Chapter 2faculty.kfupm.edu.sa/CHE/shammakh/files/My files/300_Chapter_2.pdf · Chapter 2 Chee 318 2...

Chapter 2 Chee 318 1

Chapter 2

ONE DIMENSIONAL STEADY STATE CONDUCTION


HEAT CONDUCTION THROUGH COMPOSITE RECTANGULAR WALLS

→∆Χ←→∆Χ←→∆Χ← CBA

A B

C

Temperature profile

q q

T1 X1

T2 X2

T3 X3

T4 X4


Thermal conductivity

• Fourier’s law (k is constant)

)( 12 TTx

kAq

dxdTkAq

−∆

−=

−=

• Fourier’s law (k is function of T)

⎥⎦⎤

⎢⎣⎡ −+−

∆−=

+=

−=

21

2212

0

0

(2

)(

)1(

TTTTxAkq

TkkdxdTkAq

ββ



• Energy balance

CBA qqqq ===⇒

Α∆Χ

−=

Α∆Χ

−=

Α∆Χ

−=

Β

Α

Α

c

C

B k

TT

k

TT

k

TTq 433221

Solving these three equations simultaneously

CB

C

C

A

RRRTT

kkk

TTq++

−=

Α∆Χ

+Α

∆Χ+

Α∆Χ

−=

Α

Β

ΒΑ

4141



∑∆

==R

Tq overall

Resistancediffrence potentioal thermal



(a) Surfaces normal to the x-direction are isothermal

(b) Surfaces parallel to x-direction are adiabatic

For resistances in series: Rtot=R1+R2+…+Rn

For resistances in parallel:Rtot=1/R1+1/R2+…+1/Rn


Radial Systems-Cylindrical CoordinatesConsider a hollow cylinder

Temperature distribution


Radial Systems-Cylindrical Coordinates

?resistance Thermal)ln(

)(2

?conditionsBoundary

)2(

2

12lrrTTkLq

Solution

drdTrLkq

rLAdrdTkAq

oi −=

−−=

=

−=

π

π

π

∑∆

==R

Tq overall

Resistancediffrence potentioal thermal


Composite Walls

? Express the following geometry in terms of a an equivalent thermal circuit.

Α∆Χ++

−=

C

CAA k

krrkrr

TTLq/)/ln(/)/ln(

)(2

1212

41π

kLrrR io

π2)/ln(

=

CB

C

C

A

RRRTT

kkk

TTq++

−=

Α∆Χ

+Α

∆Χ+

Α∆Χ

−=

Α

Β

ΒΑ

4141


Composite Walls


CBA krrkrrkrrTTLq

/)/ln(/)/ln(/)/ln()(2

342312

41

++−

=π

kLrrR io

π2)/ln(

=


Radial Systems-Spherical Coordinates

• Home exercise


Example 2.2

• A thick tube of steel (k= 19 W/m.C) with 2 cm inner diameter and 4 cm outer diameter is covered with 3 cm layer of insulation (k= 0.2 W/m.C)

• Given that the inside T=600 C and the outside T=100 C) calculate the heat loss per meter of length.

• Also calculate the interface T.


2.5 Overall Heat Transfer Coefficient

In terms of overall temperature difference:qx

1,∞T

1,sT

2,sT

2,∞T

xx=0 x=L

11, ,hT∞

22, ,hT∞

Hot fluid

Cold fluid

AhkAL

AhR

RTT

q

tot

totx

21

2,1,

11++=

−= ∞∞

AhTT

kALTT

AhTT

q ssssx

2

2,2,2,1,

1

1,1,

/1//1∞∞ −

=−

=−

=



AhTT

kALTT

AhTT

q ssssx

2

2,2,2,1,

1

1,1,

/1//1∞∞ −

=−

=−

=

TUAqhkLh

U

AhkALAhTT

qx

∆=++

=

++−

= ∞∞

21

21

2,1,

/1//11

/1//1


2.5 Overall Heat Transfer Coefficient? Express the following

geometry in terms of a an equivalent thermal circuit.



Alternatively

UAqTRR

TUAq

ttot

x

1=

∆==

∆=

∑where U is the overall heat transfer coefficient and ∆T the overall temperature difference.

)]/1()/()/()/()/1[(11

41 hkLkLkLhARU

CCBBAAtot ++++==


2.5 Overall Heat Transfer CoefficientConsider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures

Temperature distribution


2.5 Overall Heat Transfer CoefficientBased on the previous solution, the conduction hear transfer rate can be calculated:

( ) ( ) ( )condt

ssssssx R

TTLkrr

TTrr

TTLkq

,

2,1,

12

2,1,

12

2,1,

)2/()/ln()/ln(2 −

=π

−=

−π=

• Fourier’s law: constdrdTrLk

drdTkAqr =π−=−= )2(

In terms of equivalent thermal circuit:

)2(1

2)/ln(

)2(1

22

12

11

2,1,

LrhkLrr

LrhR

RTT

q

tot

totx

π+

π+

π=

−= ∞∞



where U is the overall heat transfer coefficient. If A=A1=2πr1L:

44

1

3

41

2

31

1

21

1

1lnlnln11

hrr

rr

kr

rr

kr

rr

kr

h

U

CBA++++

=

alternatively we can use A2=2πr2L, A3=2πr3L etc. In all cases:

∑====

tRAUAUAUAU 1

44332211


Example 2.4

• Water flows @ 50 C inside a 2.5 cm inside diameter tube such that h=3500 w/m^2. C. The thickness of the tube is 0.8 cm and k= 16 w/m^2. C . The outside of the tube loses heat by free convection with h=7.6 w/m^2. C. Calculate the overall heat transfer coefficient and the heat loss per unit length to surrounding air at 20 C.

WTUAq

CmWRA

U

WCAh

R

WCkL

ddR

WCAh

R

o

ooo

iot

iii

19

/577.71

/575.1)1)(0266.0(6.7

11

/00062.0)1)(16(2

)025.0/0266.0ln(2

/ln(

/00364.0)1)(025.0(3500

11

2

)

=∆=

==

===

===

===

∑

π

ππ

π


2.6 Critical thickness of insulation

• If A, is increased, q will increase. When insulation is added to a pipe, the outside surface area of the pipe will increase. This would indicate an increased rate of heat transfer

• The insulation material has a low thermal conductivity, – it reduces the conductive heat transfer.

– This contradiction indicates that there must be a critical thickness of insulation.

– The thickness of insulation must be greater than the critical thickness, so that the rate of heat loss is reduced as desired.


2.6 Critical thickness of insulation

hkr

drdq

hrkrr

TTLqio

i

=

=

+

∞−=

0

0

0

0

q maximize To

1)/ln()(2π

As the outside radius, ro, increases, then in the denominator, the first term increases but the second term decreases.


Example 2.5

• Calculate the critical radius of insulation for asbestos [k=0. 17 W/m. C] surrounding a pipe and exposed to room air at 20oC with h=3.0 W/m2.oC. Calculate the heat loss from a 200 C, 5.0-cm-diameter pipe when covered with the critical radius of insulation and without insulation.

From Equation (2-18) we calculate ro as cmm

hkro 67.50567.0

0.317.0

====

The inside radius of the insulation is 5.0/2=2.5 cm, so the heat transfer iscalculated from Equation (2-17) as mW

InLq \7.105

)0.3)(0567.0(1

17.0)5.2/67.5(

)20200(2=

+

−=

π


Example 2.5

mWInL

q \7.105

)0.3)(0567.0(1

17.0)5.2/67.5(

)20200(2=

+

−=

π

Without insulation the convection from the outer surface of the pipe is mWToTirh

Lq /8.84)20200)(025.0)(2)(0.3())(2( =−=−= ππ

So, the addition of 3.17 cm (5.67-2.5) of insulation actually increases the heattransfer by 25 percent.


2.7 Heat Source system. (Heat Conduction Equation)

Energy Conservation Equation stst

outgin Edt

dEEEE &&&& ==−+

dzzdyydxxout

zyxin

qqqE

qqqE

+++ ++=

++=&

&

where from Fourier’s law

zTdxdyk

zTkAq

yTdxdzk

yTkAq

xTdydzk

xTkAq

zz

yy

xx

∂∂

−=∂∂

−=

∂∂

−=∂∂

−=

∂∂

−=∂∂

−=

)(

)(

)(

(2.1)

z

y

x

xy

z



• Thermal energy generation due to an energy source:– Manifestation of energy conversion process (between

thermal energy and chemical/electrical/nuclear energy)Positive (source) if thermal energy is generatedNegative (sink) if thermal energy is consumed

) (

dV

dzdydxq

qEg

&

&&

=

=

q& is the rate at which energy is generated per unit volume of the medium (W/m3)

• Energy storage term– Represents the rate of change of thermal energy

stored in the matter in the absence of phase change.

) ( dzdydxtTcE pst ⎥⎦⎤

⎢⎣⎡

∂∂

ρ=&

is the time rate of change of the sensible (thermal) energy of the medium per unit volume (W/m3)

tTcp ∂∂ρ /



tTcq

zTk

yyTk

yxTk

x p ∂∂

ρ=+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

&

Heat Equation

Net conduction of heat into the CVrate of energy generation per unit volume

time rate of change of thermal energy per unit volume

At any point in the medium the rate of energy transfer by conduction into a unit volume plus the volumetric rate of thermal energy generation must equal the rate of change of thermal energy stored within the volume


2.7 Heat Conduction Equation- Other forms

• If k=constant

tT

kq

zT

yT

xT

∂∂

α=+

∂∂

+∂∂

+∂∂ 1

2

2

2

2

2

2 &pc

kρ

=α is the thermal diffusivity

• For steady state conditions

0=+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂ q

zTk

yyTk

yxTk

x&

• For steady state conditions, one-dimensional transfer in x-direction and no energy generation

0or 0"

==⎟⎠⎞

⎜⎝⎛

dxdq

dxdTk

dxd x Heat flux is constant in

the direction of transfer


2.7 Heat Conduction Equation

tT

kq

zT

yT

xT

∂∂

α=+

∂∂

+∂∂

+∂∂ 1

2

2

2

2

2

2 &

02

2

=+∂∂

kq

xT &

take integral twice (assume tconsq tan

*= )

212

*

2CC

kqT +Χ+Χ−=

solve with TWO Boundary conditions ( )( )22

11

,,TT

ΧΧ


2.7 Heat Conduction Equation take integral twice (assume tconsq tan

*= )

212

*

2CC

kqT +Χ+Χ−=

solve with TWO Boundary conditions ( )( )22

11

,,TT

ΧΧ

2

.

0

02

1

2

0

xk

qTT

TCC

−=−

∴==


2.8 Heat Conduction Equation

• In cylindrical coordinates:

tTcq

zTk

zTk

rrTkr

rr p ∂∂

ρ=+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛φ∂

∂φ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

&2

11

• In spherical coordinates:

tTcqTk

rTk

rrTkr

rr p ∂∂

ρ=+⎟⎠⎞

⎜⎝⎛

θ∂∂

θθ∂∂

θ+⎟⎟

⎠

⎞⎜⎜⎝

⎛φ∂

∂φ∂∂

θ+⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂

&sinsin1

sin11

2222

2


2.8 Heat Conduction Equation In cylindrical coordinates

tTcq

zTk

zTk

rrTkr

rr p ∂∂

ρ=+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛φ∂

∂φ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

&2

11

01=+⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂ q

rTkr

rr&

w

w

TkRqT

rdrdT

RrTT

+=

∴

==

==

4

0 @ 0

@

2.

0


2.8 Heat Conduction Equation In cylindrical coordinates

tTcq

zTk

zTk

rrTkr

rr p ∂∂

ρ=+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛φ∂

∂φ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

&2

11

01=+⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂ q

rTkr

rr&

)ln(

4/)(

)ln((4

@ @

0

.22

001

01

220

.

0

00

rr

krrqTTC

rrCrr

kqTT

rrTTrrTT

i

ii

ii

−+−=

+−=−

∴====


Example 2.6

• A current of 200 A is passed through a stainless-steel wire [k=19W/m.oC]3 mm in diameter. The resistivity of the steel may be taken as 70 .cm, and the length of the wire is 1 m. The wire is submerged in a liquid at 110 C and experiences a convection heat-transfer coefficient of 4 kW/m^2. C. Calculate the center temperature of the wire.


All the power generated in the wire must be dissipated by convection to theliquid: )(2 ∞−=== TTwhAqRP The resistance of the wire is calculated from

Ω=×

==−

099.0)15.0(

)100)(1070(2

6

πρ

ALR

Where ρ is the resistivity of the wire. The surface area of the wire is ,dLπ sofrom Equation (a), ( ) ( ) ( )( )( ) WTw 396011011034000099.0200 32 =−×= −π and ]419[215 FCTw oo= The heat generated per unit volume q is calculated from LrqVqP 2π&& == so that

( ) ( )]./1041.5[/2.560

1105.13960 373

23fthBtumMWq ×=

×=

−π&

Finally, the center temperature of the wire is calculated from Equation (2-26):

( )( ) ]449[6.231215)19)(4(

105.110602.54

382

FCTwkrq

T oooo =+

××=+=

−&


2.11 THERMAL CONTACT RESISTANCE

• Imagine two solid bars brought into contact with the sides of the bars insulated so that heat flows only in the axial direction

→∆← Αx →∆← Βx

→q →q

3T

2T 1T

T

1x 2x 3x x

Α∆ΧΒ

+Α

∆ΧΑ−

=⇒

Α∆ΧΒ−

=

Α∆ΧΑ−

=

ΒΑΒΑ kk

TTq

k

TT

k

TTq 313221

A B



The temperature drop across the interface between materials may be appreciable, due to surface roughness effects, leading to air pockets.

No real surface is perfectly smooth, and the actual surface roughness is believed to play a central role in determining the contact resistance. This factor can be extremely important in a number of applications because of the many heat-transfer situations which involve mechanical joining of two materials.



3T

AT2 1T

T

1x 2x 3x x

BT2

Α∆ΧΒ−

=−

=

Α∆ΧΑ−

=

Β

ΒΑ

Α

Α

k

TT

Ah

TT

k

TTq B

C

322221

1

Α∆ΧΒ+Α

+Α∆ΧΑ

−=⇒

ΒΑ khk

TTq

C

131



3T

AT2 1T

T

1x 2x 3x x

BT2

Α∆ΧΒ−

=−

=

Α∆ΧΑ−

=

Β

ΒΑ

Α

Α

k

TT

Ah

TT

k

TTq B

C

322221

1

Α∆ΧΒ+Α

+Α∆ΧΑ

−=⇒

ΒΑ khk

TTq

C

131

ΑCh1 : thermal contact resistance

Ch : contact coefficient Ffth

BtuCm

woo ..

,. 22

For design purposes the contact conductance values given in Table 2-2 may be used in the absence of more specific information.


Example 2.11

• Tow 3 cm diameter 304 stainless steel bars, 10 cm long, have ground surface and are exposed to air with a surface roughness of about 1 µm. If the surfaces are pressed together with a pressure of 50 atm and the two bar combination is exposed to an overall temperature difference of 100 C, calculate the axial heat flow and the temperature drop across the contact surface.

)hfor 2.2 (TableC/W 747.0)03.0(

)4)(1028.5(1

/679.8)03.0()3.16(

)4)(1.0(

1

c2

4

2

31

=×

==

==∆

=

Α∆ΧΒ+Α

+Α∆ΧΑ

−=

−

ΒΑ

π

π

AhR

WCkA

xR

khk

TTq

cc

th

C


Example 2.11

C 13.41.18

)100)(747.0(resistance thermal total the toresistancecontact theof ratio theby taking found be contactcan the

across drop re temperatuThe

W52.5

1.18747.0)679.8)(2(is resistance thermal totalthe

)hfor 2.2 (TableC/W 747.0)03.0(

)4)(1028.5(1

/679.8)03.0()3.16(

)4)(1.0(

1

c2

4

2

31

==∆=∆

=∆

=

=+=

=×

==

==∆

=

Α∆ΧΒ+Α

+Α∆ΧΑ

−=

∑

∑

∑

−

ΒΑ

TR

RT

RTq

R

AhR

WCkA

xR

khk

TTq

th

cc

th

th

cc

th

C

π

π


2.9 Conduction-convection systems

A

xxq ∆+→

x→

→xq

∞T

oT

∞T

Steady state OperationNo Heat GeneratedConstant Thermal Conductivity (k)Constant Convection Coefficient (h)Neglect Radiation



0)(

)( convectionby lost Energy

faceleft in theEnergy

faceleft in theEnergy

2

2

2

2

=−−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−==

−==

∞

∞

TTkAhp

dxTd

TThpdx

dxdx

TddxdTkAq

dxdTkAq

x

x



• CASE 1: The fin is very long, and temp. at the end of the fin is essentially that of the surrounding fluid.

• CASE 2: The fin is of finite length and loses heat by convection from its end.

• CASE 3: The end of the fin is insulated

0)(2

2

=−− ∞TTkAhp

dxTd



Taking 'A' to be Pdx

Where P 2 w h+( )⋅ (perimeter)

Qconv hp T Tinf−( ) dx

Qcond x ∆x+ Qcond−,

∆xhp T Tinf−( )+

in the limit as ∆x goes to infinity we get

xQcond

dd

hp T Tinf−( )+ 0

xkAc−

dTdx

⎛⎜⎝

⎞⎠

dd

hp T Tinf−( )+ 0if 'k' and 'A' are constant then

kAc−d2T

d x2⋅hp T Tinf−( )+ 0 let θ T Tinf−



a2 hpkAc

d2θ

dx2a2θ− 0

general solutionθ x( ) C1 eax⋅ C2 e ax−⋅+

boundary conditions

θ 0( ) θb Tb Tinf−

θ L( ) T L( ) Tinf− 0 assuming at L, the temperature of the fin equals the temperature of the surrounding

our solution becomes

θ x( )θb

e ax− T x( ) Tinf−

Tb Tinf−e

hpkAc

⎛⎜⎝

⎞⎠

x−


2.10 Fins


Purpose of a Fin

• Fins are extended surfaces that are utilized in the removal of heat from a body

• One can increase heat transfer by increasing the heat transfer coefficient or increasing the surface area

• Finned surfaces are manufactured by extruding, welding, or wrapping a thin metal sheet on a surface

• Fins enhance heat transfer from a surface by exposing a larger surface area to convection and radiation


The fin efficiency

( )

( )∞−Α=⎪⎭

⎪⎬

⎫=

TTohtempbaseatwere

finentireifdtransferrewouldbwhichheat

qtransferedheatactualhefficiencyFin

f

actualf

.

,

( )∞−Α=⇒ TTohq ffactual η

Total surface area of the fin

The fin efficiency


• Fin efficiency is defined as the ratio of actual heat transfer rate to the maximum possible heat transfer rate

• The surface temperature of the fin decreases from the base to the tip direction. The degree of variation depends on the dimensions and the thermal conductivity of the fin. If the thermal conductivity is very large, the surface temperature may be constant and equals to Tb.

• The heat transfer rate from the elemental area, pdx

• The total heat transfer rate from the entire fin

• The maximum possible heat transfer rate of the entire fin T = Tb = constant

• The fin efficiency

( )dQ hpdx T T∞= −&

0

( )L

fin aQ hp T T dx= −∫&

max b fin bQ hpL hAθ θ= =&

max

finQQ

η =&

&


The fin efficiency (case 3)

For CASE 3 above (The end of the fin of uniform cross-sectional area is insulated)

mLmL

ftanh

=η (2-38)

232 Lk

hmLmΑ

= Ltm =Α (2-39)

Profile area


The fin efficiency (case 3)

It has been shown that the efficiency equation for case 3 (insulated tip) can be used for case 2 (finite length) without insulation) if a corrected length is used

mLmL

ftanh

=η

Corrected length ( Case 2)


-Convection tip boundary condition

The total fin surface area At including the tip area is

Dividing the above equation by the perimeter, p, the corrected fin length is

If L is replaced by Lc, the tip surface area of the fin subjected to convection boundary condition is considered

(1) For a rectangular fin

(2) For a circular fin

t cA Lp A= +

cc

AL LP

= +

2( ) 2 2cwt wt tL L L Lw t w

= + ≈ + = ++

4cDL L= +

t t/2

Lc


Temperature Distribution• Heat conduction equation in the r-direction for steady-state conditions,

with no energy generation:

01=⎟

⎠⎞

⎜⎝⎛

drdTkr

drd

r

• Boundary Conditions: 2,21,1 )(,)( ss TrTTrT ==



• Temperature profile, assuming constant k:

2,221

2,1, ln)/ln()(

)( sss T

rr

rrTT

rT +⎟⎟⎠

⎞⎜⎜⎝

⎛−= Logarithmic temperature distribution

(see above)


Thermal ResistanceBased on the previous solution, the conduction heat transfer rate can be calculated:

( ) ( ) ( )condt

ssssssx R

TTLkrr

TTrr

TTLkq

,

2,1,

12

2,1,

12

2,1,

)2/()/ln()/ln(2 −

=π

−=

−π=



In terms of equivalent thermal circuit:

kLrrR

RTT

q

tot

totx

π2)/ln( 12

2,1,

=

−= ∞∞

Chapter 2faculty.kfupm.edu.sa/CHE/shammakh/files/My files/300_Chapter_2.pdf · Chapter 2 Chee 318 2...

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Transcript of Chapter 2faculty.kfupm.edu.sa/CHE/shammakh/files/My files/300_Chapter_2.pdf · Chapter 2 Chee 318 2...