1D STEADY STATE HEAT CONDUCTION (1)CONDUCTION (1)
Prabal TalukdarPrabal TalukdarAssociate Professor
Department of Mechanical EngineeringDepartment of Mechanical EngineeringIIT Delhi
E-mail: [email protected]
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Convection Boundary ConditionConvection Boundary Condition
Heat conductionat the surface in a selected direction
=Heat convection at the surface in the same direction
In writing the equations for convection boundary conditions, we have selectedboundary conditions, we have selectedthe direction of heat transfer to be the positive x-direction at both surfaces. Butthose expressions are equally applicable
h h t t f i i th it
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when heat transfer is in the oppositedirection
Radiative Boundary ConditionRadiative Boundary Condition
Heat conductionat the surface in a selected direction
=Radiation exchangeat the surface in the same direction
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Interface Boundary ConditionsInterface Boundary Conditions
The boundary conditions at an interface are based on the requirements that(1) two bodies in contact must have the
h fsame temperature at the area of contactand (2) an interface (which is a surface) cannot store any energy, and thusy gy,the heat flux on the two sides of an interface must be the same
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Generalized Boundary Conditions
Heat transferto the surfacein all modes
Heat transferfrom the surfacein all modes
=
in all modes in all modes
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Solution of steady heat conduction equation
1D CartesianDifferential Equation: Boundary Condition:
0dx
Td2
2= ( ) 10 TT =
Integrate:
CdT
=
Applying the boundary condition to the general solution:
( ) 21 CxCxT +=1C
dx=
Integrate again:
00
( ) 21 CxCxT +=
G l S l ti A bit C t t
1TSubstituting:
211 C0.CT +=12 TC =
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General Solution Arbitrary Constants 211 12 TC
It cannot involve x or T(x) after the boundary condition is applied.
Cylindrical - SphericalCylindrical SphericalDifferential Equation: Differential Equation:
0)drdT
r(drd
=0)
drdT
r(drd 2 =
Integrate:
1CdrdT
r =
Integrate:
12 C
drdT
r =dr
Divide by r :)0( ≠rCdT 1=
dr
Divide by r2 :)0( ≠r
1CdTrdr
Integrate again:( ) 21 CrlnCrT +=
21
rdr=
Integrate again:C
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( ) 21 CrlnCrT +
which is the general solution.
( ) 21 C
rC
rT +−=
During steady one-dimensionalheat conduction in a spherical (orheat conduction in a spherical (orcylindrical) container, the total rateof heat transfer remains constant,but the heat flux decreases withi i di
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increasing radius.
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Heat GenerationHeat Generation
Under steady conditions, the energy balance for this solid can be expressed as
Rate of heat Rate of energy =transfer
from solidhAs(Ts‐T∞)
generation within the solid
=
Vg&s( s ∞) g
gV•
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ss hA
gVTT ∞ +=
A large plane wall of thickness 2L (A = 2A and V = 2LA )A large plane wall of thickness 2L (As = 2Awall and V = 2LAwall), A long solid cylinder of radius ro (As = 2πro L and V= πr2
o L), A solid sphere of radius r0 (As = 4πr2
o L and V= 4/3πr3o )
•
ss hA
gVTT
•
∞ +=
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Under steady conditions, the y ,entire heat generated within the medium is conducted through the outer surface of the cylinder
The heat generated within this inner cylinder must
the outer surface of the cylinder.
g ybe 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 temperature in a symmetrical solid with uniform heat generation occurs at its center
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1-D plane wall1 D plane wall
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Energy balanceEnergy balanceRate of heat transfer into the =
Rate of change of energy of the wall
Rate of heat transfer out of the -
wallgy
wall
dtdE
QQ walloutin =−
••
dt
0dt
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.
dT•Fourier’s law of heat conduction for the wall
t t
dxdT
kAQ wall,cond −=•
kAdTdQ2TL •∫∫
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constantkAdTdxQ1TT
wall,cond0x ==
∫−=∫
Temp profileTemp profileTT
kAQ 21 −•(W)
LkAQ 21
wall,cond = (W)
The rate of heat conduction through a plane wall is proportional to the average thermal conductivity theaverage thermal conductivity, the wall area, and the temperature difference, but is inversely
i l h ll hi kproportional to the wall thickness
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Temp profile
1 D steady state heat conduction equation 0)dT
k(d1 D steady state heat conduction equation
Integrate the above equation twiceBoundary conditions
0)dx
k(dx
=
( ) 21 CxCxT +=
T)0(T and T)L(TBoundary conditions
Apply the condition at x = 0 and L1,sT)0(T = and 2,sT)L(T =
21s CT = 21,s C
1,s1212,s TLCCLCT +=+=
12 TT −1
1,s2,s CL
TT=
11,s2,s Tx
TT)x(T +
−=
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1,sTxL
)x(T +=
Thermal Resistance ConceptThermal Resistance Concept
Analogy between thermal and electrical resistance concepts
(W)wall
21wall,cond R
TTQ
−=&
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R wall = (oC/W)
Convection ResistanceConvection Resistance •
)TT(hAQ ssconvection ∞−=
si
TTQ ∞• −
= (W)convection
convection RQ =
convection hA1
R =
(W)
(oC/W)s
convection hA
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Radiation ResistanceRadiation Resistance (W)
rad
surrssurrssrad
4surr
4ssrad R
TT)TT(Ah)TT(AQ
−=−=−εσ=
•
(K/W)srad
rad Ah1
R =
Combined convection and radiation
(W/m2K))TT)(TT()TT(A
Qh surrs
2surr
2s
surrss
radrad ++εσ=
−=
•
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Possible when T∞ = Tsurr(W/m2K)radconvcombined hhh +=
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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Network subjected to convection on both sidesNetwork subjected to convection on both sides
Rate of heat convection into =
Rate of heat convection from the
Rate of heat conduction =
the wall wallthrough the wall
)()( 22221
111 ∞∞
•
−=−
=−= TTAhLTTkATTAhQL
AhTT
kALTT
AhTTQ
2
2221
1
11
11∞∞
• −=
−=
−=
Adding the numerators and denominators yields2,
2221
1,
11
convwallconv RTT
RTT
RTT ∞∞ −
=−
=−
=
g y
totalRTTQ 21 ∞∞
• −= (W)
PTalukdar/Mech-IITDAhkA
LAh
RRRR convwallconvtotal21
2,1,11
++=++=
TTQ 21 ∞∞• − (W)
totalRQ 21 ∞∞= (W)
The ratio of the temperature drop to the thermal resistance across any layer is constant, and thus the temperature drop
l i ti l t thacross any layer is proportional to the thermal resistance of the layer. The larger the resistance, the larger the temperaturedrop.p
RQT•
=Δ (oC)
This indicates that the temperature drop across any layer is equal to the rate of heat transfer times the thermal resistance across that layer
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times the thermal resistance across that layer
It is sometimes convenient to express heat transferto express heat transfer through a medium in an analogous manner to Newton’s law of cooling as
TQ
Δ&
TUAQ Δ=•
(W)
1UA =
totalRQ =
totalR
The surface temperature of the wall can be determined using the thermal resistance TTTTQ 1111 −
=−
= ∞∞•
concept, but by taking the surface at which the temperature is to be determined as one of the terminal surfaces.
Known
AhRQ
conv1
1,1
==
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Known
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