1D STEADY STATE HEAT CONDUCTION (1)CONDUCTION (1)

Prabal TalukdarPrabal TalukdarAssociate Professor

Department of Mechanical EngineeringDepartment of Mechanical EngineeringIIT Delhi

E-mail: prabal@mech.iitd.ac.inp

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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