The derivation of the equation of motion for various fluids is similar to the d derivation of Eular’s equation. However ,the tangential stresses arise during the motion of a real viscous fluid, must be considered

1. SUBJECT :- APPLIED FLUID MECHANICS
NAVIER-STOKES
EQUATION

2. NAVIER-STOKES EQUATION OF MOTION
•The derivation of the equation of motion for various
fluids is similar to the d derivation of Eular’s equation.
However ,the tangential stresses arise during the motion
of a real viscous fluid, must be considered
•Let , R is the body force per unit mass of fluid having
components X,Y,Z in the x,y,z directions respectively.
•Let, 𝑇𝑋,𝑇𝑌,𝑇𝑍 be the components of shear forces per unit
mass set up by the viscous effects.
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4. •As per newtons second law of motion, in X-directions
•Total force = mass * acceleration ( inertia force )
•X. p.dx. dy. dz -
𝜕p
𝜕x
. dx. dy. dz + Tx = p.dx. dy. dz
du
dt
…(1)
(body force) (pressure force) (shear force) (inertia force)
• mass = density * volume
• m = p.dx. dy. dz
•The shear stress due to viscosity on a particular surface
equals the rate of change of velocity in a direction normal
to that surface.
• 𝜏 = 𝜇 .
dv
dn
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5. •The resistance force acting on the face AEHD is
• = - 𝜇 (dy. dz)
du
d𝑥
•The resistance force acting on the face BFGC is
• = 𝜇 (dy. dz)
𝜕
𝜕𝑥
(u +
𝜕𝑢
𝜕𝑥
d𝑥)
• = 𝜇 (dy. dz) (
𝜕𝑢
𝜕𝑥
+
𝜕2 𝑢
𝜕𝑥2 . d𝑥)
•Therefore, the resultant force acting in the x-direction on
faces AEHD and BFGC is :
•= - 𝜇 (dy. dz)
du
d𝑥
+ 𝜇 (dy. dz) (
𝜕𝑢
𝜕𝑥
+
𝜕2 𝑢
𝜕𝑥2 . d𝑥)
• = 𝜇
𝜕2 𝑢
𝜕𝑥2 dx. dy. dz …...(2)
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6. •Similarly, the X-component of resistance force on
face EFGH is
• = - 𝜇 dx. dy
du
d𝑧
•the X-component of resistance force on face ABCD
is
• = 𝜇 (dy. dz) (
𝜕𝑢
𝜕𝑥
+
𝜕2 𝑢
𝜕𝑥2 . dz)
•The net – X-component of resistance force on face
EFGH and ABCD is
•= 𝜇
𝜕2 𝑢
𝜕𝑧2 dx. dy. dz …..(3)
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7. • Similarly, the X-component of resistance force on faces
HGCD and EFBA are
• = 𝜇
𝜕2 𝑢
𝜕𝑦2 dx. dy. dz ……..(4)
• The total viscous resistance parallel to x-axis is given by
the sum of (2), (3), and (4).
• = 𝜇 (
𝜕2 𝑢
𝜕𝑥2 +
𝜕2 𝑢
𝜕𝑦2 +
𝜕2 𝑢
𝜕𝑧2) dx. dy. dz
• The analogous equation for components 𝑇𝑌 and 𝑇𝑍
• 𝑇𝑌 = 𝜇 (
𝜕2 𝑣
𝜕𝑥2 +
𝜕2 𝑣
𝜕𝑦2 +
𝜕2 𝑣
𝜕𝑧2) dx. dy. dz
• 𝑇𝑍 = 𝜇 (
𝜕2 𝑤
𝜕𝑥2 +
𝜕2 𝑤
𝜕𝑦2 +
𝜕2 𝑤
𝜕𝑧2 ) dx. dy. dz
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8. •Substitute value of 𝑇𝑋 in equation (1)
•X. p.dx. dy. dz -
𝜕p
𝜕x
. dx. dy. dz + 𝜇 (
𝜕2 𝑢
𝜕𝑥2 +
𝜕2 𝑢
𝜕𝑦2 +
𝜕2 𝑢
𝜕𝑧2) dx. dy. dz
= p.dx. dy. dz
du
dt
•Dividing throughout by p.dx. dy. dz, we get
• X -
1
𝑝
𝜕p
𝜕x
=
du
dt
-
𝜇
𝑝
(
𝜕2 𝑢
𝜕𝑥2 +
𝜕2 𝑢
𝜕𝑦2 +
𝜕2 𝑢
𝜕𝑧2) but ν =
𝜇
𝑝
•X -
1
𝑝
𝜕p
𝜕x
=
du
dt
- ν (
𝜕2 𝑢
𝜕𝑥2 +
𝜕2 𝑢
𝜕𝑦2 +
𝜕2 𝑢
𝜕𝑧2)
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9. •Similarly for y and z directions
• Y -
1
𝑝
𝜕p
𝜕𝑦
=
du
dt
- ν (
𝜕2 𝑣
𝜕𝑥2 +
𝜕2 𝑣
𝜕𝑦2 +
𝜕2 𝑣
𝜕𝑧2)
• Z -
1
𝑝
𝜕p
𝜕𝑍
=
du
dt
- ν (
𝜕2 𝑤
𝜕𝑥2 +
𝜕2 𝑤
𝜕𝑦2 +
𝜕2 𝑤
𝜕𝑧2 )
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10. Applications
•Laminar flow in circular pipes.
•Laminar un-directional flow between stationary parallel
plates.
•Laminar uni-directional between parallel plates having
relative motion.
•Laminar flow between concentric rotating cylinders.
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