Continuity Equation Net outflow in x direction Continuity Equation net out flow in y direction,
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Transcript of Continuity Equation Net outflow in x direction Continuity Equation net out flow in y direction,
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
z
wdydxw dx dydz
z
ww
)( -
)(
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 z
w
y
v
x
u
)(
)(
)(
Continuity Equation
sec3m
kgm 0
z
)w(
y
)v(
x
)u(
t
0 .
Vt
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
z
w
y
v
x
u
t
m
kgm
z
w
y
v
x
u
t
If the density is constant
0 )(
)(
)(
0 )(
)(
)(
z
w
y
v
x
u
z
w
y
v
x
u
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.
Dt
Du
Dt
Du
u ).V( t
u
Dt
Du
z
u w
y
u v
x
u .u
t
u
v
ma
z
uw
y
uv
x
uu
t
u
Dt
Dua
v
m
Dt
Du
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
Dt
Dv
z
v w
y
v v
x
v u
t
v
Inertial force / volume in z direction Dt
Dw
z
w w
y
w v
x
w u
t
w
Dt
DuInertial force / volume in x direction
z
uw
y
uv
x
uu
t
u
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
dxdydz
dxdydz
v
mg
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
dydzdxx
pP
dxdydzx
pdydzdx
x
pPP
x
p
dxdydz
dxdydz
x
p
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
x
p
y
p
z
p
z
p
y
p
x
p
z
p
y
p
x
p
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
x
uexx
y
veyy
z
wezz
zzyyxxe e e
z
w
y
v
x
u
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 divx
u2xx
V divy
v2yy
V divz
w2zz
[ 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
x
v
y
uyxxy
x
w
z
uzxxz
y
w
z
vzyyz
z
y
x
F xzxyxxvx
z
y
x
F xzyyyxvy
z
y
x
F zzzyzxvz
z
u
x
w
zy
u
x
v
xFvx
y .V
x
u 2
.V 2
z
v
y
w
zy
w
yy
u
x
v
xFvy
.V. 2
y
w
zz
v
y
w
yx
w
z
u
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 x
p
Dt
Du zxyxxx
x
z
y
x
g y
p
Dt
Dv
yzyyyxy
zyx
gz
p
Dt
Dw zzzyzxz
X direction:-
Y direction:-
Z direction:-
x
w
z
u
z
y
u
x
v
y
.V x
u 2
x g
x
p
z
u w
y
u v
x
u u
t
u x
X direction:-
z
v
y
w
z .V
y
v 2
y
y
u
x
v
x g
y
p
z
v w
y
v v
x
v u
t
v y
Y direction:-
.V z
w 2
z
z
v
y
w
y
x
w
z
u
x g
y
p
z
w w
y
w v
x
w u
t
w z
Z direction:-
z
y
x g
x
t
Du 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
y
xy
xxx xg
x
t
Du
uV. . . VuuV
VuuVu . . V.
Divergence of the product of scalar times a vector.
t
u t
u
t
u
t
u t
u
t
u
Taking RHS of N-S Equation we have
u.V
t
u u.V
t
u
z
u w
y
u v
x
u u
t
u
Dt
Du
V . u uV . t
u t
u
V . t
u uV . t
u
0 u uV . t
u
Dt
Du
CONTINUITY z
w
y
v
x
u
t
since
Is equal to zero
zxz
y
xy
xxx xg
x
p uV .
t
u
z
yz
y
yy
x
yx yg
y
p uV .
t
v
zzz
y
zy
xzx zg
z
p uV .
t
w
CONSERVATION FORM:-
zxz
y
xy
xxx xg
x
p
z
uw
y
uw
x
2u
t
u
z
yz
y
yy
x
yx yg
x
p
z
vw
y
2v
x
uv
t
v
zzz
y
zy
xzx zg
z
p
z
2w
y
vw
x
uw
t
w
xg x
w
z
u
z
y
u
x
v
y
x
u 2 .V
x
x
P
z
uw
y
uv
x
2u
t
u
SIMPLICATION OF NAVIER STOKES EQUATION
xg xz
w2
2z
u2
2y
u
xy
v2
x
u 2
z
w
y
v
x
u 3
2 x
x
P
z
uw
y
uw
x
2u
t
u
If is constant
xg xz
w2
2z
u2
2y
u
xy
v2
x
u 2
z
w
y
v
x
u 3
2 x
x
P
z
uw
y
uw
x
2u
t
u
xg xz
w2
2z
u2
2y
u2
xy
v2
2x
u2 2
zx
w2 3
2 yx
v2 3
2 2x
u2 3
2
x
P
z
uw
y
uw
x
2u
t
u
xz
w2 3
1 xy
v2 3
1 2z
u2
2y
u2
x
u2 3
11 x
P
z
uw
y
uv
x
2u
t
u
z
w
x 3
1 y
v
x 3
1 x
u
x 3
1
2y
u2
2x
u2
x
P
z
uw
y
uv
x
2u
t
u
z
w
y
v
x
u 3
1 2z
u2
2y
u2
2x
u2
x
P
z
uw
y
uv
x
2u
t
u
V. 31
2z
u2
2y
u2
2x
u2
x
P
z
uw
y
uv
x
2u
t
u
2z
u2
2y
u2
2x
u2
x
P
z
uw
y
uv
x
2u
t
u
For Incompressible flow
0 V .
Energy Equation
Energy 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
y
yx.u
xxxu
x
up
for Y direction
V.vF
dxdydz yzu y
yyv
x
yxv
y
vp
for Z direction
dxdydz xxw
y
zyw
xzxw
z
wp
V.vF
Body force is given by dxdydz V.g
wzg vyg uxg
Total work done
dxdydz V.f
dxdydz
zzzw
y
yzw
xxzw
z
zyv
y
yyv
x
xyv
zzxu
y
yxu
xxxu
z
wp
y
vp
x
up
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 z
Tk
z
y
Tk
y
x
Tk
x q
z
wp
y
vp
x
up
z
Tk
z
y
Tk
y
x
Tk
x q
2
2V e
Dt
D
f.V zzzw
y
yzw xzw
z
zyv
y
yyv
x
xyv
zzxu
y
yxu
xxxu
z
wp
y
vp
x
up
z
Tk
z
y
Tk
y
x
Tk
x q
2
2V e
Dt
D
Energy EquationNonconservation form
z
wp
y
vp
x
up
z
Tk
z
y
Tk
y
x
Tk
x q
2
2V e
Dt
D
f.V
zzzw
y
yzw
xxzw
z
zyv
y
yyv
x
xyv
zzxu
y
yxu
xxxu
Non conservation:-
z
wp
y
vp
x
up
z
Tk
z
y
Tk
y
x
Tk
x
q V 2
2V e .
2
2V e
Dt
D
f.V
zzzw
y
yzw
xxzw
z
zyv
y
yyv
x
xyv
zzxu
y
yxu
xxxu
Conservation:-
.V p 2z
T2k
2y
T2k
2x
T2k q
z
Tpc w
y
Tpcu
x
Tpcu
x
Tpc
.V p 2z
T2k
2y
T2k
2x
T2k q
z
wT
y
vT
x
uT pc
x
Tpc
xfzxz
yxy
xxx
x
p
tD
uD
fyz
yzyyy
xyx
y
p
tD
vD
fzz
zzy
zyx
zxz
p
tD
wD
Momentum Equation Non conservation form
X direction
Y direction
Z direction
Momentum Equation
Conservation form
X direction
Y direction
Z direction
xfzxz
yxy
xxx
x
pVu
tD
uD
)(.
fyz
yzyyy
xyx
y
pVv
tD
vD
).(
fzz
zzy
zyx
zxz
pVw
tD
wD
)(.
Vfz
zzw
yzyw
xzxw
zxz
zyzv
yyyv
xyxv
zxzu
yxyu
xxxu
z
wp
y
vp
x
up
z
Tk
zy
Tk
yx
Tk
xq
Ve
tD
D
.)()()(
)()()()()()(
)()()(
2
2)()()()(
Energy Equation
Non conservation form
Vfz
zzw
yzyw
xzxw
zxz
zyzv
yyyv
xyxv
zxzu
yxyu
xxxu
z
wp
y
vp
x
up
z
Tk
zy
Tk
yx
Tk
xq
Ve
Ve
tV
.)(
)()()()()(
)()()()()()(
2
2.
2
2)()()(])([])([
Energy equation
Conservation form
FORMS OF THE GOVERNING EQUATIONS PARTICULARLY SUITED FOR CFD
energytotalofFluxVV
e
energyInternalofFluxVe
momentumofcomponentzofFluxVw
momentumofcomponentyofFluxVv
momentumofcomponentxofFluxVu
fluxMassV
)(2
2
Solution vectar
)(2
2Ve
w
v
u
U
Variation in x direction
xzwxyvxxux
Tkupu
Ve
xzuwxyuv
xxpu
u
F
)(2
2
2
Variation in y direction
zywyyvxyuy
Tkvpv
Ve
zyvwyypv
yxvu
v
G
)(2
2
2
Variation in z direction
zzwyzvxxzuz
Tkwpw
Ve
xzzpwxyzwv
xzwu
w
H
)(2
2
2
Source vectar
qzfwyfvxfu
zfyfxf
J
)(
0
Time marching
Jz
H
y
G
x
F
t
U
Types of time marching
1. Implicite time marching
2. Explicite time marching
Explicit FDM
Implicit FDM
Crank-Nicolson FDM
Space marching
Jz
H
y
G
x
F