The document discusses the derivation of the Navier-Stokes equations. It considers an elementary fluid mass and analyzes the forces acting on it, including pressure, gravity, and viscous forces. Shear stresses on the faces of the elementary volume are calculated. Equating the total forces gives equations (5), (6), and (7), which are the Navier-Stokes equations relating velocity, pressure, viscosity, and body forces. The Navier-Stokes equations are then applied to problems involving laminar fluid flow.

2.  The fluid element is acted upon by gravity force,
pressure force and viscous force is the case of Navier-
Stokes equation.
 Consider an elementary small mass of fluid of size dx*
dy* dz in x, y, z three directions respectively as
shown in figure.

4. Pressure force on the face ABCD= p dxdy
Pressure force on the face EFGH=(p+∂p/∂x dx) dy dz
Hence net pressure force in x-direction is equal to
= p dxdy -- (p+∂p/∂x dx) dydz = ∂p/∂x dxdydz
Let R is the body force per unit mass of fluid having components X,Y,Z in the
x,y,z directions respectively. Therefore the body force acting on the
element of fluid in the X-direction will be Xρ dxdydz.
Let tx,ty,tz are the component of shear forces in x,y,z directions
Total force=mass * acceleration
Xρ dxdydz---∂p/∂x dxdydz—Tx= ρ dxdydz *du/dt --------------(1)

5. Shear stress caused due to viscosity on a particular surface equal to the rate of
change of velocity in a direction normal to that surface
ѓ=μ dv/ dn
Let us consider any two faces of parllelopiped say ABCD and EFGH
The shear force acting on face ABCD
= μ (dydz) du/dx
The shear force acting on face EFGH
= μ (dydz) ∂/ ∂x [u+ ∂u/∂x dx]
= μ (dydz) *[∂u/∂x + ∂²u/∂x² dx]
The net shear force along the x-axis on faces ABCD and EFGH
= μ (dydz) du/dx -- μ (dydz) *[∂u/∂x + ∂²u/∂x² dx]
= - μ ∂²u/∂x² dx dy dz -----------------------------------(2)

6. Similarly x-component of shear force on face BCFG
= μ (dx dy) ∂u/∂z
X-component of shear force on face ADEH
= μ (dx dy) ∂/ ∂z [u+ ∂u/∂z dz]
= μ (dx dy) *[∂u/∂z + ∂²u/∂z² dz]
The net x-component of shear force on face BCFG and ADEH
= μ (dx dy) ∂u/∂z --_ μ (dx dy) *[∂u/∂z + ∂²u/∂z² dz]
=- μ ∂²u/∂z² dx dy dz -------------------------------(3)
Similarly net x-component of shear force on face CDEF and ABGH is
=- μ ∂²u/∂y² dx dy dz --------------------------------------------------(4)

7.  The total viscous resistance parallel to x-axis on the all six faces of the
element is given by sum of eq (2), (3) , (4)
Tx= - μ [∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² ] dx dy dz
Similarly for Ty and Tz are
Ty = -μ [∂²v/∂x² + ∂²v/∂y² + ∂²v∂z² ] dx dy dz
Tz = - μ [∂²w/∂x² + ∂²w/∂y² + ∂²w/∂z² ] dx dy dz

8. Substituting the values of Tx in eq 1
X ρ dx dy dz--- ∂p/∂x dx dy dz + μ [∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² ]
dx dy dz= ρ dx dy dz du/dt
Dividing throughout by ρ dx dy dz , we get
X - μ/ρ ∂p/∂x = du/dt –μ/ρ [∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² ]
X- μ/ρ ∂p/∂x = du/dt –v [∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z²] [ μ/ρ=v] -
----(5)
Similarly for y and z direction
Y- μ/ρ ∂p/∂x = du/dt – v [∂²v/∂x² + ∂²v/∂y² + ∂²v∂z² ]--------(6)
Z -μ/ρ ∂p/∂x = du/dt – v [∂²w/∂x² + ∂²w/∂y² + ∂²w/∂z² ] -----(7)

9. Navier-Stokes equations help in
I. The design of aircraft and cars
II. Study of blood flow
III. The design of power stations
IV. Analysis of pollution

10. Applications of Navier-Stokes Equation are
1. Laminar flow in circular pipes.
2. Laminar flow between two fixed plates.
3. Laminar flow between parallel plates having relative
motion.

11. Thank You