3. Navier Stokes Equations

7/18/2019 3. Navier Stokes Equations

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Navier

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equations

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Differential equations of motionThe Navier -Stokes equations

The Navier‐Stokes equations describe momentum balances across a differential element of incompressible fluid in a particular co‐ordinate system.

Newton’s second law of motion states that the rate of change of momentum of an element of fluid

is equal to the net force acting on the fluid

Consider a differential element of flowing fluid. Let the fluid be Newtonian (viscosity μ is constant for

constant T) and incompressible (density ρ is constant for constant T). In a rectangular co‐ordinate

system, the element may be characterised by length δx, height δy and depth δz. By Newton’ssecond

law, the change of momentum of that element in the x‐direction with time can be described by

· ∑

·

· · · · ·

It can be shown that this may be expressed mathematically as

x x x x x

z x

y x

x x g

z

u

y

u

x

u

x

P

z

uu

y

uu

x

uu

t

u ρ µ ρ +⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂2

2

2

2

2

2

.............(66)

Where the LHS is the differential change in momentum of the element in the x‐direction. The first

term on the RHS is due to differences of pressure upstream and downstream (in the x‐direction) of

the element; the second term is due to shear stresses acting on the surfaces of the element in the x‐

direction, and the final term is due to the force of gravity on the element.

Similar balances can be written for the y‐ and z‐directions to give a complete set of differential

equations in rectangular co‐ordinates:



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In rectangular co-ordinates

x‐component x x x x x

z x

y x

x x g

z

u

y

u

x

u

x

P

z

uu

y

uu

x

uu

t

u ρ µ ρ +⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂2

2

2

2

2

2

y‐component y

y y y y

z

y

y

y

x

yg

z

u

y

u

x

u

y

P

z

uu

y

uu

x

uu

t

u ρ µ ρ +

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂2

2

2

2

2

2

z‐component z z z z z

z z

y z

x z g

z

u

y

u

x

u

z

P

z

uu

y

uu

x

uu

t

u ρ µ ρ +⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂2

2

2

2

2

2

These are the differential equations of motion, or the Navier‐Stokes equations that describe the

changes in velocity (direction and magnitude) of a package of incompressible fluid as a result of the forces acting on that fluid.

These may also be written in cylindrical co‐ordinates:

In cylindrical co-ordinates

r‐component

( ) r

zr

r

r

z

r r

r

r

g

z

uu

r

u

r

ru

r r r

r

P

z

uu

r

uu

r

u

r

uu

t

u

ρ

θ θ

µ

θ ρ

θ

θ θ

+⎟⎟

⎠

⎞⎜⎜

⎝

⎛

∂

∂+

∂

∂−

∂

∂+⎟

⎠

⎞⎜

⎝

⎛

∂

∂

∂

∂+

∂

∂−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂

∂+−

∂

∂+

∂

∂+

∂

∂

2

2

22

2

2

2

211

θ‐component

( ) θ θ θ

θ

θ θ θ θ θ θ

ρ θ θ

µ

θ θ ρ

g z

uu

r

u

r ru

r r r

P

r z

uu

r

uuu

r

u

r

uu

t

u

r

zr

r

+⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂−

∂

∂+⎟

⎠

⎞⎜⎝

⎛

∂

∂

∂

∂+

∂

∂−=⎟

⎠

⎞⎜⎝

⎛

∂

∂+−

∂

∂+

∂

∂+

∂

∂

2

2

22

2

2

211

1

z‐component

z

z z z

z z

zr

z

g z

uu

r r

ur

r r

z

P

z

uu

u

r

u

r

uu

t

u

ρ θ

µ

θ ρ θ θ

+

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

∂

∂+

∂

∂+⎟

⎠

⎞⎜⎝

⎛

∂

∂

∂

∂+

∂

∂−=⎟

⎠

⎞⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂

2

2

2

2

2

11

There are no general solutions to the Navier‐Stokes equations. However, it is possible to reduce

these equations to forms that can be handled manually by identifying which of the terms in the

above expressions are negligible and can be eliminated.



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The full set of equations can only be used by finite element computing programmes (computational

fluid dynamics) that are used for predicting fluid dynamics in complicated geometries.

Boundary Conditions

Boundary conditions are needed to solve differential equations in fluid flow.

A number of commonly used boundary conditions are included below:

1.

The velocity of a fluid particle in contact with a boundary at rest is zero.

2. The velocity of a fluid particle in contact with a moving boundary is the same as the velocity

of the moving boundary.

3. For two immiscible liquids flowing through a pipe:

a.

the fluids particles have a common velocity at the liquid‐liquid interface; ie

1 2liquid liquid v v=

b. the fluids particles have a common momentum flux perpendicular to the surface; ie

1 21 2liquid liquid

dv dv

dx dxµ µ =

at the interface

where μliquid1 is the viscosity of liquid 1 and μliquid2 is the viscosity of liquid 2

v1 is the velocity distribution for liquid 1

v2 is the velocity distribution for liquid 2

is the velocity gradient in x‐direction (it can be in y‐direction, r‐direction,

θ, etc – depending on the system of coordinates)

4.

At liquid‐gas interfaces the momentum flux (hence the velocity gradient) in the liquid phase

is very nearly zero and can be assumed to be zero in most calculations.



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Applications of Navier-Stokes equations

1. Steady Laminar flow between fixed parallel plates (infinite plates)

Consider laminar flow between two horizontal infinite parallel plates as shown in the figure below

x

y

U x

h

h

The flow is in the x‐direction and we therefore need the x‐component of the Navier Stokes equations

i.e.

x x x x x

z x

y x

x x g

z

u

y

u

x

u

x

P

z

uu

y

uu

x

uu

t

u ρ µ ρ +⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂2

2

2

2

2

2

Simplifications

•

There are no flows in the y and z directions and therefore there is no velocity in the y and z

directions. I.e. uy and uz = 0.

•

There is no change in velocity along the x‐axis (uniform flow) i.e. 0

•

There is no variation of ux in the z‐direction. I.e.

0, but ux = f(y)

•

the flow is steady state i.e 0.

•

The plates are horizontal and therefore there is no effect of gravity. I.e. 0

Reduce the Navier –stokes equation by applying the above conditions:

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂−=

2

2

0 y

u

x

P xµ or ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂=

∂

∂2

2

y

u

x

P xµ

The velocity ux is a function of only y (i.e. ux=f(y)) and therefore.

i.e

which can now be integrated to obtain



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i.e.

Where C1 is a constant of integration.

Further integration yields

i.e.

Where C2 is a constant of integration.

(Note: in this case is regarded as constant for integration since the integration is with respect to y

direction and P is not a function of y.)

C1 and C2 can be determined with the help of the boundary conditions.

The two plates are fixed;

At y=+h; ux=0;

0

At y= ‐h; ux=0;

0

Thus, C1=0 and

Therefore

and the velocity profile is parabolic.

Volumetric flow rate Q

The volumetric flow rate, Q between the two parallel plates (for a unit width in the z direction) is

obtained from the relationship

· · ∆ (=velocity × cross‐sectional area [=] m3/s)

∆ 1 (for a unit width)

Therefore

The pressure gradient is negative as the pressure decreases in the direction of flow.

If we make the approximation



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Navier

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∆

ℓ

Then

∆ℓ

The volumetric flow rate is directly proportional to the pressure gradient, inversely proportional to

the viscosity and strongly dependent on the distance between the plates.

Average velocity, uave

For a unit depth between the plates i.e. ∆ 1, the overall volumetric flow may be described by

Q = cross sectional area × average velocity

i.e. · ∆ · ∆ · 2 · 1

Therefore the average velocity can be described by

∆ℓ

The maximum velocity, umax

The maximum velocity umax occurs midway (i.e. at y=o) between the two plates.

i.e consider the original equation for ux.

When y = 0,

,

0

∆

ℓ

Note that we can calculate directly the relationship between umax and uave: i.e.

,

∆ℓ

∆ℓ



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Navier

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equations

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2. Steady Laminar flow of a liquid (with constant density and viscosity)

through a horizontal circular pipe of radius, R and length, L

Consider steady laminar flow of an incompressible Newtonian fluid in a circular pipe:

r

z

R

The flow is in the z‐direction and we therefore need the z‐component of the Navier‐Stokes equations

in cylindrical coordinates i.e.

z

z z z z

z

z

r

z g z

uu

r r

ur

r r z

P

z

uu

u

r

u

r

uu

t

u ρ

θ µ

θ ρ θ θ +⎟

⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂+⎟

⎠

⎞⎜⎝

⎛

∂

∂

∂

∂+

∂

∂−=⎟

⎠

⎞⎜⎝

⎛

∂

∂+

∂

∂+

∂

∂+

∂

∂2

2

2

2

2

11

Simplifications

• There is no flows in the r direction and no flow around the pipe axis (θ‐direction); therefore θ 0

•

There is no change in velocity along the z‐axis (uniform flow) i.e. 0

•

There is no variation of uz in the θ‐direction. I.e.

0, but uz = f (r)

•

The flow is steady state i.e 0.

•

The pipe is horizontal and therefore there is no effect of gravity. I.e. 0

Reduce the Navier –stokes equation by applying the above conditions:

The equation reduces to

0

The velocity uz is a function of only r (i.e. uz = f (r))and therefore

.



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So the equation becomes

Or

This may be integrated to obtain

Setting r = 0, it can be seen that C1+0 = 0, or C1 = 0;

Therefore

And integrating for a second time gives

The boundary condition for this problem is that at r=R, uz = 0 (i.e. no slip at the walls)

Thus 0

Gives

And finally

1

The pressure gradient is negative as the pressure decreases in the direction of flow.

As before, If we make the approximation

∆

ℓ

Then

∆

ℓ

1

i.e., the velocity profile is parabolic

As before, the pressure gradient , is treated as a constant as far as integration is concerned

because the integration is with respect to r direction and also P is not a function of r.



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Navier

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The maximum velocity (umax)

From the velocity profile, the velocity is maximum when r=0

i.e ,

∆ℓ

The volumetric flow rate, Q

As before, the volumetric flow rate is

Therefore integrating with respect to r and θ gives

∆ℓ

1

∆ℓ

1

Integrating wrt θ:

∆ℓ

2 1

Integrating by parts gives

∆ℓ

2

Thus the solution is

∆ℓ

Which is the Hagen‐Poiseuille equation for laminar flow in a circular pipe, which was originally

determined using an energy balance (equation 15). Clearly the two approaches are equivalent.

Mean velocity, uave

Mean velocity, uave is given by Q/Area, i.e.

Thus 4

8∆ℓ

18

∆ℓ 2

In this case, it can be seen that the volumetric average velocity uave is exactly ½ × the maximum

velocity (umax).

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3. Navier Stokes Equations