STEADY HEAT CONDUCTION IN PLANE WALLS - RCME … · STEADY HEAT CONDUCTION IN PLANE WALLS Heat flow...

1

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FIGURE 3–1

dt

dEQQ walloutin =− &&

STEADY HEAT CONDUCTION IN PLANE WALLS

Heat flow through a wall is one dimensional when the temperature of the wall varies in one direction only

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

wall theof energy theof change of Rate

wall theofout ferheat trans

of Rate

wall theintoferheat trans

of Rate

The energy balance for the wall can be expressed as

or


FIGURE 3–2

Under steady conditions, the temperature distribution in a plane wall is a straight line.

21, (W)

LTTkAQ wallcond

−=&

Separating the variables in the above equation and integrating from where

, to , where we get

0=x 0=x 0=x

( ) 10 TT =

0=xLx = ( ) 2TLT =

∫∫==

−=2

1,0

. T

TTwallcond

L

x

kAdTdxQ

Performing the integrations and rearranging gives

2


FIGURE 3–2

(W)R

TTQwall

wallcond 21,

−=&

thermal resistance

0=x 0=x 0=x

Where

The Thermal Resistance Concept

( )WCkALRwall / °=


FIGURE 3–3

Analogy between thermal and electrical resistance concepts.

3


FIGURE 3–4

conv

sconv R

TTQ ∞−

=&

( )WChA

Rs

wall / 1°=

Schematic for convection resistance at a surface.

Where

thermal resistance

Consider convection heat transfer from a solid surface of area and temperature to a fluid whose temperature sufficiently far from the surface is , with a convection heat transfer coefficient h. Newton’s law of cooling for convection heat transfer rate can be rearranged as

sA

sA sT

∞T

( )∞−= TThAQ ssconv

.


FIGURE 3–5

( ) ( )

sradrad

rad

surs

surssradsurrssrad

AhR

RTT

TTAhTTAQ1

44

=⇒−

=

−=−= εσ&

Schematic for convection and radiation resistance at a surface.

( ) ( )( )surrssurrssurrss

radrad TTTT

TTAQ

h ++=−

= 22εσ&

thermal resistance

or, the radiation resistance

4


FIGURE 3–6

The thermal resistance network for heat transfer through a plane wall subjected to convection on both sides, and the electric analogy.


FIGURE 3–6

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

wall thefrom convectionheat

of Rate

wallethrough thconductionheat

of Rate

wall theintoconvectionheat

of Rate

The Resistance Network

Under steady conditions we have

or

( ) ( )22221

111

.

∞∞ −=−

=−= TTAhL

TTkATTAhQconv

5


FIGURE 3–6

which can be rearranged as

Adding the numerators and denominators yields (Fig.3-7)

AhTT

kALTT

AhTTQ

2

2221

1

11.

/1//1∞∞ −

=−

=−

=

2,

2221

1,

11

convwallconv RTT

RTT

RTT ∞∞ −

=−

=−

=

)( 21.

WR

TTQtotal

∞∞ −=


FIGURE 3–7

A useful mathematical identity.

Thermal Resistance Network.

6


FIGURE 3–8

The temperature drop across a layer is proportional to its thermal resistance.

WCAhkA

LAh

RRRR convwallconvtotal /11 0

2121 ++=++=


FIGURE 3–9

The thermal resistance network for heat transfer through a two-layer plane wall subjected to convection on both sides.

AhTT

RTTQ

conv 1

11

1,

11

/1−

=−

= ∞∞&

7


FIGURE 3–10

Q&

jitotal

ji

RTT

Q−

−=

,

&

The evaluation of the surface and interface temperatures when T∞1and T∞2 are given and is calculated.

Once is known, an unknown surface temperature Tj at any surface or interface j can be determined from

Q&

( ) ( )AkLAhTT

RRTT

Qwallconv 111

21

1,1,

21.

//1 +−

=+−

= ∞∞


FIGURE 3–11

Schematic for Example 3-1.

8


FIGURE 3–11


Assumptions

1.steady state heat transfer

2. (1-D) heat transfer

3. thermal conductivity is constant

Properties

1. the thermal conductivity is given to be k=0.9 W/m*C

Analysis

A = 3m x 5m = 15 m2


FIGURE 3–11

( )WCkALRwall / °=

( ) ( ) 630W 3.0216 15./ 9.0 221 =⎟

⎠⎞

⎜⎝⎛ °−

°=−

=m

CmCmWL

TTkAQ&

Example 3-1.

SOLUTION

wall

wall

RT

QΔ

=.

where

( )( ) WCmCmW

m /02222.0 15./ 9.0

3.02 °=

°=

( ) WWC

CQ 630/02222.0

216.=

°°−

=

9


FIGURE 3–12


Assumptions

1.steady state heat transfer

2. (1-D) heat transfer

3. thermal conductivity is constant

Properties

1. the thermal conductivity is given to be k=0.78 W/m*C


• solution

WCRRRR

WCmCmWAh

RR

WCmCmW

mkALR

WCmCmWAh

RR

convglassconvtotal

convo

glass

convi

/1127.002083.000855.008333.0

/02083.0)2.1)(/40(

11

/00855.0)2.1)(/78.0(

008.0

/08333.0)2.1)(/10(

11

021,

0202

22

020

0202

11,

=++=++=

====

===

====

( )[ ] WR

TTQ

total

2661127.0

102021 =−−

=−

= ∞∞&

Then Steady rate of heat transfer through the window become

10


CRQTT conv0

1,1 2.2)08333.0)(266(20 −=−=−= ∞&

1,

11

convRTTQ −

= ∞&

CRQTT conv0

1,1 2.2)08333.0)(266(20 −=−=−= ∞&

Knowing rate of heat transfer , the inner surface temperature of the winder can be determine from

Discussion – Note that the inner surface temperature of the window glass will be even though the temperature of the air in the room is maintained at Such low surface temperature are highly undesirable since they cause the formation of fog or even frost on the inner surfaces of the glass when the humidity in the room is high.

C°− 2.2 C°20


FIGURE 3–13


Assumptions1.steady state heat transfer2. (1-D) heat transfer3. thermal conductivity is constant

Properties1. the thermal conductivity of glass and air space is given to be k=0.78 ,0.026 W/m*C

11


WC /4332.002083.000427.03205.000427.008333.0 0=++++=

22,1,1,

0202

22

020

2

22

020

1

131

0202

11,

/02083.0)2.1)(/40(

11

/3205.0)2.1)(/026.0(

01.0

/00427.0)2.1)(/78.0(

004.0

/08333.0)2.1)(/10(

11

convglassairglassconvtotal

convo

air

glass

convi

RRRRRR

WCmCmWAh

RR

WCmCmW

mAk

LRR

WCmCmW

mAk

LRRR

WCmCmWAh

RR

++++=

====

====

=====

====

solution


( )[ ] WR

TTQ

total

2.694332.0

102021 =−−

=−

= ∞∞&


Knowing rate of heat transfer , the inner surface temperature ofthe winder can be determine from

CRQTT conv0

1,11 2.14)08333.0)(2.69(20 =−=−= ∞&

12


FIGURE 3–14

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


FIGURE 3–15

A typical experimental setup for the determination of thermal contact resistance.

13


FIGURE 3–16

Effect of metallic coatings on thermal contact conductance.


FIGURE 3–17



Properties1. the thermal conductivity of aluminum at room temperature is k=237 W/m*C

14


FIGURE 3–11

kLR =

ckRL =

WCmxCmWh

Rc

c /.10909.0./11000

11 024−=°

==

Example 3-4. SOLUTION

( )( )WCmxCmW /.10909.0./237 0240 −=

cmm 15.20215.0 ==

The thermal contact resistance is

For a unit surface area, the thermal resistance of a flat plate is defined as

where L is the thickness of the plate and k is the thermal conductivity, Setting ,the equivalent thickness is determined from the relation above to be

crRR =


FIGURE 3–18

Properties-The thermal conductivity of copper is given to be k=386 W/m*C. -The contact conductance

2/ 000,42 mWhc =


15


FIGURE 3–18

Solution

WC

RRRR

WCmCmWAh

R

WCmCmW

mkALR

WCmxCmWAh

R

ambientplateerfacetotal

conv

plate

ccerface

/0.40026.003.0

/0.4)01.0)(/25(

11

/0026.0)01.0)(/386(

01.0

/03.0)108)(/42000(

11

0

int

0202

0

020

02402int

++=

++=

===

===

=== −


FIGURE 3–18

( ) WWCC

RTQ

total

4.12/0326.4

2070=

°°−

=Δ

=

⋅

Solution

The rate of heat transfer is determined to be

The temperature jump at the interface is determined from

( )( ) CWCWRQT erfaceerface °=°==Δ⋅

37.0/03.04.12int

.

int

16


FIGURE 3–19

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

−+

−=+=

2121

2

21

1

2121

11RR

TTR

TTR

TTQQQ &&&

Thermal resistance network for two parallel layers.

21

21

1

21

21. 11

RRRR

RRR

RTT

Q totaltotal +

=⎟⎟⎠

⎞⎜⎜⎝

⎛+=⇒

−=

−


FIGURE 3–20

Thermal resistance network for combined series-parallel arrangement.

convconvtotal

total

RRRR

RRRRRR

RTT

Q

+++

=++=

−= ∞

321

21312

1.

and

333

33

22

22

11

11

1,,hA

RAk

LR

AkL

RAk

LR conv ====

17


FIGURE 3–21



Properties1. the thermal conductivity are given to be k=0.72 W/m*C for plaster layer, and k=0.026 W/m*C for the rigid foam


WCmxCmWAh

RR

WCmxCmW

mkALRR

WCmxCmW

mkALRRR

WCmxCmW

mkALRRR

WCmxCmW

mkALRR

WCmxCmWAh

RR

convo

brick

centerplaster

sideplaster

foam

convi

/16.0)125.0)(/25(

11

/01.1)122.0)(/72.0(

16.0

/48.48)1015.0)(/22.0(

16.0

/36.0)125.0)(/22.0(

02.0

/6.4)125.0)(/026.0(

03.0

/4.0)125.0)(/10(

11

0202

22,

0204

020,53

020,62

0201

0202

11,

====

====

=====

=====

====

====

solution

18


solution

The three resistance R3, R4 and R5 in the middle are parallel, and their equivalent resistance is determined from

WCRRRRRRR

WCR

CWRRRR

midimid

mid

mid

/85.616.036.097.036.06.44.0

is resistance total theand series,in are resistance theall Now/97.0

giveWhich

/03.148.48

101.11

48.4811111

0621

543

°=+++++=+++++=

°=

°=++=++=


solution

( )[ ] ( )

( )( ) WW

WWCC

RTT

tatal

total

26315mm/5.17Q

becomes wallentire gh thefer throuuheat trans of rate Then the .15m3mx5mA

is wall theof area totalThe area. mper W 17.54.38/0.25or

area surfacem 0.25per 38.4/85.6

1020Q

becomes wallh thefer througheat trans of ratesteady Then the

22.

2

2

221.

==

==

=

=°

°−=

−= ∞∞

19


FIGURE 3–22

Alternative thermal resistance network for example 3-6 for the case of surfaces parallel to the primary direction of heat transfer being adiabatic.


FIGURE 3–23

Heat is lost from a hot water pipe to the air outside in the radial direction, and thus heat transfer from a long pipe is one-dimensional.

20


FIGURE 3–24

)/ln(2

2

12

21,

,

,

2

1

2

1

rrTT

LkQ

rLAkdTdrA

QdrdTkAQ

cylcond

TT

TT

rr

rr

cylcond

cylcond

−=

=→−=

−=

∫∫=

=

=

=

π

π

&

&

&

A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T1and T2.

Rcyl=ln(r2/r1)/2πLk

cylcylcond R

TTQ 21,

−=&


FIGURE 3–25

The thermal resistance network for a cylindrical (or spherical) shell subjected to convection from both the inner and the outer sides.

totalRTTQ 21 ∞∞ −

=&

21


FIGURE 3–25

totalRTT

Q 21 ∞∞ −=&

The rate of heat transfer under steady condition can be expressed as

( ) ( ) ( )

( )( )

( ) ( )( )LrhLk

rrLk

rrTT

RRRTTQ

TLk

rrLrh

TTRRTTQ

LrALrAAhLk

rrLk

rrLk

rrAh

RRRRRR

o

conv

cylconv

convcylcylcylconvtotal

43

34

2

23

22

2,32

22.

2

1

12

11

21

1,1,

21.

4411

423

34

2

23

1

12

11

2,3,2,1,1,

21

2/ln

2/ln

from calculate also could We2

/ln21

2 and 2 where

12

/ln2

/ln2

/ln1

where

πππ

ππ

πππππ

++

−=

++−

=

+

−=

+−

=

==

++++=

++++=

∞∞

∞∞


FIGURE 3–26

The thermal resistance network for heat transfer through a three-layered composite cylinder

22


FIGURE 3–27

The ratio ΔT/R across any layer is equal to Which remains constant in one-dimensional steady conduction.

Q&


FIGURE 3–28


Assumptions1.steady state heat transfer2. (1-D) heat transfer and

thermal symmetry at midpoint

3. thermal conductivity is constantProperties1.the thermal conductivity of

steel is given to be k=15W/m*C2.Heat of fusion of water at

atmospheric is hif=333.7 kj/kg3.Emissivity outer surface of tank is

ε=1

23


22222

22211

0.29)04.3(

3.28)3(

mDA

mDA

===

===

ππ

ππ

))(( 2222

22 ∞∞ ++= TTTThrad εσ

solution

[ ][ ] CmWhrad02228 /34.5278295)278()295()10*67.5)(1( =++= −

Radiation heat transfer coefficient is given by

(a)

Inner and outer surface area of the tank are


Then the individual thermal resistance become

WCRR

RRR

WCmCmWAh

R

WCmCmWAh

RR

WCmmCmWrkr

rrRR

WCmCmWAh

RR

radoitotal

radrad

convo

sphere

convi

/00274.000646.0

100345.0

1000047.0000442.011

/00646.0)0.29)(/34.5(

11

/00345.0)0.29)(/10(

11

/000047.0.0)50.1)(52.1)(/15(4

5.152.14

/000442.0)3.28)(/80(

11

011

1

0202

2

0202

222

00

21

121

0202

11,

=⎟⎠⎞

⎜⎝⎛ +++=⎟⎟

⎠

⎞⎜⎜⎝

⎛+++=

===

====

=−

=−

==

====

−−

ππ

24



[ ] skJQorWR

TTQtotal

/027.8 802900274.0

02212 ==−

=−

= ∞∞ &&

equivRTTQ 22 −= ∞&

00225.000646.0

100345.0

111 11

=⎟⎠⎞

⎜⎝⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

−−

radoequiv RR

R

CRQTT equiv0

22 4)00225.0)(8029(22 =−=−= ∞&

We now determine the outer surface temperature from

Which give


Solution (b)

The total amount of heat transfer during a 24-h period is

kJsxskJtQQ 700,673)360024)(/029.8( ==Δ= &

kgkgkJ

kJhQm

ifice 2079

/7.333700,673

===

C°Note that it take 333.7kJ of energy to melt 1 kg of ice at 0 , the amount of ice that will melt during a 24-h period is

Discussion ( )( ) WCmCmWAh

Rcombined

combined /00225.029./34.15

11 022

2

=°

==

25


FIGURE 3–29



Assumptions1. steady state heat transfer2. (1-D) heat transfer and

thermal symmetry at centerline and variation in the axialdirection

3. thermal conductivity is constant4. thermal contact resistance at interface is negligible

Properties1.the thermal conductivity of cast iron is given to be k =80W/m*C

and k = 0.05W/m*C for glass wool insulation

26


222

211

361.0)1)(0575.0(22

157.0)1)(025.0(22

mmmLrA

mmmLrA

===

===

ππ

ππ

The area of the surface exposed to convection are determine to be

Then the individual thermal resistance become

WCmCmWLk

rrRR

WCmCmWLk

rrRR

WCmCmWAh

RR

insulation

pipe

convi

/35.2)1)(/05.0(2

)75.2/75.5ln(2

)/ln(

/0002.0)1)(/80(2

)5.2/75.2ln(2

)/ln(

/106.0)157.0)(/60(

11

00

1

122

00

1

121

0202

11,

====

====

====

ππ

ππ


WCRRRRR

WCmCmWAh

RR

oitotal

convo

/61.2154.035.20002.0106.0

/154.0)361.0)(/18(

11

021

0202

322

=+++=+++=

====

[ ] WR

TTQtotal

12116.2

532021 =−

=−

= ∞∞&

CWCWRQT pipepipe00 02.0)/0002.0)(121( ===Δ &

Then Steady rate of heat loss from the steam become

Temperature drop across the pipe and the insulation are determine

CWCWRQT insulationinsulation00 284)/35.2)(121( ===Δ &

27


FIGURE 3–30

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

)2(1

2)/ln(

2

12

11

LrhLkrr

TTRRTTQ

convins

ππ+

−=

+−

= ∞∞&


FIGURE 3–31

)( , mhkr cylindercr =

)( 2, m

hkr spherecr =

Critical radius of insulation for a cylindrical body to be

Critical radius of insulation for a spherical shell is

tyconductivi thermal theis where k

28


FIGURE 3–32



Assumptions1. steady state heat transfer2. (1-D) heat transfer and

thermal symmetry at centerline and variation in the axialdirection

3. thermal conductivity is constant4. thermal contact resistance at interface is negligible5. heat transfer coefficient incorporates the radiation effectProperties1.the thermal conductivity of plastic is given to be

k=0.15W/m*C

29


The rate of heat transfer become equal to the heat generate within the wire, which is determine to be

WAVVIWQ e 80)10)(8( ==== &&

The values of these two resistances are determine to be

WCRRRtherefore

WCmCmWkL

rrR

WCmCmWhA

R

mmmLrA

plasticconvtotal

plastic

conv

/94.018.076.0

/18.0)5)(/15.0(2

)5.1/5.3ln(2

)/ln(

/76.0)110.0)(/12(

11110.0)5)(0035.0(22

0

00

12

0202

2

222

=+=+=

===

===

===

ππ

ππ


We now determine the interface temperature from

totaltotal

RQTTR

TTQ && +=⇒−

= ∞∞

11

CWCW 00 105)/94.0)(80(30 =+=

The critical radius of insulation of the plastic cover is determine from

mCmWCmW

hkrcr 0125.0

/12/15.0

02

0

===

30


FIGURE 3–33

)( ∞−= TThAQ SSconv&

The thin plate fins of a car radiator greatly increase the rate of heat transfer to the air (photo by Yunus Cengeland James Kleiser).

HEAT TRANSFER FROM FINNED SURFACES

The rate of heat transfer from a surface at a temperature to the surrounding medium at is given by Newton’s law of cooling as

∞TsT


FIGURE 3–34

Some innovative fin designs.

31


FIGURE 3–35

Volume element of a fin at location x having a length of Δx, cross-sectional area of Ac, and perimeter of p.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

Δ+

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

elementthe

fromconvection

heatofRate

xxatelement

thefromconduction

heatofRate

xatelementthe

toinconduction

heatofRate

.

convxxcondxcond QQQ &&& += Δ+,,

Fin Equation

The energy balance on this volume element can be expressed as

where

( )( )∞−Δ= TTxphQconv&


FIGURE 3–35

dxdTkAQ ccond −=&

xΔ

0)(,, =−+Δ

−∞

Δ+ TThpx

QQ xcondxxcond&&

0)( =−+ ∞TThpdx

Qd cond&

0)( =−−⎟⎠⎞

⎜⎝⎛

∞TThpdxdtkA

dxd

c

Fin EquationSubstituting and dividing by ,we obtain

Taking the limit as gives

From Fourier’s law of heat conduction we have

Gives the differential equation governing heat transfer in fins,

xΔxΔxΔ

0→Δx

32


FIGURE 3–36

Boundary conditions at the fin base and the fin tip.

∞−== TTbbθθ )0(

0)()( =−= ∞TLTLθ ∞→L;

Boundary conditions at the fin tip.

Boundary conditions at the fin base


FIGURE 3–37

ckAhpxax

b

eeTTTxT /)( −−

∞

∞ ==−−

)(0

∞=

−=−= TThpkAdxdTkAQ bc

xcfinlong

&

Very long fin:

A long circular fin of uniform cross section and the variation of temperature along it.

∫∫ =−= ∞

finfin Afinfin

Afin dAxhdATxThQ )(])([ θ&

33


FIGURE 3–38

Under steady conditions, heat transfer from the exposed surfaces of the fin is equal to heat conduction to the fin at the base.

0==Lxdx

dθ

aLxLa

TTTxT

b cosh)(cosh)( −

=−−

∞

∞

0=−=

xctipinsulated dxdTkAQ&

aLTThpkA bc tanh)( ∞−=

Adiabatic fin tip:

Boundary condition at fin tip:


FIGURE 3–39

Corrected fin length Lc is defined such that heat transfer from fin of length Lc with insulated tip is equal to heat transfer from the actual fin of length L with convection at the fin tip

pA

LL cc +=

2tan,tLL fingularrecc +=

4,DLL fincylinderc +=

Corrected fin length:

and

34


FIGURE 3–40

Fins enhance heat transfer from a surface by enhancing surface area.

( )∞−= TThAQ bfinfin max,&

The heat transfer from the fin will be maximum in this case and can be expressed as


FIGURE 3–41

finthefromratetransferheatIdealfinthefromratetransferheatActual

QQ

fin

finfin ==

max,&

&η

Temperature distribution in a fin.

Fin efficiency

or

( )∞−== TThAQQ bfinfinfinfinfin ηη max,&&

35


FIGURE 3–41

Fin efficiency

For the cases of constant cross section of very long fins and fins with insulated tips, the fin efficient can be expressed as

( )( )

( )( ) aL

aLTThA

aLTThpkAQ

Q

aLhpkA

LTThATThpkA

QQ

bfin

bc

fin

fintinsulated

c

bfin

bc

fin

finfinlong

tanhtanh

and

11

max,ip

max,

=−

−==

==−−

==

∞

∞

∞

∞

&

&

&

&

η

η


FIGURE 3–42

Efficiency of circular, rectangular, and triangular fins on a plain surface of width w

36


FIGURE 3–43

Efficiency of circular fin of length L and constant thickness t.


FIGURE 3–44

( ) area surface the

from ratefer Heat trans area base offin the from ratefer Heat trans

b

b

bb

fin

finno

finfin

A

ATThA

QQQ

=−

==∞

&

&

&ε

( )( )

( ) finb

fin

bb

bfinfin

bb

fin

finno

finfin A

ATThA

TThATThA

QQQ

ηη

ε =−

−=

−==

∞

∞

∞

&

&

&

Fin Effectiveness

The effectiveness of such a long fin is determined to be

( )( ) cbb

bc

finno

finfinlong hA

kpTThA

TThpkAQQ

=−−

==∞

∞

&

&ε

37


FIGURE 3–45

Various surface areas associated with a rectangular surface with three fins.

( )( )( )∞

∞

−

−+==

TThATTAAh

QQ

bfinno

bfinfinunfin

nofintotal

fintotaloverallfin

,,

,,

ηε

&

&


FIGURE 3–46

Because of the gradual temperature drop along the fin, the region near the fin tip makes little or no contribution to heat transfer.

38


FIGURE 3–47

C°85

C°85


Assumptions1. Steady operating conditions

exist2. The transistor case is isothermal

at PropertiesThe case-to-ambient thermal resistance is given to be WC /20°

C°85

( ) WWC

CR

TTRTQ

ambientcase

c

ambientcase

3/20

2585=

°°−

=−

=⎟⎠⎞

⎜⎝⎛Δ=

−

∞

−

&


FIGURE 3–48


39


Assumptions1. steady operating conditions exist2. The heat transfer coefficient is uniform over the entire fin surfaces3. thermal conductivity is constant4. heat transfer by radiation is negligibleProperties1.the thermal conductivity of the fin is given to be k=180W/m*C


In the case of no fin, heat transfer from the tube per mater of its length is determined from Newton’s law of cooling to be

WTThAQ

mmmLDA

bfinnofinno

finno

537)25120)(0942.0)(60()(

0942.0)1)(03.0( 21

=−=−=

===

∞&

ππ

)002.0)(/180(/60*)002.0*

21015.0()

21(

07.2015.0

)002.0*2103.0(

21

0

02

1

2

mCmWCmW

kthtL

r

tr

+=+

=+

=+

The efficiency of the circular fins attached to a circular tube is plotted in fig.3-43 Noting that L=1/2(D2-D1)=1/2(0.06-0.03)=0.015m in this case we have

207.0=

(1)

(2)

40


From (1),(2) 95.0=finη

trrrAfin 22

12

2 2)(2 ππ +−=

[ ]2

22

00462.0)002.0)(03.0(2)015.0()03.0(2

mmmmm

=

+−= ππ

)(max, ∞−== TThAQQ bfinfinfinfinfin ηη &&

WCmCmW

0.25)25120)(00462.0)(/60(95.0 0202

=−=


)(

000283.0)003.0)(03.0( 21

∞−=

===

TThAQ

mmmSDA

bunfinunfin

unfin

&

ππ

WCmCmW 60.1)25120)(000283.0)(/60( 0202 =−=

WQQnQ unfinfinfintotal 5320)6.10.25(200)(, =+=+= &&&

Heat transfer from the unfinned portion of the tube is

Noting that there are 200 fins and thus 200 interfin spacings per meter length of the tube, the total heat transfer from the finned tube becomes

41


) ( 47835375320,, lengthtubemperWQQQ finnofintotalincrease =−=−= &&&

Therefore, the increase in heat transfer from the tube per meter of its length as a result of the addition of fins is

Discussion The overall effectiveness of the finned tube is

9.9537

5320

,

,, ===

WW

QQ

finnototal

fintotaloverallfin &

&ε


FIGURE 3–49


( )DzLS/4ln

2π=

( ) mx

mxS 9.621.0/5.04ln

302==

π

The shape factor for this configuration

42


FIGURE 3–50



FIGURE 3–51


43


FIGURE 3–52

Ventilation paths for a naturally ventilated attic and the appropriate size of the flow area around the radiant barrier for proper air circulation.


FIGURE 3–53

Three possible locations for an attic radiant barrier.

44


FIGURE 3-54

Thermal resistance network for a pitched roof-attic-ceiling combination for the case of an unventedattic.

STEADY HEAT CONDUCTION IN PLANE WALLS - RCME … · STEADY HEAT CONDUCTION IN PLANE WALLS Heat flow...

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