1/49 Chapter 6 Equations of Continuity and...

1/49

Chapter 6 Equations of Continuity and Motion

Session 6-2 Equation of motion

2/49

Chapter 6 Equations of Continuity and Motion

Contents

6.1 Continuity Equation

6.2 Stream Function in 2-D, Incompressible Flows

6.3 Rotational and Irrotational Motion

6.4 Equations of Motion

6.5 Examples of Laminar Motion

6.6 Irrotational Motion

6.7 Frictionless Flow

6.8 Vortex Motion

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4/49

• Apply Newton's 2nd law of motion

• External forces = surface force + body force

〮 Surface force:

~ normal force + tangential force

〮 Body forces:

~ due to gravitational or electromagnetic fields, no contact

~ act at the centroid of the element → centroidal force


(A)F ma=∑

x xF ma∆ = ∆

5/49

Consider only gravitational force


x y zg ig jg kg= + +

LHS of (A):

( )x xF gx y zρ∆ = ∆ ∆ ∆

xx xy z y zxx

σσ σ ∂ − ∆ ∆ + ∆ ∆+ ∆ ∂

yxyx yxx z x zyy

ττ τ

∂ − ∆ ∆ + ∆ ∆+ ∆ ∂

zxzx zxx y x yzz

ττ τ ∂ − ∆ ∆ + ∆ ∆+ ∆ ∂

body force (B)

normal force

tangential force

tangential force

6/49


Divide (B) by volume of element

yxx x zxx

F gx y z x y z

τσ τρ∂∆ ∂ ∂

= + + +∆ ∆ ∆ ∂ ∂ ∂

(C)

RHS of (A):

xx

ma ax y z

ρ∆ =∆ ∆ ∆

(D)

7/49


Combine (C) and (D)

yxx zxx xg ax y z

τσ τρ ρ∂∂ ∂

+ + + =∂ ∂ ∂

xy y zyy yg ax y z

τ σ τρ ρ

∂ ∂ ∂+ + + =

∂ ∂ ∂

yzxz zz zg ax y z

ττ σρ ρ∂∂ ∂

+ + + =∂ ∂ ∂ (6.21)

8/49


6.4.1 Navier-Stokes equations– Eq (6.21) ~ general equation of motion

– For Newtonian fluids (with single viscosity coeff.), use stress-strain

relation given in (5.29) and (5.30)

→ Navier-Stokes equations

Eq. (5.29):

( )223

pressure normal stress due to fluid deformation and viscosity

xup qx

σ µ µ∂ = − + − ∇ ⋅ ∂

9/49


( )223y

vp qy

σ µ µ∂ = − + − ∇ ⋅ ∂

( )223z

wp qz

σ µ µ∂ = − + − ∇ ⋅ ∂

Eq. (5.30):

yx xyv ux y

τ τ µ∂ ∂ += = ∂ ∂

yz zyw vy z

τ τ µ∂ ∂ += = ∂ ∂

zx xzu wz x

τ τ µ ∂ ∂ = = + ∂ ∂

(6.23)

(6.22)

10/49

Substitute Eqs. (5.29) & (5.30) into (6.21)

Assume constant viscosity (neglect effect of pressure and temperature on

viscosity variation)


( )223x x

v up u u wg aq x yx x y zx z xρ µ µ ρµ µ

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

( )223x x

v up u u wg aq x yx x y zx z xρ µ µ µ ρ

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

u v wx y z

∂ ∂ ∂+ +

∂ ∂ ∂

11/49


Expand and simplify

u v wx y zx

∂ ∂ ∂∂ + + ∂ ∂ ∂∂

2 2 2 2 2 2 2 2

2 2 2 2

2. . 23x

p u u v w v u u wL H S gx x x x y x z x y y z x z

ρ µ µ µ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

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

2 2 22 2 2

22 2 2

13x

u v wp u u ugx x y x zx x y z

ρ µ µ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + += − + ++ + ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂

( )2 2 2

2 2 2

13x

p u u ug qx xx y z

ρ µ µ ∂ ∂∂ ∂ ∂

= − + ++ + ∇ ⋅ ∂ ∂∂ ∂ ∂

normal stress shear stress

normal stress + shear stress

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( )2 2 2

2 2 2

13x x

p u u ug aqx xx y z

ρ µ µ ρ ∂ ∂∂ ∂ ∂

− + + =+ + ∇ ⋅ ∂ ∂∂ ∂ ∂

( )2 2 2

2 2 2

13y y

p v v vg aqy yx y z

ρ µ µ ρ ∂ ∂∂ ∂ ∂

− + + =+ + ∇ ⋅ ∂ ∂∂ ∂ ∂

( )2 2 2

2 2 2

13z z

p w w wg aqz zx y z

ρ µ µ ρ ∂ ∂∂ ∂ ∂

− + + =+ + ∇ ⋅ ∂ ∂∂ ∂ ∂

→ Navier-Stokes equation for compressible fluids with constant viscosity

(6.24)

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1) For inviscid (ideal) fluid flow, (µ = 0) → viscous forces are neglected.


♦ Vector form

( ) ( )23

qg p q qq qt

µρ µ ρ ρ∂− ∇ + ∇ + ∇ = +∇ ⋅ ⋅∇∂

( )dq qa qqdt t

∂= = + ⋅∇

∂

where --- Eq. (2.5)

( )qg p qqt

ρ ρ ρ∂− ∇ = + ⋅∇∂

→ Euler equations for ideal fluid

14/49

2) For incompressible fluids, (Continuity Eq.)

Define acceleration due to gravity as


0q∇ ⋅ =

( )2 qg p q qqt

ρ µ ρ ρ∂− ∇ + ∇ = + ⋅∇∂

(6.25)

xhg gx

∂= −

∂

yhg gy

∂= −

∂g g h= − ∇

zhg gz

∂= −

∂

15/49

where h = vertical direction measured positive upwardFor Cartesian axes oriented so that h and z coincide

→ minus sign indicates that acceleration due to gravity is in the negative

h directionThen, N-S equation for incompressible fluids and isothermal flows are


(6.26)0 , 1x yhg gz

∂= = =

∂

zg g= − (6.27)

16/49


2 2 2

2 2 2

1u u u u h p u u uu v w gt x y z x x x y z

µρ ρ

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

2 2 2

2 2 2

1v v v v h p v v vu v w gt x y z y y x y z

µρ ρ

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

2 2 2

2 2 2

1w w w w h p w w wu v w gt x y z z z x y z

µρ ρ

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

(6.28)

Local acceleration

Convective acceleration

Body force per mass

Pressure force per mass

Viscosity force per mass

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Eq. (6.28): unknowns – u, v, w, p→ We need one more equation to obtain a solution when the boundary

conditions are specified.

→ Eq. of continuity for incompressible fluid

♦ Boundary conditions

1) kinematic BC: velocity normal to any rigid boundary (wall) equal the

boundary velocity (velocity = 0 for stationary boundary)

2) physical BC: no slip condition (continuum stick to a rigid boundary)

→ tangential velocity relative to the wall vanish at the wall surface


0u v wx y z

∂ ∂ ∂+ + =

∂ ∂ ∂

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♦ General solutions for Navier-Stocks equations are not available because

of the nonlinear, 2nd-order nature of the partial differential equations.

→ Only particular solutions may be obtained by simplifications.

→ Numerical solutions are usually sought.

r - component:2

r r r rr z

v vv v v vv vt r r r z

θ θρθ

∂ ∂ ∂ ∂+ + − +

∂ ∂ ∂ ∂

( ){ } 2 22 2 2 21 21 r rr rp v v vg rvr r r r zr r θρ µ θ θ ∂ ∂ ∂ ∂ ∂∂= − + + − + ∂ ∂ ∂ ∂ ∂∂

♦ Navier-Stocks equations in cylindrical coordinates for constant density

and viscosity

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θ - component:

z - component:


rr z

v v v v v v vv vt r r r zθ θ θ θ θ θρ

θ∂ ∂ ∂ ∂ + + − + ∂ ∂ ∂ ∂

( ){ } 2 22 2 2 21 1 21 rp v v vg rvr r r r zr r θ θθ θρ µθ θ θ ∂ ∂ ∂ ∂ ∂∂= − + + + + ∂ ∂ ∂ ∂ ∂∂ z z z z

r zvv v v vv v

t r r zθρ

θ∂ ∂ ∂ ∂ + + + ∂ ∂ ∂ ∂

2 2

2 2 21 1 z zz

zp v vvg rz r r r zr

ρ µθ

∂ ∂ ∂ ∂∂ = − + + + ∂ ∂ ∂ ∂∂ (6.29)

20/49


Continuity eq. for incompressible fluid

( ) ( ) ( )1 1 0r zvrv vr r r zθθ∂ ∂ ∂

+ + =∂ ∂ ∂

(6.30)

Normal & shear stresses for constant density and viscosity

2 rrvpr

σ µ ∂= − +∂

12 rv vpr r

θθσ µ θ

∂ = − + + ∂

2 zzvpz

σ µ ∂= − +∂

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1 rr

vvrr rr

θθτ µ θ

∂ ∂ = + ∂ ∂

1 zz

v vz rθ

θτ µ θ∂ ∂ = + ∂ ∂

r zzr

v vz r

τ µ ∂ ∂ = + ∂ ∂

22/49


- N-S equations are important in viscous flow problems.

♦ Laminar motion

~ orderly state of flow in which macroscopic fluid particles move in layers

~ viscosity effect is dominant

♦ Laminar flow through a tube (pipe) of constant diameter

~ instantaneous velocity at any point is always unidirectional (along the

axis of the tube)

~ no-slip condition @ boundary wall

~ apply concept of the Newtonian viscosity

~ velocity gradient gives rise to viscous force within the fluid

~ low Re

23/49


[Re] Reynolds number = inertial force / viscous force = destabilizing force /

stabilizing force

♦ Viscous force

~ dissipative

~ have a stabilizing or damping effect on the motion

~ use Reynolds number

[Cf] Turbulent flow

~ unstable flow

~ instantaneous velocity is no longer unidirectional

~ destabilizing force > stabilizing force

~ high Re

24/49

2-D flow (x, z)

steady flow

parallel flow

z-axis coincides with h

6.5.1 Laminar flow between parallel platesConsider the two-dimensional, steady, laminar flow between parallel plates

in which either of two surfaces is moving at constant velocity and there is

also an external pressure gradient.

♦ Assumptions:


( )0 ; 0vy

∂→ = =

∂( ) 0t

∂→ =

∂

( )0 ; 0ww ∂→ = =

∂

0 ; 1h h hx y z

∂ ∂ ∂→ = = =

∂ ∂ ∂

25/49


a

y

zaz/a

0px

∂<

∂0p

x∂

=∂0

px

∂<

∂

p1 p2

26/49


♦ External pressure gradient

1 2p p>

0px

∂< →

∂i) pressure gradient assists the viscously induced motion to

overcome the shear force at the lower surface

0px

∂> →

∂ii) pressure gradient resists the motion which is induced by the

motion of the upper surface

1 2p p

27/49


Continuity eq. for two-dimensional, parallel flow:

0u wx z

∂ ∂+ =

∂ ∂

( )

2

2

only

0ux

u f z

∂=→ ∂

=

28/49


N-S Eq.:

. : ux dirt

∂−

∂uux

∂+

∂uvy

∂+

∂uwz

∂+

∂

hgx

∂= −

∂

2

2

1 p ux x

µρ ρ

∂ ∂− +

∂ ∂

2

2u

y∂

+∂

2

2u

z ∂

+ ∂

2

2

10 p ux z

µρ ρ

∂ ∂∴ = − + ∂ ∂

Steady flowContinuity eq. for incompressible fluid

2D flow

parallel flow

(6.31a)

29/49


. : wz dirt

∂−

∂wux

∂+

∂wvy

∂+

∂wwz

∂+

∂

2

2

1h p wgz z x

µρ ρ

∂ ∂ ∂= − − +

∂ ∂ ∂

2

2w

y∂

+∂

2

2w

z∂

+∂

10 pgzρ

∂∴ = − −

∂ (6.31b)

(6.31b): p gz

ρ γ∂ = − = −∂

( )p z f xγ∴ = − + (6.32)

30/49

→ hydrostatic pressure distribution normal to flow

→ For any orientation of z -axis. in case of a parallel flow, pressure is distributed hydrostatically in a direction normal to the flow.

(6.31a): ~ independent of z


p dpx dx

∂→

∂

(A)2

2dp udx z

µ ∂∴ =∂

Pressuredrop

Energy loss due toviscosity

31/49

Integrate (A) twice w.r.t. z

Use the boundary conditions,


2

2dp udzdz dzdzdx z

µ ∂=∂∫∫ ∫∫

1dp uz dz dz C dzdx z

µ ∂= +∂∫ ∫ ∫

2

1 22dp z u C z Cdx

µ= + + (6.33)

i) ( ) 2 20, 0 0 00dpz u C Cdx

µ= = → × = + ∴ =

32/49


ii) 2

1, 2dp az a u U U C adx

µ= = → = +

2

11

2dp aC U

a dxµ

∴ = −

∴ (6.33) becomes

2 212 2

dp z dp au zUdx a dx

µ µ = + −

2

2 2z dp az zu Ua dx

µ µ ∴ = − −

33/49


( ) 12

U a dp zu z u z za dx aµ

= = − −

(6.34)

Velocitydriven

Pressuredriven

i) If 0dpdx

= Couette flow (plane Couette flow)→

Uu za

= (6.35)

→ driving mechanism = (velocity)U

34/49


ii) If 0U = 2-D Poiseuille flow (plane Poiseuille flow)→

( ) parabolic1 ~2

dpu z a zdxµ

= − (6.36)

→ driving mechanism = external pressure gradient, dpdx

max @ 2au z =

2

max 8a dpu

dxµ= − (6.37)

35/49


V = average velocity

(6.38)2

max23 12

Q a dpuA dxµ

= = = −

[Re] detail

( )2 30 0

1 12 12

a a dp dpQ u dz z az dz adx dxµ µ

= = − = −∫ ∫2

max21

12 3Q a dpA a V uA dxµ

= × ∴ = = − =

36/49


[Re] Dimensionless form

2

2

12

2

1

u z a dp z zU a U dx a a

a dpPU dx

u z z zPU a a a

µ

µ

= − −

= −

= + −

37/49


[Cf] Couette flow in the narrow gap of a journal bearing

Flow between closely spaced concentric cylinders in which one

cylinder is fixed and the other cylinder rotates with a constant

angular velocity, ω

i

o i

U ra r r

Ua

ω

τ µ

=

= −

≈

38/49

→ Hagen-Poiseuille flow

→ Poiseuille flow: steady laminar flow due to pressure drop along a tube

Assumptions:

– use cylindrical coordinates


6.5.2 Laminar flow in a circular tube of constant diameter0p

x∂

<∂

39/49


parallel flow →

Continuity eq. →

paraboloid →

steady flow →

00

rvvθ

==

0zvz

∂=

∂

0zvθ

∂=

∂

0zvt

∂=

∂

0zv ≠

0zvr

∂≠

∂

40/49

Eq. (6.29c) becomes

By the way,

independent of r


10 zzp vg rz r r r

ρ µ∂ ∂ ∂ = − + + ∂ ∂ ∂ (A)

( ) ( )zp dg p h p hz z dz

ρ γ γ∂ ∂

− + = − = −+ +∂ ∂

zhg gz

ρ ρ ∂ = − ∂

( )

comp. Eq.

01

0

r

rp gr

p rr

ρ

γ

→

= −

−∂

+∂

∂→ + =

∂

41/49


Then (A) becomes

( ) 1 zd vp h rdz r r r

µγ∂ ∂ =+ ∂ ∂

( )1 zd vrp h rdz r r

γµ

∂ ∂ =+ ∂ ∂ (B)

Integrate (B) twice w.r.t. r

( )2

11

2zd r vr Cp h

dz rγ

µ∂

= ++∂

(C)

42/49


( ) 112

zd v Crp hdz r r

γµ

∂= ++

∂

( )2

1 21 ln

2 2 zd r v C r Cp hdz

γµ

= + ++ (D)

Using BCs

→ (C) :

→ (D) :

max0 , z zr v v= = 1 0C =

0 , 0zr r v= = ( )2

02

12 2

rdC p hdz

γµ

= + (D1)

43/49


Then, substitute (D1) into (D) to obtain νzpiezometricpressure

( ) ( )2 201

4zdv r rp hdz

γµ

∴ = − −+

( )22

0

0

14zrd rv p h

dz rγ

µ = − + −

(6.39)

→ equation of a paraboloid of revolution

44/49

(1) maximum velocity,

(2) mean velocity, Vz


maxzv

(6.40)max @ 0zv r =

( )max

20

4zd rv p hdz

γµ

= − + (6.41)

zdQ v dA=

( ) ( )2 201 2

4d rdrr rp hdz

πγµ

= − −+

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( ) ( )0 2 2001 2

4r dQ rdrr rp h

dzπγ

µ = − −+ ∫

( )02 4

202 2 4

r

r

d r rp h rdz

πγ

µ = − + −

( )4

0

8r d p h

dzπ

γµ

= − +

( ) max2

02

0 8 2z

z

vQ Q r dV p hA r dz

γπ µ

∴ = = = =− + (E)

[Cf] For 2 - D Poiseuille flow max23

V u=

46/49


(3) Head loss per unit length of pipe

Total head = piezometric head + velocity head

Here, velocity head is constant.

Thus, total head loss = piezometric head change

( ) 2 20

1 8 32f z zh V Vd p hL r Ddz

µ µγ

γ γ γ ≡ = =− +

(E)where 0 diameter2D r= =

(6.42)

47/49


[Re] Consider Darcy-Weisbach Eq.

(F)21

2f z

h VfL D g

=

head loss due to friction= fh

friction factorf =

Combine (6.42) and (F)

2

232 1

2z zV Vf

D D gµ

γ= (6.43)

48/49

Differentiate (6.39) w.r.t. r


64 64 64/ Rez z

fV D V D

νν

= = = → For laminar flow (6.44)

(4) Shear stress

rzr

vz

τ µ ∂=∂

zz vvrr

µ ∂∂ =+ ∂∂

(G)

( ) 12

zv d rp hr dz

γµ

∂= +

∂ (H)

49/49


Combine (G) and (H)

( )12zr

d rp hdz

τ γ= +

Linear profile

(6.45)

At center and walls

0 , 0zrr τ= =

( )max0 0

1,2zr zr

dr r rp hdz

τ τγ= = =+

Chapter 6 Equations of Continuity and MotionChapter 6 Equations of Continuity and Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.4 Equations of Motion6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion 6.5 Examples of Laminar Motion

1/49 Chapter 6 Equations of Continuity and...

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