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