2nd session - dr. s. p. asok
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1.2 Advantages of CFD
Possible to see simultaneously the effect of various parameters
and variables on the behaviour of the system. To study the same in an
experimental setup is not only difficult and time-consuming but in
many cases, impossible.
It is much cheaper than setting up big experiments or building
prototypes of physical systems.
Numerical modeling is versatile. A large variety of problems with
different levels of complexity can be simulated on a computer.
Numerical experimentation allows models and is similar to
conducting experiments.
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In some cases, it is the only feasible substitute for experiments, for
example, modeling loss of coolant accident (LOCA) in nuclear
reactors, numerical simulation of spread of fire in a building and
modeling of incineration of hazardous waste.
However, all problems can not be solved by CFD. Experiments arestill required to get an insight into the phenomena that are not well
understood and also to check the validity of the results of computer
simulation of complex problems.
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1.3 Applications of Fluid Flow and Heat Transfer
Coupled and individual Fluid flow and heat transfer play a very
important role in nature.The various applications of fluid flow and heat
transfer are:
All methods of power production, e.g. thermal, nuclear, hydraulic,
wind, and solar power plants.
Heating and air-conditioning plants.
Chemical and metallurgical industries, e.g. furnaces, heat
exchangers, condensers and reactors.
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Design of IC engines.
Optimization of heat transfer from cooling fins.
Aircraft and spacecraft.
Design of electrical machinery and electrical circuits.
Cooling of computers.
Weather prediction and environmental pollution.
Materials processing such as solidification and melting, metal
cutting, welding, rolling, extrusion, plastics and food processing in
screw extruders, laser cutting of materials.
Oil exploration.
Production of chemicals such as cement and aluminium oxide.
Drying
Processing of solid and liquid wastes.
Bio-heat transfer as in human and animal bodies.
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We can find from classical textbook on fluid dynamics and heat
transfer that there are only a handful of analytical or exact solutions. In
actual situations, problems are lot more complex as in those involving non-
linear governing equations and / or boundary conditions, and irregular
geometry which do not allow analytical solutions to be obtained. Therefore, it
is necessary to use numerical techniques for most problems of practical
interest. Furthermore, to design and optimize thermal processes and
systems, numerical simulation of the relevant transport phenomena is a
must, since experimentation is usually too involved and expensive. However,
necessary experimentation must still be done in checking the accuracy and
validity of numerical results. Sometimes, numerical model can be refined by
input from results of a companion experimental set-up for the same problem.
1.4 Necessity of CFD in Fluid Flow and Heat Transfer
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Suppose we wish to obtain the temperature field in a domain. We
imagine that the domain is filled by a grid, and seek the values of
temperature at the grid points. Therefore, the energy equation which is the
governing differential equation for the problem is valid at all the grid points.
The governing differential equation is then transformed into a system of
difference equations resulting in a set of simultaneous algebraic equations
which means that if there are 100 grid points there will be 100 equations to
solve per variable. The simplification inherent in the use of algebraic
equations rather than differential equations is what makes numerical
methods so powerful and widely applicable.
1.5 Basic Approach in Solving a Problem by Numerical Method
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There are mainly two methods of discretising a given differential
equation as briefed below:
Finite Difference method: The usual procedure for deriving finite-
difference equations consists of approximating derivatives in the differential
equation via a truncated Taylor series. The method includes the assumption
that the variation of the unknown to be computed is somewhat like a
polynomial in x, y or z so that higher derivatives are unimportant. The
popularity of finite-difference methods is mainly due to their straight-
forwardness and relative simplicity.
1.6 Methods of Discretization
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Linear second-order partial differential equation in two independent
variables is further classified into three canonical forms elliptic, parabolic, and
hyperbolic. The general form of this class of equations is
where coefficients are either constant or functions of the independent
variables only. The three canonical forms are determined by the following
criteria:
b24ac < 0 elliptic
b24ac = 0 parabolicb24ac > 0 hyperbolic
2.2 Elliptic, Parabolic and Hyperbolic Equations
02
22
2
2
gfy
ex
dy
cyx
bx
a (2.5)
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(a) Elliptic PDE: Typical examples are
Note that in both of the equations (2.6) and (2.7), b = 0, a = 1, c = 1 which
makes b24 ac =-4 which is
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In elliptical problems, the function (x,y) must satisfy both, the differential
equation over a closed domain and the boundary conditions on the closed
boundary of the domain. This is depicted pictorially in Fig. 2.1
Fig. 2.1 Pictorial representation of an elliptic problem
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(b) Parabolic PDE: A typical example is
Where is a positive, real constant,
Note that in equation (2.8), b = 0, c = 0, a = which makes
b24 ac = 0. In parabolic problems, the solution advances outward
indefinitely from known initial values, always satisfying the known
boundary conditions as the solution progresses. This is also called
marching type of problem. The open-ended type of solution domain is
pictorially illustrated in fig. 2.2.
2
2
xt(Heat conduction or diffusion equation) (2.8)
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(c) Hyperbolic PDE: A typical example is
Fig. 2.2 Pictorial representation of a parabolic problem
2
22
2
2
tx(wave equation) (2.9)
Where 2 is a real constant (always positive).
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Note that in equation (2.9), b = 0, a = 2 , c = -1 which makes b24ac = 4 2
which
is > 0.
The solution domain of hyperbolic PDE (Fig. 2.3) has the same
open-ended nature as in parabolic PDE. However, two initial conditions are
required to start the solution of hyperbolic equations in contrast with
parabolic equations where only one initial condition is required.
Fig. 2.3 Pictorial representation of a hyperbolic problem
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The initial and boundary conditions must be specified to obtain unique
numerical solutions to the partial differential equations. This is explained
next using the one-dimensional transient heat condition equation.
2.3 Initial and Boundary Conditions
2
2
xT
tT (2.10)
Equation (2.10) is the mathematical representation of a physical problem in
which for example, temperature within a large solid slab having finite
thickness changes in the x-direction (thickness direction) as a function of
time till steady state (corresponding to t ) is reached.
Following are the three categories of boundary conditions.
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The values of the dependent variables are specified at the boundaries(Fig.2.4).
Boundary conditions (abbreviated as B.C.) of the first kind can be
expressed as
2.3.1 Dirichlet Conditions (First Kind)
B.C.1 T = f(t) or T1 at x = 0
B.C.2 T = T2 at x = L
Initial conditions (abbreviated as I.C.)
0 x L
T = f(x) at t = 0
or T = T0
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Fig. 2.4 Dirichlet condition
The derivative of the dependent variable is given as a constant or as a
function of the independent variable on one boundary.
2.3.2 Neumann Conditions (second kind)
For example 0x
Tat x=L and t 0
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This condition specifies that the temperature gradient at the right
boundary is zero (insulation condition).
Cauchy conditions: A problem which combines both Dirchlet and Neumann
conditions is considered to have Cauchy conditions (Fig. 2.5)
Fig. 2.5 Cauchy condition (Dirichlet and Neumann)
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The derivative of the dependent variable is given as a function of
the dependent variable on the boundary (see Fig. 2.6). For the heat
conduction problems, this may correspondent to the case of cooling of a
large steel slab of finite thickness L by water or oil, the heat transfer
coefficient. hbeing finite.
2.3.3 Robbins Conditions (Third kind)
Fig. 2.6 Robbins condition
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On the basis of their initial and boundary conditions, PDEsmay be further
classified into initial value or boundary value problems.
2.4 Initial and Boundary Value Problems
2.4.1 Initial Value Problems
In this case, at least one of the independent variables has an open
region. In the unsteady state conduction problems the time variable has the
range 0 t , where no condition has been specified at t = ; therefore thisis an initial value problem.
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When the region is closed for all independent variables and conditions are
specified at all boundaries, then the problem is of the boundary value type.
An example of this is the three-dimensional steady state heat conduction
(with no heat generation) problem mathematically represented by the
equation:
2.4.2 Boundary Value problems
with the boundary conditions given at all the boundaries.
02
2
2
2
2
2
z
T
y
T
x
T (2.11)
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Fluid flow is described mathematically by Navier-Stokes equations
and equation of continuity. A two-dimensional, laminar, incompressible flow
with constant viscosity is described by:
3.2 Governing Equations
xMomentum: y
uvx
uut
u
=x
p
y
u
x
u2
2
2
2
(3.5)
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Equations (3.5 and 3.6) are called the NavierStokes equations
and Eq. (3.7) is the equation of continuity. It should be noted that pdenotes
the difference between the total pressure and hydrostatic pressure. This
causes the body forces to cancel, as they are in equilibrium with the
hydrostatic pressure.
yMomentum:y
vv
x
vu
t
v
=y
p
y
v
x
v2
2
2
2
(3.6)
and continuity 0u
y
v
x (3.7)
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1. Nonlinearity: A close look at Eqns. (3.5) and (3.6) reveal that the
convective part of the momentum equations involve nonlinear terms. For
example, in Eqn. (3.5), the convection coefficient u is a function of the
dependent variable u. However, this can be treated like the conductivity k
being a function of temperature T. Starting with a guessed velocity field, one
could iteratively solve the momentum equation to arrive at the converged
solution for the velocity components. Therefore, nonlinearity poses no
problems as such. It only makes the computations more involved.
3.3 Difficulties in Solving the Navier Stokes (NS) Equations
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2. Pressure gradient: The main hurdle to overcome in the calculation ofvelocity field is the unknown pressure field. The pressure gradient behaves
like a source term for a momentum equation. But, there is no particular
difficulty in solving the momentum equations. So, the challenging task is to
determine the correct pressure distribution.
The pressure field is indirectly linked with the continuity equation.
When the correct pressure field is plugged into the momentum equations the
resulting velocity field satisfied the continuity equation. Therefore, the
calculation of pressure field is the main problem facing the numerical analyst
in solving fluid flow problems.
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3. Turbulent Flow: In turbulent flows, there will be higher wall shear stressand pressure drops. Turbulence delays the point of separation in external
flows and reduces form drag. The unsteady velocity fluctuations present
can generate flow induced oscillations leading even to structural failure.
Time dependent NS equations are needed to define the flow along with
the smallest time and length scales since the eddies formed in turbulent
flow are available in a variety of sizes. Grids should be refined needing
higher computational effort and higher round of errors can result. Direct
Numerical Simulation (DNS) type of analysis is more advisable.
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4. FINITE DIFFERENCE
METHOD
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4.1 Finite Difference Discretisation :
1. Replace the derivatives in the governing equations by algebraic
difference quotients, resulting in a system of algebraic equations. In the
below mesh, uniform spacing is followed. x need not be equal to y.
Non uniform spacing is covered under automatic grid generation.
y
x
i-1,j+1 i,j+1 i+1,j+1
i+1,j
i+1,j-1
i,j
i,j-1i-1,j-1
i-1,j P
y
x
Fig. 4.1 Finite Difference Discretion 2D
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2. Let ui,j = the x component velocity at P(i,j); using Taylor series:.........
6
)(
2
)( 3
,
3
32
,
2
2
,
,,1
x
x
ux
x
ux
x
uuu
jijiji
jiji (4.1)
;2
)(~2
,
2
2
,
,,1
x
x
ux
x
uuu
jiji
jiji Second order accurate eqn. (4.2)
;~,
,,1 xx
u
uuji
jiji First order accurate eqn. (4.3)
The neglected higher order terms Truncation errors.
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3. The truncation error for (4.2) and (4.3) are :
2 ,3 ,!
)(&;
!
)(
n
n
ji
n
n
n
n
ji
n
n
n
x
x
u
n
x
x
u
Lesser the magnitude of x, lesser is the Truncation error.
4. Rearrange terms on (4.1)
........6
)(2
2
,
3
3
,
2
2
,,1
,
xxux
xu
x
uu
xu
jiji
jiji
ji
)(,,1
,
xox
uu
x
u jiji
ji
, (4.4)
The symbol o ( x) terms of order of x and imply the order of magnitude of the
truncation error. (4) first order accurate difference representation ofjix
u
,
called
the first order forward difference
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5. In a similar way, write ui-1, jfrom ui, j:
..............6
)(
2
)()(
3
,
3
32
,
2
2
,
,,1
x
x
ux
x
ux
x
uuu
jijiji
jiji
)(,1,
,
xox
uu
x
u jiji
ji
, (4.5)
(5) = First order backward or rearward difference expression for ( u/ x) at (i,j).
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6. ..............3
)(2
3
,
3
3
,
,1,1
x
x
ux
x
uuu
jiji
jiji
2,1,1
,
)()(2
xox
uu
x
u jiji
ji
, (4.6)
(6) Second order central difference for ( u/ x)i,j
we can get,
2
2
,1,,1
,
2
2
)()(
2xo
x
uuu
x
u jijiji
ji
, (4.7)
(4.7) second order central difference form forji
x
u
,
2
2
in the same way, the
expressions for the y direction can also be got.
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8. Consider the Laplace equation2
2
2
2
2 0yT
xTT
0)(
2
)(
22
1,,1,
2
,1,,1
y
TTT
x
TTT jijijijijiji
Puty
x = mesh aspect ratio
0)1(2)( ,2
1,1,
2
,1,1 jijijijiji TTTTT
)2
1,1,
2
,1,1
,1(2
)( jijijijiji
TTTT
T
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4.2 Burgers Equation
1. Fluid mechanics problems are usually non linear. Consider an equation having
convective, diffusive and time dependent terms. Burgers introduced a simple nonlinear equation to meet the conditions:
,2
2
xxu
t= any transferable property (4.8)
2. On (4.8), if the r.h.s term is neglected, the Eulers equation (or) the inviscid
Burgers equation is got.
0x
ut
3. Conservative property: a finite difference equation passes conservative
property if it preserves integral conservation relation to the continuum.
Preserving the conservative property is of special importance in the finite
volume approach - a special form of finite difference equations.
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4.3 Upwind Scheme
1. 0x
ut
, discretizing it
011
xu
t
n
i
n
i
n
i
n
i , (4.9)
Applying the Von Neumann stability analysis, it can be found that both the finite
difference equations are unconditionally unstable. The remedy is the upwind scheme.
Equation (4.9) can be made stable by substituting the forward space difference by a
backward space difference, provided u is +ve.
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2. Considering the full Burgers equation,
2
2
xxu
t;
,0),(11
ufortermviscousx
uu
t
n
i
n
i
n
i
n
i
,0),(1 ufortermviscousx
uu n
i
n
i
The upwind method of discretization is very essential for convection dominated
problems. The upwind bias retains transporative property of flow equation.
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4.4 Upwind Differencing and Artificial Viscosity
1. For the Burgers equation
0...,..........11
uforx
uu
t
n
i
n
i
n
i
n
i
(4.10)
2. Using Taylors expansion:
n
i
n
i
n
i
n
it
t
tt
2
221
2
)(+ (4.11)
n
i
n
i
n
i
n
ix
x
xx
2
22
12
)(. (4.12)
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3. substitute (4.11) and (4.12) on (4.10)
,0),(11
ufortermviscousx
uu
t
n
i
n
i
n
i
n
i
Dropping the superscript n and subscript i
tdiffusivexx
x
xx
x
ut
tt
tt
t
3
2
223
2
22 )(0
2
)()(0)(
2
11
2
2
2
2
2
)(012
1x
xxx
tuxu
xu
t,
;2
2
2
2
termsorderhigherxxx
ut
e (4.13)
Where )1(2
1Cxue
C = Courant number =x
tu
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4. While deriving (4.13) ,2
2
t was taken as
2
22
xu . However, the non
physical coefficient e leads to a diffusion like term which is dependent on the
discretisation procedure, e is called artificial/numerical viscosity. C should be
less than 1, so that e is a non zero +ve quantity, Normally Cx= Cy . Though
upwinding schemes suffer from false diffusion, finite differences provide stable
solutions.
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4.5 Consistency
1. A finite difference representation of a PDE is consistent if: mesh Lt 0 (PDEFDE)
= meshLt(TE) 0,
Consider ,2
2
x
ua
t
u
Using the DufortFrankel FD scheme for the r.h.s
2
111
111
)(2 x
uuuua
t
uu nin
in
in
in
in
i ,
The TE = )(6
1)(
12 3
32
2
22
4
4
tt
u
x
t
t
uax
x
ua n
i
n
i
n
i. The TE will be meaningful
forx
ttx
x
t;0&0,0
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2.0,
)(
0,
)(
xt
TELt
xt
FDEPDELt= ,
2
22
t
ua This r.h.s term is not vanishing
when the mesh vanishes.
Now reconstitute the PDE from the FDE2
2
2
22
xua
tua
tu we have
started from a parabolic equation and ended up in a hyperbolic equation. Thus the Dufort
Frankel scheme is not consistent unless 0
x
ttogether with t, x 0.
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4.6 Explicit and Implicit Methods
1. We have:2
11
1
)(
2
x
uuua
t
uu n
i
n
i
n
i
n
i
n
i , using the forward time and central space
(FTCS) scheme, the unknown dependent variable uin+1
can be explicitly got from the
known values of ui+1n, ui
n, ui-1
n and this is a typical example for explicit finite
difference method.
2. Now let us try a different discretion of ut = a2u, express the spatial differences on the
r.h.s. in terms of averages between n and (n+1) time level:
2
1
1
1
1
1
1
1
1
)(
22
2 x
uuuuuua
t
uu nin
i
n
i
n
i
n
i
n
i
n
i
n
i (4.14)
This is called the Crank-Nicolson implicit scheme; i.e. the unknown u in+1
has been
expressed in terms of the known quantities at time level n and the unknown quantities
at time level n+1; hence equation (4.14) at a grid point i cannot itself result in a
solution for uin+1
. We will need the equation to be written for all grid points and later
solve simultaneously.
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4.7 Error and Stability Analysis
1. Discretization error = TE + any error introduced by the numerical treatment of the
b.cs. If A = Analytical solution, D = exact solution of the FDE and N = numerical
solution from a real computer with finite accuracy, then discretization error = A-D.
Round off error = = N-D. The solution will be stable if ivalues shrink or at least
stay the same, as the solution progresses from n to n+1. If ivalues grow larger, then
the solution becomes unstable for a solution to be stable : 11 n
i
n
i
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4 9 Example for Unstable Calculation
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4.9 Example for Unstable Calculation
1. Consider the heat condition equation2
2
x
ua
t
u;
The simple explicit finite difference representation is:
2
11
1
)(
)2(
x
uuua
t
uu n
i
n
i
n
i
n
i
n
i 211
1
)(;)21()(
x
taruruuru
n
i
n
i
n
i
n
i
It can be found that the solution is stable for r 1/2
2. Consider a case where
r1/2 say r =1, ui
n+1 =
1(100+100) + (1-2) 0 =
200C; which cannot be
correct.
Fig. 4.2 Unstable Calculation
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5. FINITE DIFFERENCE
APPLICATIONS
5.1 Heat Dissipation Through a Straight Fin
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p g g
1. The governing differential equations of the problem are approximated at the nodes to
generate finite difference algebraic equations for the nodal values of (x,y,z,t). In the
FV method the governing equation is integrated over chosen CVsaround each node
of the grid before the derivatives are approximated.
1 2 i-1 i i+1
N
N+1
x
x=L
Tbx
Tf
dT/dx=0
a
P
Fig. 5.1 Heat Dissipation Through a Straight Fin
Governing equation: ,0)(2
2
fTTKa
hP
dx
Tdneglecting the losses from the end
face x = L, the b.csare; T = Tbat x = 0, 0
dx
dtat x = L
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2. Consider a finite difference grid of N equal steps having N+1 grids, x = L/N.
For any mode i in the interior
,0)()(
22
11fi
iii TTKahP
xTTT 2 i N (5.1)
T1(2+A) T2+ T3= -ATf, i = 2, where A =2)( x
Ka
hP
T2(2+A) T3+T4= -ATf, i = 3, TN(2+A) TN+ TN+1 = -ATf, i = N;3. For the boundary nodes, nodal equations are derived from the b.c s : At the fi
base i=1, T1=Tb, the zero heat flux condition at the end i = N+1 should be
discretized by introducing a hypothetical node (N+2) called a image node. The
central difference expression for the derivativedxdT at the node (N+1) is:
NNNN TT
x
TT
dx
dT2
2 02
(5.2)
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4. Using eqn. (5.1) for the end node i=N+1:
TN(2+A) TN+1+ TN+2 = -ATf, use equ. (5.2) on this 2TN(2+A) TN+1= -ATf
5. The final system of algebraic equations for the fin problem for a grid of N=4 is:
f
f
f
f
b
AT
AT
AT
ATT
T
T
T
TT
A
A
A
A
5
4
3
2
1
)2(2000
1)2(100
01)2(10
001)2(100001
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T
h
Example:A slab of 0.1m thickness with K=40W/mC generates energy at the rate o
106W/m
3. The surface at x = 0 is insulated and the end at x = 0.1m is subjected to
convection with a heat transfer coefficient of 200W/m2C with Tatm= 150C. Obtain the
temperature distribution.
Solution
Take M=5, x = L/M = 0.1/5 = 0.02m 1 2 3 4 5 6
For the internal nodes m = 2 to 5:
;0)(
)2(
2
11K
gxTTT mmmm
040
10)02.0(2
62
11
xTTT mmm
5&4,3,2,0102 11 mforTTT mmm
At x = 0,
022 12
12K
gxTT
1,01022 12 mforTT
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At x = L,
02
2
22 221
2
1
2
TK
xh
k
gx
Tk
xh
T
M
MM
,015040
20002.02
40
1002.0
40
20002.0222
62
65 xxxx
Txx
T
60402.22 65 mforTT
40
1010
10
10
10
2.220000
121000012100
001210
000121
000022
6
5
4
3
2
1
T
TT
T
T
T
5 3 One Dimensional Steady Conduction Through Cylinder
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5.3 One Dimensional Steady Conduction Through Cylinder
1.
b
H
r
rT1
T1T2
T3Tm-1
TmTm+1
TMTM+1
1 2 3 m-1 m m+1 M M+1
r
2
2
r r
r =M
b
b
Fig. 5.3 One Dimensional Steady Conduction Through Cylinder
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2. Governing equation: broorgkdr
rdTr
dr
d
r,)(
1)(1
3. [Rate of heat entering by condition] + [rate of energy generation] = 0
Applying this for a cylindrical volume element and simplifying:
1,02
)(
2
11
2
11 1
2
1 Mmfork
grT
k
rhT
k
rh
MT
M
MMM
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Example:A 10cm dia. Solid chrome-Nickel rod with k= 20 W/mC is electricity heated
by passing an electric current generating heat at the rate of 107
W/m3
. The surface of the
rod is subjected to convection with a heat transfer coefficient h=200 W/mC into an
ambient at 30C. Determine the finite difference equations.
Solution:
b = 0.05m, k =20 W/mC, g = 107W/m3, T = 30C, h =200W/m2C, Take M=5
,01.05
05.0m
M
br 1.0
20
20001.0,50
20
10)01.0()( 722 x
k
rh
k
gr
the f.d. equation for the central node m =1: 1;0)(
)(4 12
12 mfork
grTT
.1;050)(4 12 mforTT
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Now we have six algebraic equations for the six nodes from m = 1 to m = 6:
-4T1 + 4T2= -50, for m = 1,
0.5T1 - 2T2+1.5 T3= -50, for m = 2,
0.75T2 - 2T3+1.25 T4 = -50, for m = 3,
0.8333T3 - 2T4+1.666T5= -50, for m = 4,
0.875T4 - 2T5+1.125T6= -50, for m = 5,
0.9T5 - T6= -28, for m = 6
in the matrix form:
28
50
50
50
50
50
19.00000
125.12875.0000
01666.128333.000
0025.1275.00
0005.125.0
000044
6
5
4
3
2
1
T
T
T
T
T
T
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Example:Steady one dimensional heat conduction is taking place in a slab of thickness
L=1 cm, where energy is generated at a construction rate of 7.2 x 107 W/m3. The
boundary surface of the slab at x=0 is maintained at fo=50C, while the boundary surface
at the other end dissipated heat by convection with a heat transfer coefficient h =
200W/m2C into the surroundings at T = 100C. Taking the K value of the slab as
18W/mC, determine the nodal temperatures, taking five subdivisions.
Solution:
cmM
LxLxg
kx
T2.0
5
1;0,0
12
2
LxathThTdx
xdTK
CTxatCxTCB s
,)(
,50;050)(:.1
LxatTThdx
dTk )(
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we know : ;0)(
22
11
k
gxTTT mmmm
for the interior nodes and for the node 6, use the following:
02)(2
22 221
2
12 T
k
xh
k
gxT
k
xhT MMM
Now,
,1618/)102.7()102()( 723
2
xxk
gx
0444.018/200)102(22 3
xk
xh
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,44.4100
0444.02
k
xhTSubstituting, getting the nodal equations and bringing to
the matrix form
44.20
16
16
16
66
044.22000
12100
01210
00121
00012
6
5
4
3
2
T
T
T
T
T
Solving: T2 = 119.045C, T3 = 172.09C, T4 = 209.135C, T5 = 230.18C,
T6= 235.225C
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ydiffusivitwhere
X
T
t
T2
2
One-diml. Transient Analysis
The Governing Equation for the transient
analysis is given by
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Explicit Scheme
The FD equation for 1D transient equation
is given by
2
11
1
211
1
)(
)21()(
)(2
x
tr
where
TrTTrT
xTTT
tTT
p
m
p
m
p
m
p
m
p
m
p
m
p
m
p
m
p
m
Example: A marble slab of k=2 w/mC =1 x 10-6
m2/s L = 2cm thick is initially at a
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Example:A marble slab of k 2 w/m C, 1 x 10 m /s, L 2cm thick is initially at auniform temperature Ti= 200C. One of its surfaces is suddenly lowered to 0C and the
other surface is insulated. Develop an one dimensional explicit f.d. scheme to determinethe temperature distribution in the slab as a function of position and time.
Solution: 0,0,2
2
tLxx
T
t
T
B.C: 0x
T, at x = 0, t > 0
T = 0C, at x = L, t > 0
I.C: T = 200C, for t = 0
For the insulated side h1= 0,
For the x = L side hM+1= , because there is a sudden heat transfer.
Insulated
K,
2 cm
Ti= 200C
0C
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Also T M+1= 0C
We know: Tmi+1= r Tm-1
i+ (1-2r)Tmi+ r imT 1 ; for m = 2, 3, . . . ., M and r = stability
criterion
1
1
i
mT = 0C, at m = M+1; Tm= 200C, for i = 0; m = 1, 2, . . . ., M+1
Taking M = 5; x = LM
= 25= 0.4cm = 4mm
Take r =1
2 r =
1
2 =
t
( x)2 t = 8 sec
Now the equation for 1imT becomes (Tm-1i+ Tm+1i), for m = 2, 3, 4, 5
In addition we have: T1i+1
= T2i+1
for m=1
T6i+1= 0C for m = 6 and Tm
= 200C for m = 1 to 6
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Now we march on time and obtain the following values:
m=1 2 3 4 5 6Time
stept sec
x=0 4mm 8mm 12mm 16mm 20mm
0 0 200 200 200 200 200 200
1 8 200 200 200 200 200 0
2 16 200 200 200 200 100 0
3 24 200 200 200 150 100 0
4 32 200 200 175 150 75 0
5 40 187.5 187.5 175 125 75 0
6 48 181.2 181.2 156.2 125 62.5 0
7 56 168.7 168.7 153.1 109.4 62.5 0
8 64 160.9 160.9 139.1 107.8 54.7 0
9 72 150 150 134.4 96.9 53.9 0
10 80 142.2 142.2 123.5 94.2 48.5 0
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The labyrinth seal should have a minimum gap not
below0.2 mmand achieve a high pressure drop ratio
(Pr) of above 10at a low rated flow of2.5 m3/hof water
at 70C
The seal length is restricted to be as small as
possible above a minimum stipulatedlength of 120 mm.
Limiting Cavitation to tolerable levels over the seal
length.
Straight Annular Portion
Outer Sleeve
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Fig.2 Annular Seal (AS)
b
c
d
D/2
Axis of the Labyrinth
CavityChamber
aV1
V2
Fig.4 Circular-grooved Squarecavity Labyrinth Seal (CSLS)
Fig.3 Circular-grooved square cavity labyrinth
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Fig.5 Domain discretisation of CSLS named LS1
Fig.6 Domain discretisation of CTLS 1
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Fig.7 Velocity distribution for AS at Re = 2500 predicted by FEA
Fig.8 Velocity distribution of CSLS - LS1 at Re = 2500 predicted by FEA
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Fig.9 Velocity distribution of CTLS 1 at Re = 2500 predicted by FEA
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Fig.10 Pressure distribution for AS at Re = 750 predicted by FEA
ig.11 Pressure distributions for CSLS named LS1 at Re = 750 predicted by
Fig.12 Pressure contours for CTLS 1 at Re = 750 predicted by FEA
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Fig.13 CFD flow visualisation for CCLS-LS6
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Fig.14 CFD predicted flow visualisation view for CCLS-LS7
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THANK
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