(2) Heat Conduction Equation S2 2014-2015

8/9/2019 (2) Heat Conduction Equation S2 2014-2015

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(2) H ea t ondu c t ion Equa t ionby

Assoc Prof Leong Kai ChoongSchool of Mechanical and Aerospace Engineering

MA3003 Heat TransferSemester 2, AY 2014-2015

Read Chapter 2 of textbook

before these lecture slides


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At the end of these lectures, you should be able to

understand multidimensionality and time dependence of

heat transfer, and the conditions under which a heattransfer problem can be approximated as being one-

dimensional,

derive the heat conduction equation in various

coordinate systems and simplify it for the steady-stateone-dimensional case,

identify the thermal conditions at the boundaries of solids

and express them mathematically as boundary or initial

conditions, and appreciate the different solution methods for simple and

complex heat conduction problems.

Learning Objectives


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Scope of these lectures

• Basic Concepts• Fourier’s Law as a Vector Equation

• Derivation of the Heat Conduction Equation

• Boundary and Initial Conditions

• Formulation of Heat Conduction Problems

• Introduction to Solution Methods

Note that unless otherwise indicated, figures are taken from the prescribed

textbook, Çengel, Y.A., and Ghajar, A.J., Heat and Mass Transfer:Fundamentals and Applications, 5th Ed. (SI Units), McGraw-Hill, 2015.


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Basic Concepts

Temperature & Heat Transfer – Scalar & Vector

quantities


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Coordinate Systems

or Cartesian coordinates


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Steady (or steady-state) versus Transient Heat

Transfer


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Fourier’s Law as a Vector Equation

For an isotropic material, for which the thermal

conductivity is the same in all directions, Fourier’slaw in Cartesian coordinates is

)1(....;;

z

T k q

y

T k q

x

T k q z y x

∂

∂−=

∂

∂−=

∂

∂−=

where

x x,y,z,t T x

T

xq x

torespectwith)(of derivative partial

directionincomponentfluxheat

=∂

∂

=


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T k ∇−=q

=∇T Gradient of scalar temp. field

zyx qkq jqiq ++=

In vector form, Eq. (1) can be written more

compactly as

where =q Conduction heat flux vector

In Cartesian coordinates

z

T

y

T

x

T T

∂∂

+∂∂

+∂∂

=∇ k ji

where i, j and k are the unit vectors in the x, y

and z directions, respectively.


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Vector representation of Fourier’s Law

q yq

q

q


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Heat Transfer Vectors and Isotherms


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Derivation of Heat ConductionEquation in Cartesian coordinates

Source: Incropera et al. (2013) with change of symbols

xQ

dx xQ +

dz zQ +

dy yQ +

yQ

zQ


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Energy balance:

Rate of heat rate conducted into solid + rate of

heat rate generated inside solid - rate of heatconducted out of solid = rate of increase of

internal energy (or rate of heat stored)

( )

( ) ( ) .......(2)

of formsimplified

dxdydz

t

T cQQQ

dxdydzeQQQdt

dE E E E E

pdz zdy ydx x

gen z y x

system

st out genin

∂

∂ρ=++−

+++

=−+

+++

genewhere = rate of heat generation per unit

volume (W/m3)


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Rate of heat conducted into solid across face at x is

Using Taylor series expansion and neglecting

higher order terms, the rate of heat conducted outof the solid at x + dx is

( ) ( ) ...3

3

32

2

2

+∂

∂

+∂

∂

+∂

∂

+=+ dx x

Q

dx x

Q

dx x

Q

QQ

x x x

xdx x

( )dxdydz xT

k xdx x

Q

QQ x

dx x x

∂

∂

−∂

∂

−=∂

∂

−=− +

Using Fourier’s law, net heat inflow in x direction

xQ

Negligible


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Do the same for y and z directions.

Net heat conduction into volume =

( )dxdydz z

T k

z y

T k

y x

T k

x

∂∂

∂∂

+

∂∂

∂∂

+

∂∂

∂∂

gene

( )dxdydzegen

Rate of heat generation within elemental

volume =

where = rate of heat generation per unit

volume (W/m3)


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)3...(t

T ce

z

T k

z y

T k

y x

T k

x pgen ∂

∂ρ=+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

Substituting in Equation (2) and dividing

throughout by dxdydz,

- general form of heat conduction equation in

Cartesian coordinates.If k is not a function of temperature and position,

ydiffusivit thermalwhere

.....(4) 1

2

2

2

2

2

2

p

gen

c

k

t

T

k

e

z

T

y

T

x

T

ρ=α

∂∂

α=+

∂∂

+∂∂

+∂∂


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Thermal Diffusivityα

Table 1: Thermal diffusivity of typical materials

Averagetemperature

°C

Diffusivity

α × 106

m

2

/sCopper 0 114.1Zinc 0 41.3

Brick, fireclay 204 0.516Rubber, soft 0.077


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Table 2: Effect of thermal diffusivity on the

rate of heat propagation

Material Silver Copper Steel Glassα × 106

m2/s

170 103 12.9 0.59

Time 9.5min

16.5min

2.2 h 2.00days

Semi-infinite medium initially at T i = 100°C.

Surface x = 0 suddenly lowered to 0°C. What

is the time taken for x = 30 cm to reach 50°C ? x = 30 cm


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Expressing the heat conduction equation for

constant k in compact form using the Laplacianoperator ∇2, we have

......(5) 12

t

T

k

eT

gen

∂∂

α=+∇

In Cartesian coordinates,

2

2

2

2

2

22

z y x ∂∂+

∂∂+

∂∂=∇

EquationsFourier'(6)..... 1

,0For

2

2

2

2

2

2

t

T

z

T

y

T

x

T

egen

∂∂

α=∂∂

+∂∂

+∂∂

=


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const

0 i.e.

0

(3),Eq.fromcaseD-1For

Equations Laplace'(7).... 0

,conditionsgenerationheatnoand state-steadyFor

2

2

2

2

2

2

=⇒

=

=

=

∂

∂+

∂

∂+

∂

∂

x

x

q

dx

qd

dx

dT k

dx

d

z

T

y

T

x

T

Pi Si L l


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Pierre-Simon LaplaceBorn: 23-Mar-1749 (son of a farm-

labourer )

Birthplace: Beaumont-en-Auge,

Normandy, FranceFather: Farmer

Died: 5-Mar-1827

Location of death: Paris, France

Cause of death: unspecified

Occupation: Mathematician,

Politician.

Jean le Rond d'Alembert was his

“mentor”.

Fourier was his student. At 27, he was called “the Newton of

France”

Source: http://www.nndb.com/people/871/000031778/

http://www.nndb.com/people/871/000031778/http://www.nndb.com/people/871/000031778/


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Cylindrical Coordinate System

dz zQ +

zQ

r Q

φQ

φ+φ d Q

dr r Q +


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Spherical Coordinate System

θ+θ d Q

θQ

r Q

φQ

φ+φ d Q

dr r Q +

General Heat Conduction Equations


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General Heat Conduction Equations

Cartesian coordinates:

Cylindrical coordinates:

Spherical coordinates:

t

T

ce z

T

k z y

T

k y x

T

k x pgen ∂

∂

ρ=+

∂

∂

∂

∂

+

∂

∂

∂

∂

+

∂

∂

∂

∂

112

t

T ce

z

T k

z

T k

r r

T kr

r r

pgen

∂

∂ρ=+

∂

∂

∂

∂+

φ∂

∂

φ∂

∂+

∂

∂

∂

∂

t

T ce

T k

r

T k

r r

T kr

r r pgen ∂

∂ρ=+

∂θ

∂θ

θ∂

∂

θ+

∂φ

∂

∂φ

∂

θ+

∂

∂

∂

∂sin

sin

1

sin

11222

2

2


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Boundary ConditionsBoundary conditions for the heat conduction

equation at surface x = 0

1. Constant surface

temperature

(First-kind or Dirichlet *)

sT t T =),0(

* Johann Peter Gustav Lejeune Dirichlet Born: 13 Feb 1805 in Düren, FrenchEmpire (now Germany), Died: 5 May 1859 in Göttingen, Hanover (now Germany).Studied under Georg Ohm and provided proof of Fermat’s Last Theorem.

Specified or prescribed

temperature

No three positive integers a, b, and c can satisfy the equation

a n + b n = c n for any integer value of n greater than 2.

2 C t t f h t f l

http://en.wikipedia.org/wiki/Positive_numberhttp://en.wikipedia.org/wiki/Integerhttp://en.wikipedia.org/wiki/Integerhttp://en.wikipedia.org/wiki/Positive_number


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2. Constant surface heat f lux

(Second kind or Neumann )

(a) Finite heat flux

s

x

q x

T k =∂∂

−=0

(b) Adiabatic surface

0

0

=

∂

∂

= x x

T

sq

John von Neumann (December 28, 1903 – February 8,

1957) was a Hungarian American mathematician who made

major contributions to a vast range of fields.


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3. Convection surface condit ion

(Third kind or mixed)

( )[ ]t T T h x

T k

x

,00

−=∂∂

− ∞=

Note that T at x = 0 is unknown and hence we do not

write it as T s.

4 R di ti b d diti (F th ki d)


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4. Radiation boundary condition (Fourth kind)

( )[ ]44 1,10

,0 t T T x

T k surr

x

−σε=∂

∂−

=

Note that T at x = 0 is unknown and hence we do not write

it as T s. Temperatures must be in kelvins!

Homework: Write down the four kinds of boundary

conditions for the surface x = L of the plane wall.


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5. Interface boundary conditions

00 x x

B B

x x

A A

x

T k x

T k == ∂

∂−=∂∂−

( ) ( )t xT t xT B A ,, 00 =


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For block of ice melting in warm water,

( )

fs

us

us

su

hm xT k

xT k

hhm x

T k

x

T k

''

''

+∂∂=∂∂

−=

∂∂

−−∂∂

−

iceof fusionof heat)latent(orenthalpy- us fs hhh =

skg/mof unitshasfluxMass 2'' ⋅m

6. Phase change boundary condition


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Th l d t i t


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Thermal and geometric symmetry

Boundary conditions & geometrical shape are

symmetrical about a common axis or plane.

Hence, temp. distribution is symmetrical about

the same axis and heat conduction analysis is

greatly simplified.

planeor axissymmetryat0=

∂

∂

x

T

Thermal & geometric symmetry


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Thermal & geometric symmetryabout a plane wall


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Solve for just half the wall from x = 0 to x = L/2 withthe following boundary conditions

[ ]∞∞=

=

−=∂

∂

−

=∂∂

T LT h x

T

k

x

T

L x

x

)2/(

0

2/

0

Surface at x = 0 may also be considered to beadiabatic (or perfectly insulated).


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Initial conditions

For transient heat conduction problems,temperature of the entire solid must be known at

some instant of time (usually t = 0) before thesubsequent variation in temperature with time can

be determined.For example,

0)0,( T xT =


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Solution :


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Solution :

Start with general heat conduction equation

b

ss

x A

Qq

dx

dT k

==−=0

02

2

=dx

T d where T = T ( x)

t T ce

zT k

z yT k

y xT k

x pgen ∂∂ρ=+ ∂∂∂∂+ ∂∂∂∂+ ∂∂∂∂

( )[ ]∞=

−=− T LT hdx

dT k

L x

2nd Order ODE in space ⇒ two (2) boundary

conditions

44

Example 2: Heating of a copper bar


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Example 2: Heating of a copper bar

Assumptions: (1) 1-D in x since W >> L and bar is long (2)Uniform heat generation, and (3) Constant properties (k , α).

gene

Source: Incropera et al. (2007) with change of symbols

Mathematical formulation to f ind T ( x,t)?

oi T T =

Model

Solution :


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Solution :

Start with general heat conduction equation

2nd Order PDE in space ⇒ two (2) boundary conditions

1st Order PDE in time ⇒ one (1) initial condition

[ ]∞=

−=∂∂

− T t LT h x

T k

L x

),(

t

T

k

e

x

T gen

∂∂

α=+

∂∂ 1

2

2

where T = T ( x, t )

oT xT =)0,(

oT t T =),0(

t T ce

zT k

z yT k

y xT k

x pgen ∂∂ρ=+ ∂∂∂∂+ ∂∂∂∂+ ∂∂∂∂

46

gene

Example 3: Steady-state heat


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Example 3: Steady state heatconduction across a circular pipe

∞T

Inside surface of circular pipe of inside radius r 1,outside radius r 2, and thermal conductivity k is

maintained at T i by flowing steam. The outsidesurface dissipates heat by convection with a heat

transfer coefficient h into ambient air attemperature

Develop the mathematical formulation of the

problem of determining the steady-state heatconduction across the pipe.

St t ith l h t d ti ti


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For 1-D, steady-state, constant k and no heat generation,

constantor 01

=∂∂

=

∂∂

∂∂

r

T r

r

T r

r r

Constant temperature boundary condition on inner

surface 1at r r T T i ==

( )[ ] 22 at2

r r T r T hdr

dT k

r r

=−=− ∞=

Convection boundary condition on outer surface

t

T ce

z

T k

z

T k

r r

T kr

r r

pgen

∂

∂ρ=+

∂

∂

∂

∂+

φ∂

∂

φ∂

∂+

∂

∂

∂

∂

2

11

Start with general heat conduction equation,

Example 4: Transient heat


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Example 4: Transient heatconduction across a plane wall

A plane wall of thickness L with constant thermal

properties is initially at uniform temperature T 0 . At

time t = 0, the surface at x = L is subjected toheating by a flow of hot gas at temperature

while the surface at x = 0 is kept perfectly insulated(adiabatic). The heat transfer coefficient between

the hot gas and the surface is h. There is no energygeneration within the wall.

∞T


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Schematic of plane wall example


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Develop the mathematical formulation of the

problem of determining the unsteady heatconduction across the plane wall.

Do not solve the problem.

Solution:


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1-D, time-dependent problem with constant

thermal properties and no heat generation.

[ ]

( ) xT xT ndition Initial co

t L xT LT h

x

T k

t x x

T

onditions Boundary c

t L xt

T

x

T

equational Differenti

L x

x

allfor0,:

0 ,at)(

0,0at0

:

0,0for1

:

0

0

2

2

=

>=−=∂

∂−

>==∂∂

>


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radiation from a metal ball

A spherical metal ball of radius r o is heated to a

certain temperature T i and is then taken out of theoven to cool in ambient air at temperature . The

thermal conductivity of the ball material is k and the

heat transfer coefficient on the outer surface is h.The emissivity of the ball surface is ε and the

temperature of the surroundings is T surr . There is no

energy generation within the wall. Develop the

mathematical formulation of the problem.

∞T

Start with general heat conduction equation,

TTTT ∂ ∂∂ ∂∂ ∂∂ 111


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( )[ ] ( )[ ]44 surr oor r T r T T r T hr T

k o

−εσ+−=∂

∂− ∞

=

Convection & radiation boundary

condition on outer surface

t

T ce

T k

r

T k

r r

T kr

r r pgen ∂

∂ρ=+

∂θ

∂θ

θ∂

∂

θ+

∂φ

∂

∂φ

∂

θ+

∂

∂

∂

∂sin

sin

1

sin

11222

2

2

For transient conduction, constant k and no heatgeneration,

t

T

r

T r

r r ∂

∂

α=

∂

∂

∂

∂ 11 22

At the inner surface 00=∂∂ =r r

T

Initial condit ion: iT r T =)0,(

Thermal & geometric

symmetry

Boundary

Conditions:

Example 6: Combined convection,radiation and heat flux


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radiation and heat flux

( )[ ] ( )[ ] solar sky L x qT LT T LT hdxdT k α−−σε+−=− ∞= 44

222

( )[ ]0110

T T hdx

dT k

x

−=− ∞=

Boundary condit ions on inner and

outer surfaces of house?

T sky = effective sky temperature (here it

includes sky, ground and surrounding

structures); = incident solar heat flux

Steady-state, one-dimensional problem

with boundary conditions

solar q

Basic Steps in the Solution of HeatTransfer Problems


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Transfer Problems

Solution Methods


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Mainly grouped under

(1) Analytical methods

e.g. separation of variables & transform methods

Requires

1.simple geometry

2.boundary conditions to be of simple mathematical

forms, and

3.constant thermophysical properties.

(2) Numerical methods e.g. finite difference & finite element methods


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Note the limitations of analytical methods.


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You may wish to read more about the finite difference

method in Chapter 5 of Çengel and Ghajar (2011).

Some Review Questions


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1. How does transient heat transfer differ from steady state heat

transfer?

2. How does one-dimensional heat transfer differ from two-dimensional heat transfer?

3. Which property of a material determines (a) the amount of

heat it can store per unit volume, (b) the heat it can conduct

under steady-state conditions, and (c) the rate at which it willreact to transient temperature changes?

4. What physical principles are represented by the general heat

conduction equation?

5. What are the typical boundary conditions for heatconduction? Read Chapter 2: Sections 2-5, 2-6 and 2-7 and

Chapter 3: Sections 3-1 to 3-5 of the textbook

before the next lecture!

(2) Heat Conduction Equation S2 2014-2015

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Transcript of (2) Heat Conduction Equation S2 2014-2015