1. . { }= { } + { } 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

2. { }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

3. 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

4. (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
x y 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.