Assignment 8 Solution

Assignment 8 Solution - ∫ ∫ (5) (6) Prob. 2. y x g v0...

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Assignment 8 Solution g 0 0 ( ) 1 ( ) , ( ) 1 ( ) 2 rz m L rz z r r rz R rz m P P d dP r K K r dr dz L P p g z p gz d r K rdr K r R τ ρ - = = - = ≡ - = + = - = - - ∫ ∫ z r p 0 p L R 2 R 1 Prob. 1. V Rm where Rm is the radial position of the maximum in the flow field where the shear stress is zero. 1 1/ 1/ 1/ 1/ For , (since 0) 1 Then ( ) 2 ( ) 2 ( ) 2 2 z m n z z rz n z rz m n n z m n v r n z m V R z r R dv dv m dr dr dv m K R r dr dv K R r dr m K dv R r dr m K v V < = ÷ = = - ÷ = - ÷ = - ÷ = + 1/ 1 1 1 ( ) ( ) 1 n n n n n m m n R r R R m n + + - - - ÷  ÷ + (1) (2)
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1/ 1/ 1/ 1/ 0 1/ 1 1 2 ˆ For , 0 ˆ 1 ( ) 2 ˆ ( ) 2 ˆ ( ) 2 1 ˆ ( ) ( ) 2 z m z m n z rz m n n z m n v r n z m R n n n n n z m m dv r R dr dv m K r R dr dv K r R dr m K dv r R dr m K n v r R R R m n τ + + < = - - = - - ÷ - = - ÷ = - - ÷ + = - - - - ÷  ÷ (3) Assignment 8 Solution (Cont.) Equating (2) and (3) at r = Rm gives an equation to calculate Rm . ( 29 1/ ( 1)/ ( 1)/ 2 1 1 2 ( ) n n n n n m m n m R R R R V n K + + +  - + - = ÷ ÷  (4) 2 1 1 1 1 1 0 ˆ 2 2 2 ( ) m m R R z z R R L D rz m r R Q v rdr v rdr F R dz K R R R L π = = + = = -
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Unformatted text preview: ∫ ∫ (5) (6) Prob. 2. y x g v0 0, , at y = Y, or Here Bingham plastic model: , yx yx yx x x yx y yx y yx yx Y Y yx x x Y yx Y p p p d g g g dy d g dy d g dy gy gY Y g dv dv g g dy dy τ ρ μ = = + = = -= = = = = = = + = = -+ ∫ ∫ ( ) ( ) Also, ( ) ( ) x v y y Y x p p V y B y B Y x p p Y Y x y Y p p Y B Y x Y g dv dy dy g v V y B y B g v v V Y B Y B Q v wdy v wdy = = = = -+ =--+-= =--+-= + ∫ ∫ ∫ ∫ ∫ Y...
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This note was uploaded on 10/09/2011 for the course ENGINEERIN 3133 taught by Professor Sarkozi during the Spring '11 term at Texas A&M.

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Assignment 8 Solution - ∫ ∫ (5) (6) Prob. 2. y x g v0...

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