Heat transfer - University of Saskatchewankasap13.usask.ca/EE271/files/09_Heat_transfer-08.pdf ·...

Heat = amount of energy that is transferred from one system to another (or between system and surroundings) as a result of temperature difference

Heat is transferred by - thermal conduction - convection- radiation

Heat transfer

The origin of energy transfer is the random motion of molecules

T

x

HOT COLD

dQdtHEAT A

Fig. 2.19: Heat flow in a metal rod heated at one end. Consider the rateof heat flow, dQ/dt, across a thin section x of the rod. The rate ofheat flow is proportional to the temperature gradient T/ x and thecross sectional area A.From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002)http://Materials.Usask.Ca

x

T(x)

dT/dx

Jx

Fouriers Law of Thermal Conduction

dtdQ

AJ xx

1

Jx

= heat flux, dQx

/dt

= the rate of heat flow A = cross-sectional area

dxdT

dtdQJ x ==

dT/dx

= temperature gradient

= thermal conductivity[] = W m-1

K-1

or W m-1

0C-1

dxdCD= Ficks First LawReminder:

Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet

Thermal Conductivities of various materials



Weak Van-der-Waalsbonding

Strong metallicbonding

Strong covalentbonding

0

100

200

300

400

450

0 10 20 30 40 50 60 70Electrical conductivity, , 106 -1 m-1

Al

Ag

Au

Cu

Brass (Cu-30Zn)

Be

Pd-40Ag

Bronze (95Cu-5Sn)

Ag-3Cu

Ag-20Cu

WMo

Hg

Mg

Steel (1080)

Ni

= T CWFL

Ther

mal

cond

uctiv

ity,

(W

K-1

m-1

)

Fig. 2.20: Thermal conductivity, vs. electrical conductivity forvarious metals (elements and alloys) at 20 C. The solid linerepresents the WFL law with CWFL 2.44108 W K-2. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002)http://Materials.Usask.Ca

Wiedemann Franz - Lorenz Law

T

=CWFL =2.45108 W K 2

= thermal conductivity = electrical conductivityT

= temperature

CWFL = Lorenz number

0

100

200

300

400

450

0 10 20 30 40 50 60 70Electrical conductivity, , 106 -1 m-1

Al

Ag

Au

Cu

Brass (Cu-30Zn)

Be

Pd-40Ag

Bronze (95Cu-5Sn)

Ag-3Cu

Ag-20Cu

WMo

Hg

Mg

Steel (1080)

Ni

= T CWFL

Ther

mal

cond

uctiv

ity,

(W

K-1

m-1

)

Fig. 2.20: Thermal conductivity, vs. electrical conductivity forvarious metals (elements and alloys) at 20 C. The solid linerepresents the WFL law with CWFL 2.44108 W K-2. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002)http://Materials.Usask.Ca

Wiedemann Franz - Lorenz Law

T

=CWFL =2.45108 W K 2

= thermal conductivity = electrical conductivityT

= temperature

CWFL = Lorenz number

Copper

Brass (70Cu-30Zn)

Al-14%Mg

Aluminum

10

100

1000

10000

50000

1 10 100 1000Th

erm

alco

nduc

tivity

, (W

K-1

m-1

)Temperature (K)

Thermal conductivity vs. temperature for two pure metals (Cu and Al)and two alloys (brass and Al-14%Mg). Data extracted fromThermophysical Properties of Matter, Vol. 1: Thermal Conductivity,Metallic Elements and Alloys, Y.S. Touloukian et. al (Plenum, NewYork, 1970).



Weak Van-der-Waalsbonding

Strong metallicbonding

Strong covalentbonding

Thermal conduction in metals and some insulators

MetalsAg, Cu, Al ...

Insulators with very strong covalent bondingC (diamond), BeO (beryllia), ...

Heat is transferred by conduction electrons Heat is transferred as atomic vibrations due to strong bonding between atoms

Energetic atomic vibrations

Hot Cold

Equilibrium

Fig. 2.22: Conduction of heat in insulators involves the generationand propogation of atomic vibrations through the bonds that couplethe atoms. (An intuitive figure.)From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002)http://Materials.Usask.Ca

Vibrating Cu+ ionsElectron Gas

HOT COLDHEAT

Parabolic heat equation

tT

xTDth

=

2

2thermal diffusitivity

c

Dth =where

= densityc

= specific heat capacity

Heat flow in

Heat flow out == Rate of heat accumulation in volume x

Heat flow in

Heat flow out = Jx

(x)-

Jx

(x+x) = xxTx

xJ x 2

2

=

Rate of heat accumulation in volume x

=tTcx


ttxC

xtxCD

=

),(),(2

2

Ficks Second LawReminder:

Fouriers Law Q = A

TL

=T

(L /A)

Q = rate of heat flow or the heat current, A

= cross-sectional area, = thermal conductivity (material-dependent constant of

proportionality), T

= temperature difference between ends of component, L

= length of component

Ohms Law

I

= electric current, V

= voltage difference across the conductor, R

= resistance, L

= length, = conductivity, A

= cross-sectional area

I =VR

=V

(L /A)

dxdT

dtdQJ x ==

dtdQ

AJ xx

1

Fouriers Law Q = A

TL

=T

(L /A)

Q = rate of heat flow or the heat current, A

= cross-sectional area, = thermal conductivity (material-dependent constant of

proportionality), T

= temperature difference between ends of component, L

= length of component

Ohms Law

I

= electric current, V

= voltage difference across the conductor, R

= resistance, L

= length, = conductivity, A

= cross-sectional area

I =VR

=V

(L /A)

=

= R

Definition of Thermal Resistance

Q

= rate of heat flow, T

= temperature difference, = thermal resistance

Q =T

Thermal Resistance

= thermal resistance, L

= length, A

= cross-sectional area, = thermal conductivity

=L

A

Current supplyHeat generator

GroundAbsolute zero

EMF (Electromotive Force)Heat reservoir

R = resistance = thermal resistance

V = bias (voltage)T = temperature difference

I = CurrentQ = rate of heat flow

ELECTRICAL PHENOMENATHERMAL PHENOMENA


GroundAbsolute zero






Analogy between thermal and electrical phenomena

Fig. 2.23: Conduction of heat through a component in (a) can bemodeled as a thermal resistance shown in (b) where Q = T/.From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002)http://Materials.Usask.Ca

Hot

L

TT

Q

(a) (b)

Q = T/

Q

Cold

AQ

TQ =

RVI =


GroundAbsolute zero







GroundAbsolute zero









GroundAbsolute zero







GroundAbsolute zero








Ti T0

IR2

Q=(Ti

-T0

)/

C = capacitanceC = thermal capacitance


GroundAbsolute zero








GroundAbsolute zero







tTCQ

=

TCQ =

tVCI

=

VCQ =

Analogy between thermal and electrical phenomena. Equivalent circuit of transistor



GroundAbsolute zero








GroundAbsolute zero








BJT(2N3716) Pd = 15 W

jc

= 1.17 0C/W

Tj

= 110 0C

cs

= 0.5 0C/W

sa

= ??

T0

= 25 0C

Transistor specifications: estimation of required heat sink

ja

= jc

+cs

+ sasacsjc

j

ja

jd

TTTTQP

++

=

== 00'

WCW

CCP

TT

d

jja /67.515

25110 0000 ==

= ca

= ja

-jc

- cs =

5.67- 1.17 0.5 = 4 0C/W


drdTrLQ )2(' =

dTLr

drQ )2(' =

=0

)2('T

T

b

a i

dTLrdrQ

)(2)ln( 0' TTL

abQ i =

=)ln(

)(2 0'

ab

TTLQ iL

ab

QTTi

2

)ln('

0 =

=

a = 5 mmb = 3 mm

= 27 n

m

aluminum

= 0.3 W m-1

K-1

polyethyleneI = 500 AL = 1 m

WaLIQ 9.852

2' ==

WCL

ab

/25.02

)ln(0==

CQT 0' 5.21== CTi05.41=


jc

= ?

Tj

= 150 0C

cs

= 0

sa

= 5 0C/W

T0

= 25 0C

Transistor specifications: derated power

Pmax

=?


Transistor specifications: non-steady-state regime


Stefans laws = 5.6710-8 Wm-2K-4 Stefans constant, = emissivity of the surface, S

= surface area emitting the radiation, T

= temperature of the surface, T0

= ambient temperature

)( 404 TTSP sradiated =

Effective thermal resistance

T>>T0


Emissivities of different materials

120 V

40 W

0.333 A

Power radiated from a light bulb at 2408 C is equal to the electricalpower dissipated in the filament.

What is the temperature of the filament?

P

= 40 WV

= 120 VL

= 38.1 cmD

= 33 m (273K) = 5.5110-8 m (T)~

T1.2


P

= 40 WV

= 120 VL

= 38.1 cmD

= 33 m___________________

s = 5.6710-8 Wm-2K-4 = 0.35 - 0.39

)( 404 TTSPP sradiated == 2526 1095.3381.0103314.3 mmDLS ===

))293()(1095.3)(1067.5)(35.0(40 4458 KTW = KT 2673=

TW = 3680 K

What is the temperature of filament in electric bulb ?

120 V

40 W

0.333 A

Power radiated from a light bulb at 2408 C is equal to the electricalpower dissipated in the filament.

T

= ??

T=2673 K

T=6050 K

Emission spectra of heated bodies

Sun spectrum--

in outer space--

on equator--

slanting sunlight

Wavelength, nm250 500 750 1000 1250 1500


tf

= 0.042 s = 42 ms

How long does it take to light an electric bulb ?


Convection

Fig. 1.23: Solid in equilibrium in air. During collisions between thegas and solid atoms, kinetic energy is exchanged.

M V

m

v

Gas Atom

SOLID

GAS


Newtons law of cooling

Q = heat flow, Ts

= temperature of the surface, h

= coefficient of convective heat transfer, T0

= ambient temperature S

= surface area ,

)( 0' TThSQ s =

Convection : Newtons law of cooling


1000 -


P

= 750 WS

= 1m

0.75mh

6 W m-2

0C-1____________________

T

= ??

ThSTQPconvection

=

== '

Solution

0102 3.83)75.01(2)6(750

=

== mmCWmW

hSPT CT 03.108253.83 =+==>


Ti

= 250C

To

= -400CT2

T1

= 0.76 Wm-1 0C-1 S

= 1m 0.75ml

=

10 mm hi

=

15 W m-2

0C-1

ho

= 25 W m-2

0C-1________________________________

T1

=

??

T2

=

??

Q

=

??


Ti

= 250C

To

= -400CT2

T1

= 0.76 Wm-1 0C-1 S

= 1m 0.75ml

=

10 mm hi

=

15 W m-2

0C-1ho

=

25 W m-2

0C-1________________________________

T1

=

??

T2

=

??

Q

=

??

T2

=

-18.5 0C

T1

=

-11.2 0C

ho

=

25 W m-2

0C-1

hi

=

15 W m-2

0C-1


= 18 n

mI

= 700 Aa

=

5 mm b-a

=

1.5 mm c-b

=

2 mm 1 = 0.3 W m-1 0C-12

= 0.25 W m-1 0C-1

T0

= 200Ch

= 25 W m-2 K-1T

=

??

Th

=

??


= 18 n

mI

= 700 Aa

=

5 mm b-a

=

1.5 mm c-b

=

2 mm 1 = 0.3 W m-1 0C-12

= 0.25 W m-1 0C-1

T0

= 200Ch

= 25 W m-2 K-1T

=

??

Th

=

??

ASSUMPTIONS


drdTrLQ )2(' =

dTLr

drQ )2(' =

=0

)2('T

T

b

a i

dTLrdrQ

)(2)ln( 0' TTL

abQ i =

=)ln(

)(2 0'

ab

TTLQ iL

ab

QTTi

2

)ln('

0 =

=

a = 5 mmb = 3 mm

= 27 n

m

aluminum

= 0.3 W m-1

K-1

polyethyleneI = 500 AL = 1 m

WaLIQ 9.852

2' ==

WCL

ab

/25.02

)ln(0==

CQT 0' 5.21== CTi05.41=


= 18 n

mI

= 700 Aa

=

5 mm b-a

=

1.5 mm c-b

=

2 mm 1 = 0.3 W m-1 0C-12

= 0.25 W m-1 0C-1

T0

= 200Ch

=

T

=

??

Th

=

??

Lab

2

)ln(=

Thermal Resistance

Ti

=

58.9 0C


= 18 n

mI

= 700 Aa

=

5 mm b-a

=

1.5 mm c-b

=

2 mm 1 = 0.3 W m-1 0C-12

= 0.25 W m-1 0C-1

T0

= 200Ch

= 25 W m-2 K-1T

=

58.90C

Th

=

??

h =25 W m-2

K-1h =


= 18 n

mI

= 700 Aa

=

5 mm b-a

=

1.5 mm c-b

=

2 mm 1 = 0.3 W m-1 0C-12

= 0.25 W m-1 0C-1

T0

= 200Ch

= 25 W m-2 K-1T

=

58.90C

Th

=

??

TCT0

!!!


T0

= 250CTj

= 1000CS

=

100 cm2

=0.01 m2

= 0.75 h

=

10 W m-2 K-1jc

= 15 0C/Wcs

= 1 0C/W___________________________________________________________________________________________________________________

Pd

=

??

radiation

= 15.4 0C/W convection

= 10 0C/Wconvective transfer is more important

> sink

6.0

0C/W

=

5W

Slide Number 1Slide Number 2Slide Number 3Slide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Fouriers LawFouriers LawDefinition of Thermal ResistanceSlide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Slide Number 27Slide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Slide Number 40

Heat transfer - University of Saskatchewankasap13.usask.ca/EE271/files/09_Heat_transfer-08.pdf ·...

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