1D STEADY STATE HEAT CONDUCTION (2)CONDUCTION (2)

Prabal TalukdarPrabal TalukdarAssociate Professor

Department of Mechanical EngineeringDepartment of Mechanical EngineeringIIT Delhi

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

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Thermal Contact ResistanceThermal Contact Resistance

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Temperature distribution and heat flow lines along two solid plates pressed against each other for the case of perfect and imperfect contact

Consider heat transfer through two metal rods of cross-sectional area A thatare pressed against each other Heat transfer through the interface of these twoare pressed against each other. Heat transfer through the interface of these tworods is the sum of the heat transfers through the solid contact spots and thegaps in the noncontact areas and can be expressed as

Most experimentally determinedMost experimentally determinedvalues of the thermal contact resistance fall between 0.000005 and 0.0005 m2∙°C/W (the corresponding range of thermal contact 

gapcontact QQQ•••

+=

•

where A is the apparent interface area (which is the same as the cross-sectional area of the rods) and ΔT is the effective temperature

conductance is 2000 to 200,000 W/m2∙°C).erfaceintc TAhQ Δ=

sectional area of the rods) and ΔTinterface is the effective temperature difference at the interface. The quantity hc, which corresponds to the convection heat transfer coefficient, is called the thermal contact conductance and is expressed as

It is related to thermal contact resistance byerfaceint

c TAQh

Δ=

•

(W/m2 oC)

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It is related to thermal contact resistance by

AQ

Th1R erfaceint

cc •

Δ== (m2 oC/W)

Importance of considerationImportance of consideration

Th th l t t i tThe thermal contact resistance range:

between 0.000005 and 0.0005 m2·°C/W

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Two parallel layersTwo parallel layers

⎞⎛ 11TTTT⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

−+

−=+=

•••

2121

2

21

1

2121

11)(RR

TTR

TTR

TTQQQ

L L

totalRTTQ 21 −

=•

'11

11 Ak

LR = '22

22 Ak

LR =

PTalukdar/Mech-IITD21

111RRR total

+=21

21

RRRRR total +

=where

Combined series-parallelCombined series parallelTTQ ∞

• −= 1

totalRQ =

RRconvconvtotal RR

RRRRRRRR +++

=++= 321

21312

'11

11 Ak

LR = '22

22 Ak

LR = '33

33 Ak

LR =

1

3

1hA

R conv =

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Series and parallel composite wall and its thermal circuit

RA

RD

T∞1

AR∞1

RB

RC RE

RF

R∞2

T∞2

T T TT2C1 R

1111R

111RR ∞∞ +

⎟⎞

⎜⎛

++⎟⎞

⎜⎛

+=∑T1 T3 T4T2

FEDBA R1

R1

R1

R1

R1

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

TUAQ Δ=•

(W)

where U is the overall heat transfer coefficient

PTalukdar/Mech-IITD totalRUA 1

=

Complex multi-dimensional problems as 1-D problems

1 Any plane wall normal to the x-axis is isothermal1. Any plane wall normal to the x-axis is isothermal

2. Any plane parallel to x-axis is adiabatic

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Heat conduction in cylinderHeat conduction in cylinderdTkAQ cylcond −=

•

L2AdrQ cyl,cond rL2A π=

kdTdrQ

22 Tcyl,condr

•

∫=∫

Substituting A = 2πrL and performing the integrations give

kdTdrA

2

1

2

1 TTrr == ∫−=∫

)rrln(TTLk2Q

12

21cyl,cond

−π=

•cyl,condQ

•

=constant at steady state

cyl

21cyl,cond R

TTQ −=

•(W)

PTalukdar/Mech-IITD Lk2)rrln(R 12

cyl π= ln(outer radius/inner radius)

2π(length)(thermal conductivity)=

Heat conduction in sphereHeat conduction in sphere

For sphere sph

21sphere,cond R

TTQ −=&

krr4rrR21

12sph π

−= = outer radius - inner radius

4π(outer radius)(inner radius)(thermal conductivity)

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Resistance NetworkResistance Networkcylindrical

2,convcond1,convtotal RRRR ++=

( )( )12 1rrln1

++=

spherical

( ) 2211 h)Lr2(Lk2hLr2 π+

π+

π

2,convsph1,convtotal RRRR ++=

( )12 1rr1

+−

+

The thermal resistance network for a cylindrical (or spherical) shell subjected to convection from

( ) 22

221

12

12

1 h)r4(krr4hr4 π+

π+

π=

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spherical) shell subjected to convection fromboth the inner and the outer sides.

Multilayered cylinderMultilayered cylinder

2,conv3,cyl2,cyl1,cyl1,convtotal RRRRRR ++++=

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( ) ( ) ( )423

34

2

23

1

12

11 Ah1

Lk2rrln

Lk2rrln

Lk2rrln

Ah1

+π

+π

+π

+=

Radial heat conduction through cylindrical systems

tTCg

zT.k

zTr.k

r1

rTr.k

rr1

2 ∂∂ρ=+⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛φ∂∂

φ∂∂+⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂

&

0drdTr

drd

=⎟⎠⎞

⎜⎝⎛

Integrating the above equation twice T=C ln r + C

⎠⎝

Subject to the boundary conditions, T=T1 at r = r1 and T=T2 at r = r2

Integrating the above equation twice, T=C1ln r + C2

( )⎟⎟⎠

⎞⎜⎜⎝

⎛−

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

1

2

1221

1

2

12

rrln

rlnTrlnTrln

rrln

TTT

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⎠⎝⎠⎝ 11 rr

Cd

1

11rrr r

C.Lr2.k|drdTkAQ

1π−=−= =

⎟⎟⎠

⎞⎜⎜⎝

⎛−π−=

1

21

121

rrlnr

1).TT.(Lr2.kQ

⎠⎝ 1

⎟⎟⎠

⎞⎜⎜⎝

⎛−π

=2

21rln

)TT(kL2

⎟⎠

⎜⎝ 1r

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Critical Radius of InsulationCritical Radius of Insulation

1. Steady state conditions

2. One-dimensional heat flow only in the radial direction

r2 Insulation

T h3. Negligible thermal resistance due to

cylinder wall

4 Negligible radiation exchange

r1 Thin wallT∞ , h

T4. Negligible radiation exchange between outer surface of insulation and surroundings

Ts

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Critical Radius of InsulationCritical Radius of Insulation • Practically, it turns out that adding insulation in cylindrical and spherical

exposed walls can initially cause the thermal resistance to decrease, thereby p y , yincreasing the heat transfer rate because the outside area for convection heat transfer is getting larger. At some critical thickness, rcr, the thermal resistance increases again and consequently the heat transfer is reduced.

• To find an expression for rcr, consider the thermal circuit below for an insulated cylindrical wall with thermal conductivity k:

r InsulationT1 T s T∞Q&

r2

r1

Insulation

Thin wallT∞ , h hr2

12π

( )k2rrln 12

π1 Thin wall

TsRt’

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An insulated cylindrical pipe exposed to convection from the outer surface and the thermal resistance networkand the thermal resistance network associated with it.

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• To find rcr, set the overall thermal resistance dRt’/dr = 0 and solve for r:

( )rh2

1k2rrlnR i

t π+

π=′ ri = inner radius

0hr2

1kr2

1drRd

2t =

π−

π=

′

k k2

• For insulation thickness less that r the heat loss increases

hkrr cr ==

hk2rcr =Similarly for a sphere

• For insulation thickness less that rcr the heat loss increases with increasing r and for insulation thickness greater that rcr the heat loss decreases with increasing r

• If k = 0.03 W/(m∙K) and h = 10 W/(m2∙K):– cylinder mm3m003.0

·K)W/(m10W/(m·K)03.0

hkr 2cr ====

– sphere

)(

mm6hk2rcr ==

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Values of r1, h and k are constant

To see the condition maximizes or minimizes the total resistance

11Rd2

At r =k/h

hr1

kr21

drRd

32

22

22

total2

π+

π−=

Total thermal resistance per unit length

At r2=k/h

rln 2⎟⎟⎞

⎜⎜⎛ ( )

0hk2

1k21

k1

hk1

drRd

23222

total2

>π

=⎟⎠⎞

⎜⎝⎛ −

π=

Heat transfer per unit length

Always positive, total resistance at k/h is minimumhr2

1k2r

lnR

2

1total π

+π

⎟⎟⎠

⎜⎜⎝=

iTTQ −= ∞

•

kr = (m)p g

Optimum thickness is associated with r2,

totalRL

02

=dr

dRtotal

hr cylinder,cr = (m)

2dr0

hr21

kr21

222

=π

−π h

kr2 =

( )( ) cm1m01.0

CmW5CmW05.0

hk

r o2

o

min

insulationmax,max,cr ==≈=

We can insulate hot water pipes and steam lines without worrying the critical radius of insulation

Insulation of electric wires:

-Radius of electric wires may be smaller than the critical radius

Addition of insulation material increases heat transfer-Addition of insulation material increases heat transfer

Critical radius of insulation for spherical shell: hk2r sphere,cr =

Summary• Table 3 3

SummaryTable 3.3

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1D Conduction with Heat Generation

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02

+

•

qTd 02 =+kdx

( ) 2 CxCxqxT ++=

•

Boundary conditions:

( ) 212CxCx

kxT ++−=

y

( ) 1,sTLT =−L

TTC ss

21,2,

1

−=

( ) 2,sTLT =

L2

221,2,2

2ss TT

Lk

qC+

+=

•

•

PTalukdar/Mech-IITD 221

2)( 1,2,1,2,

2

22ssss TT

LxTT

Lx

kqLxT

++

−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

•

sss TTT ≡= 2,1,

xqL ⎟⎞

⎜⎛

•22

1)( sTLx

kqLxT +⎟⎟

⎠

⎞⎜⎜⎝

⎛−= 21

2)(

L•

2

If the surface temperature of the heat generating body is

Put x = 0so Tk

qLTT +=≡2

)0(2

If the surface temperature of the heat generating body is unknown and the surrounding fluid temperature is T∞

Using energy balanceFind temperature gradientfrom the above Eq. at x = L)(| ∞= −=− TTh

ddTk sLx •g gy

We can obtain the surface temperature

q

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dx

hLqTTs

•

∞ +=

01=+⎟

⎠⎞

⎜⎝⎛

•

kq

drdTr

drd

r ⎠⎝ kdrdrr

1

2

2C

krq

ddTr +−=

•

12kdr

21

2

ln4

)( CrCkrqrT ++−=

•

Boundary conditions:

4k

dT TT )( so TrkrqrT +⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

•

2

22

14

)(0| 0 ==rdr

dT so TrT =)(

rqTC o2

•

ork ⎟⎠

⎜⎝4

( ) ))(2(2∞

•

−Π=Π TTLrhLrq soo

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01 =C kqTC o

s 42 +=

hrqTT o

s 2

•

∞ +=