2-Heat Conduction Equation
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Transcript of 2-Heat Conduction Equation
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Heat transfer has direction as well as
magnitude, and thus it is a vector quantity
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The various distances and angles involved when describing the
.
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Fouriers law of heat conduction
for one-dimensional heat conduction:
)Watt(dx
dTkAQcond =&
If n is the normal of the isothermal surface atpoint P, the rate of heat conduction at that point
can be expressed by Fouriers law as
The heat transfer vector is
)Watt(n
TkAQn
=&
always normal to an isothermal
surface and can be resolved
into its components like any
kQjQiQQ zyxn
r&
r&
r&
r& ++=
P.Talukdar/Mech-IITD x
T
kA
Q xx
=&
y
T
kAQ yy
=&
z
T
kAQ zz
=&
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Steady versus Transient Heat
Transfer
The term stead im lies no
change with time at any pointwithin the medium, while
with time or time dependence.
Therefore, the temperature orheat flux remains unchanged
with time during steady heat
transfer throu h a medium at
any location, although both
quantities may vary from one
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Multidimensional Heat
Transfer
Heat transfer roblems are also classified as bein one-
dimensional, two-dimensional, or three-dimensional,depending on the relative magnitudes of heat transfer rates in
Ex:1Dheattransfer:
Heattransfer
through
the
glass
of
a
w n owcan econs ere to eone
dimensionalsinceheattransferthrough
theglasswilloccurpredominantlyinone
direction(thedirectionnormaltothe
surfaceof
the
glass)
and
heat
transfer
in
otherdirections(fromoneside
edgetotheotherandfromthetopedge
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A medium through which heat is conducted may involve the
, ,
(or thermal) energy. In heat conduction analysis, suchconversion processes are characterized as heat generation.
Heat generation is a volumetric phenomenon. That is, it occurs
throughout the body of a medium. Therefore, the rate of heat
whose unit is W/m3
The rate of heat generation in a
as position within the medium.
When the variation of heat
generation with position is known,= dVgG && Watt
P.Talukdar/Mech-IITD
g p
the total rate of heat generation in
a medium of volume V can bedetermined from
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1 D Heat Conduction Equation
Assume the density of the wall is , the specificheat is C, and the area of the wall normal to thedirection of heat transfer is A.
An energy balance on this thin element during
t
GQQ elementelementxxx
=+ + &&&
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)TT(x.A.C)TT(mCEEE tttttttttelement === +++
x.A.gVgG elementelement == &&&
E&&&
t
elementxxx
=+ +
TT &&
Dividin b
tx..x..gxxx
=+ +
&&Ax gives
tCg
xA
tttxxx
=+
++ &
Taking the limit as x0 and t0 yields and since from Fouriers Law:
=
=+
x
TkA
xx
Q
x
QQ xxxlim
&&&
t
Cg
x
kA
xA
=+
&
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Plane wall: A is constant
Variable conductivity:t
Cgx
kx
=+
&
2&
Constant conductivity:
where the property k/C is the thermal
tk
g
x2 =+
diffusivity
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Heat Conduction Equation in a
L C li dong Cylinder
ons er a n cy n r ca s e e emen o
thickness r in a long cylinder
The area of the c linder normal to the
direction of heat transfer at any location is A =
2rL where r is the value of the radius at that
location. Note that the heat transfer area A,
location.
&&&
tGQQ e ementelementrrr
=+ +
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)TT(r.A.C)TT(mCEEE tttttttttelement === +++
r.A.gVgG elementelement == &&&
t
r.A.Cr.A.gQQ tttrrr =+ ++ &
&&dividing by Argives t
CgrA
tttrrr
=+
++ &
T&&&
== rrrrrrr
0rm
TT1 &
tg
rrA =
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TT1
&
tr
..
rr
=
&Constant Conductivity:
tkrr
rr =+
dTd1 &
krdr
drr=+
t
T1
r
Tr
rr
1
=
0rd
dTr
dr
d=
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.
A = 4r2
Variable conductivity:
t
TCg
r
T.k.r
rr
1 22
=+
&
Constant Conductivity:t
T1
k
g
r
Tr
rr
1 22
=+
&
P.Talukdar/Mech-IITD
CombinedOneDimensional
Heat
Conduction
Equation t
TCg
r
T.k.r
rr
1 nn
=+
&
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General Heat Conduction
Equation
&&&&&&&
tGQQQQQQ e ementelementzzyyxxzyx
=+++ +++
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)TT.(z.y.x.C)TT(mCEEE tttttttttelement === +++
z.y.x.gVgG elementelement == &&&
t
EGQQQQQQ elementelementzzyyxxzyx
=+++ +++ &&&&&&&
t
TT
z.y.x.Cz.y.x.gQQQQQQttt
zzyyxxzyx
=+++ +
+++ &
&&&&&&
TTQQ1QQ1QQ1 tttzzzyyyxxx ++++ &&&&&&&
tzy.xyz.xxz.y
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TT1QQ11 &&&&&&&
tg
zy.xyz.xxz.y =+
&&&
=
=
=
+
x
k
xx
z.y.k
xz.yxz.yxz.y
lim xxxx
0x
=
=
=+ T
kT
z.x.k1Q1QQ1
limyyyy
&&&
...
=
=
=
+
z
Tk
zz
Ty.x.k
zy.x
1
z
Q
y.x
1
z
QQ
y.x
1lim zzzz
0z
&&&
t
TCg
z
Tk
zy
Tk
yx
Tk
x
=+
+
+
&
Under what condition?
222&
tkzyx222
=+
+
+
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0gTTT 222=+
+
+
&
zyx
T1TTT2
2
2
2
2
2 =
+
+
0z
TT
x
T2
2
2
2
2
2
=
+
+
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TTT1T1
&tz
.z
.rr
.rr 2
TT1T1T1 2 &
t.
sinrsinrr.
rr 2222
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Boundary and Initial Conditions
The temperature distribution in a medium depends on the
conditions at the boundaries of the medium as well as the heat
transfer mechanism inside the medium. To describe a heat
transfer problem completely, two boundary conditions must be
heat transfer is significant.
T ere ore, we nee to spec y two
boundary conditions for one-dimensional
problems, four boundary conditions for
- ,
boundary conditions for three-dimensional
problems
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con on, w c s usua y spec e a me = , s ca ethe initial condition, which is a mathematical expression for
the temperature distribution of the medium initially.
)z,y,x(f)0,z,y,x(T =
Note that under steady conditions, the heat conduction
equation does not involve any time derivatives, and thus we do
not need to s ecif an initial condition
The heat conduction equation is first order in time, and thus the initial
condition cannot involve any derivatives (it is limited to a specified
temperature).However, the heat conduction equation is second order in space
coordinates, and thus a boundary condition may involve first
er vat ves at t e oun ar es as we as spec e va ues o temperature
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Specified Temperature Boundary
C di iondition
The temperature of an exposed surface can
usually be measured directly and easily.
Therefore, one of the easiest ways to
specify the thermal conditions on a surface
is to specify the temperature. For one-
dimensional heat transfer through a planewall of thickness L, for example, the
specified temperature boundary conditions
can be expressed as
1T)t,0(T =Tt,LT =
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Specified Heat Flux Boundary
ondition
determined by inspection: positive ifthe heat flux is in the positive
rec on o e coor na e ax s, an
negative if it is in the opposite
direction.
Note that it is extremely important to
have the correct si n for the s ecified
heat flux since the wrong sign will
invert the direction of heat transfer
interpreted as heat loss
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2medium from both sides, for example, the specified heat flux boundary
conditions can be expressed as
50x
)t,0(Tk =50
x)t,L(Tk =
and
Special Case: Insulated Boundary
0x
)t,0(Tk =
0x
)t,0(T=
or
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t,L
T
0x
=