# ME2135 - ME2135 FLUID MECHANICS II Part 2: Viscous Flow...

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1 ME2135 FLUID MECHANICS II Part 2: Viscous Flow Tutorial Solutions 11. The boundary layer velocity profile: ) 4 η 2 η 4 η ( λ 2 η η 2 3 U u 3 2 3 + + = where δ y η = and λ is a pressure gradient parameter. At the separation point: 0 y u 0 y = = + + = 3 2 2 3 2 δ 4 y λ 3 δ y λ δ 4 λ δ y 2 3 δ 1 2 3 U y u 6 λ 4 λ 2 3 0 δ 4 λ δ 1 2 3 U y u 0 y = = = + = = Hence: 3 2 3 2 3 η 2 η 3 η 2 3 η 3 η 2 3 2 η η 2 3 U u = + = () 2 δ 2 η η η δ d η η 2 3 η 1 δ dy U u 1 δ * 1 0 4 3 1 0 3 2 δ 0 = + = + = = ( since η = y/ δ Æ dy = δ d η ) () () δ 1286 . 0 δ 2 7 4 5 9 2 1 1 η 2 η 7 4 η 5 9 η 2 1 η δ d η ) 12 η 4 η (9 η η 2 η 3 δ d η ) η 2 (3 η 2 η η 3 δ dy U u 1 U u θ 1 0 6 7 5 4 3 1 0 5 6 4 3 2 1 0 2 3 2 3 2 δ 0 = + = + = + = = = Hence, the boundary layer shape factor at the separation point 888 . 3 1286 . 0 5 . 0 θ δ H * = = = Note : H = 3 for a linear velocity profile.

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2 12. Free-stream velocity U = 0.1 m/s. For water at 20°C : ν = 1.0 x 10 – 6 m 2 /s From Blasius Solution: Ux/ ν 5.0 Re 5.0 x δ x = = Æ U ν x 5 Ux/ ν x 5.0 δ = = For point 1 (x 1 , y 1 ): At x 1 = 1 m: Re x1 = U x 1 / ν = 0.1 x 1/1 x10 – 6 = 1x 10 5 < Re xtr where Re xtr = 5 x 10 5 Hence, the b. l. flow is laminar at x 1 and therefore: mm 15.81 cm 1.581 m 10 1.581 m 40 10 x 5 m 10 x 4 1 x 5.0 Ux/ ν x 5.0 δ 2 2 5 1 = = × = = = = From Blasius Solution: x ν U y η = and U u ' f = Hence at point 1 (x 1 , y 1 ): 162 . 3 10 10 10 1 10 1 0.1 10 10 x ν U y η 5 2 6 3 1 1 1 = = = × × × = = From the Table of Blasius Solution: η f ' 3.0 0.8460 η 1 Æ 3.162 f 1 ' = ? Æ use linear interpolation 3.5 0.9130 b a c 3.0 3.162 3.5 0.8460 d 0.9130 f η ' 1 f
3 From the sketch above: b d c a d c b a = = This will give: f 1 ' = 0.8460 + a = 0.8460 + b d c = 0.8460 + 0.8460) - (0.9130 x 3.0) - (3.5 3.0) - (3.162 = 0.8460 + 0.067 x (0.5) (0.162) = 0.8677 Hence, at point 1(y 1 = 10 mm): u 1 = f 1 ' U = 0.8677 x 0.1 m/s = 0.08677 m/s 87 mm/s ( Answer ) For point 2 (x 2 , y 2 ): At x 2 = 4 m: Re x2 = U x 2 / ν = 0.1 x 4/1 x10 – 6 = 4 x 10 5 < Re xtr Hence, the b. l. flow is laminar at x 2 and therefore: mm 31.62 cm 3.162 m 10 3.162 m 40 10 x 20 m 10 x 4 4 x 5.0 Ux/ ν x 5.0 δ 2 2 5 2 = = × = = = = Following the previous method: 581 . 1 5 . 2 10 5 . 2 10 4 10 1 0.1 10 10 x ν U y η 4 2 6 3 2 2 2 = = × = × × × = = From the Table of Blasius Solution: η f ' 1.5 0.4868 η 1 Æ 1.581 f 2 ' = ? 2.0

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## This note was uploaded on 05/05/2009 for the course ME ME2135 taught by Professor Me2135 during the Spring '09 term at National University of Juridical Sciences.

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ME2135 - ME2135 FLUID MECHANICS II Part 2: Viscous Flow...

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