chapter_3_1_cond 1 dim, steady-state

8/17/2019 Chapter_3_1_Cond 1 Dim, Steady-State

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Chapter 3 Perpindahan Panas 1

Chapter 3

One-Dimensional Steady-State Conduction


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One-Dimensional Steady-State Conduction

• Conduction problems may involve multiple directions and time-

dependent conditions

• Inherently complex – Difficult to determine temperature distributions

• One-dimensional steady-state models can represent accurately

numerous engineering systems

• In this chapter e ill

!earn ho to obtain temperature profiles for common geometries

ith and ithout heat generation"

Introduce the concept of thermal resistance and thermal circuits


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The Plane Wall

Consider a simple case of one-

dimensional conduction in a plane

all# separating to fluids of different

temperature# without energy

generation

•$emperature is a function of x

• %eat is transferred in the x-direction

&ust consider

– Convection from hot fluid to all

– Conduction through all

– Convection from all to cold fluid

'egin by determining temperature

distribution ithin the all

(x

1,∞T

1, sT

2, sT

2,∞T

x

x)* x)!

11, ,hT ∞

22, ,hT ∞

Hot fluid

Cold fluid


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Temperature Distribution

• %eat diffusion e(uation +e(" ,". in the x-direction for steady-state

conditions# ith no energy generation/

0=

dx

dT k

dx

d

• 'oundary Conditions/

• Solution of the differential e(uation

2,1, )(,)0( s s T LT T T ==

• $emperature profile# assuming constant 0/

1,1,2, )()( s s s T L

xT T xT +−=

$emperature varies linearly ith x

(x is constant

(3.1)


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Thermal Resistance

'ased on the previous solution# the conduction hear transfer rate can

be calculated/

( ) ( )

kA L

T T T T

L

kA

dx

dT kAq

s s s s x

/

2,1,2,1,

−=−=−=

1ecall electric circuit theory - Ohm2s la for electrical resistance/

Similarly for heat convection# 3eton2s la of cooling applies/

Resistance

eDifferencotentialcurrent!lectric =

hAT T T T hAq S S x

/1)()( ∞

∞

−=−=

4nd for radiation heat transfer/

Ah

T T

T T Ahq r

sur s

sur sr rad /1

)(

)(

−

=−=


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Chapter 3 Perpindahan Panas "

Thermal Resistance

Compare ith e(uations 5",a-5",c

$he temperature difference is the 6potential7 or driving force for the

heat flo and the combinations of thermal conductivity# convection

coefficient# thic0ness and area of material act as a resistance to thisflo/

• 8e can use this electrical analogy to represent heat transfer problems

using the concept of a thermal circuit +e(uivalent to an electrical circuit."

Ah R

hA R

kA

L R

r rad t convt cond t

1,

1, ,,, ===

∑∆

== R

T q overall

Resistance

#orceDri$in%&$erall


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Chapter 3 Perpindahan Panas '

Thermal Resistance for Plane Wall

In terms of overalltemperature difference/(x

1,∞T

1, sT

2, sT

2,∞T

xx)* x)!

11, ,hT ∞

22,

,hT ∞

Hot fluid

Cold fluid

AhkA

L

Ah R

R

T T q

tot

tot

x

21

2,1,

11++=

−= ∞∞

Ah

T T

kA L

T T

Ah

T T q

s s s s x

2

2,2,2,1,

1

1,1,

/1//1

∞∞ −=−

=−

=


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Chapter 3 Perpindahan Panas

Composite Walls

? Express the following

geometry in terms ofa an equivalentthermal circuit.


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Composite Walls? What is the heat transfer rate for this system?

4lternatively

UAq

T R R

T UAq

t tot

x

1=

∆==

∆=

∑

here 9 is the overall heat transfer coefficient and∆

$ the overalltemperature difference"

)*/1()/()/()/()/1+(

11

41 hk Lk Lk Lh A RU

C C B B A Atot ++++==


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Composite Walls

+a. Surfaces normal to the x-

direction are isothermal

+b. Surfaces parallel to x-

direction are adiabatic


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Example

Consider a composite all that includes an :-mm thic0 hardood

siding +4.# *-mm by ;5*-mm hardood studs +'. on *"m5. and a ;,-mm

layer of gypsum +vermiculite. all board +C."

8hat is the thermal resistance associated ith a all that is ,"= m

high by


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Contact Resistance

$he temperature drop

across the interfacebeteen materials may be

appreciable# due to surface

roughness effects# leading

to air poc0ets" 8e can

define thermal contactresistance/

,

x

B Act

q

T T R

−=

#ee ta!les $.%" $.& fortypical values of ' t"c


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Chapter 3 Perpindahan Panas 1"


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Chapter 3 Perpindahan Panas 1'

lternati!e Conduction nalysis

• @or steady-state conditions# no heat generation# one-dimensional heat

transfer# (x is constant"

dx

dT x AT k q x )()(−=

8hen area varies in the x direction and 0 is a function of temperature#

@ourier2s la can be ritten in its most general form/

∫ ∫ −=∴ T

T

x

x

x

oo

dT T k

x A

dxq )(

)(


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Example 3"#

Consider a conical section fabricated from pyroceram" It is of circular

cross section# ith the diameter D)αx# here α)*",=" $he small endis at x;)=* mm and the large end at x,),=* mm" $he endtemperatures are $;)** A and $,)


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Radial Systems-Cylindrical Coordinates

Consider a hollo cylinder# hose inner and outer surfaces are

exposed to fluids at different temperatures

(emperature distri!ution


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Temperature Distribution

• %eat diffusion e(uation +e(" ,"=. in the r-direction for steady-state

conditions# ith no energy generation/

01

=

dr

dT kr

dr

d

r

• 'oundary Conditions/ 2,21,1 )(,)( s s T r T T r T ==

• $emperature profile# assuming constant 0/

2,221

2,1,ln

)/ln(

)()( s

s sT

r

r

r r

T T r T +

−= !ogarithmic temperature distribution

+see previous slide.

• @ourier2s la/ const dr

dT rLk dr

dT kAqr =π−=−= )2(


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Thermal Resistance

'ased on the previous solution# the conduction hear transfer rate can

be calculated/

( ) ( ) ( )

cond t

s s s s s s

x R

T T

Lk r r

T T

r r

T T Lk

q ,

2,1,

12

2,1,

12

2,1,

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

2 −

=π

−

=

−π

=

In terms of e(uivalent thermal circuit/

)2(

1

2

)/ln(

)2(

1

22

12

11

2,1,

Lr hkL

r r

Lr h R

R

T T

q

tot

tot x

π+

π+

π=

−=

∞∞

• @ourier2s la/ const dr

dT rLk

dr

dT kAqr =π−=−= )2(


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Composite Walls

? Express the followinggeometry in terms ofa an equivalent

thermal circuit.


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Composite Walls? What is the heat transfer rate?

here 9 is the overall heat transfer coefficient" If 4)4;),πr ;!/

44

1

3

41

2

31

1

21

1

1lnlnln

1

1

hr

r

r

r

k

r

r

r

k

r

r

r

k

r

h

U

C B A

++++=

alternatively e can use 4,),πr ,!# 45),πr 5! etc" In all cases/

∑====

t R AU AU AU AU

144332211


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Chapter 3 Perpindahan Panas 2'


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Spherical Coordinates

• Starting from @ourier2s la# ac0noledging that (r is constant#independent of r# and assuming that 0 is constant# derive the e(uation

describing the conduction heat transfer rate" 8hat is the thermalresistance?

• @ourier2s la/

dr dT r k

dr

dT kAqr

)4( 2π−=

−=


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Summary


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Extended Surfaces $%ins&

4n extended surface +also 0no as a combined conduction-

convection system or a fin. is a solid ithin hich heat transfer by

conduction is assumed to be one dimensional# hile heat is also

transferred by convection +and>or radiation. from the surface in a

direction transverse to that of conduction

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Extended Surfaces $%ins&Bxtended surfaces may exist in many situations but are

commonly used as fins to enhance heat transfer by increasing

the surface area available for convection +and>or radiation."

$hey are particularly beneficial hen h is small # as for a 'as

and natural con!ection"

Solutions for various fin geometries can be found in the literature

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-nifor Crossectional rea Ac = C(dAc /dx) = 0

urface rea A s = P x (dA s /dx) = P


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Ch t 3 Perpindahan Panas 3'

chapter_3_1_cond 1 dim, steady-state

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