NS applied to special cases

. ij

DVg

Dtρ ρ σ= + ∇

�

�

Class 10

• The fluid at rest: Hydrostatics

• Newtonian fluid

• Incompressible flow

• Inviscid flow: The Euler and Bernoulli equations

Special casesThe fluid at rest: Hydrostatics

If the fluid is at rest:

• Viscous stresses vanish:

0 ( )ij for i jσ = ≠

• Normal stresses equal to the hydrostatic pressure.

pσ σ σ= = = −xx yy zz pσ σ σ= = = −

p gρ∇ =�

Hence the NS equations become

In expended form: (we take the z coordinate as up)

0,p

x

∂=

∂0,

p

y

∂=

∂

pg

zρ

∂= −

∂

Newtonian fluid

Stokes’ postulates for Newtonian fluid:

• The fluid is continuous, and its stress tensor is at most a linear

function of the strain rate.

• The fluid is isotropic, i.e., its properties are independent of direction,

and therefore the deformation law is independent of the axes in

which it is expressed.

• When the strain rate is zero, the deformation law must reduce to the

hydrostatic pressure condition.

, :

1

0

ij ij ij

ij

ij

p

if i j

if i j

σ δ δ

δ

δ

= −

= =

= =

Kronecker delta function.

Newtonian fluid...

General deformation law for a Newtonian viscous fluid (Stokes (1845)):

.ji

ij ij ij

j i

uup V

x xσ δ µ δ λ

∂∂= − + + + ∇ ∂ ∂

� �

:λ Coefficient of bulk viscosity, it is independent of This

comes only in the compressible flow.

.µcomes only in the compressible flow.

Stokes’ hypothesis:2

03

λ µ+ =

. 0V∇ =�

Incompressible flow:

Navier-Stokes Equation for a Newtonian Fluid

: 2 .x

u u u u p ux u v w g V

t x y z x x x

u v w u

y y x z x z

ρ ρ µ λ

µ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + = − + + ∇ + ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ + + + ∂ ∂ ∂ ∂ ∂ ∂

�

: y

v v v v p v uy u v w g

t x y z y x x yρ ρ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

2 .

yt x y z y x x y

v v wV

y y z z yµ λ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂+ ∇ + +

∂ ∂ ∂ ∂ ∂

�

:

2 .

z

w w w w p w uz u v w g

t x y z z x x z

v w wV

y z y z z

ρ ρ µ

µ µ λ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + = − + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ + + + ∇

∂ ∂ ∂ ∂ ∂

�

For incompressible flow if the viscosity is constant, the NS equations reduce to:

2DVg p V

Dtρ ρ µ= − ∇ + ∇

�

�

�

But in general viscosity could be a function of temperature and pressure.

Inviscid flow ( )0µ =

The NS equations reduce to the Euler’s equation: DV

g pDt

ρ ρ= − ∇

�

�

Euler’s equation for unidirectional flow: ρ ρ∂ ∂ ∂

+ = − ∂ ∂ ∂

�w w pw g

t z z

Further for steady flow:

( )

w pw g

z zρ ρ

∂ ∂ = − −

∂ ∂

( )2

1

2

w pg

z zρ ρ

∂ ∂⇒ = − −

∂ ∂Integrating w.r.t. z (we take the z coordinate as up) along a streamline

between points 1 and 2, we obtain

2 2

1 1 1 2 2 2

1 1

2 2p w gz p w gzρ ρ ρ ρ+ + = + +

Bernoulli’s Equation for steady, inviscid and incompressible flow.

Bernoulli’s Equation...

21

2p w gz Constρ ρ+ + =

Pressure

Energy

Kinetic

Energy

Potential

Energy

Divide by we get ,gρ2

p w2

2

p wH z Const

g gρ= + + =

Total head = Pressure+ Kinetic + Static

head head head

Bernoulli’s Equation for Real Fluid

However, all the real fluids are viscous and hence offer resistance to flow. Thus

there are always some losses in the flows.

Assumptions in deriving the Bernoulli’s equation:

• Fluid is inviscid

• Incompressible

• Flow is steady

• Flow is irrotational

The Bernoulli’s equation for real fluids between points 1 and 2 is given by

2 2

1 1 2 2

1 2 ,2 2

L

p w p wz z h

g g g gρ ρ+ + = + + +

where hL is the loss of energy between points 1 and 2.

Exact solution of the NS equations

Objective: To find out the exact solution of the NS equations for some

laminar, Newtonian viscous flows.

Contents:

• Conditions at the solid boundary

• Laminar flow velocity profile• Laminar flow velocity profileI. Steady pressure-driven flow through two parallel plates (Channel

flow).

II. Steady wall-driven flow through two parallel plates (Couette flow)

III. Steady pressure-driven flow through circular pipe (Poiseuille flow).

Conditions at a solid boundary

Normal component of velocity: Obviously the normal component of velocity must be zero! This is called no-

penetration condition.

Tangential component of velocity: Most fluids ‘stick’ to boundary. The tangential component of velocity is same

as that of the boundary. If the boundary is at rest, this is zero. This is called

no-slip condition.no-slip condition.

Some complex fluids (suspensions, pastes, etc.) show apparent slip at the wall.

Shear stress at the wall From Newton’s law of viscosity:

0y

du

dyτ µ

=

=

y

u(y)

u=0

Pressure-driven flow through two parallel plates (Channel flow/Poiseuille flow)

2 2

2 2

2 2

2 2

u u u p u uu v

t x y x x y

v v v p v vu v g

t x y y x y

ρ µ

ρ µ ρ

∂ ∂ ∂ ∂ ∂ ∂+ + = − + + ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂+ + = − + + − ∂ ∂ ∂ ∂ ∂ ∂

Assumption: wallτAssumption:

• Steady

• Incompressible

• Neglecting the gravity

• Fully developed flow, i.e., v=0, u=u(y)

2

2,

0 ( ).

p u

x y

pp p x

y

µ ∂ ∂

= ∂ ∂

∂= ⇒ =

∂

x

y

y=hwallτ

Hence,2

2

dp d u

dx dyµ=

Integrating the above equation w.r.t y, we obtain

1(1)

du dpy c

dy dxµ= +

Integrating again, we obtain21

(2)2

dpu y cy d

dxµ= + +

2 dxµ

The constants are obtained using the no-slip boundary conditions at the walls

0, 0,

0, .

u at y

u at y h

= =

= =

1,

2

0

dpc h

dx

d

µ= −

=Substituting the values of these constants,

we get( )21

2

1

2

dpu y hy

dx

du dp hy

dy dx

µ

µ

= −

= −

Velocity parabolic for this case.

Ratio of maximum velocity to average velocity:

Velocity is maximum, when centerline of the channel.0 ,2

du hy

dy= ⇒ =

2

max

1

8

dpu h

dxµ= −

Volume flow rate, Q, is given by

31

12

hdp

Q udy hdxµ

= = −∫0

12 dxµ∫

Thus, average velocity,3

2

1

112

1 12avg

dph

Q dpdxu h

A h dx

µ

µ

−

= = = −×

max 3

2avg

u

u=Hence,

Drop of pressure head for a given length:

2

2 1 2 12

12

. . . ,

12( )

avg

avg

udp

dx h

Integrating w r t x we get

up p x x

h

µ

µ

µ

= −

− = − −

1 2 2 12

1 2

2

12( )

,

12

avg

f

avg

f

u Lp p where L x x

h

If h is the drop of pressure head then

u Lp ph

g gh

µ

µ

ρ ρ

⇒ − = = −

−= =

21

Shear stress distribution:

1( 2 )

2

1

2wall

du dph y

dy dx

dph

dx

τ µ

τ

= = − −

⇒ = −

See how we are getting this expression at both the walls??

2 2

2 2

2 2

2 2

u u u p u uu v

t x y x x y

v v v p v vu v g

t x y y x y

ρ µ

ρ µ ρ

∂ ∂ ∂ ∂ ∂ ∂+ + = − + + ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂+ + = − + + − ∂ ∂ ∂ ∂ ∂ ∂

Assumption: U

Wall-driven flow between two parallel plates (Couette flow)

Assumption: • Steady

• Incompressible

• Neglecting the gravity

• Fully developed flow, i.e., v=0, u=u(y)

• No pressure gradient2

20

u

yµ ∂

= ∂

x

y

y=h0

U

duc

dy=

u cy d= + 0

0, 0,

, .

u at y

u U at y h

= =

= =

Boundary conditions

0 , 0U

c dh

= =

0

0

0

Uu y

h

Udu

dy h

Udu

dy hτ µ µ

=

=

= =dy h

Pressure-driven flow through circular pipe (Hagen-Poiseuille flow)

Assumption: • Steady

• Incompressible

• Neglecting the gravity

• Fully developed flow, i.e., 0, 0, ( )r z zu u u u rφ= = =

1 1zd du dpr

r dr dr dzµ

=

2

2

. . . ,

1

2

. . . ,

1ln

4

z

z

rr dr dr dz

Integrating w r t r weobtain

du dp rr c

dr dz

Integrating w r t r weobtain

dpu r c r d

dz

µ

µ

µ

=

= +

= + +

0, 0,

0, .

z

z

u at r

duat r R

dr

= =

= =

Boundary conditions

2 21( )

4

1

2

1

2

z

z

z

dpu R r

dz

du dpr

dr dz

du dpr

dr dz

µ

µ

τ µ

= − −

=

= − = −

R

Volume flow rate,

0

2

R

zQ u rdrπ= ∫

max( )

2( )

z

z avg

u

u=

( )

( )

2

max

2

1

4

1

8

z

z avg

dpu R

dz

dpu R

dz

µ

µ

= −

= −

( )2

1 2

2

8

32

z avg

avg

f

udp

dz R

u Lp ph

g gD

µ

µ

ρ ρ

= −

−= =

Hagen-Poiseuille Formula

Navier - Stokes equation:

We consider an incompressible , isothermal Newtonian flow (density ρ =const, viscosity μ =const), with a velocity field ))()()(( x,y,z, w x,y,z, vx,y,zuV =

r

Incompressible continuity equation:

0=∂∂

+∂∂

+∂∂

zw

yv

xu eq1.

Navier - Stokes equation:

vector form: VgPDt

VD rrr

2∇++−∇= μρρ

x component:

)( 2

2

2

2

2

2

zu

yu

xug

xP

zuw

yuv

xuu

tu

x ∂∂

+∂∂

+∂∂

++∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂ μρρ eq2.

y component:

)( 2

2

2

2

2

2

zv

yv

xvg

yP

zvw

yvv

xvu

tv

y ∂∂

+∂∂

+∂∂

++∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂ μρρ eq3.

z component:

)( 2

2

2

2

2

2

zw

yw

xwg

zP

zww

ywv

xwu

tw

z ∂∂

+∂∂

+∂∂

++∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂ μρρ eq4.

Cylindrical coordinates ),,( zr θ : We consider an incompressible , isothermal Newtonian flow (density ρ =const, viscosity μ =const), with a velocity field .uuuV zr ),,( θ=

r

Incompressible continuity equation:

0)(1)(1

=∂∂

+∂

∂+

∂∂

zuu

rrru

rzr

θθ eq a)

r-component:

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

−∂∂

+−⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

++∂∂

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+−∂∂

+∂∂

+∂∂

2

2

22

2

22

2

211zuu

ru

rru

rur

rrg

rP

zuu

ruu

ru

ruu

tu

rrrrr

rz

rrr

r

θθμρ

θρ

θ

θθ

eq b)

θ -component:

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+−⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

++∂∂

−=

⎟⎠⎞

⎜⎝⎛

∂∂

++∂∂

+∂∂

+∂∂

2

2

22

2

22

2111zuu

ru

rru

ru

rrr

gPr

zu

uruuu

ru

ru

ut

u

r

zr

r

θθθθθ

θθθθθθ

θθμρ

θ

θρ

eq c)

z-component:

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

++∂∂

−=

⎟⎠⎞

⎜⎝⎛

∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

11zuu

rrur

rrg

zP

zuuu

ru

ruu

tu

zzzz

zz

zzr

z

θμρ

θρ θ

eq d)