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D. Keffer - ChE 240: Heat Transfer and Fluid Flow−+( v )∂ ρ∂ t⎛∂τ⎜⎝ ∂ xxxx+⎡ ∂ x⎢ρvx +∂ x⎢⎢⎢∂= + ρvx∂ y⎢⎢ ∂ z⎢+ ρvx⎣ ∂ z∂τyx∂ y+( v ) ∂ρ ( v )∂τ∂ z( vy) ∂( ρv)( v ) ∂ρ ( v )zx⎞⎟+⎠v++xvvyz∂ p−∂ x∂ xx∂ y∂ zρgxxx⎤⎥⎥⎥⎥⎥⎥⎥⎦Recall, the substantial derivative of the x-momentum component has the definition:( ρv) ∂( ρv) ∂( ρv) ∂( ρv) ∂( ρv)DDtx=∂ tx+∂ xxvx+∂ yxvy+∂ zxvz−⎡⎢ρv⎣( v ) ∂( ρv) ∂( ρv) ∂( ρv)⎡∂ ρ⎢⎣ ∂ txx( v ) ∂( v y ) ∂( v )∂∂ xx++∂ xρvxxvx∂ y++∂ yρvxxvyz∂ z+⎤⎥ +⎦∂ zx⎛∂τ⎜⎝ ∂ xxxv+z⎤⎥⎦=∂τyx∂ y+∂τ∂ zzx⎞⎟+⎠∂ p−∂ xρgx−( ρv)DDtx= ρvx⎛∂τ⎜⎝ ∂ x∂τ∂τ∂ z∂ p∂ xxx yx zx( ∇ ⋅ v) + ⎜ + + ⎟+− ρgx∂ y⎞⎟⎠but using the product rule for differentiation again:( ρv) D( v ) D( ρ)DDtx= ρDtx+vxDtDρDtFrom the mass balance: − = ρ( ∇ ⋅ v)so( ρv) D( v ) D( ρ)DDtx= ρDt( ρv) D( v )DDtx= ρDtxx−+vxvxDt( ⋅ v)ρ ∇3-8
D. Keffer - ChE 240: Heat Transfer and Fluid FlowAnd the momentum balance becomes:−( )D vρDtx+vx⎛∂τ⎜⎝ ∂ x∂τ∂ z∂ p∂ xxx yx zx( v) − ρvx( ∇ ⋅ v) = ⎜ + + ⎟+− ρgxρ ∇ ⋅∂τ∂ y⎞⎟⎠−( )D vρDtx⎛∂τ=⎜⎝ ∂ xxx+∂τyx∂ y+∂τ∂ zzx⎞⎟+⎠∂ p−∂ xρgxThis is the equation of motion in the x-direction. It is a momentum balance.The y- and z-components of the momentum are obtained in a precisely analogous manner.They look like:( )D v y− ρDt( )D v− ρDtz⎛∂τ=⎜⎝ ∂ x⎛∂τ=⎜⎝ ∂ xxyxz+∂τyy∂ y∂τyz∂ ywhich in vector notation looks like:( )+++∂τzy∂ z∂τ∂ zzz⎞⎟+⎠⎞⎟+⎠∂ p−∂ y∂ p−∂ zρgρgD v− ρ = ∇ ⋅τ + ∇p− ρgGeankoplis, 3.7-13DtThis momentum balance is true for any continuous medium.B. Equations for stresses using Newton’s Law of viscosity (page 172-173)1. Shear stress components for Newtonian fluids in rectangular coordinatesSee Geankoplis page 172, eqns (3.7-14 to 3.7-20)2. Shear stress components for Newtonian fluids in cylindrical coordinatesSee Geankoplis page 172-173, eqns (3.7-21 to 3.7-27)3. Shear stress components for Newtonian fluids in spherical coordinatesSee Geankoplis page 173, eqns (3.7-28 to 3.7-34)For the qualitative explanation of the basis of these equations, you must go to “An Introduction toFluid Dynamics” by Stanley Middleman, Wiley, New York, 1998, page 142-143Examine rectangular case: obtain 1-D results from 3-D case by looking at equation 3.7-17 whenthe y component of the velocity is zero.yz3-9
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tâ Ïâ tâ Ïâ = Ï dt d