Differential Equation: Conservation of Momentum { } = { } + { } Sum of the External Forces Net Rate of Linear Momentum Efflux Accumulation of Linear Momentum within C.V.. (x,y,z) (x+  x, y+  y, z+  z) x y z xx zz yy  F =  F x i +  F y j +  F z k Let’s focus on the x-component of the external forces (Note that normal forces have been combined  xx = -p +  xx ):  F x = (  xx  y  z)  x+  x - (  xx  y  z)  x + (  yx  x  z)  y+  y - (  yx  x  z)  y + (  zx  x  y)  z+  z - (  zx  x  y)  z +  g x  x  y  z Component of g in the x-direction Similar expressions exist for  F y and  F z. Now we look at the momenturm flux for the cubic element. Volumetric flow across the x-plane at x+  x: (v x  y  z )  x+  x The momentum flux then is: (  v v x  y  z )  x+  x Note:First velocity is a vector for momentum. And for the x-plane at x: (  v v x  y  z )  x

For all of the faces: [(  v v x )  x+  x - (  v v x )  x ]  y  z + [(  v v y )  y+  y - (  v v y )  y ]  x  z + [(  v v z )  z+  z - (  v v z )  z ]  x  y Rate of accumulation of linear momentum The momentum of the cube is:  v  x  y  z (i.e. momentum per unit volume times the volume) and its rate of accumulation:  (  v)  x  y  z  t Substituting into the word equation: { } i + { } j + {  F z } k = [(  v v x )  x+  x - (  v v x )  x ]  y  z + [(  v v y )  y+  y - (  v v y )  y ]  x  z + [(  v v z )  z+  z - (  v v z )  z ]  x  y +  (  v)  x  y  z  t Expression for  F x  F y

lim { } = lim { (  xx )  x+  x - (  xx )  x + (  yx )  y +  y - (  yx )  y + (  zx )  z +  z - (  zx )  z +  g x } x0x0 y0y0 z0z0  F x xyzxyz  x  y x0x0 y0y0 z0z0 Divide by  x  y  z and take the limit as  x,  y, and  z  0. Focus on RHS: lim { i + j + } = x0x0 y0y0 z0z0  F x xyzxyz  F y xyzxyz  F z xyzxyz  (  vv x )  (  vv y )  (  vv z )  (  v)  x  y  z  t Reorganize the RHS by the product rule for differentiation. = v [ ]  (  v x )  (  v y )  (  v z )  (  )  x  y  z  t  v  v  x  y  z  t vxvx vyvy vzvz  0 (Why?) x0x0 y0y0 z0z0  F x xyzxyz  F y xyzxyz  F z xyzxyz Divide by  x  y  z and take the limit as  x,  y, and  z  0. Focus on RHS: lim { i + j + } = Our equation becomes:  Dv Dt We can use our short hand notation, the substantial derivative. x-component is  Dv x Dt { This is a vector equation so components must be equal. Let’s look at the x-component of the LHS: z z

 (  xx ) +  (  yx ) +  (  zx ) +  g x  x  y  z lim { } = xyzxyz  F x x0x0 y0y0 z0z0 Combining the RHS & LHS, we obtain:  Dv x = Dt Basically, this is Newton’s second law of motion If we use our viscosity relationship for the stresses, assume constant viscosity and an incompressible fluid, this becomes:  (  xx ) +  (  yx ) +  (  zx ) +  g x  x  y  z  Dv x = Dt  g x -  P  x  2 v x  2 v x  2 v x  x 2  y 2  z 2 +  { + + } Laplacian or  2  Dv x = Dt  g x -  P +   2 v x  x x-component If we multiply the x-equation by i, the y-equation by j and the z-equation by k, then add:  Dv = Dt  g -  P +   2 v Navier-Stokes equation (vector equation) Even though derived in rectangular coordinates, it holds for cylindrical and other coordinate systems. Components are in tables.