+Conduction Heat Transfer

HANNA ILYANI ZULHAIMI

+OUTLINE

uCONDUCTION: PLANE WALL

uCONDUCTION: MULTI LAYER PLANE WALL (SERIES)

uCONDUCTION: MULTI LAYER PLANE WALL (SERIES AND PARALLEL)

uMULTIPLE LAYERS WITH CONDUCTION AND CONVECTION

uCONDUCTION: CYLINDER

uCONDUCTION: SHPERE

uCRITICAL RADIUS OF INSULATION

+Conduction: Plane Wall

)( 12 TTxkAQ

=

A plane wall

First consider the plane wall where a direct application of Fouriers law may be made. Integration yields:

T1

T2 Q = heat rate in direction normal to surface

x = Wall thickness

T1, T2 = the wall face temperature

A = surface area

k = thermal conductivity

x

Q

+Conduction: Plane Wall

u If k varies with T according linear rela3on :

+composite wall

cc

BB

AA x

TTAkxTTAk

xTTAkQ

=

=

= 342312

The temperature gradients in the three materials are shown, and the heat flow may be written:

AkxAkxAkxTTQ

ccBBAA ///41

++

=

Solving these three equations simultaneously, the heat flow is written

Note: the heat flow must be the same through all sections.

Heat flow through multilayer plane walls

+composite wall

Note: the heat flow must be the same through all sections.

A relation quite like Ohms law in electric-circuit theory

Rth = the thermal resistances of the various materials

CBA RRRTTQ++

= 41

AkxRn

ncond

=

=th

overall

RTQ

Heat flow through multilayer plane walls

+EXAMPLE 1

+EXAMPLE 1

+ QUIZ 1 A composite wall is formed of a 2.5-cm copper plate, a 3.2-mm layer of asbestos, and a 5-cm layer of glass wool. The wall is subjected to an overall temperature difference of 560C. Calculate the heat flux through the composite structure.

kcopper = 385 W/m.C

kasbestos = 0.166 W/m.C

Kglass wool = 2.22 W/m.C

THERMAL RESISTANCE NETWORKS uTHE GENERALIZED

FORM FOR THE THERMAL RESISTANCE NETWORK IS BASED ON THE ELECTRICAL ANALOGY

uFOR PARALLEL PATHS, THE DRIVING FORCES ARE THE SAME FOR THE SAME TERMINAL TEMPERATURES, AS PER FIGURE (3-19)

THERMAL RESISTANCE NETWORKS uTOTAL HEAT

TRANSFER

uRESISTANCE THROUGH EACH LAYER

uOVERALL EQUATION

uOVERALL RESISTANCE FOR PARALLEL FLOWS:

+A

B

C

D

q = ?

Construct the electrical analog

HEAT FLOW THROUGH PLANE WALL

+A

B

C

D

1.1 Heat flow through plane wall

DCB

cBA RRR

RRR

TTQ+

++

= 41

AkxRn

ncond

=

Akx

Akx

Akx

Akx

Akx

Akx

TTQ

D

D

C

C

B

B

C

C

B

B

A

A +

+

+

=

2/2/

2/2/

41

+Dr. aziye Balku

14 NEWTONS LAW OF COOLING FOR CONVECTION HEAT TRANSFER RATE

)(

= TThAQ SSconv

conv

Sconv R

TTQ

=

Sconv hAR 1=

convR

h

Convection resistance of surface

(W)

(0C / W)

Convection heat transfer coefficient

MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION uFOR A SERIES OF LAYERS WHERE

SYSTEM THE FLUX THROUGH EACH LAYER IS CONSTANT

MULTIPLE LAYERS

uTHE FLUX THROUGH EACH LAYER IS THE SAME, SO:

uIN TERMS OF RESISTANCE THIS RELATIONSHIP BECOMES:

MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION

MULTIPLE LAYERS

uIN OVERALL TERMS, CONSIDER THE DRIVING FORCE TO BE T1 - T2 AND THEN EXPRESS THE OVERALL RESISTANCE AS

uSO THE OVERALL HEAT TRANSFER CAN THEN BE EXPRESSED AS

MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION

+Dr. aziye Balku

18

HEAT CONDUCTION IN CYLINDERS

Steady-state heat conduction

Heat is lost from a hot-water pipe to the air outside in the radial direction.

Heat transfer from a long pipe is one dimensional

+Dr. aziye Balku

19

A LONG CYLINDERICAL PIPE STEADY STATE OPERATION

drdTkAQ cylcond =

,

Fouriers law of conduction

=

cylcondQ , constant

==

=2

1

2

1

, T

TT

r

rr

cylcond kdTdrA

Q

rLA 2=

)/ln(2

12

21, rr

TTLkQ cylcond

=

cylcylcond R

TTQ 21,

=

LkrrRcyl 2)/ln( 12=

+Heat flow through radial system

Multilayer cylinder

CBA krr

krr

krr

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

342312

41

++

=

Note: the heat flow, q must be the same through all layers!

+EXAMPLE: Combination Conduction and Convection

A thickwalled tube of stainless steel (A) having k= 21.63 W/m.K with dimension of 0.0254 m ID and 0.0508 m OD is covered with a 0.0254-m thick layer of insulation (B), k= 0.2423 W/m.K . The inside wall temperature of the pipe is 811 K and outside surface of the insulation is at 310.8 K. For a 0.305 m length of pipe, calculate the heat loss and also the temperature at the interface between metal and insulation.

+ANSWER

+ANSWER

+ 24 CONDUCTION IN SPHERES

24 rA =

Rsph =r2 r14r1r2k

sphsphcond R

TTQ 21,

=

FOR A SPHERICAL SYSTEM (HOLLOW BALL) THE SAME METHOD IS USED:

+Dr. aziye Balku

25

CRITICAL RADIUS OF INSULATION

)2(1

2)/ln(

2

12

11

LrhLkrr

TTRRTTQconvins

+

=

+

=

0/ 2 =

drQd

hkr cylindercr =,

Thermal conductivity

External convection heat transfer coefficient

show

CYLINDER

+Dr. aziye Balku

26

CHOSING INSULATION THICKNESS

cr

cr

cr

rrrrrr

>

=