navier-stokes equation
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IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Navier-Stokes Equation
Newtonian Fluid Constant Density, Viscosity Cartesian, Cylindrical, spherical coordinates
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Cartesian Coordinates gVPDt
DV 2
xxxxx
zx
yx
xx g
z
V
y
V
x
V
x
P
z
VV
y
VV
x
VV
t
V
2
2
2
2
2
2
yyyyy
zy
yy
xy g
z
V
y
V
x
V
y
P
z
VV
y
VV
x
VV
t
V
2
2
2
2
2
2
zzzzz
zz
yz
xz g
z
V
y
V
x
V
z
P
z
VV
y
VV
x
VV
t
V
2
2
2
2
2
2
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Cylindrical Coordinates gVPDt
DV 2
rrr
r
rz
rrr
r
gV
rz
VV
rrV
rrr
r
P
z
VV
r
VV
r
V
r
VV
t
V
22
2
2
2
2
2
211
zzzzz
zzz
rz g
z
VV
rr
Vr
rrz
P
z
VV
VV
r
VV
t
V
2
2
2
2
2
11
gV
rz
VV
rrV
rrr
P
rz
VV
r
VVV
r
V
r
VV
t
V
r
zr
r
22
2
2
2
2
211
1
Centrifugal force
Coriolis force
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Spherical Coordinates gVPDt
DV 2
rrr
r
rrrr
r
gV
r
V
rVr
rr
r
P
r
VVV
r
VV
r
V
r
VV
t
V
2
2
222
2
2
2
22
sin
1sin
sin
11
sin
gV
r
V
r
V
r
Vr
Vrrr
P
rr
VVVV
r
VV
r
V
r
VV
t
V
r
rr
sin
cot22
sin
1
sinsin
111
1cot
sin
222
2
2
22
2
2
2
2
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Spherical Coordinates gVPDt
DV 2
gV
r
V
r
V
r
Vrr
Vr
rr
P
rr
VVVVV
r
VV
r
V
r
VV
t
V
r
rr
sin
cot2
sin
2
sin
1
sinsin
111
sin
1cot
sin
222
2
2
22
2
BSL has g here instead of g
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Spherical Coordinates (3W) gVPDt
DV 2
rrr
rrrr
r
gV
rV
r
V
rV
rV
r
P
r
VVV
r
VV
r
V
r
VV
t
V
sin
2cot
222
sin
22222
22
2
2
222
22
sin
1sin
sin
11
rrr
rrr
3W &R have the formula in terms of However, the expression for is incorrect in the book2 2
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Continuity 0.
Vt
0
zyx Vz
Vy
Vxt
011
zr Vz
Vr
Vrrrt
0sin
1sin
sin
11 22
V
rV
rVr
rrt r
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Newton’s law of viscosity
VVV .3
2
Constant Density, zero dilatational viscosity
y
V
x
Vxy
xyyx
rrr
V
rr
V
rr
1
z
VVr z
zz
r
V
z
V zrrzrz
Cylindrical coordinates
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Newton’s law of viscosity
VVV .3
2
Spherical coordinates:Constant Density, zero dilatational viscosity
rrr
V
rr
V
rr
1
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Examples
•ODE vs PDE•Spherical and cylindrical coordinates•Eqn for pipe flow (Hagen Poiseulle)•Flow between rotating cylinders (not solved in class)•Thin film flow with temp variation (not solved in class, steps were discussed briefly. BSL ‘worked out’ example)•Radial flow between circular plates (BSL 3B.10)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example problems
1. Pressure driven steady state flow of fluid
between two infinite parallel platesinside a circular tube
2. Steady state Couvette flow of a fluidbetween two infinite parallel plates with top plate moving
at a known velocitybetween two circular plates of finite radius, with the top
plate rotating at a known angular velocitybetween two circular cylinders with outer cylinder
rotating at a known angular velocity (end effects are negligible)between a cone and plate (stationary plate and cone is
rotating at a known angular velocity). Angle of cone is very small (almost a parallel plate with almost zero gap)3. Coutte Poisseuille flow
between two parallel plates
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Examples
•PDE• Please refer to the book “Applied Mathematical Methods for Chemical Engineers” by Norman W Loney (CRC press), pages 330 to 342 for “worked out” examples for Momentum Transfer problems involving PDE.• Either multi dimensional or time dependent (however multidimensional and time dependent cases are not discussed in detail)• Steady state in Rectangular channel: pressure driven , coutte flow• Plan suddenly moving with constant velocity (or stress) from time t=0 (Stokes problem)• Sudden pressure gradient in a cylindrical tube (unsteady flow , converging to Hagen-Poisseuille’s flow (Bessel functions)• Flow between two (non rotating) cylinders, caused by boundary movement (coutte flow). Unsteady vs steady (not discussed in class or covered in tutorial)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Guidelines for solving PDE in Momentum Transfer
Method: If the problem involves finite scales, “separation of variable” method should
be tried If the problem involves infinite (or semi-infinite) distances, “combination of
variables” method should be tried
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Guidelines for solving PDE in Momentum Transfer
Solution forms, for finite scales: Applying the separation of variables directly may not always give proper
results If the equation is non-homogenous
For time dependent problems, first try to get steady state solution (and try that as the ‘particular solution’ for the equation). Unsteady state solution may be the ‘general solution’ for the corresponding homogenous equation
For multi dimensional problems, first try to get solution for ‘one dimensional’ problem and try that as particular solution. The ‘correction term’ may be the ‘general solution’ for corresponding homogenous equation.
Even if the equation is homogenous, you can try the above methods of obtaining ‘steady state’ or ‘one dimensional’ solution. The ‘complete solution’ will be the sum of ‘steady state + transient’ solution OR ‘one dimensional solution + correction for presence of plates’ (for example).
Always make sure that the ‘correction term’ goes to zero in the appropriate limit (eg time --> infinity, or the ‘width of the channel --> infinity)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Guidelines for solving PDE in Momentum Transfer
Other relevant Information: Problems in Cartesian coordinates tend to give Cosine/ Sine series
solution. In cylindrical coordinates, Bessel functions. In spherical coordinates, Legendre functions
When you attempt a ‘complete solution’ as ‘steady state+ transient’ (OR ‘one dimensional + correction’), make sure that you also translate the boundary conditions correctly
While solving for the ‘transient’ or ‘correction’ terms, you may encounter a situation where you have to choose an arbitrary constant (either positive or negative or zero). Usually the constant will not be zero. Choose the constant as positive or negative, depending on the boundary conditions (otherwise, you will proceed only to realize that it will not work).
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Stoke’s first problem (Please refer to BSL for solution)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
xxxxx
zx
yx
xx g
z
V
y
V
x
V
x
P
z
VV
y
VV
x
VV
t
V
2
2
2
2
2
2
N-S Equation: Example: Steady state flow in Rectangular channel
• Steady state in Rectangular channel: pressure driven flow, incompressible fluid
h2
h
hbb2
h2• Vy = Vz =0•Vx is function of y and z• gravity has no component in x direction
x
y
z
2
2
2
2
0z
V
y
V
x
P xx
• Method employed: Find a particular solution satisfying above equation; then find a general solution satisfying following differential eqn
2
2
2
2
0z
V
y
V generalxgeneralx particularxgeneralxx VVV
1
23
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Example; Rectangular channel
• Hint: To obtain a physically meaningful format, we can take particular solution to resemble one dimensional flow (when b goes to infinity)
h2
h
hbb2
h2x
y
z
2
22
12 h
yh
x
PV particularx
• Note: Check that the above solution is a valid particular solution• Before trying to get general solution, write down the boundary conditions for the over all solution Vx
0, zhyVx
0, bzyVx
000
z
x
y
x
z
V
y
V
4
5
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Examples; rectangular channel
• Translate that to get the boundary conditions for Vx-general
0, zhyV generalx
particularxgeneralx VbzyV ,
000
z
generalx
y
generalx
z
V
y
V
particularxxgeneralx VVV
2
22
12 h
yh
x
PV particularx
• We know
•Hence, from equation 5,
• Use separation of variables method
zgyfVAssume generalx
2
2
2
2
0z
V
y
V generalxgeneralx implies 0 fggf
3
2
Eqn
6
7b
7a
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Examples; rectangular channel
• Since LHS is only a function of y and RHS is fn of z, both must be equal to a constant•We say
• Note: Why do we say , why not ? What will happen if you try that? Or if we say ?
implies0 fggfg
g
f
f
2
g
g
f
f
• In any case, the chosen constant leads to
zz eCeCzg 43 )sin()cos( 21 yCyCyf
00
z
generalx
z
V implies43 CC
implies 02 C00
y
generalx
y
V
8
22
0
9
From 6
and
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Examples; rectangular channel
Hence, substituting in
zzgeneralx eeyCCzgyfV )cos(31
Using superposition principle
zz
ngeneralx eeyCV )cos(
0, zhyV generalx
2
22
12 h
yh
x
PV particularx
Now, from
)cos(12 2
22
yeeCh
yh
x
P bbn
9 7a
6
particularxgeneralx VbzyV ,
h
n2
12
Again, from 6
implies
implies
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Examples; rectangular channel
• Using Fourier cosine expansion for an even function
we can find Kn
2
22
12 h
yh
x
PV particularx
)cos(12 2
22
yKh
yh
x
Pn
•Equating the co-efficients, we get
nbb
n KeeC bbn
n ee
KC
Hence, general solution part is
zz
bbn
generalx eeyee
KV
)cos(
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Examples; rectangular channel
2
22
12 h
yh
x
PV particularx
“Complete” solution for the original problem is given by
• Note: When “b” goes to infinity, the ‘correction’ part goes to zero
zzbb
nx eey
ee
K
h
yh
x
PV
)cos(12 2
22
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Other examples
• To determine the velocity profile in a rectangular channel, where the top plate is moving at a constant velocity of V-zero, under steady state conditions
• Try out a solution of the form “V-parallel-plate + V-correction”• Use separation of variable techniques, to determine V-correction• What happens if you try separation of variable in the first place?
• To determine the unsteady state solution for a flow in a cylindrical pipe, caused by sudden application of pressure
• Try a solution of the form ‘Steady state + Transient’, just like the one we saw for flow between parallel plates• You will get Bessel Equations. Just like we represented functions in rectangular coordinates by sine and cosine functions, we can represent functions in cylindrical co ordinates by Bessel functions, because they are orthogonal.
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