Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 1

Chapter 2

Heat Transfer Principles

Update: 9/3/2019

The scope of heat transfer covers the following items.

The amount of energy transfer

The rate of energy transfer

How to enhance or to depress energy transfer

(2.1). Heat transfer modes

Ways of Energy Transfer

Direct contact between atoms and molecules

Absorption and emission of electromagnetic waves

Categories of Heat Transfer

Conduction

Convection

Phase Transition：Condensation, Boiling, Two phase flow

Radiation

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 2

Figure 1 shows how heat flows from inside of chip to the environment.

Fig. 1: The heat transfer from chip to its environment

The route of heat transfer for a laptop:

Heat source → Conduction → Convection → Boiling → Two phase flow →

Condensation → Conduction → Convection →Ambient

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 3

The route of heat transfer for an LED lamp:

Heat source → Conduction → Convection → Conduction → Convection →

Ambient

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 4

熱對流：固體與流體之間的熱傳，熱通量與溫度差成正比

熱對流之現象：牛頓冷卻定律(Newton’s cooling law)

熱由高溫處往低溫處傳遞，熱傳量與溫度差成正比。

比例常數稱為熱傳係數(heat transfer coefficient)，一般以 h 表示。

q h T

Q Aq hA T

熱傳係數不是流體的特性，與流場有關。

常用的熱傳係數

h(W/m2K)

Free Convection Gas 1 - 2 window, room to outside air Free Convection Gas - Forced liquid water 5 - 15 radiator central heating Free Convection Gas - Condensing Vapor Water 5 - 20 steam radiator Forced Convection Gas - Free Convection Gas 3 - 10 Super heater Forced Convection Gas - Forced Convection Gas 10 - 30 heat exchanger Forced Convection Gas - Forced liquid water 10 - 50 gas cooler Forced Convection Gas - Condensing Vapor Water 10 - 50 gas boiler Liquid Free Convection - Forced Convection Gas 25 - 500 Liquid Free Convection - Free Convection Liquid 50 - 100 heating coil in vessel water Liquid Free Convection - Free Convection Liquid 500 - 2000 water with steering Liquid Free Convection - Condensing vapor water 300 - 1000 steam jackets around vessels Forced liquid water - Free Convection Gas 10 - 40 combustion chamber Forced liquid water - Free Convection Liquid 500 - 1500 cooling coil - stirred Forced liquid water - Forced liquid water 900 - 2500 heat exchanger Forced liquid water - Condensing vapor water 1000 - 4000 condensers steam water Boiling liquid water - Free Convection Gas 10 - 40 steam boiler Boiling liquid water - Forced Liquid flowing 300 - 1000 evaporation of refrigerator Boiling liquid water - Condensing vapor water 1500 - 6000 evaporators steam/water

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 5

Dimensionless Numbers Commonly Used in Heat Transfer

Name Symbol Definition Applications

Nusselt number Nu hD/k Natural or forced

convection, boiling

or condensation

Reynolds number

Re ρVD/μ Forced convection,

boiling or

condensation

Prandtl number

Pr cpμ/k (ν/α) Natural or forced

convection, boiling

or condensation

Stanton number

St Nu/Re Pr Forced convection

Grashof number

Gr gβρ2ΔtL3/μ

2 Natural convection

Peclet number

Pe Re Pr Forced convection

Graetz number

Gz Re Pr D/L Laminar forced

convection

Rayleigh number

Ra Gr Pr Natural convection

Biot number Bi hD/k Transient

conduction

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 6

(2.2). Conduction Heat Transfer

The primary energy transfer mode occurs inside electronic chips.

Chip inside → chip surface → fin

Conduction heat transfer inside IC chip

Conduction heat transfer along the fin

(2.2.1). Energy Equation

The energy equation describes how energy flows inside the conveying medium.

( , , , )T T t x y z

.

'''T T T T

c cu cv cw q ut x y z

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 7

.

'''T

c cV T q ut

Energy stored + Energy conveyed = Energy transferred + Energy generated

(2.2.2). Fourier’s Law

Proposed by a French physicist at 1804.

Heat is transferred from a high temperature media to a low temperature media,

and the rate of transfer is proportional to the temperature gradient.

q k T

The proportional constant is called the thermal conductivity.

.

( ) '''T T T T

c cu cv cw k T ut x y z

(2.2.3). Heat Conduction Equation

.

( ) '''T T T T

c cu cv cw k T ut x y z

If the media is not moving, just like the case heat transfer in chips, we have

0, 0, 0u u w

.

( ) '''T

c k T ut

If the thermal conductivity is a constant，.

2 '''T

c k T ut

.

2 ''T k uT

t c c

k

c

, thermal diffusivity

At steady state 0T

t

2 2 2

2 2 20

T T T u

x y z k

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 8

(2.2.4). Thermal Conductivity

Thermal conductivity is a positive figure. It could not be of negative value.

A negative value of thermal conductivity would violate the second law of

thermodynamics.

Materials with high thermal conductivity are conductors.

Conductor： k = 10~1000 W/m-℃. Most metals are conductors.

Material Thermal Conductivity

Silver 406

Copper 390

Gold 310

Aluminum 200

Iron 55

Lead 35

Diamond 1000~2500

Boron nitride 1500

Diamond has the highest thermal conductivity. (1000~2500 W/ m-℃)

Boron nitride (BN): Cubic boron nitride, diamond-like allotrope(同分異構物),

c-BN, β-BN, or z-BN. Density (g.cm-3

) 2.3, k = 1500 W/m.K

Ceramic materials have thermal conductivity in the range 0.1~10 W/m-K.

Materials with low thermal conductivity are insulators.

Insulator：k = 0.001~0.1 W/m-℃

Cotton, fiberglass, cork are examples of insulators.

Air a very good insulator. The thermal conductivity of air is 0.026 W/m-K.

Aerogel is a silica-based substance, composed of 90-99.8% air, with densities

1.9 ~ 150 mg/cm³. It has extremely low thermal conductivity (0.003 W/m·K).

Its melting point is 1,200 °C.

Thermal conductivity is temperature dependent.

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 9

人工石墨片 iGS (Intelligent Graphite Sheet) 具有薄型化、輕量化、與高熱

傳導等特性，適合用在具有空間局限性的 3C 電子產品，作為熱管理元件。iGS

人工石墨片可以單獨使用在智慧型手機與平板電腦中，或者與傳統散熱元件整

合成為散熱模組。iGS 人工石墨片熱傳導係數最高可達 1500W/mk，是銅的四倍

以上，重量約為銅的四分之一，可繞折，適合用在複雜的 3D 機構設計。透過簡

易的加工貼附雙面膠與絕緣材料，可以裁切成不同形狀，量產性佳。具有 99.9wt%

以上的碳含量，符合各種歐盟環保指令。應用包括智慧型手機，平板電腦，筆

記型電腦，數位相機等。

石墨烯(Graphene)目前是世上最薄卻也是最堅硬的奈米材料，導熱系數高

達 5300 W/m·K，高於奈米碳管和金剛石，而電阻率只約 10-6

Ω·cm，比銅或銀

更低，為目前世上電阻率最小的材料。由於它的電阻率極低，電子的移動速度

極快，因此被期待可用來發展出更薄、導電速度更快的新一代電子元件或電晶

體。石墨烯實質上是一種透明、良好的導體，也適合用來製造透明觸控螢幕、

光板、甚至是太陽能電池。

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 10

(2.2.5). One Dimensional Heat Conduction

2

20

d T

dx

1 2T c x c

T

x

Temperature distribution is linear inside the plate.

The constants, 1c and 2c , are determined by the boundary conditions.

Temperatures at both sides are given

0, ; ,L Hx T T x L T T

( )L H L

xT T T T

L

( )H LT Tq k

L

thermal resistance：L

Rk

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Example，Determine the heat transfer rate if the surface temperature at both sides of

a flat plate are given.

TH=100℃，TL=25℃，L=5mm，k=3 W/m-℃ (ceramics)

Comment: Tremendous heat flux would occur if the surface temperatures at both

sides of a flat plate are given. The thermal conductivity of the plate is the

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 11

determining factor.

-----------------------------------------------------------------------------------------------------

Heat flux at one side is given, and the other side is convective.

0x ，dT

x k qdx

； x L ， ( )dT

k h T Tdx

1

dTc

dx

0x ， 1kc q ， 1

qc

k

x L ， 2( )q

q h L c Tk

2

1( )

Lc q T

h k

1( )

q LT x q T

k h k

0x ， max

1( )

LT q T

h k

x L ， min

qT T

h

thermal resistance：1L

Rk h

-----------------------------------------------------------------------------------------------------

Example，Determine the heat transfer rate if heat flux at one side is given, and the

other side is convective.

qH=100W/cm2，Ta=25℃，L=5mm，k=3 W/m-℃，h = 20 W/m

2-℃

Comment: The heat transfer coefficient is the determining factor. It is practically

not possible to keep the surface temperature at room temperature.

-----------------------------------------------------------------------------------------------------

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 12

Temperature at one side is given, and the other side is convective.

0x ，HT T ； x L ， ( )

dTk h T T

dx

( )H H

hxT T T T

k hL

H

H

T T hx

T T k hL

x L ， ( )L H

kT T T T

k hL

( ) ( )

1L H

hq h T T T T

hL

k

thermal resistance：1L

Rk h

1/

hL L

k k h = ratio of inside and outside thermal resistance

-----------------------------------------------------------------------------------------------------

Example，Determine the heat transfer rate if the surface temperature at one side is

given, and the other side is convective.

TH=100℃，Ta=25℃，L=5mm，k=3 W/m-℃，h = 20 W/m2-℃

Comment: The heat transfer coefficient is the determining factor. It is practically

not possible to keep the surface temperature at room temperature.

-----------------------------------------------------------------------------------------------------

Both sides are convective.

1 20, ( ); , ( )H L

dT dTx k h T T x L k h T T

dx dx

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 13

1 2

( )

1 1H LT T T

qL R

h k h

thermal resistance：1 2

1 1LR

h k h

1 1

1 1

1H

q TT T T

h R h

2 1 2

L T LT T T q

k R k

3 2

2 2

1L

q TT T T

h R h

-----------------------------------------------------------------------------------------------------

Example，Determine the heat transfer rate if the heat transfer coefficients at both

sides of a flat plate are given.

TH=100℃，TL=25℃，hH=30 W/m2-℃，hL=10 W/m

2-℃

L=5cm，k=390 W/m-℃ (copper)；L=5cm，k=3 W/m-℃ (limestone)

Comment: Moderate heat flux would occur if the heat transfer coefficients at both

sides of a flat plate are given. The thermal conductivity of the plate is NOT the

determining factor.

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Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 14

(2.2.6). Thermal resistance of multi-layer material

1

( )

1 1H L

n

i

iH i L

T T Tq

L R

h k h

-----------------------------------------------------------------------------------------------------

Example，Determine the heat transfer rate if two types of plates are placed in series

and the heat transfer coefficients at both sides of a flat plate are given.

TH=100℃，TL=25℃，hH=30 W/m2-℃，hL=10 W/m

2-℃

L=5cm，k=390 W/m-℃ (copper)；L=5cm，k=3 W/m-℃ (limestone)

-----------------------------------------------------------------------------------------------------

Example，An IC chip is made of three parallel layers. The top layer is molding

compound with thermal conductivity of 0.9 W/m-K. The middle layer is ceramic

material and the bottom layer is copper film. The associated thickness and thermal

conductivity for each layer are listed on the table below. Heat is generated between

the top and the middle layer with the intensity of q”H. Find the maximum

temperature within the chip.

top middle bottom

t(mm) 3 0.1 0.2

k (W/m-K) 0.9 3.0 390

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 15

hH=30 W/m2-℃，TL=25℃，q”H=1000 W/m

2

-----------------------------------------------------------------------------------------------------

This illustration shows a simplified network of thermal resistances in an LED system.

The total thermal resistance of the system can be divided into internal and external

thermal resistances. The internal thermal resistance is from the junction to the

package of the LED. The external thermal resistance includes the selection of the

cooling mode, heat sink design, selection of substrate material, and attachment

process.

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 16

(2.2.7)、Internal Heat generation

.2 ''' 0k T u

2 .

2''' 0

d Tk u

dx

The general solution is as follows.

.

2

1 2

1 '''

2

uT x c x c

k

Temperature inside is parabolic

T

x

Temperatures at the surface of both sides are given

To TL

L

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 17

0x ，OT T

x L ，LT T

.

2

1

1 '''

2L O

uT L c L T

k

.

2

1

1 1 '''( )

2L O

uc T T L

L k

. .

2 21 ''' 1 1 '''( )

2 2L O o

u uT x T T L x T

k L k

.

21 '''( ) ( )

2o L O

u xT T Lx x T T

k L

The maximum temperature occurs at

.

1 ''' 1( 2 ) ( ) 0

2L O

dT uL x T T

dx k L

max ( )2

L O

L kx T T

Lu

If L OT T ,

max2

Lx

If L OT T ,

max2

Lx

The maximum temperature will be close to the side of high temperature.

If ( )2

L O

k LT T

Lu

,

maxx L

If ( )2

L O

k LT T

Lu

,

max 0x

The maximum temperature is outside of the heating element.

.

maxmax max max

1 '''( ) ( )

2o L O

u xT T x L x T T

k L

-----------------------------------------------------------------------------------------------------

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 18

Example，A ceramic layer with inner heat generation is placed between two plates of

fixed temperature. The left side is kept at 100℃. The right side is kept at 30℃.

The thickness of the layer is 1 mm, and the heat generation rate is 107 W/m

3. Find

the maximum temperature within the layer.

-----------------------------------------------------------------------------------------------------

Heat transfer coefficients on both sides are given

To TL

ho hL

.

2

1 2

1 '''

2

uT x c x c

k

.

1

'''dT ux c

dx k

0x ，1( )o O

dTk h T T

dx

x L ，2( )L L

dTk h T T

dx

0x ，1 2( )o Okc h T c ，

1 2 2( )o o oO O

h h hc T c c T

k k k

x L ，

. .

2

1 1 2

''' 1 '''( ) ( )

2L L

u uk L c h L c L c T

k k

..

2

1 1 2

1 ''''''

2L L L L L

uu L kc h Lc h c L h T h

k

L

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 19

..

2

2

1 '''[( ) ] ''' ( )

2

o oL L L L L L O

h u hh L k h c u L L h T h h L k T

k k k

.

2

1''' (1 ) ( )

2

( )

L oL L L O

oL L

Lh hu L T h h L k T

k kch

h L k hk

.

1

1( ) ''' (1 )

2

( )

LL O L

o

oL L

Lhh T T u L

h kchk

h L k hk

..

2

1( ) ''' (1 )

1 ''' 2

2( )

LL O L

o

oL L

Lhh T T u L

u h kT x xhk k

h L k hk

. 1''' (1 ) ( )

2

( )

L oL L L O

oL L

Lh hu L T h h L k T

k kh

h L k hk

0x ，

.

1

''' (1 ) ( 1)2

( 1)

L Lo O L L

Lo L

Lh h Lu L h T T h

k kTh L

h hk

x L ，

..

2

2

1( ) ''' (1 )

1 ''' 2

2( )

LL O L

o

oL L

Lhh T T u L

u h kT L Lhk k

h L k hk

. 1''' (1 ) ( )

2

( )

L oL L L O

oL L

Lh hu L T h h L k T

k kh

h L k hk

.

2

( 1) ''' ( 1)2

( 1)

o oL L o O

Lo L

Lh h Lu L h T h T

k kTh L

h hk

Once 1T and 2T are obtained, the maximum temperature and the location that

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 20

the maximum temperature occurs are thus can be obtained.

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Example，If the ceramic layer is placed in the atmosphere instead of between two

plates, and the heat transfer coefficients on both sides are 30 W/m2-℃ and 100

W/m2-℃ respectively. The ambient temperature is 25℃. Find the maximum

temperature within the layer.

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Multi layers heating element

If other layers of different material are covered on top of the heating element,

just like a chip made of several layers, the calculation procedure is as the following.

To TL

TA T1 T2 TB

1 ( )AA A A O A

A

T Tk h T T q

L

1

1O

AA

A A

T Tq

L

k h

2 ( )BB B B L B

B

T Tk h T T q

L

L

LB

LA

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 21

2

1L

BB

B B

T Tq

L

k h

.

2

2 1 1

1 '''( ) ( )

2

u xT Lx x T T T

k L

.

2 1

1 ''' 1( 2 ) ( )

2

dT uL x T T

dx k L

.1

2 10

1''' ( )

12

OA

x A

A A

dT k T Tq k u L T T

Ldx L

k h

.2

2 1

1''' ( )

12

LB

x L B

B B

dT k T Tq k u T T

Ldx L

k h

.1 1 2 1

'''1 1 2

O

A A

A A A A

T T T Tu L

L L L L

k h K K k h

.1 2 2 1

'''1 1 2

L

B B

B B B B

T T T Tu

L L L L

K k h K k h

.

1

.2

1 1 1 1'''

1 1 2

1 1 1 1'''

1 1 2

O

A A

A A A A

L

B B

B B B B

Tu L

L L L L

k h K K k hT

T Tu

L L L L

K k h K k h

1.

1

.2

1 1 1 1'''

1 1 2

1 1 1 1'''

1 1 2

O

A A

A A A A

L

B B

B B B B

Tu L

L L L L

k h K K k hT

T Tu

L L L L

K k h K k h

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 22

max 2 1( )2

L kx T T

Lu

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Assignment2.1

An IC chip is made of three parallel layers. The top layer is molding compound

with thermal conductivity of 0.9 W/m-K. The middle layer is ceramic material and

the bottom layer is copper film. The associated thickness and thermal conductivity

for each layer are listed on the table below. Heat is generated inside the middle

layer with the rate of 107 W/m

3. Find the maximum temperature within the chip.

top middle bottom

t(mm) 3 1 0.2

k (W/m-K) 0.9 3.0 390

hH=30 W/m2-℃，TL=25℃

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Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 23

(2.3). Forced convection Heat Transfer

Energy Equation

.

( ) '''T T T T

c cu cv cw k T ut x y z

(2.3.1). Convective Heat Transfer in electronic equipments

Flow over a plate: PCB, chip surface.

Flow along parallel plates: PCBs

Flow over a cylinder: copper column, heat pipe, fin.

Flow over cylinder banks: fin arrays.

Flow inside a tube: water tube.

Natural convection along a vertical plate: PCB, chip.

Natural convection over a horizontal plate: PCB, chips

Natural convection between vertical plates: PCBs

Jet impingement on a plate: fin cooling.

Hydraulic diameter (水力直徑)

4h

Ad

p ， A : cross section area, p : wet peripheral length

Circular annulus elliptic rectangular square

Circular pipe：2

4A d

， p d ， hd d

Annulus pipe：2 2( )

4o iA d d

， ( )e ip d d ， h e id d d

Elliptic pipe: A ab , 3( ) ( 3 )(3 )p a b a b a b ,

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 24

4

3( ) ( 3 )(3 )h

abd

a b a b a b

Rectangular pipe： A ab ， 2( )p a b ，2

h

abd

a b

Square pipe： 2A a ， 4p a ， hd a 。

It is noted that hydraulic diameter is not equivalent diameter.

Equivalent diameter(等效直徑).

4e

Ad

Annulus pipe：2 2( )

4o iA d d

，

2 2

e o id d d

Rectangular pipe： A ab ， ed ab 。

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Example : Determine the hydraulic diameter and the equivalent diameter of a square

air duct with the width of 30 cm.

-----------------------------------------------------------------------------------------------------

Example : Determine the hydraulic diameters of four kinds of pipe with the same

cross sectional area, a circular pipe, an annulus pipe, a rectangular pipe, and a

triangular pipe. The flow area is 10 cm2.

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Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 25

(2.3.2). Flow over a flat plate

Boundary layer developed along the surface of a plate:

Flow over a flap plate

Single flat plate

For flow over a flat plate, all properties are evaluated at mean film temperature.

1( )

2m wT T T

The flow is composed of laminar region and turbulent region.

0

1( )

crit

crit

x L

la tu

x

h h dx h dxL

, where 5Re 5 10critcrit

u x

55 10critxu

For example, air at 300 K:

= 1.8462×10-5

kg/m-sec, =1.1774 kg/m3,

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 26

u= 1 m/s,

critx = 7.841 m,

u= 5 m/s,

critx = 1.568 m,

u= 10 m/s,

critx = 0.784 m,

For most fins, the length is less than 10 cm, and the flow velocity is less than 5 m/sec.

It is appropriate to assume that the flow over a flat plate is laminar.

In the laminar region:

1 1

3 20.332Pr Rex xNu , 50 Re 5 10x

1 1

3 20.664Pr ReL LNu

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Example: Calculate the heat transfer coefficient and the associated heat transfer

rate for a flat plate fin with the width of 3cm, and length of 1 cm. The surface

temperature is 100℃. The air velocity is u= 1 m/s, and the air temperature is 300

K.

= 1.8462×10-5

kg/m-sec, =1.1774 kg/m3, k=0.02624 W/m-K

Re= 1272, Nu= 21.1, h= 27.7 W/m2-℃

If velocity is raised to u= 10 m/s, Re= 12718, Nu= 66.7, h= 87.7 W/m2-℃

If the width of fin is increased to w= 4 cm, Re= 2544, Nu= 29.8, h= 19.6 W/m2-℃

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Example: If the flat plate in the above example is an IC chip with heat dissipation

rate of 1W, find the surface temperature. The room temperature is 25℃.

Q=hA(Ts - Ta)

Ts = 25 +1/27.7/ 0.0002= 25 + 180.5 = 205.5℃

It is noted that the cooling is not enough fir the IC chip. If velocity is raised to u=

10 m/s, surface temperature would drop to an acceptable value of 82℃.

-----------------------------------------------------------------------------------------------------

Assignment2.2

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 27

An IC chip with the dimension of 2×2 cm is composed of three parallel layers.

The top layer is made of molding compound. The middle layer is made of ceramic

material and the bottom layer is made of copper film. The associated thickness and

thermal conductivity for each layer are listed in the table below. The power

dissipation rate of the IC chip is 4 watts. Heat is generated between the top and the

middle layer. Air flows over the chip with velocity of 2 m/sec, and the air

temperature is 25℃.

Top middle bottom

t(mm) 3 0.1 0.2

k (W/m-K) 0.9 3.0 390

(a). Find the heat transfer coefficient on the surface of chip.

(b). Find the maximum temperature within the chip.

(c). Find the heat transfer on the top side and the bottom side.

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Parallel flat plates

Flow between parallel plates may seem to be a simple heat transfer case. There are

nevertheless many flow types to consider. Laminar or turbulent flow is one example

but the difference between the fully and the developing flow regions is also important.

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 28

The correlations used must in addition converge towards the single plate case when

the air gap is large.

wT T , temperature difference between air and wall

i i wT T , inlet temperature

e e wT T , exit temperature

ln

i em

i

e

, mean temperature

s : wall to wall distance

w : wall width

4 42

2( )h

A wsd s

p w s

: hydraulic diameter, for w s

Re Pr hdGz

L , Graetz number

Laminar plates: Re 2500hd u

1.37

0.87

0.02897.54

1 0.0438

m hm

Gz h dNu

Gz k

, the averaged value.

4

0 14

mNu

GzGz

Nu e

, the Nusselt number at the exit.

Turbulent plates:

0.3

0.550.407Re hm

dNu

L

, 2500 Re 7000d , 3 20

h

L

d

0.2

0.80.0358Re hm

dNu

L

, 7000 Re 20000d , 3 20

h

L

d

0.8 0.40.023Re PrmNu , Re 2300d , 20h

L

d

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 29

Ti Tw Te

dx

a i im Au

( )a a wm c dT h T T pdx

( )a a w

dTm c hp T T

dx

( ) 0w

a a

dT hpT T

dx m c

wT T , 1

a a

hp

m c

0d

dx

0x , iT T , i

x

i

e

x L , eT T , e

L

e

i

e

( ) ( ) (1 ) ( )L

a a e i a a e i a a i w m mQ m c T T m c m c e hpL T T hpL

(1 )L

i m m

a a

hpL Le

m c

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 30

1

/

L

m

i

e

L

ln

i em

i

e

This is the definition of bulk mean temperature.

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Example: Two parallel fins are separated with spacing of 0.5 cm. The length and

the width of fin are 1cm and 3 cm. The surface temperature of fin is 100℃. Find

the heat transfer rate if air velocity is u= 1 m/s. The room temperature is 25℃.

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Assignment 2.3: Two parallel fins are separated with spacing of 0.5 cm. The length

and the width of fin are 1cm and 3 cm. The heat transfer rate of fin is 1W. Find the

wall temperature if air velocity is u= 1 m/s. The room temperature is 25℃.

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Example: If the heat transfer rate of fin is 2W and the surface temperature of fin is

confined to 80℃, find the air velocity.

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Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 31

(2.3. 3). Pipe Flow

Laminar: (Re<2300)

Developing flow

0.141/ 3RePr

1.86/ s

NuL D

,

0.42

RePr

8 s

L

D

2 / 3

0.065( / )RePr3.66

1 0.04[( / )RePr]

D LNu

D L

Fully developed flow

3.66Nu , uniform surface temperature

4.36Nu , uniform heat flux

Turbulent flow:

Developing flow

2 / 3

1/ 2 2 / 3

( / 2)(Re 1000)Pr1

1 1.27( / 2) (Pr 1)

s

s

f LNu

f D

,

2

1

1.58ln Re 3.28sf

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 32

Fully developed flow

0.8 0.40.023Re PrNu , heating fluid

0.8 0.30.023Re PrNu , cooling fluid

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Example: Water flows along a tube with the diameter of 2mm. Tube length is 20

cm. The velocity is 0.1 m/s. The inlet temperature is 25℃, and the wall temperature

is kept at 100℃. Find the exit temperature.

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Example: A copper plate is used to dissipate 1000W of heat. The dimension of

plate is 10cm×10cm×1cm. Water is used to cool this plate. Five circular channels

with diameter of 6mm are drilled inside the plate. Water velocity is 0.2 m/sec and

the inlet temperature is 25℃. Find the plate temperature.

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Assignment 2.4: A copper plate is used to dissipate 500W of heat. The

dimension of plate is 20cm×15cm×1cm. The plate temperature should not be higher

than 100℃. Design a water cooling system to remove the heat of plate. The ambient

temperature is 25℃

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 33

The amount of heat transfer for one tube is

( ) ( ) ( )(1 )L

e i e i w iq mc T T mc mc T T e

2

4

4

L hpL h dL hL

mc ducd uc

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Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 34

(2.4). Natural Convection

(2.4.1). Natural convection of a vertical plate

The fluid motion is driven by density variations caused by temperature

difference. The density of air is inversely proportional to temperature. Air in the

vicinity of hot wall is heated and gets lighter. The cold air outside pushes the light

air up.

Laminar flow on a vertical wall at constant temperature

11/ 2

4

1/ 41/ 2

0.75Pr

4 0.609 1.221Pr 1.238Pr

xx

hx GrNu

k

3

2

( )wx

g T T xGr

, Grashof number, the ratio of the buoyancy force to the viscous

force.

1 1

PT T

, volumetric thermal expansion coefficient.

11/ 2

4

1/ 41/ 2

0.75Pr

4 0.609 1.221Pr 1.238Pr

xk Grh

x

11/ 2

4

1/ 41/ 2

4 4 0.75Pr

3 3 4 0.609 1.221Pr 1.238Pr

LL

k Grh h

L

PrRa Gr , Rayleigh number, transition from laminar to turbulence occurs at

910Ra

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Example:

A vertical plate with 10 cm width and 20 cm height, is suspended in air. The surface

temperature is 70℃. Calculate the heat transfer rate if the air temperature is 25℃.

wT = 70℃, 1

( )2

f wT T T = 320.5 K

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 35

= 17.95×10-6

m2/s, Pr= 0.7,

1

fT = 3.12×10

-3 K

-1, k = 2.77×10

-2 W/m-K

3

2

( )wL

g T T LGr

= 3.41×107, PrRa Gr = 2.39×10

7< 10

9

11/ 2

4

1/ 41/ 2

4 0.75Pr

3 4 0.609 1.221Pr 1.238Pr

Lk Grh

L

= 8.64×0.499 = 4.31 W/m

2-K

( )wQ Ah T T = 3.6 W

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Example:

A vertical plate with 10 cm width and 20 cm height is suspended in air. The heat

dissipation rate is 5W. Calculate the surface temperature if the air temperature is 25

℃.

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Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 36

(2.4.2). Natural Convection Heat Transfer for Arrays of Vertical Parallel Flat Plates

0.7535

124

Rahb Ra

Nu ek

3

2

( )Pr Prw

b

g T T b bRa Gr

L

If the width of a fin array is fixed, the gap between two neighboring fins is

determined by the number of fins.

1

f

f

W N tb

N

2 fQ hA T N HLh T

It is noted that increasing the number of fins would decrease the gap between fins.

However, the heat transfer rate would not increase proportionally because the heat

transfer coefficient would be depressed if the spacing between fins is too crowded.

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 37

-----------------------------------------------------------------------------------------------------

Example:

A vertical fin array has the following dimensions: L = 100 mm, W = 100 mm, H = 50

mm, t = 2mm. Calculate the heat transfer coefficient if the air temperature is 30℃.

-----------------------------------------------------------------------------------------------------

Example:

Heat Transfer Technology in Electronic System NCHU ME Dept. 頁 38

A vertical fin array has the following dimensions: L = 100 mm, W = 100 mm, H = 50

mm, t = 2mm. Calculate the surface temperature if the dissipation power is 30W and

the air temperature is 30℃.

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