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Test_1.key.2004 - PGE 323 Test#1 Key Pope 1(a The mass flux...

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PGE 323 Test #1 Key February 20, 2004 Pope 1 (a) The mass flux in each direction is given as follows: For x-direction: [ ] ( 29 [ ] ( 29 z , y , x x x z , y , x x z y u t Out Mass z y u t In Mass + ρ = ρ = For y-direction [ ] ( 29 [ ] ( 29 z , y y , x y z , y , x y x z u t Out Mass x z u t In Mass + ρ = ρ = For z-driection [ ] ( 29 [ ] ( 29 z z , y , x z z , y , x z y x u t Out Mass y x u t In Mass + ρ = ρ = Mass in - Mass out = Accumulated Mass ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ]} y x u y x u [ ] z x u z x u [ ] y z u y z u {[ t ] [ z y x z z , y , x z z , y , x z z , y y , x y z , y , x y z , y , x x x z , y , x x t t t + + + + ρ - ρ + ρ - ρ + ρ - ρ = φρ - φρ .(1) Dividing both the sides by t z y x and taking limits as x, y, z and t go to zero: ρ + ρ + ρ - = ρφ x ) uz ( x ) uy ( x ) ux ( t ) ( ----------------------(2) From Darcy's law: x k u x x Φ μ - = ----------------------------------------------(3) y k u y y Φ μ - = ----------------------------------------------(4) z k u z z Φ μ - = ----------------------------------------------(5) Substituting equations (3), (4) and (5) into equation (2), gives the diffusivity equation 1.5-3 in the course notes: z x y ρ x u u z ρ u y ρ
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Φ μ ρ + Φ μ ρ + Φ μ ρ = ρφ z ) z k ( y ) y k ( x ) x k ( t ) ( z y x ------------------(6) (b) Assuming steady state flow, equation (6) simplifies to: Φ μ ρ + Φ μ ρ + Φ μ ρ = z ) z k ( y ) y k ( x ) x k ( 0 z y x Assuming constant density and viscosity, gives: Φ + Φ + Φ = z ) z k ( y ) y k ( x ) x k ( 0 z y x Assuming homogeneous and anisotropic medium (that is kx, ky, and kz are
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