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

{ }i + { }j + { Fz}k For all of the faces: [(v vx)x+x - (v vx)x ] yz + [(v vy)y+y - (v vy)y ] xz + [(v vz)z+z - (v vz)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 + { Fz}k = [(v vx)x+x - (v vx)x ] yz + [(v vy)y+y - (v vy)y ] xz + [(v vz)z+z - (v vz)z ] xy + (v) xyz Expression for  Fx  Fy

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

(xx) + (yx) + (zx) + gx x y z lim { } =  Fx (xx) + (yx) + (zx) + gx x y z xyz x0 y0 z0 Combining the RHS & LHS, we obtain: Basically, this is Newton’s second law of motion  Dvx = Dt (xx) + (yx) + (zx) + gx x y z If we use our viscosity relationship for the stresses, assume constant viscosity and an incompressible fluid, this becomes:  Dvx = Dt gx - P x 2vx 2vx 2vx x2 y2 z2 + { + + } x-component  Dvx = Dt gx - P +  2vx x Laplacian or 2 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 +  2v Navier-Stokes equation (vector equation) Even though derived in rectangular coordinates, it holds for cylindrical and other coordinate systems. Components are in tables.