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Tata Technologies Limited

A Report on

1D and 2D heat transfer project(including a general overview of company)

Aerospace DepartmentTATA Technologies Limited, Hinjewadi

By:

Vishesh Gupta 2011A4PS273P

At

TATA Technologies Limited, Hinjewadi, Pune

A Practice School I station of

Birla Institute of Technology and Science,

Pilani, Rajasthan

July, 2013


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A Report on

1D and 2D heat transfer project

(including a general overview of company)

Aerospace Department

TATA Technologies Limited, Hinjewadi

By:

Vishesh Gupta 2011A4PS273P

Prepared in partial fulfillment of the

Practice school I Course

At

TATA Technologies Limited, Hinjewadi, Pune

A Practice School I station of

Birla Institute of Technology and Science,

Pilani, Rajasthan

July, 2013


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Acknowledgement

I am grateful to Tata Technologies Limited for giving me this

opportunity to gain experience at the organization and forgiving me invaluable guidance right at the start.

I would also like to thank Mr Sumeet Dhar (Project Manager),

Mrs Prachi Dhadheech (HR) for their valuable support

throughout our tenure. I would like to make a special mention

of Vinay Shekhar (employee at the aerospace department) for

providing me deep insight at crucial junctures of my project

and small helps in debugging the code. I would also like toextend my gratitude towards Mr Manoj (aerospace project

manager) for giving me the freedom to work through my

project and patiently listening to my queries and explanations

also giving worthy advice. I would further like to express my

gratitude towards Dr Biswadip Shome for giving me worthy

suggestions and sufficient time to extend my project to 2D

analysis in the short time I worked under him.

I would finally like to thank Dr Ranjit Patil( PS Instructor

from BITS Pilani) for being a constant source of motivation for

me throughout my internship at Tata Technologies.


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Birla Institute of Technology and Science

Pilani (Rajasthan)

Practice School Division

Station: Tata Technologies Ltd. Centre: Hinjewadi,

Pune

Duration: 53 Days Date of Start: 22/05/2013Date of Submission: 07/07/2013

Title of Project: Simulation tool for a floor fire test

and the behavior of the material in the test by modeling

of a transient heat transfer process through a series of

slabs.

Name of Student I.D.No. Discipline

Vishesh Gupta 2011A4PS273P B.E. Mechanical

Name and designation of experts:

Dr Biswadip Shome (Head Aerospace Design &

Validation)

Manoj Radle (Project Manager)Name of PS Faculty: Dr. Ranjit Patil

Key words: heat transfer, finite difference analysis

Project Areas: transient heat transfer, partial

differential equations, fully implicit FDM


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Abstract

This project report has been primarily divided into four

chapters. Chapter 1 deals with company information. Chapters

2, 3 and 4 give detailed analysis of my project related work.

Chapter 1 aims at giving a general overview of the company,

mainly information pertaining to what was observed during

the orientation sessions.

Chapter 2 deals with 1 dimensional heat transfer solved by the

analytical method using techniques like transformation of non-

homogenous boundary conditions to homogenous boundary

conditions by superposition principle, separation of variables

and Greens function.

Chapter 3 concerns with the numerical approach adopted to

solving the 1D heat transfer problem. The fully implicit finite

difference scheme has been utilized here which is

unconditionally stable and convergent. The matrix formed has

been solved both directly using in-built MATLAB methods as

well as Gauss-Seidel iterative procedures.

Finally, Chapter 4 showcases the 2D heat transfer problem

solved by a numerical approach which uses a slight variant of

the fully implicit finite difference scheme called the alternating

direction implicit method (ADI).

Signature (Student) Signature (Project Guide) Signature (PS Faculty)

Date: Date: Date:


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Table of Contents

S No. Topic Page No.

1 CHAPTER I

Introduction 5

Orientation Related Information 11

ERM 20

Software Partners 24

2 CHAPTER II

1D Analytical approach 34

Improvements & Conclusion 46

3 CHAPTER III

1D Numerical Approach 47

Improvements & Conclusion 58

4 CHAPTER IV

2D Numerical Approach 59

Improvements and Conclusion 70

5 Appendix A 71

6 Appendix B 777 Appendix C 87

8 References 92


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List Of Illustrations

Figures

Fig. 1.1 Tata overview

Fig. 1.2 Presence of Tata Technologies over the

globe

Fig. 1.3 Product Cycle at Tata TechnologiesFig. 1.4 Historical PerformanceFig. 1.5 Global initiatives to sustain and grow

employee engagement

Fig. 1.6 Risk Management ProcessFig. 1.7 Some clients of Tata TechnologiesFig. 1.8 Aerospace Design ProcedureFig. 1.9 Airplane structure components

Fig. 2.1 Time-Temperature Curve

Fig. 2.2 Values of source temperature and time

Fig. 2.3 Transient heat transfer through a

series of blocks

Fig. 3.1 1 dimensional heat transfer through a

series of blocks

Fig. 4.1 2 dimensional heat transfer through a

series of blocks


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Graphs

Graph 2.1 Variation of source and interface 1

temperature with time

Graph 2.2 Variation of source and interface 2temperature with time

Graph 2.3 Variation of source and all interfacetemperature with time

Graph 3.1 Variation of source and interfacetemperature with time by direct

solving of matrix

Graph 3.2 Variation of source and interfacetemperature with time (Gauss-Seidel

iteration)Graph 3.3 Variation of source and interface

temperature with time (both direct by

direct matrix and Gauss-Seidel

iteration)

Graph 3.4 Convergence of temperature at

particular time by Gauss-Seidel

iterations

Graph 4.1 Variation of node 1 temperatures of

all interfaces with time

Graph 4.2 Variation of temperature of all nodes

at top surface with timeGraph 4.3 Temperature of all nodes at 30 min at

top surface


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CHAPTER I

Introduction

The origins of Tata Technologies lie with disparate

organizations started in India, Europe and the USA over 20

years ago. Gradually the European and North American

companies came together until in 1989 a single entity emerged

INCAT. Following growth and acquisition, that company

firmly established itself in North America, Europe and Japan.

And in 2004, INCAT was launched on the London StockExchange. Tata Technologies started in Singapore in 1994 as a

provider of specialized IT enabled consulting, services and

products to leading manufacturers. Establishing itself in the

USA, UK and India, the companys first major outsourcing

contract was with Tata Motors. Then, in 2005, Tata

Technologies acquired INCAT.

Today, the enlarged company serves the major automotive andaerospace OEMs and their suppliers. Tata Technologies is

active in North America, Europe, the Middle East and the

Asia-Pacific region, and currently services all of the top ten

aerospace original equipment manufacturers (OEMs) and all of

the top ten automotive OEMs.


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Tata Technologies Limited1

The Tata group

Tata Technologies is part of the Tata group, one of Indias

oldest, largest and most respected businesses. Founded by

Jamsetji Tata in 1868, the group pioneered industries of

national importance to India: steel, power, hospitality and

airlines. With revenues over US $100 billion, the group today

has 3.2 million shareholders, more than 100 operating

companies in six continents and employs more than 450,000

people. It focuses on steel production, power generation,

commercial and passenger vehicles, chemicals, hotels, textiles,

consumer goods, consultancy, information technology and

telecommunications. Earning the trust and respect of millions

of stakeholders around the world, the group and its enterprises

have adhered to a rigorous set of business ethics and a strong

commitment to corporate social responsibility for over 140

years.

Fig. 1.1: Tata overview


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Fig. 1.2: Presence of Tata Technologies over the globe


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Fig. 1.3: Product Cycle at Tata Technologies (1st half)


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Fig. 1.3: Product Cycle at Tata Technologies (2nd half)


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Fig. 1.4: Historical Performance


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Orientation Related Information

Tata Technologies enables manufacturing companies to design

and build better products through engineering services and the

application of information technology to product development

and manufacturing enterprise processes. With over 6,000

engineers, representing 17 nationalities, Tata Technologies

covers every aspect of the value chain from conceptualization,

manufacturing and aftermarket/MRO support. The three

major industry verticals that it serves include aerospace,

automotive and machinery manufacturing. Tata Technologies

supports clients through comprehensive engineering services

and IT processes and tools to manage product development

and the complete manufacturing ecosystem. It serves clients in

25 countries, with a delivery model specifically designed for

engineering and IT engagements that offers a unique blend of

deep, local expertise integrated with our nine global delivery

centers based in Detroit (USA), Coventry (UK), Pune and

Bangalore (India), Brasov, Craiova & Iasi (Romania), Stuttgart

(Germany), and Bangkok (Thailand).


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VPD Vehicle Programs & Development (VPD)

The VPD provides complete outsourced program management,

concept development, detail design, validation and

manufacturing planning services. Projects of this scale and

complexity are achieved through a combination of automotive

experts in the US and Europe, coupled with Indias most

experienced automotive engineers. Programs include electric

vehicles, or EV variants that help achieve sustainability

targets, while providing mobility at an affordable price point.

The eMO-C electric delivery vehicle concept, designed by Tata

Technologies, follows the eMO showcased at the 2012

International Auto Show in Detroit, was a first for any India-

based engineering services firm.

E&D Outsourced Engineering & Design

The services of this group include concept development,

VA/VE, CAE, detailed engineering, embedded software

development, product verification, and manufacturing process

design, tool design and validation, applied to major product

subsystems and components. Offerings include the provision of

providing services from offshore engineering centers in India,

Romania and Thailand.

PLM Product Lifecycle Management Solutions

The PLM optimizes product development processes,

implementing collaborative PLM tools, a major contributor to

ER&D investment efficiency, especially for global engineering

teams with extensive supply chains. Tata Technologies is the

worlds largest independent reseller of PLM technology. It and


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its customers are among the worlds top users of PLM

technology. The companys engineers use these products to

efficiently deliver services to our clients worldwide. It packages

the insights and best practices we have developed through ourengineering heritage into service offering templates to improve

the efficiency of engineering teams at Tata Technologies and

its clients.

ESG Enterprise Solutions Group

The ESG provides consulting and IT solutions that help

manufacturing customers in optimizing critical enterpriseprocesses through the application and data analytics of

Enterprise Resource Planning, (ERP), Manufacturing

Execution Systems, (MES), and Customer Relationship

Management, (CRM), including the use of social media and

improving manufacturing planning and performance. It also

has extensive experience in rapidly integrating the processes,

systems and data of companies acquired by manufacturers.


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Overview and Outlook

1. Manufacturing

The sectors that it serves aerospace, automotive andmachinery manufacturing, have emerged from the global

recession and engineering R&D (ER&D) spends have grown

consistently. Demand is being driven by an increasingly

competitive market and changing demographics in the west,

where availability of a skilled engineering workforce is on the

decline in many countries, adds NASSCOM. Tata Technologies

helps its partners in the manufacturing industry tacklechallenges relating to the shortage of engineering resources

and skills. It partners with the worlds top automotive OEMs

and tier one auto suppliers, top aerospace companies and

leading machinery manufacturing OEMs to supply high-end

capability and variable capacity. It can scale up rapidly,

flexing resources and specific skills to meet the project

demands of our clients, helping them move from fixed tovariable costs, yet retaining access to top talent, who

understand their products and processes.

2. Aerospace

Global air travel is poised to grow significantly over the next

few years and expand the addressable market for airlines and

their partners. While innovation in business and operatingmodels evolve at a rapid pace, product innovation needs to

gather exponential momentum for airlines to win the battle

against high operating costs. The aerospace sector is projected

to be worth over $4 trillion by 2030 according to Airbus and


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Boeing with 86,000 new aircraft expected to be delivered over

the next 20 years.

The backlog of orders for large commercial fixed-wing aircraft

stands at 8,000 and is growing, driven mainly by the

introduction of new aircraft such as Airbus A320neo and

A350, and Boeings 737 MAX and 787. The depressed market

for business jets is starting to recover and new product

development is on the rise with expectations that the market

will grow five percent annually over the next few years. Even

the global commercial helicopter market is recovering sooner

than expected. Booz & Co identifies four major challenges

confronting the civilian aerospace sector: increasing production

rates; unsustainable development cost and value distribution;

growing demand for more efficient aircraft; and digitization of

the industry. The spurt in production of aircraft has led to a

slip-up in the supply chain, as not all suppliers are geared to

meet the increased demand. The existing tooling,

infrastructure and capabilities will be pushed to the limit

across the upstream supply chain because of increased

outsourcing and the complexity of sub-systems and assemblies.


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3. Automotive

Rarely has an industry confronted the magnitude of

multidimensional change that the automotive industry faces

today. Meanwhile, 2012 sales in Europe reached the lowest

level in 17 years, due to a weak economy and low consumer

confidence. In the United States, strong pent-up demand and a

falling unemployment rate helped auto sales reach a five-year

high in 2012. The regional shift in vehicle sales opportunities

translates into a need to develop products that resonate with

new groups of consumers. The automotive industry is being

shaped by a number of forces, according to KPMG's 2013

Global Automotive Executive Survey. Among them are

environmental pressures leading to more efficient engines.

Electric mobility

To capitalize on these industry trends, Tata Technologies

formed a Vehicle Programs Development group in2011. The

group has demonstrated its capabilities to develop a complete

vehicle. In 2012, Tata Technologies' eMO (electric mobility)

vehicle study was introduced to the industry at the North

American International Auto Show in Detroit. The worlds first

complete vehicle study developed by an India-based

engineering services company, the eMO showcases innovation

in automotive packaging and design, manufacturing processes,

as well as electric vehicle engineering benchmarks, and was

developed to go to market at the disruptive price point of

$20,000.


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In 2013, the group introduced the eMO-C, a commercial

vehicle variant, profiled in another part of this report, and

opened an all-new 10,000-square foot North American

Engineering and Innovation Center in Troy, Michigan, USA.The facility opened in March with 60 engineering

professionals, with the number expected to exceed 100 by the

end of the year.

4. Machinery Manufacturing

There are five major trends that are evident in the global

industrial machinery segment: agile product development; cost

fitness/fine-tuning the cost structure to fund growth;

leveraging IT/digitization; developing the BRIC markets and

other emerging regions; and impacting technological trends

connectivity, monitoring and emissions reduction.

For many industrial companies, developing the BRIC markets

and other emerging regions represent the best opportunities

for growth. With the machinery manufacturing sector

witnessing steady growth worldwide, Tata Technologies is wellpositioned to support various OEMs and tier 1 suppliers. Its

service offerings cover product design, mechanical, electrical

and embedded electronics and manufacturing engineering. The

aim is to be a full service provider to the machinery


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manufacturing industry by leveraging our skills and

knowledge in engineering, Product Lifecycle Management

(PLM) and enterprise IT solutions. Its machinery

manufacturing domain comprises experts with extensiveknowledge in systems engineering, mechanical engineering,

product design, electrical, electronics and embedded design

and development. Its innovative and frugal engineering

approach helps organizations create products, at a faster pace

with a lower cost, delivering more value to end customers.

Human Capital

The Company continues to pursue its strategy of growing

returns on human capital. It strives to consistently sustain a

high performance culture to bring out the best in people in a

signature working experience. This supports its customers

with highly proactive and motivated teams to ensure delivery

on promises of best of breed solutions cost effectively.

Customer delight leads to Shareholder value.

Talent Acquisition

A focused branding of the Company both with Campuses and

the industry job market lies at the foundation of our strategy

for talent acquisition. The focus on acquiring skills, specific to

customer needs takes root with specialized courses designed by

us and offered to select institutions. The company addressesthe job market with several propositions for quick engagement

of exceptional talent. Its PFLE branding (Passionate Fun

Loving Engineer), a proactive team of recruiters supported by

imaginative IT enabled processes, ensures the highest


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productivity on talent acquisition. Costs of recruitment

benchmark well with industry. Recruitment lead times are

continuously on the decline.

Fig. 1.5: Global initiatives to sustain and grow employee engagement


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ERM

Definition

ERM is a process, effected by an entity's Board of Directors,management and other personnel, applied in strategy setting

and across the enterprise, designed to identify potential events

that may affect the entity, and manage risks to be within its

risk appetite, to provide reasonable assurance regarding the

achievement of entity objectives.

The Company has established a formal Enterprise Risk

Management (ERM). The Company has adopted the

recommendations on the Enterprise Risk Management

framework provided by the Committee of Sponsoring

Organizations of the Treadway Commission (COSO). As a

services focused Company, it is necessary for the Company to

manage risk at the individual transaction level and to consider

aggregate risk at the customer, industry and geographic,

where appropriate.

ERM Organization and Process

The Executive Management Team of the Company is

responsible for implementing the Risk Management

Framework under the direction of the Audit Committee of the

Company, and the Audit Committee provides periodical

updates to the Board of Directors of the Company. The Board

monitors the overall performance of the Risk Management

function.


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Risk Management Activities

A disciplined approach to risk is important in an organization

such as ours in order to ensure that we are executing according

to our strategic objectives and this process is designed to

identify potential events that, if they occur, will affect our

Company. The management has identified the following top 10

risks, classified into external risk factors and internal risk

factors, to help achieve business objectives in a robust manner.

Fig. 1.6: Risk Management Process


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Global delivery

Tata Technologies has fine-tuned its ability to gather its global

resources in combinations that reap the utmost benefit for its

clients. It can combine its scalable resourcesboth human and

physical with the proven technology and our accumulated

lengthy experience to address any client requirement. And it

can do it while providing the underlying assurance that all its

efforts are underpinned by a robust reputation for thorough

organization and first-class engineering. At each client

engagement, it balances its resources for maximum benefit.

With access to several thousand engineering and technology

specialists, it can also provide the right level of qualifications,

experience and skills to meet your exact requirements. In

addition, its global services delivery team is led by individuals

with a robust background in engineering.

It offers real-world education for:

1) Dassault Systmes

2) Siemens PLM

3) Autodesk (mechanical design applications)

4) MSC Software

Product-development IT

IT is the lifeblood of your business. And aligning your

enterprise systems, PLM, enterprise resource planning (ERP),

customer relationship management (CRM), information

lifecycle management (ILM), application lifecycle management

(ALM) and production systems to your business strategies is


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key to creating successful products, satisfied customers and

enhanced profits. Tata Technologies is unique in that it look at

this area holistically, allying your business strategies to the

most appropriate software and systems.

The company has been engaged in SAP implementation

services for product-centric organizations in a wide variety of

industries around the world. Its SAP service offerings span all

phases of the lifecycle of your enterprise from planning to

implementation, customization, development, testing and post-

manufacturing support. And thats how it keeps the lifeblood of

your company flowing. Better CRM initiatives are not limited

to software installation, they involve considering the context,

support and understanding of professionals so that they can

learn and take full advantage of your information systems.


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Software partners

Dassault Systmes

Tata Technologies is recognized as a Premier Partner withDassault Systmes, a global software alliance for the

CAD/CAM/CAE/PLM market.

1)CATIA products are based on the open and scalableDassault Systmes V5 and V6 architecture. CATIA is the

leading product development solution for all

manufacturing companies, from big OEMs to small

producers.

2)SMARTEAM brings affordable product datamanagement capabilities to small-to-medium companies.

3)ENOVIA is a leading enterprise solution for robust,collaborative product data management.

4)DELMIAs digital manufacturing solutions enable thecontinuous creation and validation of manufacturing

processes throughout the product lifecycle.

MSC Software

MSC Software is the leading global provider of integrated

enterprise simulation solutions.

1)SimOffice easy-to-use simulation enables engineers toverify design in the Microsoft Windows desktop

environment.


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2)Nastran is the powerful all-purpose finite elementanalysis solution used by the worlds most admired

manufacturers.

3)Adams, the market-leading motion-simulation software,simulates system level and loads.4)Patran, with its universal graphics user interface enables

finite element modeling, analysis and data integration,

analysis simulation and visualization capabilities.

5)Marc is ideal for the simulation of non-linear physicalbehavior of material conditions under extreme stress.

Siemens PLM

Siemens PLM provides leading software solutions that help

manufacturers turn more ideas into successful products.

1)Teamcenter powers innovation and improvesproductivity by connecting your team with the product

and process knowledge you need to make good decisions

throughout the product lifecycle. Teamcenters open PLM

foundation powers end-to-end lifecycle process excellence.

2)NX Synchronous technology from Siemens PLMSoftware can make your design process up to 100 times

faster. With this breakthrough, you no longer have to

choose between constraint-driven or history-free

modeling, and you can use data from multiple CAD

systems.

3)Solid Edge with synchronous technology is a completefeature-based 2D/3D CAD system that combines the

speed and flexibility of direct modelling with precise

control of dimension-driven design.


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Autodesk

Autodesk is a world-recognized leader in design, visualization

and documentation software products. Tata Technologies is an

Autodesk Premier Solutions Provider with the added

distinctions of being a Manufacturing Specialist and an

Authorized Training Center. Tata Technologies provides

clients with AutoCAD, AutoCAD Electrical, AutoCAD

Mechanical and Autodesk Inventor software and service-

based solutions.


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Fig. 1.7: Some clients of Tata Technologies


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Project in the Aerospace Department

Todays intelligent aerospace businesses understand

that the key to successful engineering and PLMcollaboration is finding the right strategic partner a

partner with credible experience in the aerospace

domain, integrated global design delivery centers,

innovative business processes, proven resource

scalability, and one who understands your unique

business requirements.

For over two decades Tata Technologies has been

providing the worlds foremost aerospace organizations

with complete design-to-build solutions and customized

answers to the most complex PLM challenges. Seeing

the problems better and delivering better solutions is

what we do well no matter what the challenge. Asglobal delivery and outsourcing become key strategies

for the aerospace industry, its approach helps

companies to achieve cost savings, time to market, and

gain competitive advantage. Tata Technologies brings

blended onshore and offshore delivery resources

combined with state-of-the-art design center facilities

and the largest concentration of PLM aerospace

technology expertise in the world. Its engineering

pedigree is its foundation which is better in every way


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for aircraft design. Andbetter for the customers at

35,000 feet.

Fig. 1.8: Aerospace Design Procedure


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Fig. 1.9: Airplane structure components


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Project on 1 dimensional heat transfer

Project Statement

Create a simulation tool/calculation tool that will allowus to simulate a floor fire test and the behavior of the

material in the test. See sketch below for test setup.

- The tool shall be able to calculate the temperature at

the top of the flooring material inside the vehicle floor

- The tool shall have the ability to simulate thetemperature rise at the top of the flooring material

when the under said floor is subjected to a temperature

curve per ASTM E-119. The tool has to be able to

calculate the "top of floor" temperature at least every

minute for at least 30 minutes.


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Fig. 2.1: Time-Temperature Curve


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Fig. 2.2: Values of source temperature and time


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CHAPTER II

1D analytical approach

In Simple Words

Fig. 2.3: Transient heat transfer through a series of blocks

The given problem is to be sampled for the base of an aero

plane which is linked to the engine and has a high heat

exposure. Since, exact experimental tests are not very feasible

for this problem, a theoretical study and a simulation model is

necessary for this very purpose. This model is then to be

simulated in reality and passed through required tests for the

aircraft to be considered risk-free for manufacturing purpose.


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If the heat at the top of the surface is found to exceed danger

limits, then the experimental simulation will not pass the

given test. In any case, a theoretical model for this 1

dimensional heat transfer is highly essential.

Explanation

In the given model, an unknown number of blocks (to be

decided by the user) are placed on top of one another (all of

different materials and properties). The sides of the blocks are

insulated so that heat transfer takes place only from bottom to

base in 1 direction. The bottom of the 1st block is heated on an

imaginary oven which has varying temperature according to

the plot shown in figure 2.1 (and also the values printed in

figure 2.2). The ambient temperature is 293 K.

Assumptions

Heat flow is considered to be transient in nature The interfaces of the blocks are assumed to be in perfect

thermal contact so that temperature at the interface is

equal for both the blocks and thus no convection takes

place at the interface

Rate of heat conduction at the interfaces is assumed to bethe same for both the blocks

Radiation and convection effects within the blocks areneglected (convection on the top surface is however

considered)


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The properties of the blocks like thermal conductivity,density etc. are assumed to be constant with temperature.

The heat transfer coefficient at the convection surface is

also assumed to be constant. The temperature at the bottom surface is approximated to

be a sixth order polynomial (varying with time).

The values of beta in the solution are taken only up to 10values as the effect of the latter values on the solution is

assumed to be negligible.

No heat generation in any of the blocks

Formulation of the differential equations

Boundary Conditions


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Initial Conditions

Solution

Background:

non homogeneity in equations

knowledge of Greens function transformation of non-homogenous boundary conditions to

homogenous boundary conditions by superposition

principle

separation of variables for solving partial differentialequations

In the given problem, we have a non - homogeneity at the

last boundary condition which we attempt to remove. We

do this by transforming the non homogenous boundary

conditions to homogenous boundary conditions although it


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will leave us with a non-homogenous set of differential

equations which can be solved through the greens

function.

The function will be obtained through the polyfit()function of MATLAB as a sixth order polynomial.

The functions (x) and (x) are described below.

Boundary Conditions

and


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Boundary Conditions

Solution of these set of ordinary differential equations is

pretty simple and can be easily obtained through matrixmanipulations in MATLAB.

For (x,t), the following equations hold:


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Boundary Conditions

Initial Conditions

( )

We first solve the homogenous system assuming gi(x,t)=0 .

From the method of sepration of variables,


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For a 2 block system

The eigen value problem for 2 slabs then yields

=0 where a=a1, b=a2These equations are for a 2 block system and can be extended

to multi block by extending the determinant. After finding the

values of for each iteration, values of will be known.Thus, []=[ ]will give the values of and .The solution for for a 2 block system would then be | | | |


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where

|

and The final solution is for any of the two blocks.

The code for the above solution has however been developed for

any number of slabs in MATLAB. The code developed is

printed inAppendix A.


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Input Data

Enter the number of slabs: 3

Enter material of plate 1: aluminium

Enter thickness of plate 1: 0.005

Enter material of plate 2: brass


Enter material of plate 3: stainless steel


Enter value of coefficient h for given problem: 10

Graph 2.1: Variation of source and interface 1 temperature with time

0

200

400

600

800

1000

1200

0 5 10 15 20 25 30 35

Temp

erature(K)

Time(minutes)

Source temperature

Interface 1


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Graph 2.2: Variation of source and interface 2 temperature with time

0

200

400

600

800

1000

1200

0 5 10 15 20 25 30 35

Temperature(K)

Time(minutes)

Source temperature

Interface 2


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Graph 2.3: Variation of source and all interface temperatures with time

The graphs obtained from the program have been plotted into

EXCEL and shown above. The graphs show the variation oftemperature versus time at the different interfaces. As we can

see, the trend followed is similar to the source curve but the

graphs are not easily distinguishable. The code takes about a

minute to show up the solution.

0

200

400

600

800

1000

1200

0 5 10 15 20 25 30 35

Tem

perature(K)

Time(minutes)

Source temperature

Interface 1

Interface 2

Interface 3


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Improvements

converting the program into GUI The temperatures obtained are not fully correct as they

exceed the source temperature at some locations

The time taken to solve this problem can certainly bereduced

increasing options for user to input values likesurrounding temperature etc.

creating separate functions for some part of the code forbetter reusability

validating the answer obtained through experimentaltechniques

making the problem more and more generalized byreducing number of assumptions

Conclusion

Although this model gives a fair idea of the trend followed by

the interface temperatures, it is not a very accurate or fast

model, and thus I have shifted to solving this problem by the

finite difference scheme covered in Chapter 3 which is really

fast as well as accurate.


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CHAPTER III

1D numerical approach

In Simple Words

Fig. 3.1: 1 dimensional heat transfer through a series of blocks

Explanation

In the given model, an unknown number of blocks (to bedecided by the user) are placed on top of one another (all of


insulated so that heat transfer takes place only from bottom to

base in 1 direction. The bottom of the 1st block is heated on an


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imaginary oven which has varying temperature according to

the plot shown in the figure 2.1 (and also the values printed in

figure 2.2).The ambient temperature is 293 K.

Assumptions




place at the interface


Radiation and convection effects within the blocks areneglected (convection on the top surface is however

considered)


The heat transfer coefficient at the convection surface isalso assumed to be constant.

No heat generation in any of the blocks The block is given 5 partitions and the time step is taken

to be 0.1 minute although these values can be changed

any time before the program run.


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Boundary Conditions

Initial Conditions


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Solution

Background

Fully implicit finite difference scheme of solving partialdifferential equations

Solving a matrix using Gauss-Seidel iterative methodsThe given set of equations is solved using the fully

implicit finite difference. The fully implicit scheme is

chosen due to its ease of implementation and because it isunconditionally convergent and stable, thereby

warranting any time step and number of divisions to be

chosen. In figure 3.1, each of the blocks is partitioned into

a given number of segments. The finite difference

equations would be applied at each of the segments. For

the position derivative, the central difference

approximation is used and for time derivative, thebackward difference approximation. The following

substitutions were made:


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If the number of slabs is taken to be 3 and the number of

nodes per slab to be 5 as shown in figure 3.1, then the

following system of equations will hold starting from n=0:

An imaginary node

is taken here to satisfy the

conditions. Similarly, other such nodes will be taken when

required to form the given matrix although the values

obtained for these variables will hold no meaning.

Continuing with the system of equations:


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The given system of equations has 17 variables and 17

equations to be solved which is then encoded into a matrix andsolved by MATLAB using the command A\b.

But in very high dimensional codes, the non-iterative codes

used by MATLAB may not succeed and thus an iterative

technique (the Gauss-Seidel method) has been worked out

which gives just the same solutions but takes a lot more time

inside the for loops.

In the Gauss-Seidel procedure, the above system of equations

would be written in the following manner:


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and so on till each node is explicitly written in terms of other

nodes. A set of values for all 17 nodes is then assumed and the

1st node is then calculated from the above equation. The

modified 1st node and the remaining nodes are then put into

the 2nd node equation. The procedure keeps repeating till the

consecutive values of the same node converge. For the Gauss-

Seidel method, this convergence mostly happens if the given

matrix is tri-diagonal or diagonally dominant. In this case,

even though the formed matrix is neither, it is very close to

being a tri-diagonal matrix and thus the solution does

converge.

After the 17 values for n=0 are obtained either by direct matrix

solutions or iterative procedures, the matrix is solved again forn=1 and so on till all required values for all times are obtained.

The code for the above solution has however been developed for

any number of slabs and partitions in MATLAB. The code

developed is printed inAppendix B.


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Input Data








Enter value of coefficent h for given problem: 10

Without Gauss-Seidel iteration

Graph 3.1: Variation of source and interface temperature with time by

direct solving of matrix

0

200

400

600

800

1000

1200

0 5 10 15 20 25 30 35

Temperature(K)

Time(minutes)

Source temperature

Interface 1

Interface 2

Interface 3


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With Gauss-Seidel iteration

Graph 3.2: Variation of source and interface temperature with time

(Gauss-Seidel iteration)

0

200

400

600

800

1000

1200

0 5 10 15 20 25 30 35

Temperature(K)

Time(minutes)

Source temperature

Interface 1 gs

Interface 2 gs

Interface 3 gs


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Comparison (with and without Gauss-Seidel)

Graph 3.3: Variation of source and interface temperature with time

(both direct by direct matrix and Gauss-Seidel iteration)

As we can see, both the non-iterative and the iterative

programs yield exactly the same results with the respective

graphs in both the methods overlapping. The iterative

procedure however takes up slightly more time and memoryspace although it guarantees solution for very large matrices

too.

0

200

400

600

800

1000

1200

0 5 10 15 20 25 30 35

Temperature(K)

Time(minutes)

Source temperature

Interface 1 gs

Interface 2 gs

Interface 3 gs

Interface 1

Interface 2

Interface 3


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Convergence of Gauss-Seidel procedure

Graph 3.4: Convergence of temperature at particular time by Gauss-

Seidel iterations

The above graph is shown for the convergence of temperatures

of the top-most surface of all the slabs at different time

intervals for the Gauss-Seidel method. This confirms that eventhough the formed matrix in this case does not rigorously

satisfy convergence criteria, but the implemented code does

happen to converge.

0

200

400

600

800

1000

1200

0 10 20 30 40 50 60 70 80 90

Temperatu

re(Kelvin)

Iterations

5min

10min

15min

20min

25min


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Improvements

converting the program into GUI increasing options for user to input values like

surrounding temperature etc.

creating separate functions for some part of the code forbetter reusability and optimizing code to work faster



Conclusion

The fully implicit finite difference method, both by with and

without Gauss-Seidel iteration, gives a seemingly good and

accurate solution which can now be verified by experimental

techniques. Now that a seemingly right methodology has beenaccomplished, the scope of expanding the project to higher

dimensions (2D has been included next) or including radiation

conditions always remains.


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CHAPTER IV

2D numerical approach

In Simple Words

Fig. 4.1: 2 dimensional heat transfer through a series of blocks

Explanation

In the given model, an unknown number of blocks (to be

decided by the user) are placed on top of one another (all of


now open to atmosphere so that heat transfer takes place in 2

directions, from bottom to top and through the sides. The

bottom of the 1st block is heated on an imaginary oven which


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has varying temperature according to the plot shown in figure

2.1 (and also the values printed in figure 2.2). The ambient

temperature is 293 K.

Assumptions




place at the interface.


Radiation and convection effects within the blocks areneglected (convection on the top surface and the sides is

however considered)


The heat transfer coefficient at the convection surface isalso assumed to be constant.

No heat generation in any of the blocks There are two values of interface temperatures found at

each time step which differ by a slight proportion. Thus,

the first has been taken as the valid temperature in such

a scenario.


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Boundary Conditions


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Initial Conditions

Solution

Background

The alternating direction implicit (ADI) finite differencescheme of solving partial differential equations which is

well suited for a 2D heat transfer problem

The above set of equations is solved using a slight variant

of the fully implicit FDM scheme called the alternating

direction implicit method. A usual implementation of a

fully implicit FDM in this case would result in a near

penta-diagonal matrix which would have to be coded all

over again. Instead, the ADI scheme breaks a time step

into two halves. In the first half, the x-derivative is

written as an implicit central-difference approximation

and the y-derivative as an explicit central-difference

approximation. The reverse holds true for the second half

of the time step. Each of these halves result in matrices


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very similar to tri-diagonal matrix, the results of which

are then combined to obtain the final solution.

Since the resulting matrices are close to tri-diagonal, thecode from Chapter III could be re-used and modified to

give the solution. This scheme also has the advantage of

being unconditionally stable just as the fully implicit

methods. The Gauss-Seidel iterative procedure for the

solving of the matrices has not been implemented here as

it has already been taken up in the previous chapter and

the implementation would exactly be the same.

This time the blocks were divided into segments in both x

and y directions such that it resulted in creation of point

nodes. At each of these nodes, the equations and

boundary conditions stated above are valid. The following

substitutions were made:

For (n+)th time step


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For (n+1)th time step

If the number of slabs is taken to be 3 and the number of

segments per slab to be 5 in both x and y direction as

shown in figure 4.1, then the following system ofequations will hold starting from n=0:

( )

(

)

( ) ( ) And so on for the remaining slabs.


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From the equations obtained a 1717 matrix is formed and

solved for corresponding temperatures. Once the temperatures

for j=1 have been obtained, we move on to j=2.

( ) ( ) ( ) ( )

And so on for the remaining slabs.

Thus, we solve the matrix for 1 j 5, and finally, we would

have solved the complete node system at n= .After having all the values, we move to the next half of the

solution method for finding values at n=1. The following

equations hold:

( ) ( )

( )

( ) ( )


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and so on.

Similarly, here we form the matrix at the ith value and keep

solving for each i until we have the temperatures for each node

in the mesh at n=1.

We then move on to solving for n=2 by the same procedure

described above. The difference from the previous chapters is

that the temperatures at each node here can be different even

varying along the y direction although only slightly.

The code for the above solution has been developed for any

number of slabs and partitions in MATLAB. The code

developed is printed inAppendix C.


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Input Data








Enter equal width of all plates: 0.01

Enter value of coefficent h for given problem: 10

Graph 4.1: Variation of node 1 temperatures of all interfaces with time

0

200

400

600

800

1000

1200

0 5 10 15 20 25 30 35

Tem

perature(K)

node

Source temperature

Interface 1 node 1

Interface 2 node 1

Interface 3 node 1


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The graph above has been shown for the corner nodes of all the

interfaces and follows similar trends as discussed in the

previous chapters.

Graph 4.2: Variation of temperature of all nodes at top surface withtime

The graph above shows temperature curves for all nodes of the

top surface. As we see, the temperature variation along the y

0

200

400

600

800

1000

1200

0 5 10 15 20 25 30 35

Temperature(K)

Time(minutes)

Top Surface

Source temperature

node 1

node 2

node 3

node 4

node 5


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direction is not much and the temperature curves for all these

nodes overlap.

Graph 4.3: Temperature of all nodes at 30 min at top surface

This graph shows the temperature variation along the top

surface at time t=30 min. As expected, the temperature is

symmetric with respect to the middle node with heat flowing

from both sides of the middle node to the surroundings at

ambient temperature.

1092.2

1092.4

1092.6

1092.8

1093

1093.2

1093.4

1093.6

1093.8

0 1 2 3 4 5 6

Temperature(K)

node

Top Surface

30 min


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Improvements

converting the program into GUI increasing options for user to input values like

surrounding temperature etc.

creating separate functions for some part of the code forbetter reusability and optimizing code to work faster



can be extended to the 3 dimensional case

Conclusion

The ADI finite difference technique works for the 2

dimensional case and gives good results which can now be

validated using experimental techniques.


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Appendix A

% This program executes the given project statement using methods mentioned

in Chapter II. It gives the output as the graphs of all interfaces and source

temperature varying with time

% The symbolic toolbox is must for the running of this program% Please refer to the theory mentioned for smooth understanding of thisprogram% The following user inputs are required:% 1) material of slabs (only following materials are allowed:% (i)brass (ii)alunminium (iii) stainless steel (iv) wood% 2) thickness of slabs (the user is requested to keep thickness small < 0.02

meters)% 3)heat transfer convection coefficient% "All values must be entered in SI Units"clcclearA={'*' 'k' 'rho' 'c' ; 'water' 0.617 996 4178 ; 'brass' 110 8530 380 ;

'stainless steel' 14.9 7900 477 ; 'soil' 0.4 3333.3 800 ; 'wood' 1.26 700

1700 ; 'aluminium' 237 2700 897}; %declares matrix of available materialswith physical propertiesA{1,5}='alpha';

for i=2:7A {i,5}=A{i,2}/(A{i,3}*A{i,4}); % calculates value of alpha for all

materialsendn=input('Enter the number of slabs: ');

mat=cell(1,30);L=zeros(1,30);k=zeros(1,30);rho=zeros(1,30);c=zeros(1,30);

alpha=zeros(1,30);for i=1:nfprintf('Enter material of plate %d',i);mat{i}=input(': ','s');if(i==1)

fprintf('Enter thickness of plate %d',i);

L(i)=input(': ');else

fprintf('Enter thickness of plate %d',i);L(i)=L(i-1)+input(': ');

endendh=input('Enter value of coefficent h for given problem: ');for j=1:n

confirm=0;for i=2:size(A,1)%disp('1');if (strcmp(A{i,1},mat{j})==1) %loop compares values entered with

values stored in matrix and assigns properties accordingly%disp('1');confirm=1;k(j)=A{i,2};rho(j)=A{i,3};


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c(j)=A{i,4};alpha(j)=A{i,5};

endif(confirm==1)

break;end

endendphi=zeros(2*n-1,2*n-1);phi(1,1)=L(1);phi(1,2)=L(1)*-1;phi(1,3)=-1;for i=2:n-1

phi(i,2*i-2)=L(i);phi(i,2*i-1)=1; % creates left hand matrix for solution of phi(x)phi(i,2*i)=-1*L(i);phi(i,2*i+1)=-1;

endphi(n,1)=k(1);phi(n,2)=k(2)*-1;for i=n+1:2*n-2

phi(i,2*i-2*n)=k(i-n+1);phi(i,2*i-2*n+2)=-1*k(i-n+2);

endphi(2*n-1,2*n-2)=k(n)+h*L(n);phi(2*n-1,2*n-1)=h;ans1=zeros(2*n-1,1); % creates right hand matrix for solution of phi(x)ans1(1,1)=-1;coeffphi=phi\ans1; %coeffiecients obtainedsyms x;fphi=zeros(n,1);

fphi=sym(fphi);fphi(1,1)=coeffphi(1,1).*x+1;for i=2:n

fphi(i,1)=coeffphi(2*i-2,1)*x+coeffphi(2*i-1,1); %creates function(x)

using coefficientsendzi=phi;ans2=zeros(2*n-1,1);ans2(2*n-1,1)=h;coeffzi=zi\ans2; % coefficients for zhi(x) obtainedfzi=zeros(n,1);fzi=sym(fzi);fzi(1,1)=coeffzi(1,1).*x;for i=2:n

fzi(i,1)=coeffzi(2*i-2,1)*x+coeffzi(2*i-1,1); %creates function(x)using coefficientsendF=zeros(n,1);F=sym(F);for i=1:n

F(i,1)=293*(1-fphi(i,1)-fzi(i,1)); %inital condition for transformed

solutionend


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syms x1;betamat=zeros(2*n-1,2*n-1);betamat=sym(betamat);betamat(1,1)=-1*sin((x1*L(1))/sqrt(alpha(1)));betamat(1,2)=sin((x1*L(1))/sqrt(alpha(2)));

betamat(1,3)=cos((x1*L(1))/sqrt(alpha(2)));betamat(n,1)=-

((k(1)/k(2))*sqrt(alpha(2)/alpha(1))*cos((x1*L(1))/sqrt(alpha(1))));betamat(n,2)=cos((x1*L(1))/sqrt(alpha(2)));betamat(n,3)=-(sin((x1*L(1))/sqrt(alpha(2))));for i=2:n-1

betamat(i,2*i-2)=sin((x1*L(i))/sqrt(alpha(i)));betamat(i,2*i-1)=cos((x1*L(i))/sqrt(alpha(i)));betamat(i,2*i)=-(sin((x1*L(i))/sqrt(alpha(i+1))));betamat(i,2*i+1)=-(cos((x1*L(i))/sqrt(alpha(i+1))));

endfor i=n+1:2*n-2

betamat(i,2*i-2*n)=(k(i-n+1)/k(i-n+2))*sqrt(alpha(i-n+2)/alpha(i-

n+1))*cos((x1*L(i-n+1))/sqrt(alpha(i-n+1)));betamat(i,2*i-2*n+1)=-((k(i-n+1)/k(i-n+2))*sqrt(alpha(i-n+2)/alpha(i-

n+1))*sin((x1*L(i-n+1))/sqrt(alpha(i-n+1))));betamat(i,2*i-2*n+2)=-(cos((x1*L(i-n+1))/sqrt(alpha(i-n+2))));betamat(i,2*i-2*n+3)=sin((x1*L(i-n+1))/sqrt(alpha(i-n+2)));

endbetamat(2*n-1,2*n-

2)=(((h*sqrt(alpha(n)))/(x1*k(n)))*sin((x1*L(n))/sqrt(alpha(n))))+cos((x1*L(n

))/sqrt(alpha(n)));betamat(2*n-1,2*n-

1)=(((h*sqrt(alpha(n)))/(x1*k(n)))*cos((x1*L(n))/sqrt(alpha(n))))-

sin((x1*L(n))/sqrt(alpha(n)));

y=det(betamat); %calculates determinant of matrix formed for getting values

of betay=simplify(y);y=matlabFunction(y);syms t;%a=[0 5 10 15 20 25 30];%a=[0 5 10 15 20 25 30 35 40 45 50 55 60 ];a=[0 300 600 900 1200 1500 1800 2100 2400 2700 3000 3300 3600 ];%b=[293 811 977 1033 1068 1094 1116];b=[293 811 977 1033 1068 1094 1116 1135 1151 1165 1178 1189 1200 ];f1temp=polyfit(a,b,6);%f1temp=polyfit(a,b,15);f1=f1temp(1).*(t).^6+f1temp(2).*(t).^5+f1temp(3).*(t).^4+f1temp(4).*(t).^3+f1

temp(5).*(t).^2+f1temp(6).*(t)+f1temp(7); %sixth-order polynomial

approximation, any of the other functions commented can also be usedf1(t)=f1;%f1=f1temp(1).*t.^6+f1temp(2).*t.^5+f1temp(3).*t.^4+f1temp(4).*t.^3+f1temp(5)

.*t.^2+f1temp(6).*t+f1temp(7);%put simulating time after iteration%f1=(1.7267*t)+293;%t/60 is 1.7267%f1=990.1098*exp((0.0027948556*t)/60) - 728.6*exp((-0.23*t)/60);%f1=990.10977*exp(0.0040060346*(t/60)) - 697.22292*exp(-0.25160733*(t/60));%f1=713.09*cos(0.16482*(t/60)) - 34.844*cos(0.2747*(t/60)) -

13.599*sin(0.2747*(t/60)) - 268.81*sin(0.16482*(t/60)) -

1128.9*cos(0.05494*(t/60)) + 1100.0*cos(0.10988*(t/60)) +

2.1099*cos(0.21976*(t/60)) + 2311.0*sin(0.05494*(t/60)) +

1211.9*sin(0.10988*(t/60)) - 235.36*sin(0.21976*(t/60)) - 358.39;


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%f1=(0.0001513*t^5 - 0.02684*t^4 + 1.787*t^3 - 53.74*t^2 + 1965*t + 2613)

/(t + 8.92);%f1=2163*sin(0.06295*(t/60)-0.6615) + 1875*sin(0.1206*(t/60)+0.5249) +

1207*sin(0.1647*(t/60)+2.177) + 361.1*sin(0.1984*(t/60)+4.188);%f1= 2.501*10^-5*(t/60)^5 - 0.004423*(t/60)^4 + 0.2938*(t/60)^3 -

9.088*(t/60)^2 + 134*(t/60) + 304;f1der=diff(f1);g=zeros(n,1);g=sym(g);for i=1:n

g(i,1)=(fphi(i,1)*f1der); %function g(x,t) is calculated for all slabsendsyms xdtou;g=subs(g,x,xd);g=subs(g,t,tou);%Theta(x,t)=sym('Theta(x,t)');Theta=0;green(x,xd,t,tou)=sym('green(x,xd,t,tou)');%integral1(x,t,tou)=sym('integral1(x,t,tou)');%integralgen1(x,t,tou)=sym('integralgen1(x,t,tou)');

for z=1:ngreen=0*x;p=1;i=1;beta2=0;

while(p~=11)beta=fzero(y,0.1*i); %value of beta around 0.1*i is obtained by solving

y=0if(beta


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rhsmat=zeros(2*n-2,1);rhsmat(1,1)=sin((beta*L(1))/sqrt(alpha(1)));

rhsmat(n,1)=(k(1)/k(2))*sqrt(alpha(2)/alpha(1))*cos((beta*L(1))/sqrt(alpha(1)

));solmat=coeffmat\rhsmat; %coefficients A and B for different zhi are

obtainedzhimat=zeros(n,1);zhimat=sym(zhimat);zhimat(1,1)=sin((x*beta)/sqrt(alpha(1)));for j=2:n

zhimat(j,1)=solmat(2*j-3,1)*sin((x*beta)/sqrt(alpha(j)))+solmat(2*j-

2,1)*cos((x*beta)/sqrt(alpha(j))); %zhi(x) is formed, not to be confused with

previous zhi(x) , this is part of theta(x) solutionendNn=(k(1)/alpha(1))*int(zhimat(1,1).^2,0,L(1)); %Nn evaluatedfor j=2:n

Nn=Nn+(k(j)/alpha(j))*int(zhimat(j,1).^2,L(j-1),L(j));end

green=green+((1/Nn)*(k(z)/alpha(z))*zhimat(n,1)*exp(-(beta.^2*(t-

tou)))*subs(zhimat(z,1),x,xd)); %green's function evaluatedgreen=vpa(green,6);i=i+1;p=p+1;

end%green=simplify(green);if(z==1)

integral1(x,t,tou)=int((green*subs(F(1,1),x,xd)),xd,0,L(1));integral1=integral1(x,t,0);integral1=vpa(integral1,6);integral2=int(int(green*g(1,1),xd,0,L(1)),tou,0,t);integral2=vpa(integral2,6);Theta=Theta+integral1+integral2;

Theta=vpa(Theta,6);Theta=simplify(Theta);Theta=vpa(Theta,6);

elseintegralgen1(x,t,tou)=int((green*subs(F(z,1),x,xd)),xd,L(z-1),L(z));integralgen1=integralgen1(x,t,0);integralgen1=vpa(integralgen1,6);integralgen2=int(int(green*g(z,1),xd,L(z-1),L(z)),tou,0,t);integralgen2=vpa(integralgen2,6);Theta=Theta+integralgen1+integralgen2; %final calculation of Theta(x,t)Theta=vpa(Theta,6);Theta=simplify(Theta);Theta=vpa(Theta,6);

% Theta=simplify(Theta);%input for calcualting temp. bet. a and b

endendTfinal(x,t)=Theta+fphi(n,1)*f1+fzi(n,1)*293; %The actual temperature

varying with time is finally obtained

Tfinal=vpa(Tfinal,6);%Tfinal=simplify(Tfinal);disp('Equation of temperature on surface is ');%pretty(Tfinal(L(n),t));%why is it giving 2 answers but it is 1*1 matrixezplot(f1(t*60),[0,30]);hold all;


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Tfinalatsurf=zeros(n,1);Tfinalatsurf=sym(Tfinalatsurf);for i=1:nTfinalatsurf(i)=Tfinal(L(i),t); %recording temperature at interfacesendTfinalatsurf=vpa(Tfinalatsurf,6);Tfinalatsurf(t)=Tfinalatsurf;Tfinalatsurf=Tfinalatsurf(t*60);Tfinalatsurf=simplify(Tfinalatsurf);Tfinalatsurf=vpa(Tfinalatsurf,6);%Tfinalatsurf(t)=Tfinalatsurf;%disp(Tfinalatsurf);for i=1:nezplot(Tfinalatsurf(i),[0,30]);end% This program has certain pitfalls as it does not give exact variation of

temperature of interfaces with time but merely a correct trend.% Some other improvements include:% 1) optimization of code for faster running

% 2) Ambient temperature can be input by user

% 3) User can be asked if he/she wants the trend of temperature with time fora line within the block instead of interface in which case only values

varying between the given lengths will be accepted


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Appendix B

Matrix solved directly

% This program executes the given project statement using methods mentionedin Chapter III. It gives the output as the graphs of all interfaces and

source temperature varying with time% Please refer to the theory mentioned for smooth understanding of this

program% The following user inputs are required:% 1) material of slabs (only following materials are allowed:% (i)brass (ii)alunminium (iii) stainless steel (iv) wood% 2) thickness of slabs (the user is requested to keep thickness small < 0.02

metres)% 3) heat transfer convection coefficient% "All values must be entered in SI Units"clc

clearA={'*' 'k' 'rho' 'c' ; 'water' 0.617 996 4178 ; 'brass' 110 8530 380 ;'stainless steel' 14.9 7900 477 ; 'soil' 0.4 3333.3 800 ; 'wood' 1.26 700

1700 ; 'aluminium' 237 2700 897};A{1,5}='alpha';for i=2:7

A {i,5}=A{i,2}/(A{i,3}*A{i,4}); % calculates value of alpha for all

materialsendn=input('Enter the number of slabs: ');mat=cell(1,n);L=zeros(1,n);k=zeros(1,n);rho=zeros(1,n);

c=zeros(1,n);alpha=zeros(1,n);for i=1:n

fprintf('Enter material of plate %d',i);mat{i}=input(': ','s');%if(i==1)

fprintf('Enter thickness of plate %d',i);L(i)=input(': ');

% else% fprintf('Enter thickness of plate %d',i);% L(i)=input(': ');

% endend

h=input('Enter value of coefficent h for given problem: ');for j=1:nfor i=2:size(A,1)%disp('1');if (strcmp(A{i,1},mat{j})==1) %loop compares values entered with values

stored in matrix and assigns properties accordingly%disp('1');k(j)=A{i,2};rho(j)=A{i,3};c(j)=A{i,4};


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alpha(j)=A{i,5};

endend

enddeltax=zeros(1,n);deltat=0.1;r=zeros(1,n);N=5;a=[0 5 10 15 20 25 30];b=[293 811 977 1033 1068 1094 1116];for i=1:n

deltax(i)=L(i)/(N-1);

r(i)=(alpha(i)*deltat)/(deltax(i)^2);endfdmmat=zeros(n*N+n-1,n*N+n-1);fdmmat(1,1)=1+2*r(1);fdmmat(1,2)=-r(1);for i=2:(N-1)

fdmmat(i,(i-1))=-r(1);fdmmat(i,i)=1+2*r(1);

fdmmat(i,i+1)=-r(1);end%fdmmat(N,N-1)=1+2*r(2);%fdmmat(N,N+1)=-r(2);%fdmmat(N,N+2)=-r(2);j=1;%for i=N+1:2*N-1% fdmmat(i,i-2)=-r(2);% fdmmat(i,i+1)=1+2*r(2);% fdmmat(i,i+2)=-r(2);

%endfor i=N:n*N-1

if(mod(i,N)==0)j=j+1;fdmmat(i,i+j-3)=1+2*r(j);fdmmat(i,i+j-1)=-r(j);fdmmat( i,i+j)=-r(j);

elseif(mod(i,N)==1)fdmmat(i,i+j-4)=-r(j); %creating left hand side matrixfdmmat(i,i+j-1)=1+2*r(j);fdmmat(i,i+j)=-r(j);else

fdmmat(i,i+j-2)=-r(j);fdmmat(i,i+j-1)=1+2*r(j);fdmmat(i,i+j)=-r(j);

endend

endj=N-4;p=1;for i=n*N:n*N+n-2

fdmmat(i,j+2)=-k(p)/deltax(p);fdmmat(i,j+4)=k(p)/deltax(p);fdmmat(i,j+5)=k(p+1)/deltax(p+1);fdmmat(i,j+6)=-k(p+1)/deltax(p+1);j=j+N+1;p=p+1;


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endfdmmat(n*N+n-1,n*N+n-3)=-k(n)/(2*deltax(n)*h);fdmmat(n*N+n-1,n*N+n-2)=1;fdmmat(n*N+n-1,n*N+n-1)=k(n)/(2*deltax(n)*h);rhsmat=zeros(n*N+n-1,1);rhsmat(1,1)=293+r(1)*((deltat/5)*(b(2)-b(1))+b(1));for i=2:n*N-1

rhsmat(i,1)=293; %creating right hand side matrixendrhsmat(n*N+n-1,1)=293;solmat=fdmmat\rhsmat; %first solution matrixsurfrec=zeros(301,n);for i=1:n

surfrec(1,i)=293;end%surfrec(2,2)=solmat(4.1);i=N-1;j=1;while(i=0 && q*deltat=5 && q*deltat=10 && q*deltat=15 && q*deltat=20 && q*deltat=25 && q*deltat


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solmat=fdmmat\rhsmat; %these are the temperatures

obtained at given iteration(time)%surfrec(q+1)=solmat(n*N+n-2,1);%surfrec(q+1,1)=solmat(1,1);c=N-1;d=1;while(c


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alpha=zeros(1,n);for i=1:n


fprintf('Enter thickness of plate %d',i);L(i)=input(': ');

% else% fprintf('Enter thickness of plate %d',i);% L(i)=input(': ');

% endendh=input('Enter value of coefficent h for given problem: ');% mat{1,1}='aluminium';% mat{1,2}='aluminium';% mat{1,3}='aluminium';% L(1)=0.005;L(2)=0.005;L(3)=0.005;% h=10;for j=1:n

for i=2:size(A,1)

%disp('1');if (strcmp(A{i,1},mat{j})==1) %loop compares values entered with values

stored in matrix and assigns properties accordingly%disp('1');k(j)=A{i,2};rho(j)=A{i,3};c(j)=A{i,4};alpha(j)=A{i,5};

endend

enddeltax=zeros(1,n);deltat=0.1;r=zeros(1,n);N=5;a=[0 5 10 15 20 25 30];b=[293 811 977 1033 1068 1094 1116];for i=1:n

deltax(i)=L(i)/(N-1);r(i)=(alpha(i)*deltat)/(deltax(i)^2);

endfdmmat=zeros(n*N+n-1,n*N+n-1);fdmmat(1,1)=1+2*r(1);fdmmat(1,2)=-r(1);for i=2:(N-1)

fdmmat(i,(i-1))=-r(1);fdmmat(i,i)=1+2*r(1);

fdmmat(i,i+1)=-r(1);end%fdmmat(N,N-1)=1+2*r(2);%fdmmat(N,N+1)=-r(2);%fdmmat(N,N+2)=-r(2);j=1;%for i=N+1:2*N-1% fdmmat(i,i-2)=-r(2);% fdmmat(i,i+1)=1+2*r(2);% fdmmat(i,i+2)=-r(2);


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%endfor i=N:n*N-1

if(mod(i,N)==0)j=j+1;fdmmat(i,i+j-3)=1+2*r(j);fdmmat(i,i+j-1)=-r(j); %creating left hand side matrixfdmmat( i,i+j)=-r(j);

elseif(mod(i,N)==1)fdmmat(i,i+j-4)=-r(j);fdmmat(i,i+j-1)=1+2*r(j);fdmmat(i,i+j)=-r(j);else

fdmmat(i,i+j-2)=-r(j);fdmmat(i,i+j-1)=1+2*r(j);fdmmat(i,i+j)=-r(j);

endend

endj=N-4;p=1;

for i=n*N:n*N+n-2fdmmat(i,j+2)=-k(p)/deltax(p);fdmmat(i,j+4)=k(p)/deltax(p);fdmmat(i,j+5)=k(p+1)/deltax(p+1);fdmmat(i,j+6)=-k(p+1)/deltax(p+1);j=j+N+1;p=p+1;

endfdmmat(n*N+n-1,n*N+n-3)=-k(n)/(2*deltax(n)*h);fdmmat(n*N+n-1,n*N+n-2)=1;fdmmat(n*N+n-1,n*N+n-1)=k(n)/(2*deltax(n)*h);rhsmat=zeros(n*N+n-1,1);rhsmat(1,1)=293+r(1)*((deltat/5)*(b(2)-b(1))+b(1));for i=2:n*N-1

rhsmat(i,1)=293; %creating right hand side matrixendrhsmat(n*N+n-1,1)=293;solmat=zeros(n*N+n-1,1);for i=1:n*N+n-1

solmat(i,1)=293;endlarge=1;

while(large>=0.0001)

temp=solmat;k=1;for i=1:n*N-1

sum=rhsmat(i,1);change=1;for j=1:n*N+n-1if mod(i,N)~=0

if fdmmat(i,j)~=1+2*r(k)sum=sum-fdmmat(i,j)*solmat(j,1); %getting first solution

endelseif mod(i,N)==0


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if j==1k=k+1;endif fdmmat(i,j)==-r(k)

if change==1change=0;

elseif change==0sum=sum-fdmmat(i,j)*solmat(j,1);end

endelseif fdmmat(i,j)~=-r(k)

sum=sum-fdmmat(i,j)*solmat(j,1);end

endend

endendfor j=1:n*N+n-1

if mod(i,N)~=0if fdmmat(i,j)==1+2*r(k)

sum=sum/fdmmat(i,j);end

elseif fdmmat(i,j)==-r(k)

sum=sum/fdmmat(i,j);break;

endend

endsolmat(i+k-1,1)=sum;

endsum=0;j=N-4;for i=n*N:n*N+n-2

sum=-

(fdmmat(i,j+2)*solmat(j+2,1)+fdmmat(i,j+5)*solmat(j+5,1)+fdmmat(i,j+6)*solmat

(j+6,1));solmat((i-n*N)*(N+1)+N)=sum/fdmmat(i,j+4);j=j+N+1;

endsolmat(n*N+n-1,1)=(rhsmat(n*N+n-1,1)-fdmmat(n*N+n-1,n*N+n-2)*solmat(n*N+n-2)-

fdmmat(n*N+n-1,n*N+n-3)*solmat(n*N+n-3,1))/fdmmat(n*N+n-1,n*N+n-1);

temp=solmat-temp;

large=max(temp);endsurfrec=zeros(301,n);

for i=1:nsurfrec(1,i)=293;

end%surfrec(2,2)=solmat(4.1);i=N-1;j=1;while(i


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end

for q=2:300

iter=0;j=2;

if q*deltat>=0 && q*deltat=5 && q*deltat=10 && q*deltat=15 && q*deltat=20 && q*deltat=25 && q*deltat=0.0001)iter=iter+1;

temp=solmat;k=1;for i=1:n*N-1

sum=rhsmat(i,1);change=1;for j=1:n*N+n-1if mod(i,N)~=0

if fdmmat(i,j)~=1+2*r(k)sum=sum-fdmmat(i,j)*solmat(j,1);

endelseif mod(i,N)==0

if j==1k=k+1;


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endif fdmmat(i,j)==-r(k)

if change==1change=0;

elseif change==0sum=sum-fdmmat(i,j)*solmat(j,1);end

endelseif fdmmat(i,j)~=-r(k)

sum=sum-fdmmat(i,j)*solmat(j,1);end

endend

endendfor j=1:n*N+n-1

if mod(i,N)~=0if fdmmat(i,j)==1+2*r(k)

sum=sum/fdmmat(i,j);end

elseif fdmmat(i,j)==-r(k)

sum=sum/fdmmat(i,j);break;

endend

endsolmat(i+k-1,1)=sum;

endsum=0;j=N-4;for i=n*N:n*N+n-2

sum=-

(fdmmat(i,j+2)*solmat(j+2,1)+fdmmat(i,j+5)*solmat(j+5,1)+fdmmat(i,j+6)*solmat

(j+6,1));solmat((i-n*N)*(N+1)+N)=sum/fdmmat(i,j+4);j=j+N+1;

endsolmat(n*N+n-1,1)=(rhsmat(n*N+n-1,1)-fdmmat(n*N+n-1,n*N+n-2)*solmat(n*N+n-2)-

fdmmat(n*N+n-1,n*N+n-3)*solmat(n*N+n-3,1))/fdmmat(n*N+n-1,n*N+n-1);if q==50

iter2=1;result(iter,iter2)=solmat(n*N+n-2,1);

elseif q==100iter2=2;result(iter,iter2)=solmat(n*N+n-2,1);

elseif q==150

iter2=3;result(iter,iter2)=solmat(n*N+n-2,1);



elseif q==250iter2=6;


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result(iter,iter2)=solmat(n*N+n-2,1);

endtemp=solmat-temp;

large=max(temp);end%surfrec(q+1)=solmat(n*N+n-2,1);%surfrec(q+1,1)=solmat(1,1);c=N-1;d=1;while(c


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Appendix C

% This program executes the given project statement using methods mentioned

in Chapter IV. It gives the output as the graphs of node 1 of all interfaces

and source temperature varying with time

% Please refer to the theory mentioned for smooth understanding of thisprogram% The following user inputs are required:% 1) material of slabs (only following materials are allowed:% (i)brass (ii)alunminium (iii) stainless steel

(iv)wood)% 2) thickness of slabs (the user is requested to keep thickness small < 0.02

metres)% 3) width of slabs (all slabs must be equal width)% 3) heat transfer convection coefficient% "All values must be entered in SI Units"

%"Do not take very large number of partitionsclcclear

A={'*' 'k' 'rho' 'c' ; 'water' 0.617 996 4178 ; 'brass' 110 8530 380 ;'stainless steel' 14.9 7900 477 ; 'soil' 0.4 3333.3 800 ; 'wood' 1.26 700

1700 ; 'aluminium' 237 2700 897};A{1,5}='alpha';for i=2:7

A {i,5}=A{i,2}/(A{i,3}*A{i,4});endn=input('Enter the number of slabs: ');%can ask user for outside temperature%n=3;mat=cell(1,n);L=zeros(1,n);k=zeros(1,n);rho=zeros(1,n);

c=zeros(1,n);alpha=zeros(1,n);for i=1:n


fprintf('Enter thickness of plate %d',i);%you have to introduce vpa

at many places. Check to see results for 3 or more slab right or not then use

vpa% L(i)=input(': ');

% else% fprintf('Enter thickness of plate %d',i);L(i)=input(': ');

% endendwidth=input('Enter equal width of all plates: ');h=input('Enter value of coefficent h for given problem: ');% mat{1,1}='aluminium';% mat{1,2}='brass';% mat{1,3}='stainless steel';% L(1)=0.005;L(2)=0.006;L(3)=0.007;L(4)=0.008;% h=10;% width=0.01;


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for j=1:nfor i=2:size(A,1)%disp('1');if (strcmp(A{i,1},mat{j})==1)

%disp('1');k(j)=A{i,2};rho(j)=A{i,3};c(j)=A{i,4};alpha(j)=A{i,5};

endend

endN=5;deltax=zeros(1,n);deltay=width/(N-1);deltat=0.1;rx=zeros(1,n);ry=zeros(1,n);a=[0 5 10 15 20 25 30];b=[293 811 977 1033 1068 1094 1116];

for i=1:ndeltax(i)=L(i)/(N-1);rx(i)=(alpha(i)*deltat)/(2*deltax(i)^2);ry(i)=(alpha(i)*deltat)/(2*deltay^2);

endsolmat=zeros(n*N+n-1,N);fdmmat=zeros(n*N+n-1,n*N+n-1);rhsmat=zeros(n*N+n-1,1);for w=1:Nfdmmat(1,1)=1+2*rx(1);fdmmat(1,2)=-rx(1);for i=2:(N-1)

fdmmat(i,(i-1))=-rx(1);fdmmat(i,i)=1+2*rx(1);fdmmat(i,i+1)=-rx(1);

endj=1;for i=N:n*N-1

if(mod(i,N)==0)j=j+1;fdmmat(i,i+j-3)=1+2*rx(j);fdmmat(i,i+j-1)=-rx(j);fdmmat( i,i+j)=-rx(j);

elseif(mod(i,N)==1)fdmmat(i,i+j-4)=-rx(j);fdmmat(i,i+j-1)=1+2*rx(j);fdmmat(i,i+j)=-rx(j);

elsefdmmat(i,i+j-2)=-rx(j);fdmmat(i,i+j-1)=1+2*rx(j);fdmmat(i,i+j)=-rx(j);

endend

endj=N-4;p=1;for i=n*N:n*N+n-2


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fdmmat(i,j+2)=-k(p)/deltax(p);fdmmat(i,j+4)=k(p)/deltax(p);fdmmat(i,j+5)=k(p+1)/deltax(p+1);fdmmat(i,j+6)=-k(p+1)/deltax(p+1);j=j+N+1;p=p+1;

endfdmmat(n*N+n-1,n*N+n-3)=-k(n)/(2*deltax(n)*h);fdmmat(n*N+n-1,n*N+n-2)=1;fdmmat(n*N+n-1,n*N+n-1)=k(n)/(2*deltax(n)*h);rhsmat(1,1)=293+rx(1)*((deltat/10)*(b(2)-b(1))+b(1));for i=2:n*N-1

rhsmat(i,1)=293;endrhsmat(n*N+n-1,1)=293;solmat(:,w)=fdmmat\rhsmat;endfdmmat2=zeros(N+2,N+2);rhsmat2=zeros(N+2,1);solmat2=zeros(n*N-1,N+2);

p=1;u=0;for w=1:n*N-1

if mod(w,N)==0p=p+1;

endfor i=1:N

fdmmat2(i,i)=-ry(p);fdmmat2(i,i+1)=1+2*ry(p);fdmmat2(i,i+2)=-ry(p);

endfdmmat2(N+1,1)=k(p)/(2*deltay*h);fdmmat2(N+1,2)=1;fdmmat2(N+1,3)=-k(p)/(2*deltay*h);fdmmat2(N+2,N)=-k(p)/(2*deltay*h);fdmmat2(N+2,N+1)=1;fdmmat2(N+2,N+2)=k(p)/(2*deltay*h);if w==1for i=1:N

rhsmat2(i,1)=(1-

2*rx(p))*solmat(w,i)+rx(p)*solmat(w+1,i)+rx(p)*((deltat/10)*(b(2)-

b(1))+b(1));end

elseif mod(w,N)==0rhsmat(i,1)=rhsmat(i-1,1);u=u+1;

else

for i=1:Nrhsmat2(i,1)=(1-

2*rx(p))*solmat(w+u,i)+rx(p)*solmat(w+u+1,i)+rx(p)*solmat(w+u-1,i);endend

endrhsmat2(N+1,1)=293;


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rhsmat2(N+2,1)=293;solmat2(w,:)=fdmmat2\rhsmat2;

endsurfrec=zeros(301,n);surfrec2=zeros(301,N);surfrec(1,:)=293;surfrec2(1,:)=293;i=N-1;j=1;while(i=0 && q*deltat=5 && q*deltat=10 && q*deltat=15 && q*deltat=20 && q*deltat=25 && q*deltat


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if mod(w,N)==0p=p+1;

endfor i=1:N

fdmmat2(i,i)=-ry(p);fdmmat2(i,i+1)=1+2*ry(p);fdmmat2(i,i+2)=-ry(p);

endfdmmat2(N+1,1)=k(p)/(2*deltay*h);fdmmat2(N+1,2)=1;fdmmat2(N+1,3)=-k(p)/(2*deltay*h);fdmmat2(N+2,N)=-k(p)/(2*deltay*h);fdmmat2(N+2,N+1)=1;fdmmat2(N+2,N+2)=k(p)/(2*deltay*h);if w==1for i=1:N

rhsmat2(i,1)=(1-

2*rx(p))*solmat(w,i)+rx(p)*solmat(w+1,i)+rx(p)*Tatiter;end

elseif mod(w,N)==0rhsmat(i,1)=rhsmat(i-1,1);u=u+1;

else

for i=1:Nrhsmat2(i,1)=(1-

2*rx(p))*solmat(w+u,i)+rx(p)*solmat(w+u+1,i)+rx(p)*solmat(w+u-1,i);endend

endrhsmat2(N+1,1)=293;rhsmat2(N+2,1)=293;

solmat2(w,:)=fdmmat2\rhsmat2;endc=N-1;d=1;while(c


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References

Heat conduction by M Necati Ozisik http://highered.mcgraw-hill.com/sites/dl/free/0073129305/314124/cen29305_ch04.pdf

en.wikepedia.orgwww.mathworks.inwww.wolframalpha.orgComputational Fluid Dynamics by John D. Anderson
http://highered.mcgraw-hill.com/sites/dl/free/0073129305/314124/cen29305_ch04.pdfhttp://highered.mcgraw-hill.com/sites/dl/free/0073129305/314124/cen29305_ch04.pdfhttp://www.mathworks.in/http://www.mathworks.in/http://www.wolframalpha.org/http://www.wolframalpha.org/http://www.wolframalpha.org/http://www.mathworks.in/http://highered.mcgraw-hill.com/sites/dl/free/0073129305/314124/cen29305_ch04.pdf


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Thank You!!!