(4 5) 1d ss conduction part1.pdf
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1D STEADY STATE HEAT
Associate Professor
IIT Delhi
E-mail: [email protected]
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Heat conductionat the surface in a
selected direction
=Heat convectionat the surface in
the same direction
In writing the equations for convection
boundar conditions we have selected
the direction of heat transfer to be the
positive x-direction at both surfaces. But
those expressions are equally applicable
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w en ea rans er s n e oppos e
direction
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Heat conductionat the surface in a
selected direction
=Radiation exchangeat the surface in the
same direction
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The boundary conditions at an interface arebased on the requirements that
(1) two bodies in contact must have the
same temperature at t e area o contact
and
(2) an interface (which is a surface) cannotstore an ener and thus
the heat flux on the two sides of an
interface must be the same
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Generalized Boundary
Conditions
Heat
transfertothesurface Heat
transferfromthesurface=
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Solution of steady heat
conduction equation1D Cartesian
Differential Equation: Boundary Condition:
0dx
Td2
2= ( ) 10 TT =
Integrate:
dT
Applying the boundary condition to the general solution:
( ) 21 CxCxT +=1
dx=
Integrate again:
00
( ) 21 CxCxT += 1T
Substituting:
C0.CT += =
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enera o u on r trary onstants
It cannot involve x or T(x) after theboundary condition is applied.
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-Differential Equation:
Differential Equation:
0)dr
dTr(dr
d
= 0)dr
dTr(dr
d 2 =
Integrate:
1CdTr =
Integrate:
12 CdTr =
Divide by r :)0( rCdT 1=
r
Divide by r2 :)0( r
CdT
rdr
Integrate again:
=
2rdr =
Integrate again:
PTalukdar/Mech-IITD which is the general solution.
( ) 21 C
rrT +=
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During steady one-dimensional
heat conduction in a s herical or
cylindrical) container, the total rateof heat transfer remains constant,
but the heat flux decreases with
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ncreas ng ra us.
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Under steady conditions, the energy
balance for this solid can be expressed as
Rateof
heat
Rate
of
energy
transfer
fromsolid
hA (T T )
generationwithin
thesolid
Vg&
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ss
hATT +=
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s wall wall ,A long solid cylinder of radius ro (As = 2ro L and V= r
2o L),
A solid sphere of radius r0 (As = 4r2o L and V= 4/3r
3o )
ss
hA
gVTT +=
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Under stead conditions theentire heat generated within the
medium is conducted through
The heat enerated within this inner c linder must
.
be equal to the heat conducted through the outer
surface of this inner cylinder
Integrating from r = 0 where T(0) = T0 to r = ro where T(ro) = Ts yields
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The maximum temperaturein a symmetrical solid with
uniform heat generation
occurs at its center
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Rate of heat
transfer into the =Rate of change of
energy of the wall
Rate of heat
transfer out of the-
wall wall
dEQQ walloutin =
0
dt
dEwall = for steady operation
Therefore, the rate of heat transfer into the wall must be equal to the rate
of heat transfer out of it. In other words, the rate of heat transfer through
the wall must be constant, Qcond, wall constant.
Fouriers law of heat conduction for the walldx
kAQ wall,cond =
2TL
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cons anA TxQ1TT
wall,cond0x ==
=
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TT
L
wall,cond =
The rate of heat conduction through a
plane wall is proportional to the,
wall area, and the temperature
difference, but is inversely
proport ona to t e wa t c ness
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Temp profile
dTd
Integrate the above equation twice dxdx
=
( ) 21 CxCxT +=
Apply the condition at x = 0 and L
1,s= an 2,s=
CT =,
1,s1212,s TLCCLCT +=+=
1
,s,s
CL =
1,s2,s TT =
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,sL
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Analogy between thermal and
electrical resistance concepts
(W)
wall
21wall,cond
R
TTQ
=&
PTalukdar/Mech-IITD kA
LR
wall = (oC/W)
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)TT(hAQ ssconvection
=
s TT
convectionconvect on
R
convection 1R = (oC/W)s
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(W)
rad
surrssurrssrad
4surr
4ssrad
R
TT)TT(Ah)TT(AQ
===
(K/W)srad
rad Ah
1
R =
Combined convection and radiation
(W/m2K))TT)(TT()TT(A
Qh surrs
2surr
2s
surrss
radrad ++=
=
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Possible when T
= Tsurr(W/m2K)radconvcombined hhh +=
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The thermal resistance network for heat transfer through a plane wall subjected
to convection on both sides, and the electrical analogy
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Rate of heat
convection into =Rate of heat
convection from the
Rate of heat
conduction=
the wall wallthrough the wall
)()( 22221
111
=
== TTAhTT
kATTAhQ
Ah
TT
kAL
TT
Ah
TTQ
2
2221
1
11
11
=
=
=
Addin the numerators and denominators ields
2,
2221
1,
11
convwallconv R
TT
R
TT
R
TT =
=
=
totalR
TTQ 21 = (W)
PTalukdar/Mech-IITD AhkA
L
Ah
RRRR convwallconvtotal21
2,1,
11++=++=
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TT
totalR
=
The ratio of the temperature drop to the
thermal resistance across any layer is
constant, and thus the temperature drop
across any ayer s propor ona o e
thermal resistance of the layer. The larger
the resistance, the larger the temperature
dro .
RQT
= (oC)
This indicates that the temperature drop across
any layer is equal to the rate of heat transfer
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It is sometimes convenient
through a medium in an
analogous manner to
Newtons law of cooling as
T&
TUAQ =
(W)
1UA=
totalR=
totalR
The surface temperature of the wall can be
determined using the thermal resistance TTTT 1111
concept, but by taking the surface at which thetemperature is to be determined as one of the
terminal surfaces.Ah
Rconv1
1, 1
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