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Conduction Heat Transfer
Dr. Subrahmanya S. KatteAshok Leyland Chair Professor & Associate Dean (Research)
School of Mechanical EngineeringSASTRA University
Thanjavur 613401, India
Lecture Notes Compiled by
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Based on References
Incropera F.P., and DeWitt D.P., Fundamentals of Heat and Mass Transfer, Fifth edition, John Wiley & Sons, 2001
John H. Lienhard IV, and John H. Lienhard V, A Heat Transfer Text Book, Third edition, Phlogiston Press, Cambridge, Massachusetts, U.S.A.
Other sources on web
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1. Fundamentals of Conduction Heat Transfer
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1.1 Thermophysical Properties
• Transport properties:k (thermal conductivity – conduction heat transfer) (kinematic viscosity - momentum transfer)D (diffusion coefficient - mass transfer)
• Thermodynamic properties:relating to equilibrium state of a system, such as density and specific heat c
ν
ρ
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Thermal Conductivity
Recall Fourier’s law
Km/W
mK
mW
;2
=)dx/dT(
qk
"x
x −≡
dx
dTkq x
"x −=
It is an empirical law, i.e., based on experimental evidence
Applies to all states of matter
Defines the transport property, thermal conductivity:
Is thermal conductivity different between gases, liquids and solids?
We, mostly, assume isotropic material (independent of the direction): kx=ky=kz=k
Range of thermal conductivity for various states of matter at NTP
Solid metalsLiquid metals
Nonmetalic solids
NM liquids
Gases
No theory to predict.k Metals > Non Metalsk Crystals > Non Crystalsk metals - with + Tk nonmetals + with + TPure metals - electrical and Thermal conductance correlate.
Thermal Conductivity: Solids
• Heat, in solids, is transported through– Migration of free electrons– Lattice vibrational waves
• On can disperse solid material throughout an air space to form an insulator - fiber, powder and flake type insulations, even though solids have higher k
Thermal Conductivity: Fluids• Heat transport is less effective than in
solids; k is lower• Physical mechanisms controlling k not
well understood in liquid state
• Generally k decreases with increasing temperature (exceptions glycerine and water)
• k decreases with increasing molecular weight
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Thermal Diffusivity
• The volumetric heat capacity (J/m3 K) represents the ability of a material to store heat energy– For solids and liquids, MJ/m3 K– For gases, kJ/m3 K
cρ
c
k
ρ=α
• Thermal diffusivity represents the ability of a material to conduct heat relative to its ability to store the same
• Materials of large thermal diffusivity will respond quickly to changes in their thermal environment
• Materials of small thermal diffusivity will respond more sluggishly, taking longer to reach a new equilibrium condition
1>cρ1≈cρ
Thermal diffusivity
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Density and Specific Heat
Solids – density ~ constant, cv= cp= c increases with +T(Eg. iron)
Liquids – density decreases with + T, cv= cp= c increases with +T(Eg. Water above 4oC)
Gases – density decreases with + T, cv< cp both increases with +T(Eg. air), volumetric heat capacity decreases with +T
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1.1.1 A Note on Thermophysical Property Data
• The accuracy of engineering calculations depends on the accuracy with which the thermophysical properties are known
• Numerous examples could be cited of design failures attributable to misinformation associated with key thermophysical properties
• Selection of reliable property data is an integral part of any careful engineering analysis
• The casual use of data from the literature or handbooks, which have not been well characterized or evaluated, is to be avoided
http://h
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ws.infoFourier’s law is used to determine the conduction heat
transfer rate
Temperature is a scalar field T(x,y,z)
Heat transfer in three dimensions are:
z
Tkq;
y
Tkq;
x
Tkq zzyyxx ∂
∂−=∂∂−=
∂∂−=
Heat flux is a vector quantity, q is vector sum of qx, qy and qz
Hence, to determine qx, qy, qz we require the knowledge of the manner in which temperature varies within the medium (temperature distribution)
1.2 Heat Conduction Equation
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This knowledge is obtained by developing, from fundamental principles, a partial differential equation (PDE), termed as the “Heat Conduction Equation”, which governs the temperature distribution within a medium
The solution to this PDE, subject to boundary and/or initial conditions, provides the knowledge of temperature distribution
http://h
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Control Volume
Surroundings
Boundary (Control Surface)
stst
outgin Edt
dEEEE ==−+
Accumulation (Storage)
Generation
Heat Addition Heat Rejection
inE outEgEstE
Inflow and outflow are surface phenomenaGeneration and accumulation are volumetric phenomena
http://h
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Energy conservation for differential volume, dx dy dz, gives
stgoutin QQQQ =+−
zyxin QQQQ ++= dzzdyydxxout QQQQ +++ ++=
(2.1)
where
stgdzzzdyyydxxx QQQQQQQQ =+−+−+− +++
Substituting these in Eq. (2.1)
(2.2)
QxQx+ ∆x
Qy+ ∆y
Qz+ ∆z
Qy
Qz
Qg
T(x,y,z,t)
http://h
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z
QQQ;dy
y
QQQ;dx
x
QQQ z
zdzzy
ydyyx
xdxx ∂∂+=
∂∂
+=∂
∂+= +++
Using Taylor series, neglecting higher order terms:
Substituting these in Eq. (2.2)
stgzyx QQdz
z
Qdy
y
Qdx
x
Q =+∂
∂−∂
∂−
∂∂− (2.3)
Using Fourier’s law
)dxdy(z
Tk
z
TkAqAQ
)dxdz(y
Tk
y
TkAqAQ
)dydz(x
Tk
x
TkAqAQ
zzzzzz
yyyyyy
xxxxxx
∂∂−=
∂∂−==
∂∂−=
∂∂−==
∂∂−=
∂∂−==
(2.4)
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t
Tcq
z
Tk
zy
Tk
y
x
Tk
x zyx ∂∂ρ=′′′+
∂∂
∂∂+
∂∂
∂∂+
∂∂
∂∂
dxdydzqQg ′′′=
dVm
dzdydxt
Tc)mcT(
tQst
ρ=∂∂ρ=
∂∂=Heat stored or
rate of change of energy is:
is rate at which heat is “generated” per unit volume of medium (W/m3), could be a source or sink
q ′′′
, if c is a constant
(2.6)
(2.5)Therefore heat generated is
Substituting Eq. (2.4-2.6) in Eq. (2.3), and dividing by (dxdydz),
Net conduction of heat into the CV rate of heat “generation”
or sink
time rate of change of heat energy
General Heat Equation
(2.7a)
0 for steady state
0 for no generation or sink
Terms depend on 1-D, 2-D, 3-D conduction
http://h
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t
T1
k
q
z
T
y
T
x
T2
2
2
2
2
2
∂∂
α=
′′′+
∂∂+
∂∂+
∂∂
where the thermal diffusivity
For steady state
c
k
ρ=α
02 =′′′
+∇k
qT
0T2 =∇ Laplace equation
Eq. (2.7b) can be written as
(2.7b)
(2.7c)Heat Equation
Poisson equation
t
T1
k
qT2
∂∂
α=
′′′+∇
If there is no heat generation
(2.7d)
(2.7e)
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1.2.2 Conduction Equation in Polar Coordinates
stgdzzzddrrr QQQQQQQQ =+−+−+− +θ+θθ+
zrin QQQQ ++= θ
dzzddrrout QQQQ +θ+θ+ ++=
Substituting these in Eq. (2.1)
(2.8)
dzz
rdr
drr ∂
∂+=∂
∂+=∂
∂+= +++z
zdzzdr
rdrr
QQQ;
QQQ;
QQQ θ
θθ
θθθ
Using Taylor series expansion,
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dVm
dzdrdrtT
c)mcT(t
Qst
ρ=
θ∂∂ρ=
∂∂=
dzrdrdqQg θ′′′=Heat generated
Heat stored (rate of change of energy)
, if c is a constant
Using Fourier’s law Qr=A r qr=−A r kr∂T∂r
=−k r∂T∂ r
rdθ dz
Qθ=Aθ qθ=−Aθ kθ∂Tr ∂θ
=−kθ∂Tr ∂ θ
drdz
Q z=Az qz=−Az k z∂T∂ z
=−k z∂T∂ z
rdrdθ
(2.10)
(2.12)
(2.11)
Substituting these in Eq. (2.8)
stgzr QQdz
z
Qrd
r
Qdr
r
Q =+∂
∂−∂
∂−∂
∂− θθθ (2.9)
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Substituting Eq. (2.10-2.12) in Eq. (2.9),and dividing by (rdrdθdz),
(2.13a)t
Tcq
z
Tk
z
Tk
r
1
r
Trk
rr
1z2r ∂
∂ρ=′′′+
∂∂
∂∂
θ∂∂
θ∂∂+
∂∂
∂∂
+θ
t
T1
k
q
z
TT
r
1
r
T
r
1
r
T2
2
2
2
22
2
∂∂
α=
′′′+
∂∂
θ∂∂+
∂∂+
∂∂
+
t
T1
k
qT2
∂∂
α=
′′′+∇
If material is isotropic kr=kθ=kz=k, a constant
(2.13b)t
T1
k
q
z
TT
r
1
r
Tr
rr
12
2
2
2
2 ∂∂
α=
′′′+
∂∂
θ∂∂+
∂∂
∂∂
+
(2.13c)
or
or
t
Tcq
Tk
sinr
1Tsink
sinr
1
r
Trk
rr
1222
2r2 ∂
∂ρ=′′′+
φ∂∂
φ∂∂
θ+
θ∂∂θ
θ∂∂
θ+
∂∂
∂∂
φθ
(2.14a)
1.2.3 Conduction Equation in Spherical Coordinates
t
T1
k
qT
sinr
1Tsin
sinr
1
r
Tr
rr
1222
22 ∂
∂α
=′′′
+
φ∂∂
φ∂∂
θ+
θ∂∂θ
θ∂∂
θ+
∂∂
∂∂
If material is isotropic kr=kθ=kφ=k, a constant
(2.14b)
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1.2.4 Methods of Solution
•Integration
•Separation of variables
•Superposition principle for linear problems
•Approximate analytical methods (Integral methods)
•Numerical methods
•Numerical integration
•Finite Difference Method (FDM)
•Finite Volume Method (FVM)
•Finite Element Method (FEM)
•Direct Numerical Simulation (DNC)
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1.2.5 Boundary and Initial Conditions
The temperature distribution in a medium depends on the conditions existing at the boundaries, and at some initial timeThe solution of heat equation requires a maximum of 6 B.Cs and 1 I.CInitial condition is the specification of temperature distribution at some initial time
T
x
T
1
T
2
∞T
Qcond ”Qrad ”
Qconv ”
Surface Energy Balance
For a control surface:
0QQQ radconvcond =−−
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B.C of first kind (Dirichlet condition)
Surface temperature is specified
Eg.: a surface in contact with a melting solid or a boiling liquid or condensing steam
B.C of second kind (Neumann condition)
Heat flux at the surface (temperature gradient at the surface) is specified
Heat flux is zero: approximation for situations where thermal insulation is very high
What if the surface is perfectly insulated?
Eg.: Radiative heating of a body, electrical heating
qs” W/m2
B.C. of third kind
A relationship for the temperature gradient at the surface is specifiedEg.: convection at the surface
Can be approximated to
NOTE:Specified values could be constant or a function of timeMany more types of B.C s possible: radiative, interface with another solid with a contact resistance
(Same as B.C of first kind)
http://h
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and identify the independent variables that determine T
Step 2. Write the appropriate differential equation, starting with one of the forms of Eq. (2.7c)
Step 3. Obtain the general solution of the differential equation
1.2.6 Solution Procedure
Step 4. Write the “side conditions” on the differential equation - the initial and boundary conditions
This is the step that most seriously tests your physical or “practical” understanding of problems
Never, never introduce inaccessible information in a boundary or initial condition
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Step 6. Put the calculated constants back in the general solution to get the particular solution to the problem
It would be better to express the solution in a neat dimensionless formBy nondimensionalizing the result, we can succeed in representing all situations with a simple curve while plottingThis is highly desirable when the calculations are not simple
Step 5. Substitute the general solution in the initial and boundary conditions and solve for the constants
This process gets very complicated in the transient and multidimensional problems. Fourier series methods are typically needed to solve the problem
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Step 7. Play with the solution - look it over - see what it has to tell you. Make any checks you can think of to be sure it is correct
Step 8. If the temperature field is now correctly established, you can, if you wish, calculate the heat flux at any point in the body by substituting temperature distribution back into Fourier’s law
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Fourier-Poisson Equation
Assumptions
Geometry
Boundaryand/or initialconditions
Solve the Equation for TAnalytical or Numerical
Obtain TemperatureProfile
Fourier’s Law of HeatConduction
Obtain rate of heat transfer
TYPICAL METHOD FOR SOLVING CONDUCTION PROBLEMS
COORDINATESYSTEM
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2. One-dimensional, Steady Conduction
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• Temperature is a function of only one spatial coordinate
• Steady state means that the temperature is not a function of time
• Despite their inherent simplicity, one-dimensional, steady-state models may be used to accurately represent numerous engineering systems
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Qx
1T
2T
x
x=0 x=L
A slab/plane wall, as shown, is at steady state with dissimilar temperatures on either side and no internal heat generation
2.1 Plane Wall / Slab
Step 1
T=T(X) for steady x-direction heat flow
Step 2
For steady 1-D conduction without heat generation
0dx
Td2
2
= (i) GDE
Recall the solution procedure
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Step 4
T(0)=T1 = 0 + C2
Step 5
L
TTC 12
1−=
Step 3
1Cdx
dT =
By integrating (i) twice
( ) 21 CxCxT += (ii) General solution
The B.Cs are: T(0)=T1 T(L)=T2and
; so
; so C2=T1
T(L)=T2=C1L + C2 = C1L+ T1
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Step 6
Step 7
( ) ( ) 112 TL
xTTxT +−=
( )L
x
TT
TxT
12
1 =−−
or
We note that the solution satisfies the boundary conditions and that the temperature profile is linear
Step 8
Putting constants in the general solution
( ) ( )22112112 m/W
L
TTk
L
TTkT
L
xTT
dx
dk
dx
dTkq
−=−−=
+−−=−=
( )WLT
kAL
TTkAqAQ 21 ∆=−==
kAL
TQ
∆=
2.1.1 Thermal Resistance (Electrical Analogy)
The expression for slab reminds us Ohm’s lawL
TkAQ
∆=
Thus, if we rearrange it:
condR
TQ
∆=
R
EI
∆=is like
where, the thermal resistance for conduction in a plane wall (K/W) is
kA
LR cond =
Thus, we can write
kA
LR cond =
TE ∆=
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( ) ( )kA
LTT
TTL
kAQ 2,S1,S
2,S1,Sx
−=−=
hA
1Rconv =
qx
1,∞T
1,sT
2,sT
2,∞T
x
x=0 x=L
11, ,hT∞
22, ,hT∞
Hot fluid
Cold fluid Consider a plane wall, separating two fluids of different temperature
Based on the previous solution,the conduction heat transfer rate is
For heat convection,
hA1
)TT()TT(hAQ S
Sx∞
∞−=−=
where, the thermal resistance for convection (K/W) is
Plane Wall with Convection
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1,s1,s2,s TL
x)TT()x(T +−=
tot
2,1,
2,conv
2,2,s
cond
2,S1,S
1,conv
1,S1,
R
TT
R
TT
R
TT
R
TTQ ∞∞∞∞ −
=−
=−
=−
=
Ah
1
kA
L
Ah
1RRRR
212,convcond1,convtot ++=++=
Ah1
kAL
Ah1
TT
Ah1
TT
kAL
TT
Ah1
TT
21
2,1,
2
2,2,S2,S1,S
1
1,S1,
++
−=
−=
−=
− ∞∞∞∞
Using the concept of a thermal circuit
∑∆=
R
TQ overall
Temperature distribution can be obtained by treating the slab as subject to the first kind boundary conditions on both surfaces
Find surface temperatures
http://h
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conditions on either side
Qx
x
x=0 x=L
q0 W/m2Convection, given: h, T∞
TS,1
TS,2
( )∞−=−
= TThAL
TTkAAq 2,S
2,S1,S0
hA1
TT
kAL
TTAq 2,S2,S1,S
0∞−
=−
=
h
qTT 0
2,S += ∞
Equating the first and last terms
Equating the first and second terms
++=+= ∞ h
1
k
LqT
k
LqTT 0
02,S1,S
Under steady state, by energy balance
Plane Wall with Heat flux / Convection
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Substituting for TS,1 and TS,2 , we get
The temperature distribution, as we know
( ) ( )
−++= ∞ k
xL
h
1qTxT 0
1,S1,S2,S TL
x)TT()x(T +−=
The same expression can also be obtained by integrating
0dx
Td2
2
=
Subject to following boundary conditions
00x
qdx
dTk =−
=( ){ }∞
=−=− TLTh
dx
dTk
Lxand
( )
−+
=− ∞
L
x1
hL
k
kLq
TxT
0or
Often a slab consists of two or more materials, like the walls of a house
A composite wall may involve any number of series and parallel thermal resistances due to layers of different materials
Hot fluid
Cold fluid
Q
Alternatively, Q can be related to each layer
For resistances in series:Rtot =R1+R2+…+Rn
2.1.2 Conduction Through Composite Walls
Surfaces normal to x-directionare isothermal
Surfaces parallel to x-direction are adiabatic
For resistances in parallel:1/Rtot =1/R1+1/R2+…+1/Rn
Heat transfer is actually multidimensional
Hence, both circuits give different values of Rtot with increasing GF kk −
Qx Qx
"x
BA"c,t
q
TTR
−=
No two solid surfaces will ever form a perfect thermal contact is a composite slabHence, the temperature drop across the interface between materials may be appreciable, due to surface roughness effects, leading to air pockets
For unit area of interface
2.1.3 Contact Resistance
This temperature drop is attributed to the thermal contact resistance
http://h
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1. By increasing the area of contact spots:by increasing the joint pressureby reducing the roughness of the mating surfaces
2. By selecting an interfacial fluid/ substance of larger thermal conductivity
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2.2 Temperature Dependent Thermal Conductivity
( ) 0dx
dTTk
dx
d =
1Cdx
dT)T(k = dxCdT)T(k 1=∴
Qx
1T
2T
x
x=0 x=L
For steady 1-D conduction in a plane wall
with B.Cs: T(0)=T1 T(L)=T2and
By integrating
Integrating again
[ ] LCCxCdxCdT)T(kL
01
L0211
T
T
2
1
∫ =+==∫
dT)T(kL
1C
2
1
T
T1 ∫=∴
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L
AdT)T(k
L
AAC
dx
dTA)T(kQ
1
2
2
1
T
T
T
T1 ∫=∫−=−=−=
If k is linearly dependent on T : )aT1(k)T(k 0 +=
k0 is constant
a is temperature coefficient of k
By Fourier’s law
( ) ( )
−+−=
+=∫ += 2
22
1210
T
T
20
T
T0 TT
2
aTT
L
Ak
2
TaT
L
AkdT)aT1(k
L
AQ
1
2
1
2
( ) ( )
++−= 2121
0 TT2
a1TT
L
AkQ ( ) ( )
L
TTA
2
TTa1k 2121
0
−
++=
++=2
TTa1kk 21
0m
( )L
TTAkQ 21
m
−=∴ where
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t
T1
k
q
z
TT
r
1
r
Tr
rr
12
2
2
2
2 ∂∂
α=
′′′+
∂∂
θ∂∂+
∂∂
∂∂
+
0drdT
rdrd =
For steady state, one-dimensional radial conduction
0 0
⇒0dr
dTr
dr
d
r
1 =
By integrating 1Cdr
dTr =
r
C
dr
dT 1=∴
By integrating again
Subject to boundary conditions
21 CrlnCT +=
ri
Ti
T(ri)=Ti T(ro)=Toand
2.3 Radial Conduction in Polar Coordinates
General solution
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Solving for the constants2o1o
2i1i
CrlnCT
CrlnCT
+=+=
Substituting boundary conditions in the general solution
i
o
io1
rrln
TTC
−= i
i
o
ioi2 rln
rrln
TTTC;
−−=
( ) ( )i
o
iioi
rrln
rrln
TTTrT −+=
Substituting the constants in the general solution
( ) ( )
i
o
oi1
rrln
TTLk2
r
CrL2k
dr
dTkAQ
−π=π−=−=
By Fourier’s law
http://h
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rrlnR io
cond π=( ) cond
oi
io
oi
R
TT
Lk2rrln
TTQ
−=
π
−= So, for hollow cylinders
( ) =π−=dr
dTrL2kQ
Hence, the temperature profile in a cylinder is non-linear, q falls off inversely with r
Constant
dT/dr has to decrease as r increases
For slab, A is constant and so is dT/dr
A
Comparison of temperature distributions
Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures
22
12
11
2,1,
22
2,2,s
12
2,s1,s
11
1,s1,r
Lhr21
Lk2)r/rln(
Lhr21
TT
Lhr21
TT
Lk2)r/rln(TT
Lhr21
TTQ
π+π+π
−=
π
−=
π
−=
π
−= ∞∞∞∞
Temperature profile, can be calculated using
−+=
112
1,s2,s1,s r
rln
)r/rln(
)TT(T)r(T
Hollow Cylinder with Convection
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2.3.1 Radial Conduction in Solid Cylinder
If only one B.C is given at the outer surface, what should be done for the other B.C?
0dr
dTr
dr
d =
GDE is
For 1-D radial conduction, the temperature profile is symmetric about any radial line
So, B.C at r = 0 is
T(r)
r0
dr
dT
0r
==
2.3.2 Radial Conduction in Composite Coaxial Cylinders
44
3
4
2
3
1
2
11
2,1,
44
4,4,
3
4
4,3
2
3
32
1
2
21,
11
1,1,
21
2
ln
2
ln
2
ln
21
2
1
2
ln
2
ln
2
ln2
1
LhrLk
rr
Lk
r
r
Lk
rr
Lhr
TTQ
Lhr
TT
Lk
rr
TT
Lk
r
r
TT
Lk
r
r
TT
Lhr
TTQ
CBA
r
s
C
s
BA
ssr
πππππ
π
ππππ
+
+
+
+
−=
−=
−
=
−=
−
=−
=
∞∞
∞∞
Insulation thickness
Lhr2
1R;
Lk2r
rlnR
oconv
i
o
cond π=
π=
2.4 Critical radius of insulation
Q↑ as r↑if r<rc
Q↓ as r↑if r>rc
Insulation is useful
Representative Graph
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Lhr21
Lk2r
rln
TT
RR
TTQ
o
i
o
i
convcond
i
π+
π
−=+−= ∞∞
rc is determined by
For a particular ro/ri (ro=rc), total resistance is a minimum,hence Q is a maximum as
( ) 0Lhr2
1rlnrln
Lk2
1
dr
d
dr
dR
oio
oo
=
π+−
π=
0Lhr2
1
r
1
Lk2
12
cc
=π
−
π
⇒h
krc =
When ro=rc
or
Existence of critical thickness requires that the heat transfer area change in the direction of heat transferThis idiosyncrasy is of concern, as an insulation can actually increase the heat transfer if r<rc
This principle is used in electrical insulations
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2.5 Radial Conduction in Spherical Coordinates
0dr
dTr
dr
d 2 =
( ) ( ) 2211 TrT;TrT ==
12 C
dr
dTr =
21
r
C
dr
dT =
( ) 21 C
r
CrT +−=
1-D radial conduction, subject to first kind B.Cs
By integrating
Integrating again
or
Substituting B.Cs 22
122
1
11 C
r
CT;C
r
CT +−=+−=
General solution
Solving two equations( )
12
21211 rr
TTrrC
−−−=
( )12
21212 rr
TTrTC
−−−=and
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Substituting the constants in the general solution:
( ) ( )
−
−−−=
2
1
1
211
rr1
rr1
TTTrT
( ) ( ) ( ) ( )
k4r1
r1
TT
rr
TTrrk4
r
Cr4k
dr
dTr4kQ
21
21
12
21212122
π
−
−=−
−π=
π−=π−=
cond
21
R
TTQ
−=k4
r1
r1
R 21cond π
−
=
rc, when Rcond is a minimum can be shown to beh
k2rc =
By Fourier’s law
or
(Q)
A Comparison
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3. One-dimensional, Steady, Conduction with Heat ‘Generation’
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Thermal energy may be “generated” or “consumed” due to conversion from/to some other form of energy
If thermal energy is generated in the material at the expense of some other energy form, we have a source: is +ve– Deceleration and absorption of neutrons in a nuclear
reactor– Exothermic reactions – Conversion of electrical to thermal energy– Volumetric absorption of radiation
If thermal energy is consumed we have a sink: is -ve– Endothermic reactions
q ′′′
q ′′′
Caution: Heat flux depends on spatial coordinate. Hence, it would be incorrect to use the conduction resistance concepts
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Heat diffusion equation
0k
q
dx
Td2
2
=′′′
+
General Solution
3.1 Plane WallConsider one-dimensional, steady-state conduction in a plane wall of constant k, with uniform heat generation:
By integrating
Integrating again
1Cxk
q
dx
dT +′′′
−=
( ) 21
2
CxC2
x
k
qxT ++
′′′−=
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2,s1,s T)L(T;T)L(T ==−
Consider a slab as shown:
The problem can be treated by considering that Ts,1 and Ts,2 are known, (could be calculated using surface energy balance for other B.Cs)Substituting the B.Cs in general solution
( ) 21
2
1,s CLC2
L
k
qTLT +−
′′′−==−
( ) 21
2
2,s CLC2
L
k
qTLT ++
′′′−==
Solving two equations,L2
TTC 1,s2,s
1
−=
k2
Lq
2
TTC
22,s1,s
2
′′′+
+=and
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2
TT
L
x
2
TT
L
x1
k2
Lq)x(T 2,s1,s1,s2,s
2
22 ++
−+
−
′′′=
Temperature profile is parabolic (not independent of x)
Heat flux (may be determined from Fourier’s law) also dependent on x
Substituting constants in the general solution
s2
22
TL
x1
k2
Lq)x(T +
−
′′′=
Temperature distribution is symmetrical
s
2
Tk2
Lq)0(T +
′′′=
If both surfaces are maintained at a common temperature, Ts,1 = Ts,2 = Ts
Maximum temperature exists at the mid plane, x=0
Note that at the plane of symmetry (x=0),the temperature gradient is zero:
0q" 0dx
dT0x
0x
=⇒=
==
∴Mid plane is equivalent to adiabatic surface
( ) 2,s0x
TLT;0dx
dT ===
Hence, the problem may also be treated as steady, 1-D conduction with heat generation, subject to B.Cs
Calculation of surface temperature Ts
k
Lq
dx
dT
Lx
′′′−=⇒
=
)TT(hk
Lqk s ∞−=
′′′
−−∴
)TT(hdx
dTk s
Lx∞
=
−=−
h
LqTTs
′′′+= ∞
If the B.Cs are not of the first kind, the surface temperatures may be calculated using the surface energy balance
For convective B.C,
Differentiating the symmetrical temperature distribution
−
′′′= 2
2
L
x2k2Lq
dxdT
or
Ts may be eliminated in preceding equations
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Heat diffusion equation is:
0k
q
dr
dTr
dr
d
r
1 =′′′
+
1
2
Ck2rq
drdT
r +′′′
−=
By integrating (for uniform heat generation)
k
rq
dr
dTr
dr
d ′′′−=
or
k2
rq
r
C
dr
dT 1 ′′′−=
( ) 2
2
1 Ck4
rqrlnCrT +
′′′−=Integrating again General solution
L
hT ,∞
or
( ) so0r
TrT;0dr
dT ===
B.Cs areFirst condition results from the symmetry
3.2 Radial Conduction in a Solid Cylinder
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dr
dT1
0r
=→==
First B.C gives
( )k4
rqCTrT
2o
2so
′′′−==
Second B.C gives
k4
rqTC
2o
s2
′′′+=∴
( )k4
rqT
k4
rqrT
2o
s
2 ′′′++
′′′−=
Substituting the constants in the general solution, we get
( )
−
′′′+=
2
o
2o
s r
r1
k4
rqTrT
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)TT)(Lr2(h)Lr(q so2o ∞−π=π′′′
h2
rqTT o
s
′′′+= ∞
Calculation of surface temperature:
Either surface energy balance or an overall energy balance may be used
The second approach gives
or
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k
q
dr
dTr
dr
d
r
1 22 =
′′′+
3.3 Radial Conduction in a Solid SphereHeat diffusion equation is:
132 Cr
k3
q
dr
dTr +
′′′−= or
k
rq
dr
dTr
dr
d 22 ′′′
−=
By integrating
k3
rq
r
C
dr
dT21 ′′′
−=
( ) 2
212
1 Ck6
rq
12
rCrT +
′′′−
+−=
+−
Integrating again
or
General solution( ) 2
21 C
k6
rq
r
CrT +
′′′−−=
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drdT
10r
=∴==
( ) so TrT =
First B.C
Second B.C gives
( ) 2
2o
so Ck6
rqTrT +
′′′−==
k6
rqTC
2o
s2
′′′+=∴
Substituting the constants
( )
−
′′′+=
2
o
2o
s r
r1
k6
rqTrT
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4. Extended Surfaces (Fins) of Constant Area
Heat transfer by convection:
Stegosaurus( )∞−= TThAQ S
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ApplicationsCooling of IC enginesAutomobile “radiators”Electric power transformersTube-fin and plate-fin heat exchangersProcessor of a computer
Radiating fins in satellites, space shuttle, and experimental robots on other heavenly bodies
Fins are particularly beneficial when h is small,as for a gas and natural convection
Straight finUniform c/s area
Straight finTriangular, trapezoidal
Selection of a particular fin configuration may depend on:space, weight, manufacturing and cost considerationsextent to which the fin increase the pressure drop of flow
Annular fin Pin fin
Fin designs are limited only by imagination!
Externally finned tubing
Internally finned tubing
AssumptionsConduction is one-dimensional {along the length of fin, T(x), and steadyHeat loss (gain) is only by convection (no radiation)Thermal conductivity is constantHeat transfer coefficient is uniform (this can introduce serious errors!)
0QQQ convxxx =−− ∆+
dx
dTkAQ cx −=
dxdx
dQQQ x
xxx +=∆+
( )∞−= TThPdxQconv
Energy balance gives
where
(3.1)
4.1 Fin Equation and Solution
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( ) 0TThPdx
TdkA 2
2
c =−− ∞ ( ) 0TTkA
hP
dx
Td
c2
2
=−− ∞
0mdx
d 22
2
=θ−θ
c
2
kA
hPm =∞−=θ TT
2
2
2
2
dx
d
dx
Td θ=∴
We define
We get Fin equation
If k, Ac are constants
or
General solution
( ) mx2
mx1 eCeCx +=θ − or ( ) mxsinhCmxcoshCx 43 +=θ
Substitution gives
( ) 0TThPdxdxdx
dTkA
dx
dc =−−
∞
and
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Second B.C gives 0C2 =
or
( ) 10 C0 =θ=θ
( ) mx0ex −θ=θ
First B.C gives
Substitution gives mx
00e
TT
TT −
∞
∞ =−−=
θθ
or
( ) mx0 em
dx
d −−θ=θ∴
( ) 0c
c0c0x
mx0c
0xc kA
hPkAmkAemkA
dx
dkAQ θ=θ=−θ−=θ−=
=−
=
By Fourier’s law, heat transfer by fin is
4.1.1 Long Fin
0chPkAQ θ=
( ) ( ) ( ) 0TTx;TT0 00 →−→∞→θθ=−=θ ∞∞∞
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4.1.2 Short Fin with Adiabatic Tip
( ) 0dxd
;0Lx
0 =θθ=θ=
30 C=θ
B.Cs:
First B.C gives
General solution ( ) mxsinhCmxcoshCx 43 +=θ
0dx
dTk
Lx
=−=
( ) mxsinhCmxcoshx 40 +θ=θ∴
Differentiating mxcoshmCmxsinhmdx
d40 +θ=θ
Second B.C: 0mLcoshmCmLsinhmdx
d40
Lx
=+θ=θ=
because
mLcosh
mLsinhC 04 θ−=∴
(cosh 0 = 1 ; sinh x=0)
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−θ−=θ−=
= mLcosh
mLsinhmkA
dx
dkAQ 0c
0xc
ckA
hPm =
By differentiating( ) ( )
−−θ=θ
mLcosh
xLmsinhm
dx
xd0
mLtanhmkAQ 0c θ=
or mLtanhhPkAQ 0c θ=
By Fourier’s law
or ( )
−θ=θ
mLcosh
mxsinhmLsinhmLcoshmxcoshx 0
or ( ) ( )mLcosh
xLmcoshx 0
−θ=θ
or
Substitution gives ( )
−θ=θ mxsinh
mLcosh
mLsinhmxcoshx 0
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4.1.3 Short Fin with Convective Tip
( ) ( )Lhdx
dk;0
Lx0 θ=θ−θ=θ
=
30 C=θ
B.Cs:
First B.C gives
General solution ( ) mxsinhCmxcoshCx 43 +=θ
( ){ }∞=
−=− TLThdx
dTk
Lx
( ) mxsinhCmxcoshx 40 +θ=θ∴
Differentiating mxcoshmCmxsinhmdx
d40 +θ=θ
Second B.C gives
( ) ( )mLsinhCmLcoshhmLcoshmCmLsinhmk 4040 +θ=+θ−
because
mLsinhmkhmLcosh
mLcoshmkhmLsinh
C 04 +
+θ−=∴
(cosh 0 = 1 ; sinh x=0)
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Substituting the constants
( )
+
+−θ=θ mxsinh
mLsinhmkhmLcosh
mLcoshmkhmLsinh
mxcoshx 0
or
( ) ( )mLsinhmk
hmLcosh
mxsinhmLcoshmLsinhmxcoshmkhmxsinhmLsinhmLcoshmxcoshx
0 +
−+−=
θθ
or ( ) ( ) ( )mLsinhmk
hmLcosh
xLmsinhmkhxLmcoshx
0 +
−+−=
θθ
By differentiating ( ) ( ) ( )
+
−−−−θ=θ
mLsinhmkhmLcosh
xLmcoshmmkhxLmsinhm
dx
xd0
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+
−−θ−=θ−=
= mLsinhmkhmLcosh
mLcoshmmkhmLsinhm
kAdx
dkAQ 0c
0xc
ckA
hPm =
+
+θ=
mLsinhmkhmLcosh
mLcoshmkhmLsinh
mkAQ 0c
or
+
+θ=
mLsinhmkhmLcosh
mLcoshmkhmLsinh
hPkAQ 0c
By Fourier’s law
Comparison of various finsCase Tip Condition Temp. Distribution (θ/θb) Fin heat transfer
A Convection heat transfer:
hθ(L)= -k(dθ/dx)x=L mLmkhmL
xLmmkhxLm
sinh)(cosh
)(sinh)()(cosh
+
−+−
MmLmk
hmL
mLmkhmL
sinh)(cosh
cosh)(sinh
+
+
B Adiabatic (dθ/dx)x=L=0 mL
xLm
cosh
)(cosh −
mLM tanh
C Given temperature: θ(L)= θL
mL
xLmxLmb
L
sinh
)(sinh)(sinh)( −+−θθ
mL
mLM b
L
sinh
)(cosh θθ−
D Infinitely long fin θ(L)=0
mxe− M
bCbb
C
2
hPkAM,TT)0(
kA
hPm,TT
θθθ
θ
=−==
≡−≡
∞
∞
4.2 Fin Performance
Temperature distribution and Heat transferfor fin with adiabatic tip
Fin represents a conductive resistance to heat transfer from the original surfaceThere is no assurance that the heat transfer will be enhanced through the use of fins
( )
hA1kA
L
kA
LhA
kA
LhPLL
kA
hPmL
c
cc
2
c
2
=
===
Conductive resistance
Convective resistance=
PLA =
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4.2.1 Fin efficiency
Ideally, the fin material should have a large thermal conductivity to minimize the temperature variation from its base to the tipAs k tending to infinity, the entire fin would be at the base temperature, thereby providing the maximum possible heat transfer enhancement
Efficiency: Ratio of actual heat transfer by a fin to the heat that would be transferred if the entire fin were at the base temperature (Q/Qmax )
For infinite k, T(x)=T0
The heat transfer is maximum
( ) 00max hATThAQ θ=−= ∞
T0 x
T(x)<T0 for heat transferto take place
Real situation Ideal situation
x
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For long finmL
1LhP
kA
hPL
hPkA
hA
hPkA cc
0
0c ===θ
θ=η
For fin with adiabatic tipmL
mLtanh
hPL
mLtanhhPkA
0
0c =θ
θ=η
As (mL) tends to infinity, tanh (mL) tends to 0; efficiency tends to 0
As (mL) tends to 0, tanh (mL) tends to (mL); efficiency tends to 1 (100%)
It is not generally advisable to design toward a particular value of efficiency
maxQ
Q=η
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4.2.2 Fin Effectiveness
finwithout
finwith
Q
Q=εc0c
0
A
A
hA
hA η=θθη= maxQQ η=
Fin with adiabatic tip mLtanhhA
kP
A
P
kAhP
mLtanh
A
PL
mL
mLtanh
ccc
c
===ε
ChPkA tanh( )tanh( )
( )
If the fin is long enough, mL> , tanh(mL2 ) , 1
it can be considered an infinite fin (case D of table . )3 4
In order to enhance heat tra
f ff
C b C C
fC C
q q mL kPmL
q hA T T hA hA
kP k P
hA h A
ε
ε
∞
= = = =−
→
→ =
nsfer, .1
However, will be considered justifiabl2 e
If < then we have an insulator instead 1 of a heat fin
f
f
f
εε
ε
>
≥
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To increase effectiveness, the fin’s material should have higher thermal conductivity, k
It seems to be counterintuitive that the lower convection coefficient, h, the higher effectiveness. Therefore, fins are more effective if h is low
Observation: If fins are to be used on surfaces separating gas and liquid, fins are usually placed on the gas side
P/AC should be as high as possible.
Conclusion: It is preferred to use thin and closely spaced (to increase the total number) fins
fC C
kP k P
hA h Aε
→ =
http://h
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( )b0
0b0
AANh
hAhAN
+ηθ=θ+θη=
( )
0ot
0t
t
0tt
t
hA
1A
AN1hA
A
AN1
A
ANhA
θη=
θ
η−−=
θ
−+η=
Applicable for fin array
maxQQ η=
bt ANAA += NAAA tb −=∴
( ){ }NAAANhQ t0t −+ηθ=
( )η−−=θ
=η 1A
AN1
hA
Q
t0t
to
Qbase
Qfin
Ab: base area exposed to coolantA: surface area of a single finAt: total area including base area and At=Ab+NAN: total number of fins
4.2.3 Overall Surface Efficiency
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5. Transient Conduction
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5.1 Lumped System
No Internal Resistance:
It is assumed that the temperature variation in all three spatial directions is negligible and therefore temperature varies only with time – T(t)
This assumption is reasonable if internal resistance inside the solid is small compared to the external resistance in the fluid
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Consider a hot metal that is initially at a uniform temperature, Ti , and at t=0 is quenched by immersion in a cooler fluid
The temperature of the solid will decrease for time t>0, due to convection heat transfer at the solid-fluid interface, until it reaches bulk temperature of fluid
If the thermal conductivity of solid is very high, resistance to conduction within the solid will be small compared to resistance to heat transfer between solid and fluid
Temperature gradients within the solid will be negligible, i.e., the temperature of the solid is spatially uniform at any instant
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( ) iT0T =
let
( ) ii TT θ=−=θ ∞0;
The energy balance on solid:
dt
dTVc)TT(hAs ρ=−− ∞
stout EE =−
( ) ( ) ∞−=θ TtTtdt
d
dt
dT θ=∴
dt
dVchAs
θρ=θ−⇒
τθ−=θ
ρ−=θ
Vc
hA
dt
d s
shA
Vcρ=τor where, time constant
τ=
θθ dtd
Initial Condition:
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By integrating
1ln Ct +τ
−=θ
or τ−τ−+τ− ===θtCtCt
eCeee 211
( ) 2;0 Cii =θθ=θ
τ−θ=θ∴t
ie
or τ−=θθ t
i
e
( ) ( )t
Vc
hA
i
s
eTTTtT ρ∞∞ −+=
( ) tVc
hA
i
s
eTT
TtT ρ−
∞
∞ =−−or
or
The time required for the solid to reach temperature T is:θθτ= it ln
The total energy transfer, Q, occurring up to some time t is:
∫ θ=∫ ∫ θ== τtiS
t tS dthAdthAdtqQ 0
t-
0 0 e
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Need a suitable criterion to determine validity of method. Must relate relative magnitudes of temperature drop in the solid to the temperature difference between surface and fluid
Bik
hL
R
R
hA
kAL
T
T
conv
cond
liquidsolid
solid ≡===∆
∆)/1(
)/(
)convection todue(/
)conduction todue(
The lumped capacitance method is valid when
1.0<=k
hLBi c where the characteristic length:
Lc=V/As=Volume of solid/surface area
For a large Biot number conductive resistance controls and for a small Biot number convective resistance controls
Then the error in temperature calculation is less than 5 % using lumped parameter approach
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5.2 One-dimensional, Transient Conduction
• When the lumped capacitance analysis is not valid, we must solve the partial differential equations analytically or numerically
• Exact and approximate solutions may be used
• Tabulated values of coefficients used in the solutions of these equations are available
• Transient temperature distributions for commonly encountered problems involving semi-infinite solids can be found in the literature
http://h
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q’’’=0, convective BC, uniform initial condition
2
21
x
T
t
T
∂∂=
∂∂
α
( )( )
k
hLBitan
2sin2
sin4C
L
xcos)Foexp(C
TT
T)t,x(T)t,x(
nn
nn
nn
n2n
1nn
ii
*
==ζζ
ζ+ζζ=
ζζ−=
−−=
θθ=θ ∑
∞
=∞
∞
2L
t
c
kFo
ρ=
One Term Approximations or Heisler Charts can be Used for Fo >0.2
Graphical Representation of One-Term Approximation - Heisler Charts
Temperature Distribution:
Mid plane Temperature
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6. Two-dimensional, Steady Conduction in a Slab
http://h
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directions
Solution involves partial differential equations
• Need analytical, graphical or numerical approaches
• Analytical methods involve mathematical series and functions.
– Exact solutions
– Limited types of problems can be solved
0y
T
x
T2
2
2
2
=∂∂+
∂∂
Method of Separation of Variables
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GDE and B.Cs are
0y
T
x
T2
2
2
2
=∂∂+
∂∂
12
1
TT
TT
−−≡θ
(To get homogeneous BC’s)
( ) ( )( ) ( ) 21
11
TW,xT;T0,xT
Ty,LT;Ty,0T
====
x
y
T(x,y)
L
WT2
T1T1
T1
θ=1
θ=0
θ=0 θ=0
0yx 2
2
2
2
=∂
θ∂+∂
θ∂(1)
( ) ( ) ( )yYxXy,x =θAssume the solution of the form (2)
( ) ( )( ) ( ) 1W,x;00,x
0y,L;0y,0
=θ=θ=θ=θ
GDE: B.Cs:
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2
2
2
2
dy
Yd
Y
1
dx
Xd
X
1 λ==−
Substituting (2) in (1) and dividing by XY
0Xdx
Xd 22
2
=λ+ 0Ydy
Yd 22
2
=λ−or and
y4
y3 ececY λλ− +=( )xsincxcoscX 21 λ+λ=
General solutions are
and
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4y
321 ececxsincxcoscXY λλ− +λ+λ==θ∴
( ) 0c0y,0 1 =→=θ ( )y4
y32 ececxsinc λλ− +λ=θ∴
( ) ( )432 ccxsinc00,x +λ==θ 43 cc −=→
( )y4
y42 ececxsinc λλ− +−λ=θ ( )yy
42 eexsincc λ−λ −λ=θor
n=0 precluded (unacceptable)
0Lsin =λ
L
nπ=λ∴
The only way to satisfy this is
( ) 0y,L =θ ( ) 0eeLsincc yy42 =−λ→ λ−λ
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−π=θ∴
π−πL
ynL
yn
42 eeL
xnsincc
Infinite solutions present, as the problem is linear, general solution, by superposition is
L
ynsinh
L
xnsinc)y,x( n
ππ=θ
∑ππ=θ
∞
=1nn L
ynsinh
L
xnsinc)y,x(
1L
Wnsinh
L
xnsinc
1nn =∑
ππ→∞
=
Denoting 2c2c4, new constant may depend on n
1)W,x( =θ
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{ }π
+−=+
n
1)1(2A
1n
n
Comparing with Fourier series ( ) 1L
xnsinAxf
1nn =∑
π=∞
=
{ }L
Wnsinhn
1)1(2
LWn
sinh
Ac
1nn
n ππ+−=π=∴
+
Substituting cn in the general solution
( ) ( )∑ π
ππ+−
π=θ
∞
=
+
1n
1n
LWnsinh
Lynsinh
L
xnsin
n
112y,x
1L
xnsinA
1nn =∑
π∴∞
= L
WnsinhcA nn
π=where
