steady- state-heat-transfer-in-multiple-dimensions

Lectures on Heat Transfer --Steady State Heat Conduction in

Multiple Dimensions

by

Dr. M. ThirumaleshwarDr. M. Thirumaleshwarformerly:

Professor, Dept. of Mechanical Engineering,St. Joseph Engg. College, Vamanjoor,

MangaloreIndia

Preface

• This file contains slides on Steady State Heat Conduction in Multiple Dimensions.

• The slides were prepared while teaching Heat Transfer course to the M.Tech. students in Mechanical Engineering Dept. of St. Joseph Engineering College, Vamanjoor, Mangalore, India, during Sept. – Dec. 2010.

Aug. 2016 2MT/SJEC/M.Tech.

• It is hoped that these Slides will be useful to teachers, students, researchers and professionals working in this field.

• For students, it should be particularly useful to study, quickly review the subject, useful to study, quickly review the subject, and to prepare for the examinations.

• ��

Aug. 2016 3MT/SJEC/M.Tech.

References• 1. M. Thirumaleshwar: Fundamentals of Heat &

Mass Transfer, Pearson Edu., 2006• https://books.google.co.in/books?id=b2238B-

AsqcC&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false

• 2. Cengel Y. A. Heat Transfer: A Practical Approach, 2nd Ed. McGraw Hill Co., 2003

Aug. 2016 MT/SJEC/M.Tech. 4

Approach, 2nd Ed. McGraw Hill Co., 2003• 3. Cengel, Y. A. and Ghajar, A. J., Heat and

Mass Transfer - Fundamentals and Applications, 5th Ed., McGraw-Hill, New York, NY, 2014.

References… contd.

• 4. Incropera , Dewitt, Bergman, Lavine: Fundamentals of Heat and Mass Transfer, 6th

Ed., Wiley Intl.• 5. M. Thirumaleshwar: Software Solutions to • 5. M. Thirumaleshwar: Software Solutions to

Problems on Heat Transfer – CONDUCTION-Part-II, Bookboon, 2013

• http://bookboon.com/en/software-solutions-problems-on-heat-transfer-cii-ebook


Steady State Heat Conduction in Multiple Dimensions…

Outline..

• 2-D conduction - Various methods of solution – Analytical - Graphical -


solution – Analytical - Graphical -Analogical – Numerical – Shape factors for 2-D conduction - Problems

Two-dimensional conduction.. • Practical examples of multi-dimensional heat transfer

are: heat treatment of engineering components of irregular shapes, heat transfer in I.C.Engine blocks, chimneys, air conditioning ducts etc.

• To solve multi-dimensional heat transfer problems, basically, there are four methods:


basically, there are four methods:

• Analytical methods - solutions are quite cumbersome

• Graphical methods - for two dimensional problems with isothermal and adiabatic boundaries. This is an approximate method.

Two-dimensional conduction ..

• Analogical methods - Special conducting paper (or, conducting solution in a bath) is used to make a model of the geometry being investigated and the isothermal (equi-potential) lines are traced using a probe.


probe.

• Numerical methods - Numerical methods have taken over other methods because of availability of high speed computers and the ability to analyze complex shapes and deal with complicated boundary conditions.

Analytical methods:• For 2D conduction, we have to solve:


Analytical solution becomes complicated and difficult since now we have to deal with a partial differential equation. We illustrate this with a problem:

Method used is Separation of variables, and it is briefly illustrated below for a specific problem (Ref: Incropera et al.)

• Consider the system of Fig. 4.2: three sides of a thin rect. plate maintained at a const. temp. T1 and the fourth side is at T2.


Introduce the transformation:

Substituting eqn. (4.2) in (4.1) we get:

• B.C’s are:


In (4.5) LHS depends only on x and RHS depends only on y. So, both sides are equal to the same constant. We write:


Now, applying the B.C’s and after much manipulation, we get the final solution:

• Consider one more problem (Ref: Schaum’s Series, Pitts & Leighton):


B.C’s are:

Proceeding as earlier,


Now, applying the B.C’s and simplifying, final expression for temp. distribution is:

Graphical methods:(Ref: Schaum’s Series, Pitts & Leighton):

• Consider a heated pipe with thick insulation, with inside temp. Ti and outside temp. T0 as shown. Constructing perpendiculars to isothermal lines result in heat flow lanes. Because of symmetry consider only one quadrant as shown, and in this quadrant


quadrant as shown, and in this quadrant there are 4 heat flow lanes.

• Method is to determine the heat flow in a single lane and then find the total.

Graphical methods:(Ref: Schaum’s Series, Pitts & Leighton):


• Applying Fourier’s Law to element a-b-c-d of a typical lane, heat transfer per unit depth is:


Now, if the isotherms are uniformly spaced, and if there are M such curvilinear squares in a flow lane, then temp. difference across one square is:

Then, for N flow lanes:

• And, the conductive shape factor per unit depth is:

Freehand plotting:

As shown above, a graphical plot of equally spaced isotherms and adiabatics is enough to determine the


isotherms and adiabatics is enough to determine the Shape factor. The graphical net can usually be obtained by freehand plotting. Typical one-eighth section is shown, and its shape factor is one-eighth of overall shape factor.

Shape factors for 2-D conduction:

• This is a simple method to analyse a particular type of 2-D conduction problems where steady state heat transfer occurs between two surfaces at fixed temperatures, T1 and T2, with an intervening solid medium in between.


• If Q is the rate of heat transfer between two temperature potentials T1 and T2, with the thermal conductivity of intervening material being k, with no heat generation in the medium, we write:

• Q = k S (T1 – T2)……….(4.76)where S is known as ‘Shape factor’ and has dimension of length.


• From eqn.(4.76), immediately it follows that thermal resistance of the medium is given by:

Rth = 1/(k S) ………..(4.77)• Then, since we can write: S = 1/(R.k), we get:



• In calculation of heat transfer in a furnace, separate shape factors are used to calculate the heat flow through the walls, edges and corners.

• When all the interior dimensions are greater than one–fifth of the wall thickness, we get:


fifth of the wall thickness, we get:

where, A = inside area of the wall, L = wall thickness,and D = length of edge

Ref: Cengel
