heat transfer - university of saskatchewankasap13.usask.ca/ee271/files/09_heat_transfer-08.pdf ·...
TRANSCRIPT
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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:
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
Thermal Conductivities of various materials
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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).
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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
-
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
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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
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
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 =
-
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
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
Analogy between thermal and electrical phenomena
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
-
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
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
Analogy between thermal and electrical phenomena
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
Ti T0
IR2
Q=(Ti
-T0
)/
-
C = capacitanceC = thermal capacitance
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
C = capacitanceC = thermal capacitance
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
Analogy between thermal and electrical phenomena
tTCQ
=
TCQ =
tVCI
=
VCQ =
-
Analogy between thermal and electrical phenomena. Equivalent circuit of transistor
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
Current supplyHeat generator
GroundAbsolute zero
EMF (Electromotive Force)Heat reservoir
C = capacitanceC = thermal capacitance
R = resistance = thermal resistance
V = bias (voltage)T = temperature difference
I = CurrentQ = rate of heat flow
ELECTRICAL PHENOMENATHERMAL PHENOMENA
Current supplyHeat generator
GroundAbsolute zero
EMF (Electromotive Force)Heat reservoir
C = capacitanceC = thermal capacitance
R = resistance = thermal resistance
V = bias (voltage)T = temperature difference
I = CurrentQ = rate of heat flow
ELECTRICAL PHENOMENATHERMAL PHENOMENA
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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=
-
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
jc
= ?
Tj
= 150 0C
cs
= 0
sa
= 5 0C/W
T0
= 25 0C
Transistor specifications: derated power
Pmax
=?
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
Transistor specifications: non-steady-state regime
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
tf
= 0.042 s = 42 ms
How long does it take to light an electric bulb ?
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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
-
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
1000 -
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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 =+==>
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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
=
??
-
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
= 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
=
??
-
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
= 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
-
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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=
-
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
= 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
-
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
= 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 =
-
Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
= 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
!!!
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Essential Heat Transfer for Electrical Engineers ( S.O. Kasap, 2003: v.2.02) An e-Booklet
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