Continuity Equation

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Continuity Equation. Continuity Equation. Net outflow in x direction. Continuity Equation. net out flow in y direction,. Continuity Equation. Net out flow in z direction. Net mass flow out of the element. Continuity Equation. Time rate of mass decrease in the element. - PowerPoint PPT Presentation

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Continuity Equation

Continuity Equation

dxdydz x

)u( dz dy u - dy dz dx x

)u(u

Net outflow in x direction

Continuity Equation

net out flow in y direction, 

dxdydz y

)v( dz dx v - dx dz dy y

)v(v

Continuity Equation

Net out flow in z direction   dxdydz

zwdydxw dx dydz

zww )( - )(

Net mass flow out of the element 

dxdydz z

)w( y

)v( x

)u(

Time rate of mass decrease in the element

dxdydzt

-

Net mass flow out of the element =

Time rate of mass decrease in the control volume

dxdydzt

dxdydz zw

yv

xu )( )( )(

Continuity Equation

sec3m

kgm 0 z

)w( y

)v( x

)u( t

0 . V

t

The above equation is a partial differential equation form of the continuity equation. Since the element is fixed in space, this form of equation is called conservation form.

0 )( )( )(

0

sec 0 )( )( )( 3

zw

yv

xu

t

mkgm

zw

yv

xu

t

If the density is constant

0 )( )( )(

0 )( )( )(

zw

yv

xu

zw

yv

xu

This is the continuity equation for incompressible fluid

Momentum equation is derived from the fundamental physical principle of Newton second law

Fx = m a = Fg + Fp + Fv

Fg is the gravity force Fp is the pressure force Fv is the viscous force  Since force is a vectar, all these forces will have three components.  

First we will go one component by next component than we will assemble all the components to get full Navier – Stokes Equation.

MOMENTUM EQUATION

[NAVIER STOKES EQUATION]

 

Fx – Inertial Force

Inertial Force = Mass X Acceleration derivative.  Inertial Force in x direction = m X

represents instantaneous time rate of change of velocity of the fluid element as it moves through point through space.  

DtDu

DtDu

u ).V( tu

DtDu

zu w

yu v

xu .u

tu

vma

zuw

yuv

xuu

tu

DtDua

vm

DtDu

Inertial force per unit volume in x direction =

Is called Material derivative or

Substantial derivative or

Acceleration derivative

‘u’ is variable

Inertial force / volume in y direction  

DtDv

zv w

yv v

xv u

tv

Inertial force / volume in z direction    Dt

Dw

zw w

yw v

xw u

tw

DtDuInertial force / volume in x direction

zuw

yuv

xuu

tu

Body forces act directly on the volumetric mass of the fluid element. The examples for the body forces are

Eg: gravitationalElectricMagnetic forces.

 Body force =  

Body force in y direction

Body force in z direction

xx g

dxdydzdxdydz

vmg

g

yg

zg

Body force per unit volume

Pressure on left hand face of the element

 Pressure on right hand face of the element

 Net pressure force in X direction is

 

Net pressure force per unit volume in X direction

dydzP

dydzdxxpP

dxdydzxpdydzdx

xpPP

xp

dxdydzdxdydz

xp

Pressure forces per unit volume

Net pressure force per unit volume in X direction

 

Net pressure force per unit volume in Y direction

 

Net pressure force per unit volume in Z direction  

Net pressure force in all direction

   Net pressure force in 3 direction  

xp

yp

zp

zp

yp

xp

zp

yp

xp

P

Viscous forces

Resolving in the X direction Net viscous forces 

dxdy dz z

dxdz dy y

dydz dx

dx

zxzx

zx

yxyx

yxxxxx

xx

  

  

 

 

 

 

dxdydz z

y

x

F zxyxxxv

a z

y

x

zxyxxx

b z

y

x

zyyyxy

c

zyxzzyzxz

Net viscous force per unit volume in X direction

Net viscous force per unit volume in Y direction

Net viscous force per unit volume in Z direction

UNDERSTANDING VISCOUS STRESSES

LINEAR STRESSES = ELASTIC CONSTANT X STRAIN RATE

strainlinear of rate average local x 2 x xxxx

Linear strain in X direction   

xuexx

yveyy

zwezz

 

 

zzyyxxe e e

zw

yv

xu

V divor V . Volumetric strain

Three dimensional form of Newton’s law of viscosity for compressible flows involves two constants of proportionality.  1. dynamic viscosity.

2. relate stresses to volumetric deformation.  

V divxu2xx

V divyv2yy

V divzw2zz

 

[ Effect of viscosity ‘ ’ is small in practice.

For gases a good working approximation can be obtained taking

Liquids are incompressible. div V = 0]

3/2

In this the second component is negligible

SHEAR STRESSES = ELASTIC CONSTANT X STRAIN RATE

n.deformatioangular rate average x 2 x yxxy

xv

yu

yxxy

xw

zu

zxxz

yw

zv

zyyz

  

z

y

x

F xzxyxxvx

z

y

x

F xzyyyxvy

z

y

x

F zzzyzxvz

  

zu

xw

zyu

xv

xFvx

y .V

xu 2

.V 2

zv

yw

zyw

yyu

xv

xFvy

.V. 2

yw

zzv

yw

yxw

zu

xFvz

Having derived equations for inertial force per unit volume, pressure force per unit volume body force per unit volume, and viscous force per unit volume now it is time to assemble together the subcomponents. 

vgfx F F F F

Assembly of all the components

  

  

 

z

y

x

g xp

DtDu zxyxxx

x

z

y

x

g yp

DtDv yzyyyx

y

zyx

gzp

DtDw zzzyzx

z

X direction:-

Y direction:-

Z direction:-

xw

zu

z

yu

xv

y

.V xu 2

x g

xp

zu w

yu v

xu u

tu x

X direction:-

zv

yw

z .V

yv 2

y

yu

xv

x g

yp

zv w

yv v

xv u

tv y

Y direction:-

.V zw 2

z

zv

yw

y

xw

zu

x g

yp

zw w

yw v

xw u

tw z

Z direction:-

z

y

x g

x

tDu xzxyxx

x

+

. uV u V. Vu .

CONVERTING NON CONSERVATION FORM ONN-S EQUATION TO CONSERVATION FORM

  

  

 Navier-stokes equation in the X direction is given by 

zxz

yxy

xxx xg

x

tDu

uV. . . VuuV

VuuVu . . V.

Divergence of the product of scalar times a vector.

  

t

u tu

tu

t

u tu

tu

Taking RHS of N-S Equation we have

u.V

tu u.V

tu

zu w

yu v

xu u

tu

DtDu

V . u uV . t

u tu

V . t

u uV . tu

0 u uV . tu

DtDu

CONTINUITY zw

yv

xu

t

since

Is equal to zero

zxz

yxy

xxx xg

xp uV .

tu

zyz

yyy

xyx yg

yp uV .

tv

zzz

yzy

xzx zg

zp uV .

tw

CONSERVATION FORM:-

zxz

yxy

xxx xg

xp

zuw

yuw

x

2u tu

zyz

yyy

xyx yg

xp

zvw

y

2v xuv

tv

zzz

yzy

xzx zg

zp

z

2w yvw

xuw

tw

xg xw

zu

z

yu

xv

y

xu 2 .V

x

xP

zuw

yuv

x

2u tu

SIMPLICATION OF NAVIER STOKES EQUATION

xg xzw2

2zu2

2y

u xyv2

xu 2

zw

yv

xu 3

2 x

xP

zuw

yuw

x

2u tu

If is constant

xg xzw2

2zu2

2yu

xyv2

xu 2

zw

yv

xu 3

2 x

xP

zuw

yuw

x

2u tu

xg xzw2

2zu2

2yu2

xyv2

2xu2

2 zx

w2 3

2 yxv2

32

2xu2

32

xP

zuw

yuw

x

2u tu

xzw2

31

xyv2

31

2zu2

2yu2

xu2

311

xP

zuw

yuv

x

2u tu

zw

x 3

1 yv

x 3

1 xu

x 3

1

2yu2

2xu2

xP

zuw

yuv

x

2u tu

zw

yv

xu 3

1 2zu2

2yu2

2xu2

xP

zuw

yuv

x

2u tu

V. 31

2zu2

2yu2

2xu2

xP

zuw

yuv

x

2u tu

2zu2

2yu2

2xu2

xP

zuw

yuv

x

2u tu

For Incompressible flow

0 V .

Energy EquationEnergy is not a vector

So we will be having only one energy equation which includes the energy in all the direction.

The rate of Energy = Force X velocity

Energy equation can be got by multiplying the momentum equation with the corresponding component of velocity

dQ = dE + dW  dE = dQ - dW = dQ + dW [Work done is negative] because work is done on the system.

Work done is given by dot product of viscous force and velocity vector.

for Xdirection

V.vF

dxdydz

zzxu

yyx.u

xxxu

xup

for Y direction

V.vF

dxdydz yzu yyyv

xyxv

yvp

for Z direction

dxdydz xxw

yzyw

xzxw

zwp

V.vF

Body force is given by dxdydz V.g

wzg vyg uxg

Total work done 

dxdydz V.f

dxdydz

zzzw

yyzw

xxzw

zzyv

yyyv

xxyv

zzxu

yyxu

xxxu

z

wp

yvp

xup

C

Net Heat flux into element = Volumetric Heating + Heat transfer across surface.

Volumetric heating dxdydz .q

Heat transfer in X direction = dydz dx

x

.xq

xq x q

dxdydz x

.q

dxdydz z

.zq

y

.yq

x

.xq

Heating of fluid element

dQ = B = dxdydz z

.zq

y

.yq

x

.xq

.q

dQ = B dxdydz zTk

z

yTk

y

xTk

x q

z

wp yvp

xup

zTk

z

yTk

y

xTk

x q

2

2V e DtD

f.V zzzw

yyzw

xzw

zzyv

yyyv

xxyv

zzxu

yyxu

xxxu

z

wp yvp

xup

zTk

z

yTk

y

xTk

x q

2

2V e DtD

Energy EquationNonconservation form

z

wp yvp

xup

zTk

z

yTk

y

xTk

x q

2

2V e DtD

f.V

zzzw

yyzw

xxzw

zzyv

yyyv

xxyv

zzxu

yyxu

xxxu

Non conservation:-

z

wp yvp

xup

zTk

z

yTk

y

xTk

x

q V 2

2V e . 2

2V e DtD

f.V

zzzw

yyzw

xxzw

zzyv

yyyv

xxyv

zzxu

yyxu

xxxu

Conservation:-

.V p 2zT2

k 2yT2

k

2xT2

k q z

Tpc w

y

Tpcu

x

Tpcu

x

Tpc

.V p 2zT2

k 2yT2

k

2xT2

k q z

wT y

vT x

uT pc x

Tpc

xfzxz

yxy

xxx

xp

tDuD

fyz

yzyyy

xyx

yp

tDvD

fzz

zzy

zyx

zxzp

tDwD

Momentum Equation     Non conservation form 

X direction

Y direction

Z direction

Momentum Equation 

Conservation form 

X direction

Y direction

Z direction

xfzxz

yxy

xxx

xpVu

tDuD

)(.

fyz

yzyyy

xyx

ypVv

tDvD

).(

fzz

zzy

zyx

zxzp

VwtDwD

)(.

Vfz

zzwy

zywx

zxwz

xzz

yzvy

yyvx

yxvz

xzuy

xyuxxxu

zwp

yvp

xup

zTk

zyTk

yxTk

xqVe

tDD

.)()()(

)()()()()()(

)()()(2

2)()()()(

Energy Equation

   Non conservation form

Vfz

zzwy

zywx

zxwzxz

zyzv

yyyv

xyxv

zxzu

yxyu

xxxu

zwp

yvp

xup

zT

kzy

Tk

yxT

kx

qVeVet

V

.)(

)()()()()(

)()()()()()(2

2.

2

2)()()(])([])([

Energy equation

Conservation form

FORMS OF THE GOVERNING EQUATIONS PARTICULARLY SUITED FOR CFD

energytotalofFluxVVe

energyInternalofFluxVemomentumofcomponentzofFluxVwmomentumofcomponentyofFluxVvmomentumofcomponentxofFluxVu

fluxMassV

)(2

2

Solution vectar

)(2

2Ve

wvu

U

Variation in x direction

xzwxyvxxuxTkupuVe

xzuwxyuv

xxpuu

F

)(2

2

2

Variation in y direction

zywyyvxyuyTkvpvVe

zyvwyypv

yxvuv

G

)(2

2

2

Variation in z direction

zzwyzvxxzuzTkwpwVe

xzzpwxyzwv

xzwuw

H

)(2

22

Source vectar

qzfwyfvxfuzfyfxf

J

)(

0

Time marching

JzH

yG

xF

tU

Types of time marching

1. Implicite time marching

2. Explicite time marching

Explicit FDM

Implicit FDM

Crank-Nicolson FDM

Space marching

JzH

yG

xF

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