Conduction Heat Transfer - UniMAP Portalportal.unimap.edu.my/portal/page/portal30/Lecturer...

30
+ Conduction Heat Transfer HANNA ILYANI ZULHAIMI

Transcript of Conduction Heat Transfer - UniMAP Portalportal.unimap.edu.my/portal/page/portal30/Lecturer...

+

Conduction Heat Transfer

HANNA ILYANI ZULHAIMI

+OUTLINE

u CONDUCTION: PLANE WALL

u CONDUCTION: MULTI LAYER PLANE WALL (SERIES)

u CONDUCTION: MULTI LAYER PLANE WALL (SERIES AND PARALLEL)

u MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION

u CONDUCTION: CYLINDER

u CONDUCTION: SHPERE

u CRITICAL RADIUS OF INSULATION

+Conduction: Plane Wall

)( 12 TTxkAQ −Δ

−=

A plane wall

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

T1

T2

Q = heat rate in direction normal to surface

Δx = Wall thickness

T1, T2 = the wall face temperature

A = surface area

k = thermal conductivity

Δx

Q

+Conduction: Plane Wall

u   If  k  varies  with  T  according  linear  rela3on  :                                

+

composite wall

cc

BB

AA x

TTAkxTTAk

xTTAkQ

Δ

−−=

Δ

−−=

Δ

−−= 342312

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

AkxAkxAkxTTQ

ccBBAA ///41

Δ+Δ+Δ

−=

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 Ohm’s law in electric-circuit theory

Rth = the thermal resistances of the various materials

CBA RRRTTQ++

−= 41

AkxRn

ncond

Δ=

∑Δ

=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 560ºC. Calculate the heat flux through the composite structure.

kcopper = 385 W/m.ºC

kasbestos = 0.166 W/m.ºC

Kglass wool = 2.22 W/m.ºC

THERMAL RESISTANCE NETWORKS u THE GENERALIZED

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

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

THERMAL RESISTANCE NETWORKS u TOTAL HEAT

TRANSFER

u RESISTANCE THROUGH EACH LAYER

u OVERALL EQUATION

u OVERALL 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 R

RRRRR

TTQ+⎥⎦

⎤⎢⎣

++

−= 41

AkxRn

ncond

Δ=

Akx

Akx

Akx

Akx

Akx

Akx

TTQ

D

D

C

C

B

B

C

C

B

B

A

A Δ+

⎥⎥⎥⎥

⎢⎢⎢⎢

Δ+

Δ

Δ⋅

Δ

−=

2/2/

2/2/

41

+

Dr. Şaziye Balku

14 NEWTON’S 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 u FOR A SERIES OF LAYERS WHERE

SYSTEM THE FLUX THROUGH EACH LAYER IS CONSTANT

MULTIPLE LAYERS

u THE FLUX THROUGH EACH LAYER IS THE SAME, SO:

u IN TERMS OF RESISTANCE THIS RELATIONSHIP BECOMES:

MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION

MULTIPLE LAYERS

u IN OVERALL TERMS, CONSIDER THE DRIVING FORCE TO BE T∞1 - T∞2 AND THEN EXPRESS THE OVERALL RESISTANCE AS

u SO THE OVERALL HEAT TRANSFER CAN THEN BE EXPRESSED AS

MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION

+

Dr. Şaziye 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. Şaziye Balku

19

A LONG CYLINDERICAL PIPE STEADY STATE OPERATION

drdTkAQ cylcond −=

,

Fourier’s law of conduction

=•

cylcondQ , constant

∫∫ ==

−=2

1

2

1

, T

TT

r

rr

cylcond kdTdrA

Q

rLA π2=

)/ln(2

12

21, rr

TTLkQ cylcond−

=•

π

cylcylcond R

TTQ 21,

−=

LkrrRcyl π2)/ln( 12=

+Heat flow through radial system

Multilayer cylinder

CBA krr

krr

krr

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

342312

41

++

−=

π

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

+EXAMPLE: Combination Conduction and Convection

A thick–walled 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 π=

Rsph =r2 − r14πr1r2k

sphsphcond R

TTQ 21,

−=

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

+

Dr. Şaziye Balku

25

CRITICAL RADIUS OF INSULATION

)2(1

2)/ln(

2

12

11

LrhLkrr

TTRRTTQconvins

ππ+

−=

+

−= ∞∞

0/ 2 =•

drQd

hkr cylindercr =,

Thermal conductivity

External convection heat transfer coefficient

show

CYLINDER

+

Dr. Şaziye Balku

26

CHOSING INSULATION THICKNESS

cr

cr

cr

rrrrrr

>

=

<

2

2

2

max

Before insulation check for critical radius

hkr spherecr2

, =

+EXAMPLE

An electric wire having diameter of 1.5 mm and covered with a plastic insulation (thickness = 2.5 mm) is exposed to air at 300 K and ho= 20 W/m2. K. The insulation has a k of 0.4 W/m. K. It is assumed that the wire surface temperature is constant at 400 K and it is not affected by the recovering.

a)  Calculate the value of the critical radius

b)  Calculate the heat loss per m of wire length with no insulation

c)  Repeat (b) for insulation being present

+ANSWER

+SUMMARY

u Specify appropriate form of the heat equation.

u Solve for the temperature distribution.

u Apply Fourier’s Law to determine the heat flux.

Simplest Case: One-Dimensional, Steady-State Conduction with

u Common Geometries:

Ø  The Plane Wall: Described in rectangular (x) coordinate. Area

perpendicular to direction of heat transfer is constant (independent

of x).

Ø  The Tube Wall: Radial conduction through tube wall.

Ø  The Spherical Shell: Radial conduction through shell wall

+

Thank you J