(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!
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