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Transcript of heat transfer Project
Chapter-1
INTRODUCTION
Heat transfer
Extended surface fins
Heat equation
Analytical solution
HEAT TRANSFER
Heat transfer is the transition of thermal energy from a hotter mass to a cooler mass.
When an object is at a different temperature than its surroundings or another object, transfer
of thermal energy, also known as heat flow, or heat exchange, occurs in such a way that the
body and the surroundings reach thermal equilibrium; this means that they are at the same
temperature. Heat transfer always occurs from a higher-temperature object to a cooler-
temperature one as described by the second law of thermodynamics or the Clausius
statement. Where there is a temperature difference between objects in proximity, heat transfer
between them can never be stopped; it can only be slowed.
MODES OF HEAT TRANSFER
There are three modes of Heat transfer, they are
Conduction
Convection
Radiation
CONDUCTION
Conduction is the transfer of heat by direct contact of particles of matter. The transfer of
energy could be primarily by elastic impact as in fluids or by free electron diffusion as
predominant in metals or phonon vibration as predominant in insulators. In other words, heat
is transferred by conduction when adjacent atoms vibrate against one another, or as electrons
move from one atom to another. Conduction is greater in solids, where a network of
relatively fixed spacial relationships between atoms helps to transfer energy between them by
vibration.
Heat conduction is directly analogous to diffusion of particles into a fluid, in the situation
where there are no fluid currents. This type of heat diffusion differs from mass diffusion in
behavior, only in as much as it can occur in solids, whereas mass diffusion is mostly limited
to fluids.
Metals (e.g. copper, platinum, gold, iron, etc.) are usually the best conductors of thermal
energy. This is due to the way that metals are chemically bonded: metallic bonds (as opposed
to covalent or ionic bonds) have free-moving electrons which are able to transfer thermal
energy rapidly through the metal.
As density decreases so does conduction. Therefore, fluids (and especially gases) are less
conductive. This is due to the large distance between atoms in a gas: fewer collisions between
atoms means less conduction. Conductivity of gases increases with temperature. Conductivity
increases with increasing pressure from vacuum up to a critical point that the density of the
gas is such that molecules of the gas may be expected to collide with each other before they
transfer heat from one surface to another. After this point in density, conductivity increases
only slightly with increasing pressure and density.
To quantify the ease with which a particular medium conducts, engineers employ the thermal
conductivity, also known as the conductivity constant or conduction coefficient, k. In thermal
conductivity k is defined as "the quantity of heat, Q, transmitted in time (t) through a
thickness (L), in a direction normal to a surface of area (A), due to a temperature difference
(ΔT) [...]." Thermal conductivity is a material property that is primarily dependent on the
medium's phase, temperature, density, and molecular bonding.
A heat pipe is a passive device that is constructed in such a way that it acts as though it has
extremely high thermal conductivity.
Steady-state conduction vs. Transient conduction
Steady state conduction is the form of conduction which happens when the temperature
difference driving the conduction is constant so that after an equilibration time, the spatial
distribution of temperatures (temperature field) in the conducting object does not change
any further. For example, a bar may be cold at one end and hot at the other, but the
gradient of temperatures along the bar does not change with time. The temperature at any
given section of the rod remains constant, and this temperature varies linearly along the
direction of heat transfer.
In steady state conduction, the amount of heat entering a section is equal to amount of
heat coming out. In steady state conduction, all the laws of direct current electrical
conduction can be applied to "heat currents". In such cases, it is possible to take "thermal
resistances" as the analog to electrical resistances. Temperature plays the role of voltage
and heat transferred is the analog of electrical current.
Transient conduction There also exists non-steady-state situations, in which the
temperature drop or increase occurs more drastically, such as when a hot copper ball is
dropped into oil at a low temperature. Here the temperature field within the object
changes as a function of time, and the interest lies in analyzing this spatial change of
temperature within the object over time. This mode of heat conduction can be referred to
as transient conduction. Analysis of these systems is more complex and (except for
simple shapes) calls for the application of approximation theories, and/or numerical
analysis by computer.
CONVECTION
Convection is the transfer of thermal energy by the movement of molecules from one part of
the material to another. As the fluid motion increases, so does the convective heat transfer.
The presence of bulk motion of the fluid enhances the heat transfer between the solid surface
and the fluid.
There are two types of convective heat transfer:
Natural convection: when the fluid motion is caused by buoyancy forces that result from the
density variations due to variations of temperature in the fluid. For example, in the absence of
an external source, when the mass of the fluid is in contact with a hot surface, its molecules
separate and scatter, causing the mass of fluid to become less dense. When this happens, the
fluid is displaced vertically or horizontally while the cooler fluid gets denser and the fluid
sinks. Thus the hotter volume transfers heat towards the cooler volume of that fluid.
Forced convection: when the fluid is forced to flow over the surface by external source such
as fans and pumps, creating an artificially induced convection current.
Internal and external flow can also classify convection. Internal flow occurs when the fluid is
enclosed by a solid boundary such as a flow through a pipe. An external flow occurs when
the fluid extends indefinitely without encountering a solid surface. Both of these convections,
either natural or forced, can be internal or external because they are independent of each
other.[citation needed]
The rate of convective heat transfer is given by:
q = hA(Ts − Tb)
A is the surface area of heat transfer. Ts is the surface temperature and Tb is the temperature of
the fluid at bulk temperature. However, Tb varies with each situation and is the temperature of
the fluid “far” away from the surface. h is the constant heat transfer coefficient that depends
upon physical properties of the fluid such as temperature and the physical situation in which
convection occurs. Therefore, the heat transfer coefficient must be derived or found
experimentally for every system analyzed. Formulas and correlations are available in many
references to calculate heat transfer coefficients for typical configurations and fluids. For
laminar flows, the heat transfer coefficient is rather low compared to the turbulent flows; this
is due to turbulent flows having a thinner stagnant fluid film layer on heat transfer surface.
RADIATION
Radiation is the transfer of heat energy through empty space. All objects with a temperature
above absolute zero radiate energy at a rate equal to their emissivity multiplied by the rate at
which energy would radiate from them if they were a black body. No medium is necessary
for radiation to occur, for it is transferred through electromagnetic waves; radiation works
even in and through a perfect vacuum. The energy from the Sun travels through the vacuum
of space before warming the earth.
Both reflectivity and emissivity of all bodies is wavelength dependent. The temperature
determines the wavelength distribution of the electromagnetic radiation as limited in intensity
by Planck’s law of black-body radiation. For any body the reflectivity depends on the
wavelength distribution of incoming electromagnetic radiation and therefore the temperature
of the source of the radiation. The emissivity depends on the wave length distribution and
therefore the temperature of the body itself. For example, fresh snow, which is highly
reflective to visible light, (reflectivity about 0.90) appears white due to reflecting sunlight
with a peak energy wavelength of about 0.5 micrometers. Its emissivity, however, at a
temperature of about -5°C, peak energy wavelength of about 12 micrometers, is 0.99.
Gases absorb and emit energy in characteristic wavelength patterns that are different for each
gas.
Visible light is simply another form of electromagnetic radiation with a shorter wavelength
(and therefore a higher frequency) than infrared radiation. The difference between visible
light and the radiation from objects at conventional temperatures is a factor of about 20 in
frequency and wavelength; the two kinds of emission are simply different "colours" of
electromagnetic radiation.
Extended Surfaces: Fins
Extended surfaces are often used to reduce the thermal resistance at a surface and thereby
increase the heat transfer rate from the surface to the adjacent fluid. They are also referred to
as fins. It is also not possible to increase the temperature difference (T1 – Tf) because
temperatures in the system are fixed by other constraints. The only option is to increase the
area. This is done by using extended surfaces or fins.
The use of fins is very common and they are fabricated in a variety of shapes. A familiar
application is the use of circumferential fins around the cylinder of motorcycle engine. This
geometry is shown in the figure for two shapes. In the figures shown below, one show the
fins having a rectangular cross section, while in the other figure shows them having a
triangular cross section.
Fins are most commonly used in heat exchanging devices such as radiators in cars and heat
exchangers in power plants. They are also used in newer technology such as hydrogen fuel
cells. Nature has also taken advantage of the phenomena of fins. The ears of jackrabbits and
Fennec Foxes act as fins to release heat from the blood that flows through them. Aluminum
Fins take 30% & Copper fins takes 35% less time for cooling from 350 degree centigrade to
100 degree centigrade than standard Fins.
HEAT EQUATION
The heat equation is an important partial differential equation which describes the distribution
of heat (or variation in temperature) in a given region over time. For a function u(x,y,z,t) of
three spatial variables (x,y,z) and the time variable t, the heat equation is given as follows.
or ,
where α is a constant.
Derivation in one dimension
The heat equation is derived from Fourier's law and conservation of energy (Cannon 1984).
By Fourier's law, the flow rate of heat energy through a surface is proportional to the negative
temperature gradient across the surface,
where k is the thermal conductivity and u is the temperature. In one dimension, the gradient is
an ordinary spatial derivative, and so Fourier's law is
where ux is du/dx. In the absence of work done, a change in internal energy per unit volume
in the material, ΔQ, is proportional to the change in temperature, Δu. That is,
where cp is the specific heat capacity and ρ is the mass density of the material. Choosing zero
energy at absolute zero temperature, this can be rewritten as
.
The increase in internal energy in a small spatial region of the material
over the time period
is given by[1]
where the fundamental theorem of calculus was used. Additionally, with no work done and
absent any heat sources or sinks, the change in internal energy in the interval [x-Δx, x+Δx] is
accounted for entirely by the flux of heat across the boundaries. By Fourier's law, this is
again by the fundamental theorem of calculus.[2] By conservation of energy,
This is true for any rectangle [t−Δt, t+Δt] × [x−Δx, x+Δx]. Consequently, the integrand must
vanish identically:
Which can be rewritten as:
or:
which is the heat equation. The coefficient k/(cpρ) is called thermal diffusivity and is often
noted α.
ANALYTICAL SOLUTION
To create a simplified equation for the heat transfer of a fin, many assumptions need to be
made.
Assume:
1. Steady state
2. Constant material properties (independent of temperature)
3. No internal heat generation
4. One-dimensional conduction
5. Uniform cross-sectional area
6. Uniform convection across the surface area
With these assumptions, the conservation of energy can be used to create an energy balance
for a differential cross section of the fin.
Fourier’s law states that
Where, Ac is the cross-sectional area of the differential element. Therefore the conduction rate
at x+dx can be expressed as
Hence, it can also be expressed as
.
Since the equation for heat flux is
Then dqconv is equal to
Where, As is the surface area of the differential element. By substitution it is found that
This is the general equation for convection from extended surfaces. Applying certain
boundary conditions will allow this equation to simplify.
Chapter-2
Computer simulation
Computer simulation Fluid flow and heat
transfer Methods of discretization Closure
COMPUTER SIMULATION
What is Computer Simulation???
The simulation of an industrial system on computers involving mathematical representation
of the physical process undergone by the various component of the system, by a set of
governing equations (usually differential equations) transformed to difference equations
which are in turn solved as a set of simultaneous algebraic equations.
Advantages of Computer Simulation:-
Here are some of the important advantages of computer simulation or also known as
numerical simulation:-
It is possible to see simultaneously the effect of various parameters and variables on
the behavior of the system since the speed of computing is very high. To study the
same in an experimental setup is not only difficult and time consuming but in many
cases, may be 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 level of
complexity can be simulated on a computer.
A numerical simulation allows models and hence physical understanding of the
problem to be improved. It is similar to conducting experiments.
In some cases, it is the only feasible substitute foe 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, it is to be emphasized that not every problem can be solved by computer
simulation. Experiments are still 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.
Application of Fluid Flow and Heat Transfer
Fluid flow and heat transfer plays a very important role in nature, living organism and in a
variety of practical situations. More often than not, flow and heat transfer are coupled and
rarely an engineer solves a problem of either pure fluid flow or pure heat transfer. The
various applications of fluid flow and heat transfer are:-
Power production e.g. thermal, nuclear, hydraulic, wind, and solar plants.
Heating and Air-conditioning of buildings.
Chemical and metallurgical industries, e.g. furnace, heat exchangers, condensers and
reactors.
Design of an I.C engine.
Optimization of heat transfer from cooling fins
Aircraft and spacecraft.
Design of electrical machinery and electronic circuits.
Cooling of computers.
Weather prediction and environment pollution.
Material processing such as solidification and melting, metal cutting, welding, rolling,
extrusion, plastics and food processing in screw extruders and laser cutting of
materials.
Oil exploration.
Production of chemicals such as cement and aluminium oxide.
Drying.
Processing of solid and liquid waste.
Bio-heat transfer.
A COMPARATIVE STUDY OF EXPERIMENTAL, ANALYTICAL, AND
NUMERICAL METHODS:-
1. Experimental Method : - Experimental methods are used to obtain reliable
information about the physical process that are not well understood example
combustion and turbulence. The major disadvantage this method are high cost,
measurements difficulties and probe errors. Often, small scale models do not
always simulate all the features if the full scale set up. The advantage of this
method is that it is more realistic.
2. Analytical Method :- Analytical methods or methods of classic mathematics are
used to obtain the solution of a mathematical model consisting of a set of
differential equation which represents a physical process within the limit of
assumption made. Only a handful of analytical solutions are available in heat
transfer because of the complexity involved in the handling of the complex
boundary and non-linearities in differential equation or boundary conditions.
Furthermore, the analytical solution contains infinite series, special functions,
transcendental equations foe Eigen values etc. thus the numerical evaluation
becomes quite cumbersome.
3. Numerical Method : - The major advantages of numerical solution are its abilities
to handle complex geometry and non linearity in the governing equation and
boundary condition. Other advantages of numerical method are briefly described
below :-
a. Low cost
b. High speed
c. Complete information
d. Ability to simulate realistic conditions
e. Ability to simulate ideal conditions
The disadvantage of the numerical predictions is the associated truncation error
and round off error and the difficulties in simulating the complicated boundary
condition.
METHODS OF DISCRETIZATION
A. The Finite Difference Method
Because of the importance of the heat equation to a wide variety of fields, there are many
analytical solutions of that equation for a wide variety of initial and boundary conditions.
However, one very often runs into a problem whose particular conditions have no analytical
solution, or where the analytical solutions even more difficult to implement than a suitably
accurate numerical solution. Here we will discuss one particular method for analytical
solution of partial differential equations called the finite difference method.
The usual procedure for deriving the finite difference equations consists of of approximating
derivatives in the differential equation via the 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 great popularity of the
finite difference method is mainly due to their straight-forwardness and relative simplicity by
which a newcomer in the field is able to obtain solutions of a simple problem.
Disadvantages of Finite Difference Method
Several shortcomings and limitations of the FDM were discovered when researchers tried to
solve problems with increasing degree of physical complexity such as :
1. Flow at higher Reynolds’s numbers.
2. Flows around arbitrarily shaped bodies.
3. Strong time-dependant flows.
This led to the development of superior methods, particularly in the area where finite
difference method have some disadvantage.
These Methods can be divided into two main categories
1. Finite Element Method.
2. Spectral Method.
B. Finite Element Method:-
Finite Element Method (FEM) basically seeks solution at discrete spatial regions called
elements by assuming that the governing differential equations apply to the continuum within
each element. Their introduction and ready acceptance was due to the relative ease by which
flow problems with complicated boundary shapes can be modelled, especially when
compared with finite difference method. However disadvantage of FEM arises from the fact
that more complicated matrix operation are required to solve the resulting system of
equations. Furthermore, meaningful variational formulations are difficult to obtain for high
Reynolds number flows. Hence, variation principle based FEM is limited to solutions of
creeping flow and heat conduction problems.
Glarkin’s weighted residual method, which is also another FEM is a powerful method and
circumvents the difficulties faced by variational calculus based FEM. Much research is in
progress in the use of this method.
C. Spectral Method:-
Spectral methods are generally much more accurate than simple first or second order finite
difference methods. In this method, the approximation is based on expansions of independent
variables into finite truncated series of smooth functions. A disadvantage of the spectral
method is their relative complexity in comparisons with standard finite difference methods.
Also the implementation of complex boundary conditions appears to be a frequent source of
considerable difficulty.
D. Control Volume Formulation:-
In this method, the calculation domain is divided into number of non overlapping control
volumes such that there is one control volume surrounding each grid point. The differential
equation is integrated over each control volume.
The major advantage of this method is its physical soundness. The disadvantage is that it is
not as straightforward as finite difference method.
Closure
Here, the finite difference method is chosen as a method of discretization because
For a newcomer in the field of numerical field of flow and heat transfer, this is the best
possible method to begin with simply because of its inherent straight forwardness and
simplicity.
In recent years tremendous refinements and advances have been made by numerous
researchers in the finite difference method, particularly in the areas where it was known
to be weak, such as irregular boundaries or superior accuracy.
FINITE DIFFERENCE METHOD
There are three numerical techniques:
1. Finite –difference (numerical solution)
2. Finite –element
3. Finite volume (in CFD)
This project uses finite-difference method to get temperature distribution, which is then
verified using finite-volume method in CFD.
NODES>>>
Slice the fins at right angle to the length, defining ∆X=L/M
For interior elements the node is centre having width ∆x , while for two end elements the
node is at the edge having width ∆x/2.
D2D1
0 1 2 3 4 5 M-1 M
∆X
L
Chapter-3
PROBLEM FORMULATION
Introduction
Governing equations
Boundary conditions
Discretization
Closure
ONE DIMENSIONAL STEADY STATE PROBLEM:
Considering the one-dimensional steady state heat conduction in an isolated rectangular
horizontal fin as shown figure below. The base temperature is maintained at T=T0 and the tip
of the fin is insulated. The fin is exposed to a convective environment (neglecting the heat
transfer by radiation from the tip of the fin) which is at T∞ (T∞<T0).
T=T0dTdx
= 0
(Physical domain of the rectangular fin)
The average heat transfer coefficient of the fin to the ambient is ‘h’. the length of the fin is
‘L’ and the coordinate axis begins at the base of the fin. The one dimentionality arises from
the fact that the thickness of the fin is much small as compared to its length and width can be
considered either too long or the sides to the fin to be insulated.
Governing differential equation:
The energy equation for the fin at the steady state (assuming constant k) is
d2Tdx2 −hP
kA(T−T ∞)=0
Where P = Perimeter of the fin
A= Cross sectional area of the fin
Boundary conditions:
L
h, T∞
x
The governing equation is a linear, second order ordinary differential equation, two boundary
conditions are needed to completely describe this problem( which is a boundary value
problem).
Boundary conditions are:
B.C.1: At x=0, T=T0 B.C.2: At x=L, dTdx
= 0
Non – Dimensionalisation:
Non – dimensionalising the governing equation and the boundary conditions, using the
dimensionaless variables.
Ѳ = T−T ∞
T0−T∞ and X =
xL
i.e dx=dxL
Also,
d Ѳdx
=dTdx
From above eqn,
d ѲLdX
=dTdx
d2ѲL2dx2 =
d2Tdx2
Now if, m=√ h PKA
m2= h PKA
We get
d2ѲL2dx2 −m2(T−T ∞)=0
i.e d2Ѳ
L2dx2 −m2Ѳ=0
i.e d2Ѳdx2 −m2 L2 Ѳ=0
i.e d2Ѳdx2 – (mL)2Ѳ=0
Boundary conditions are:
Ѳ(0) = 1 and Ѳ’(1) = d ѲdX |
X=1= 0
Discretization:-
The governing equation in non-dimensional form is discretized at any interior grid point i
using central difference for d2θd x2 as follows
( d2 θd x2 )
i
−{(mL)2θ }i=0
Ѳi−1−DѲi +Ѳi+1
(∆ x )2 - m2L2Ѳi = 0
Ѳi-1 – DѲi + Ѳi+1 = 0 .........(A) i=1,2,3.......
Where D = 2 + (mL)2(∆X)2
Ѳi-1 – 2Ѳi + Ѳi+1 - (mL)2Ѳi (∆X)2 = 0
Ѳi-1 – [2 + (mL)2(∆X)2]Ѳi + Ѳi+1 = 0
Handling of the boundary condition:-
At x = L i.e. i=M
The above equation reduces to
ѲM-1 – DѲM + ѲM+1 = 0
Hence ѲM+1 represents a fictious temperature at point M+1 which lies outside the
computational domain. Hence here the image point technique is used.
Image point technique:-
It is assumed that Ѳ Vs X curve extends beyond X = 1 so that at X = 1, the condition dθdx
= 0
is satisfied.
ѲM-1 ѲM+1
ѲM-1
∆X ∆X
In other words, Ѳ Vs X curve can be imagined to look as above figure. The dotted line
represents the mirror imageextension of the solid lineindicating that a minima exist at X=1.
Therefore, the boundary condition at X=1 can be aproximately satisfied by
ѲM+1 = ѲM-1
Substituting this in the equation
ѲM-1 – DѲM + ѲM+1 = 0
ѲM-1 – DѲM + ѲM-1 = 0
2ѲM-1 – DѲM = 0 ..........(B)
Therefore, we can write that
Ѳ1 = 1 for i = 1 (Known)
Ѳi-1 – DѲi + Ѳi+1 = 0 for i=2,3.......
2ѲM-1 – DѲM = 0 for i=M
Hence, we have a set of M-1 linear simultaneous algebraic equations and M-1 unknown
which can be easily solved by standard numerical method.
If M=5 we have
N=M-1=4 equations to solve
These 4 equation are,
From equation A, putting i = 1 and Ѳi-1 =Ѳ1-1= Ѳ0=1(given boundary temperature)
Ѳi-1 – DѲi + Ѳi+1 = 0
1 – DѲ1 + Ѳ2 = 0
DѲ1 - Ѳ2 = 1 ..........(1)
For i = 2,
Ѳ1 – DѲ2 + Ѳ3= 0
-Ѳ1+ DѲ2 - Ѳ3= 0 ...........(2)
-ve is multiplied to make the ‘D’ as +ve.
For i = 3,
-Ѳ2 + DѲ3 – Ѳ4= 0
For i = 4,
-Ѳ3+ DѲ4 – Ѳ5= 0 .........(3)
But Ѳ3 = Ѳ5 (Image point technique or apply equation B)
-2Ѳ3+ DѲ4 = 0 ........(4)
Now arranging these 4 equations in the matrix form.
[ D −1−1 D
0 0−1 0
0 −10 0
D −1−2 D
] Ѳ1
Ѳ2
Ѳ3
Ѳ4
=
1000
It should be noted that Ѳ1 corresponds to the temprature at grid point ‘2’ and so on for Ѳ2 ,Ѳ3,
Ѳ4.
Chapter-4
About Matlab and Ansys-Inc
Matlab Ansys-Inc Closure
MATLAB
MATLAB is a numerical computing environment and fourth generation programming
language. Developed by The Math Works, MATLAB allows matrix manipulation, plotting of
functions and data, implementation of algorithms, creation of user interfaces, and interfacing
with programs in other languages. Although it is numeric only, an optional toolbox uses the
MuPAD symbolic engine, allowing access to computer algebra capabilities. An additional
package, Simulink, adds graphical multidomain simulation and Model-Based Design for
dynamic and embedded systems.
MATLAB (meaning "matrix laboratory") was invented in the late 1970s by Cleve Moler,
then chairman of the computer science department at the University of New Mexico.[3] He
designed it to give his students access to LINPACK and EISPACK without having to learn
Fortran. It soon spread to other universities and found a strong audience within the applied
mathematics community. Jack Little, an engineer, was exposed to it during a visit Moler
made to Stanford University in 1983. Recognizing its commercial potential, he joined with
Moler and Steve Bangert. They rewrote MATLAB in C and founded The MathWorks in
1984 to continue its development. These rewritten libraries were known as JACKPAC.
[citation needed] In 2000, MATLAB was rewritten to use a newer set of libraries for matrix
manipulation, LAPACK[4].MATLAB was first adopted by control design engineers, Little's
specialty, but quickly spread to many other domains. It is now also used in education, in
particular the teaching of linear algebra and numerical analysis, and is popular amongst
scientists involved with image processing.
This project utilizes MATLAB tool for solving “n” equations found by the finite difference
method, hence plotting of temperature distribution diagram for different materials and finally
comparing the results..
ANSYS Inc.
ANSYS, Inc. is an engineering simulation software provider founded by software engineer
John Swanson. It develops general-purpose finite element analysis and computational fluid
dynamics software. While ANSYS has developed a range of computer-aided engineering
(CAE) products, it is perhaps best known for its ANSYS Mechanical and ANSYS
Multiphysics products.
ANSYS Mechanical and ANSYS Multiphysics software are non exportable analysis tools
incorporating pre-processing (geometry creation, meshing), solver and post-processing
modules in a graphical user interface. These are general-purpose finite element modeling
packages for numerically solving mechanical problems, including static/dynamic structural
analysis (both linear and non-linear), heat transfer and fluid problems, as well as acoustic and
electro-magnetic problems.
ANSYS Mechanical technology incorporates both structural and material non-linearities.
ANSYS Multiphysics software includes solvers for thermal, structural, CFD,
electromagnetics, and acoustics and can sometimes couple these separate physics together in
order to address multidisciplinary applications. ANSYS software can also be used in civil
engineering, electrical engineering, physics and chemistry.
ANSYS, Inc. acquired the CFX computational fluid dynamics code in 2003 and Fluent, Inc.
in 2006. The CFD packages from ANSYS are used for engineering simulations. In 2008,
ANSYS acquired Ansoft Corporation, a leading developer of high-performance electronic
design automation (EDA) software, and added a suite of products designed to simulate high-
performance electronics designs found in mobile communication and Internet devices,
broadband networking components and systems, integrated circuits, printed circuit boards,
and electromechanical systems. The acquisition allowed ANSYS to address the continuing
convergence of the mechanical and electrical worlds across a whole range of industry sectors.
What is ANSYS?
ANSYS is a finite-element analysis package used widely in industry to simulate the response
of a physical system to structural loading, and thermal and electromagnetic effects. ANSYS
uses the finite-element method to solve the underlying governing equations and the
associated problem-specific boundary conditions.
How to obtain a solution in ANSYS?
A solution can be obtained by following these nine steps:
1. Start-up and preliminary set-up
2. Specify element type and constants
3. Specify material properties
4. Specify geometry
5. Mesh geometry
6. Specify boundary conditions
7. Solve!
8. Postprocess the results
9. Validate the results
Results were plotted using different materials by simulating them in the ANSYS package.
The results show obtained were compared with experimental data.
Closure
Matlab is used to get numerical solutions of one dimensional fin. The temperature profiles are
hence plotted. The fin is simulated in Ansys cfd tool and hence respective plots are generated
and atudied.
Chapter-5
EXPERIMENTAL STUDY
Experimental setup
Procedure
Observation
Closure
AIM OF THE EXPERIMENT:
For a pin fin under electrical heating and subjected to natural convection with insulated tip.
We are to find out temperature distribution along the length of the fin as measured by
thermocouples and comparison with theoretical temperature distribution.
DESCRIPTION OF TEST RIG:
The test rig consists of a square duct with a pin fin attached on inside duct wall at the mid
length. Thermocouples are attached on the fins surface at 5 locations including the fin root
and tip. A sixth thermocouple to measure the temperature of the air surrounding the fin. The
base of the fin receives the heat from an electrical heater. The test rig is shown in photograph
below.
PHOTOGRAPH:
PROCEDURE:
The heater is first switched on and the system is allowed to com to steady state. The
temperature read by all the thermocouples are monitored from time to time. When the
temperature measured by all the thermocouples does not vary anymore with time, steady state
is reached. The steady state temperature of all thermocouples are measured and tabulated.
The dimensions, heat transfer coefficient and thermal conductivity of the fin are:
Fin diameter – 5 mm
Fin length – 120 mm
Heat transfer coefficient of the fin – 10.68 W/m2-K
OBSERVATION TABLE:
Thermal conductivity of fin material(copper alloy) – 386 W/m-K
TEMPERATURE T1 (Root) T2 T3 T4 T5 T∞(Ambient)
In Celcius(0C) 89.5 87.5 86 85 84.5 32
In Kelvin(K) 362.5 360.5 359 358 357.5 305
Thermal conductivity of fin material(aluminium) – 250 W/m-K
TEMPERATURE T1 (Root) T2 T3 T4 T5 T∞(Ambient)
In Celcius(0C) 89.5 86 85.5 84 83.5 32
In Kelvin(K) 362.5 359 358.5 357 356.5 305
Thermal conductivity of fin material(copper) – 401 W/m-K
TEMPERATURE T1 (Root) T2 T3 T4 T5 T∞(Ambient)
In Celcius(0C) 89.5 88 86.5 86 85.5 32
In Kelvin(K) 362.5 361 359.5 359 358.5 305
Thermal conductivity of fin material(carbon steel) – 54 W/m-K
TEMPERATURE T1 (Root) T2 T3 T4 T5 T∞(Ambient)
In Celcius(0C) 89.5 80.5 74 70 67 32
In Kelvin(K) 362.5 353.5 347 343 340 305
CONCLUSION:
The experiment was conducted sucessesfully and the temperatures were recorded.
SOURCES OF ERROR:
Probably, some heat is lost from the steel wall which goes unaccounted for.
The temperature readings from the thermocouple may not be at steady state.
There may be some energy loss in the thermocouple wire.
Impurities present in the fin material.
Closure
The observation were taken for various materials of fin under given experimental conditions.
The results were plotted using matlab tool. The results were idealized for one-dimensional
steady state conditions.
Chapter-6
Results and discussions
Plots from Ansys Programme output Experimental plots Comparative study
of the results
Results obtained from ANSYS
For the experimental setup:
By taking following data
Fin length = 120 mm
Fin width = 5 mm
Convection heat transfer coefficient = 10.68 W/m2-K
Thermal conductivity of the lab specimen = 386 W/m-K
TEMP
For Brass fin:
By taking following data
Fin length = 120 mm
Fin width = 5 mm
Convection heat transfer coefficient = 10.68 W/m2-K
Thermal conductivity of the Brass = 109 W/m-K
TEMP
For Aluminium fin:
By taking following data
Fin length = 120 mm
Fin width = 5 mm
Convection heat transfer coefficient = 10.68 W/m2-K
Thermal conductivity of the Aluminium = 250 W/m-K
TEMP
For Copper fin:
By taking following data
Fin length = 120 mm
Fin width = 5 mm
Convection heat transfer coefficient = 10.68 W/m2-K
Thermal conductivity of the Copper = 250 W/m-K
TEMP
For Carbon Steel fin:
By taking following data
Fin length = 120 mm
Fin width = 5 mm
Convection heat transfer coefficient = 10.68 W/m2-K
Thermal conductivity of the Carbon Steel = 54 W/m-K
TEMP
Programme output
Pin fin Material: BRASS
K (thermal conductivity)=109 w/m-K
h (coefficient of convection)=10.68 w/m2-K
l (length of pin-fin)=120mm
d (diameter of pin fin)=5mm
hence, m=√hpKA
= 8.85
convergence criteria =0.01
1 1.5 2 2.5 3 3.5 40
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
tem
prat
ure
nodes
1 2 3 4 5 6 7 8 90
0.05
0.1
0.15
0.2
0.25
0.3
0.35
tempra
ture
nodes
0 2 4 6 8 10 12 140
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45tem
prature
nodes
0 2 4 6 8 10 12 14 16 18 200
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5tem
pratur
e
nodes
convergence criteria = 10 -20
0 2 4 6 8 10 12 14 16 18 200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
tempra
ture
nodes
Pin fin Material: ALUMINIUM
K (thermal conductivity)=200 w/m-K
h (coefficient of convection)=10.68 w/m2-K
l (length of pin-fin)=120mm
d (diameter of pin fin)=5mm
hence, m=√hpKA
= 6.536
convergence criteria=0.01
1 1.5 2 2.5 3 3.5 40
0.05
0.1
0.15
0.2
0.25
tem
prat
ure
nodes
1 2 3 4 5 6 7 8 90
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4tem
pratur
e
nodes
0 2 4 6 8 10 12 140
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5tem
pratur
e
nodes
convergence criteria = 10 -20
0 2 4 6 8 10 12 14 16 18 200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
tempra
ture
nodes
Pin fin Material: COPPER
K (thermal conductivity)=401w/m-K
h (coefficient of convection)=10.68 w/m2-K
l (length of pin-fin)=120mm
d (diameter of pin fin)=5mm
hence, m=√hpKA
= 4.616
convergence criteria =0.01
1 1.5 2 2.5 3 3.5 40
0.05
0.1
0.15
0.2
0.25
0.3
0.35
tem
prat
ure
nodes
1 2 3 4 5 6 7 8 90
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
tempra
ture
nodes
0 2 4 6 8 10 12 140
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
tempra
ture
nodes
0 2 4 6 8 10 12 14 16 18 200
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
tempra
ture
nodes
convergence criteria = 10 -20
0 2 4 6 8 10 12 14 16 18 200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
tempra
ture
nodes
Pin fin Material: CARBON STEEL
K (thermal conductivity)=54w/m-K
h (coefficient of convection)=10.68 w/m2-K
l (length of pin-fin)=120mm
d (diameter of pin fin)=5mm
hence, m=√hpKA
= 12.579
convergence criteria=0.01
1 1.5 2 2.5 3 3.5 40
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
tem
prat
ure
nodes
1 2 3 4 5 6 7 8 90
0.05
0.1
0.15
0.2
0.25
0.3
0.35
tempra
ture
nodes
0 2 4 6 8 10 12 140
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
tempra
ture
nodes
0 2 4 6 8 10 12 14 16 18 200
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
tempra
ture
nodes
Convergence criteria= 10 -20
0 2 4 6 8 10 12 14 16 18 200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
tempra
ture
nodes
COMAPARISON OF RESULTS FOR DIFFERENT PIN FIN MATERIALS OBTAINED BY PROGRAMME OUTPUT
blue-copperRed-aluminiumBlack-brass
0 2 4 6 8 10 12 14 16 18 200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
nodes
tem
prat
ure
Green-carbon steel
EXPERIMENTAL RESULTS
0 20 40 60 80 100 12083
84
85
86
87
88
89
90
TE
MP
RA
TU
RE
(in c
els
ius)
DISTANCE(in mm)
TEMPRATURE V/S DISTANCE(aluminium)
0 20 40 60 80 100 12084.5
85
85.5
86
86.5
87
87.5
88
88.5
89
89.5
TE
MP
RA
TU
RE
(in c
els
ius)
DISTANCE(in mm)
TEMPRATURE V/S DISTANCE(COPPER ALLOY)
0 20 40 60 80 100 12085.5
86
86.5
87
87.5
88
88.5
89
89.5
TE
MP
RA
TU
RE
(in c
els
ius)
DISTANCE(in mm)
TEMPRATURE V/S DISTANCE(copper)
0 20 40 60 80 100 12065
70
75
80
85
90
TE
MP
RA
TU
RE
(in c
els
ius)
DISTANCE(in mm)
TEMPRATURE V/S DISTANCE(carbon steel)
COMAPARITIVE PLOT OF EXPERIMENTAL RESULTS FOR DIFFERENT MATERIALS
RED-CARBON-STEEL
GREEN-COPPER
BLUE-ALUMINIUM
0 20 40 60 80 100 12065
70
75
80
85
90
distance(in mm)
tem
prat
ure(
in c
elsi
us)
temprature v/s distance plot for different materials
BLACK-COPPER ALLOY
Chapter-7
Conclusion and further scope
7.1 CONCLUSION
A numerical was developed to solve the one- dimensional differential heat equation and plot
the variation of temperature along the entire length of the pin fin of different material.
An experiment was conducted to measure the temperature along the surface of the pin fin
assuming steady state heat transfer, free convection and the tip of the fin insulated. The
experiment was repeated with different materials having different values of thermal heat
conductivity (k) and a graph was plotted between temperature variance and length of the fin.
The results obtain thought the numerical coding and the experiments were verified by using
the CFD tool of ANSYS. Thus a comparative study was done for each of the materials to see
the temperature variation along the length of the pin fin in different cases.
With the above analysis we could see the difference in heat transfer of different materials in
one dimension and could predict the most suitable material for this purpose. But there were
some deficiency in the results as the following facts were ignored in the code formation
process.
The heat transfer takes place in all three dimensions
Some heat is lost from the tip of the fin which goes unaccounted for
In practical cases, free and forced convection depends on the nature of the
surrounding environment
Due to the shortage of time and the complexity of the practical case these facts were not
taken into consideration into the present investigation but can be sorted out and can even
better analyzed in the future work.
7.2 SCOPE FOR FUTURE WORK
In this analysis we have considered only one dimensional heat transfer from a pin fin an
hence used the one dimensional heat equation in the problem formulation and numerical
coding. There exists much more scope to extend this work. For example, using the two
dimensional heat equation or even the three dimensional heat equation during the problem
formulation and numerical coding, which will further optimize the solution. Moreover the fin
can be considered of nay size and shape and the heat transfer from the tip and sides of the fin
can be considered. In place of free convection forced convection can be considered
depending on the air flow around the surface of the fin.
BIBLIOGRAPHY
1. HEAT TRANSFER BY SP SUKHATME
2. GETTING STARTED WITH MATLAB BY RUDRA PRATAP
3. HEISLER,M.P 1947.TEMPARATURE CHARTS FOR INDUCTION AND
CONSTANT TEMPARATURE.
4. SCHNEIDER,P.J,1955 CONDUCTION HEAT
TRANSFER.READING,MASS:ADDISON-WESLEY PUBLISHING CO.
5. GARDNER,KA,1945.EFFICIENCY OF EXTENDED
SURFACES.TRANS.ASME,VOL 67:621
6. LEON ,OCTAVIO.,MEY,GILBERT DE., JAN VIERENDEELS,ERIK
DICK ..”COMPARISON BETWEEN THE STANDARD AND STAGGERED
LAYOUT FOR COOLING FINS IN FREE CONVECTION COOLING ”,
ASME SEPTEMBER 2003
7. BEBNIA ,MASUD,NAKAYAMA,WATARU.”CFD SIMULATIONS OF HEAT
TRANSFER FROM A HEATED MODULE IN AN AIR STREAM”
8. NAKAYAMA ,MATSOUKIS, HACHO,YAKIMA “A NEW ROLE OF CFD
SIMULATION IN THERMAL DESIGN OF COMPACT
EQUIPMENT :APPLICATON OF THE BUILD-UP APPROACH TO THERMAL
ANALYSIS OF BENCH MARK MODEL”
Appendix A
Matlab program
ml=input('enter the value of ml......>');
m=input('enter the number of nodes...>');
n=m-1;
x=1/n;
d=2+(ml^2)*(x^2);
node=zeros(n,1);
a=zeros(n,n);b=zeros(n,1);c=zeros(n,1);f=zeros(n,1);k=0;r=0;
a(1,1)=d;a(n,n)=d;f(1,1)=1;
a(1,2)=-1;a(n,n-1)=-2;
for i=2:1:n-1
a(i,i)=d;
a(i,i-1)=-1;
a(i,i+1)=-1;
end
disp(a);
if d>-2;
disp(char('scanborough criterian is sattisfied'));
end
e=input('enter the value of convergence factor i.e. E...>');
while k>=0
for i=1:1:n;
c(i,1)=b(i,1);
end
for i=1:1:n;
cal=0;
66 | P a g e
for j=i+1:1:n;
cal=cal+(a(i,j)*b(j,1));
end
for j=1:1:i-1;
cal=cal+(a(i,j)*b(j,1));
end
b(i,1)=(f(i,1)-cal)/a(i,i);
end
for i=1:1:n;
if (b(i,1)-c(i,1))<e
k=-1;
else
k=0;
end
end
r=r+1;
disp([char('iteration no.') int2str(r)]);
disp(b);
end
for i=1:1:n;
node(i,1)=i;
end
plot(node,b);
ylabel('temprature');
xlabel('nodes');
67 | P a g e
Appendix B
Program iteration results
enter the value of ml......>8.85
enter the number of nodes...>5
6.8952 -1.0000 0 0
-1.0000 6.8952 -1.0000 0
0 -1.0000 6.8952 -1.0000
0 0 -2.0000 6.8952
scanborough criterian is sattisfied
enter the value of convergence factor i.e. E...>0.0000000000000001
iteration no.1
0.1450
0.0210
0.0031
0.0009
iteration no.2
0.1481
0.0219
0.0033
0.0010
iteration no.3
0.1482
0.0220
0.0033
0.0010
68 | P a g e
iteration no.4
0.1482
0.0220
0.0033
0.0010
iteration no.5
0.1482
0.0220
0.0033
0.0010
iteration no.6
0.1482
0.0220
0.0033
0.0010
iteration no.7
0.1482
0.0220
0.0033
0.0010
iteration no.8
0.1482
0.0220
0.0033
0.0010
iteration no.9
0.1482
0.0220
0.0033
0.0010
69 | P a g e
iteration no.10
0.1482
0.0220
0.0033
0.0010
iteration no.11
0.1482
0.0220
0.0033
0.0010
iteration no.12
0.1482
0.0220
0.0033
0.0010