 2 order to the remaining inviscid terms. That is, e.g ∂ 2u ∂u ) = Ο( µ 2 ) ∂x ∂y Ο ( ρu ⇒ Ο( ρ ∞ u ∞2 c µ ∞u∞ ) δ2 µ∞ ) = Ο( δ ⇒ ( ) 2 = Ο( ) c ρ ∞u∞ c In other words, δ C 1 = Ο( Re ⇐ Major result! ) y − momentum ∂v ∂v ∂p ∂ 2v ∂ 2v ρu + ρv = − + µ 2 + µ 2 ∂x ∂y ∂y ∂x ∂y ρ ∞ u ∞2 δ 1 Ο( ρ ∞ u 2 ) Ο( ρ ∞ u 2 ) Ο( ) Ο( µ ∞ u ∞ 3 ) Ο( µ ∞ u ∞ ) δ δc c c c 2 ∞ δ 2 ∞ δ Normalizing all of these orders by and δ C = Ο( Ο( 1 Re 1 Re ρ u C 1 and substituting in for Re = ∞ ∞ 2 µ∞ ρ ∞u∞ / C ) gives: ) Ο( Clearly, except for 1 Re ) Ο( Re ) Ο( 1 Re 3 ) Ο( 2 1 Re ) ∂p , all other terms are small as Re increases. Thus, we must ∂y conclude that: ∂p is small ∂y ∂p ⇒ 0= is the y − momentum equation for a boundary layer ∂y 16.100 2002 3