Conservation Equations

From Last Time…

D" #" #" Definition of material = + uk Dt #t #xk derivative

Reynolds Transport Theorem (in combination with the Gauss Divergence Theorem) provides a way of taking the material derivative through the ! volume integral D % $" $ ( # "dV = # ' + ("Uk )* dV Dt V V& $t $xk )

When dealing with conservation laws it is expected the the rate of change ! (material derivative) of the volume integral of some quantity (ρ in this case) will be zero. Therefore… "# " + (#Ui ) = 0 "t "xi

From Last Time…

"# " + (#Ui) = 0 "t "xi

"# "# "ui + ui + # = 0 Expanding out the derivative… ! "t "xi "xi

D" #ui + " = 0 this expression satisfies the ! Dt #x i conservation of mass law

D" = 0 if we assume the fluid is ! Dt incompressible…

"u i = 0 if the fluid is incompressible then conservation of mass ! "xi requires that the divergence of the velocity be zero

1 Alternative Derivation for Continuity Equation for Constant ρ

$ "u ' u"y"z & u + #x)# y#z % "x (

! Homework: ! Assume volume flow into the cube (fluid element) has to equal to volume out flow

Show that… "u "v "w + + = 0 "x "y "z

Alternative Derivation for Continuity Equation for Constant ρ

$ "u ' u"y"z & u + #x)# y#z % "x (

% $u ( $u u"y"z #' u + "x*" y"z = # "x"y"z = net flow in the x direction & $x ) $x ! ! % $v ( $v v"x"z #' v + "y*" x"z = # "x"y"z = net flow in the y direction ! & $y ) $y

M % $w ( $w w"x"y # w + "z "x"y = # "x"y"z = net flow in the z direction

U ' * ! & $z ) $z S

$ #u #v #w' # "u "v "w& "& + + )* x*y*z = 0 % + + ( = 0 ! % #x #y #z ( $ "x "y "z '

! !

Conservation of Momentum

ρU = fluid density x velocity = momentum per unit volume so that the volume integral of ρU is the total momentum of the fluid element

# "UdV V

The conservation of momentum is a statement of Newton 2nd Law: F = ma which relates that any change in the momentum of a fluid parcel must equal the forces applied ! D # "UdV = applied forces Dt V

2 The two most common kinds of forces acting on a fluid are surface forces (e.g., viscous stress) and body forces (e.g., gravity)

D # "UdV = # Pds + # "FdV Dt V S V

F = Body force per unit mass ! rate of change of P = surface force momentum of fluid per unit area element

Note: U, P and F are all Vectors

σ33 The surface force vector P can be related to the σ32 σ31 Stress Tensor σi,j in the σ13 following manner: σ23

The subscripts of go from σ σ11 σi,j 21 σ12 1-3 and represent, respectively, σ the planes on which the 22 stresses act and the directions in which they act.

Then, the net surface force Note: P = [p , p , p ] component in the j direction 1 2 3 derived from the j-directed Therefore, describing P requires forces acting on sides 1,2,3 is all 9 components of σi,j given by:

pj = σ1,j + σ2,j + σ3,j

D # "U = # Pds + # "FdV Dt V S V Where it is assumed that stress in D the j direction is summed over all # "u = # $ ds + # "f dV three i surface elements Dt j ij j ! V S V

Recall the Reynolds Transport Theorem has the following form… ! D % $" $ ( # "dV = # ' + ("uk )* dV Dt V V& $t $xk )

By replacing α with ρu , the left hand side can be expanded… ! j $ " " ' * & (#u j ) + (#u juk )) dV = * +ijds + * #f jdV V% "t "xk ( S V

Note that we are dealing with each component of U=[u1,u2,u3] separately. !

3 $ " " ' * & #u j + #u juk ) dV = * +ijds + * #f jdV V% "t "xk ( S V

Recall: Divergence Theorem from Vector Calculus States: #u r r u n dS i dV ! " U • nˆ dS = " # •UdV " i i = " s V #xi S V The equivalent expression in tensor notation

$ " " ' "+ij ! ! * & #u j + #u juk ) dV = * dV + * #f jdV V% "t "xk ( V "xi V Since the volume choice is "$ " " ij arbitrary ! (#u j ) + (#u juk ) = + #f j "t "xk "xi

" " "$ij (#u j ) + (#u juk ) = + #f j "t "xk "xi

expanding the derivative…

! # # # # #$ij " (u j ) + u j (") + u j ("uk ) + "uk (u j ) = + "f j #t #t #xk #xk #xi

terms sum to zero because of continuity (conservation of ! mass) requirements

# # #$ij " (u j ) + "uk (u j ) = + "f j #t #xk #xi

Local rate of change !of momentum

rate of change due to advection of momentum

divergence of stress

body force (e.g., gravity)

4 # # #$ij " (u j ) + "uk ("u j ) = + "f j #t #xk #xi

σ33

σ σ 32 31 " = #P$ + % where τij is shear stress ! σ13 ij ij ij and P is the normal pressure σ23 δ =1 when i=j σ σ11 ij 21 σ12 δij=0 when i䍫㼍 σ22 !

Assume that shear stress is linearly related to deformation rate… & ) %uk %ui %u j "ij = #$ij + µ( + + %xk ' %x j %xi * "u k = 0 continuity requirement for "xk incompressible flow $ ' #ui #u j "ij = µ& + ) ! % #x j #xi ( !

"#ij " = ($%ijP + &ij ) "xi "xi

"# "P " % % "u "u (( ij ' i j * = $ + ' µ' + ** ! "xi "x j "xi & & "x j "xi ))

equals zero by continuity equation "# % "u ( ! ij "P "ui j = $ + µ' + * "xi "x j & "xi"x j "xi"xi )

2 "#ij "P " u j ! = $ + µ "xi "x j "xi"xi

2 # # #P # u j Navier-Stokes " (u j ) + "uk ("u j ) = $ + µ + "f j ! #t #xk #x j #xi#xi Equation