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