Heat transfer
Heat = amount of energy that is transferred from one
system to another (or between system and
surroundings) as a result of temperature difference
The origin of energy transfer is the random motion of molecules
Heat is transferred by
- thermal conduction
- convection
- radiation
Fourier’s Law of Thermal Conduction
T(x)
1 dQx
Jx ≡
A dt
Jx
dT/dx
x
HOT
δT
COLD
dQ
dt
HEAT
A
δx
Fig. 2.19: Heat flow in a metal rod heated at one end. Consider the rate
of heat flow, dQ/dt, across a thin section δ x of the rod. The rate of
heat flow is proportional to the temperature gradient δ T/δ x and the
cross sectional area A.
Jx
= heat flux,
dQx/dt = the rate of heat flow
A
= cross-sectional area
dQ
dT
Jx =
=− κ
dt
dx
dT/dx = temperature gradient
κ = thermal conductivity
[κ] = W m-1 K-1 or W m-1 0C-1
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Reminder:
Γ = −D
dC
dx
Fick’s First Law
Thermal Conductivities of various materials
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Thermal Conductivities of various materials
Strong metallic
bonding
Strong covalent
bonding
Weak Van-der-Waals
bonding
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Wiedemann – Franz - Lorenz Law
Thermal conductivity, κ (W K-1 m-1)
450
Ag
400
Ag-3Cu
κ
−8
−2
= CWFL = 2.45×10 W Ω K
σT
Cu
Ag-20Cu
300
κ
= T CWFL
σ
Au
κ = thermal conductivity
σ = electrical conductivity
Al
200
W
Be
Mg
Mo
Brass (Cu-30Zn)
Ni
Bronze (95Cu-5Sn)
Steel (1080)
Pd-40Ag
Hg
100
0
0
10
20
30
40
50
60
6
-1
-1
Electrical conductivity, σ, 10 Ω m
70
Fig. 2.20: Thermal conductivity, κ vs. electrical conductivity σ for
various metals (elements and alloys) at 20 °C. The solid line
represents the WFL law with CWFL ≈ 2.44×108 W Ω K-2.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
T = temperature
CWFL = Lorenz number
Wiedemann – Franz - Lorenz Law
Ag
400
Ag-3Cu
κ
−8
−2
= CWFL = 2.45×10 W Ω K
σT
Cu
Ag-20Cu
300
κ
= T CWFL
σ
Au
Al
200
W
Be
Mg
Mo
Brass (Cu-30Zn)
Ni
Bronze (95Cu-5Sn)
Steel (1080)
Pd-40Ag
Hg
100
0
0
10
20
30
40
50
60
6
-1
-1
Electrical conductivity, σ, 10 Ω m
70
Thermal conductivity, κ (W K-1 m-1)
Thermal conductivity, κ (W K-1 m-1)
450
Fig. 2.20: Thermal conductivity, κ vs. electrical conductivity σ for
various metals (elements and alloys) at 20 °C. The solid line
represents the WFL law with CWFL ≈ 2.44×108 W Ω K-2.
κ50000= thermal conductivity
σ10000= electricalCopper
conductivity
T = temperature
Aluminum
1000
CWFL = Lorenz number
Brass (70Cu-30Zn)
100
Al-14%Mg
10
1
10
100
Temperature (K)
Thermal
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill,
2002)conductivity vs. temperature for two
http://Materials.Usask.Ca
and two alloys (brass and Al-14%Mg).
1000
pure metals (Cu and Al)
Data extracted from
Thermophysical Properties of Matter, Vol. 1: Thermal Conductivity,
Metallic Elements and Alloys, Y.S. Touloukian et. al (Plenum, New
York, 1970).
Thermal Conductivities of various materials
Strong metallic
bonding
Strong covalent
bonding
Weak Van-der-Waals
bonding
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Thermal conduction in metals and some insulators
Insulators with very strong covalent bonding
C (diamond), BeO (beryllia), ...
Metals
Ag, Cu, Al ...
HOT
COLD
HEAT
Equilibrium
Hot
Cold
Energetic atomic vibrations
Fig. 2.22: Conduction of heat in insulators involves the generation
and propogation of atomic vibrations through the bonds that couple
the atoms. (An intuitive figure.)
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Electron Gas
Vibrating Cu+ ions
Heat is transferred by conduction electrons
Heat is transferred as atomic vibrations
due to strong bonding between atoms
Parabolic heat equation
∂ 2T ∂T
Dth 2 =
∂x
∂t
where
κ
Dth =
cρ
thermal diffusitivity
ρ = density
c = specific heat capacity
Heat flow in – Heat flow out =
= Rate of heat accumulation in volume δx
Rate of heat accumulation in volume δx =
∂T
δx × ρ × c
∂t
∂J x
∂ 2T
δx = −κ 2 δx
Heat flow in – Heat flow out = Jx(x)- Jx(x+δx) =
∂x
∂x
∂ 2C ( x, t ) ∂C ( x, t )
Reminder: D
=
∂x 2
∂t
Fick’s Second Law
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Fourier’s Law
ΔT
ΔT
Q′ = A κ
=
L
(L / κ A)
Jx =
dQ
dT
=− κ
dt
dx
Jx ≡
1 dQx
A dt
Q′ = rate of heat flow or the heat current, A = cross-sectional area, κ
= thermal conductivity (material-dependent constant of
proportionality), ΔT = temperature difference between ends of
component, L = length of component
Ohm’s Law
ΔV
ΔV
I=
=
R
(L / σ A)
I = electric current, ΔV = voltage difference across the conductor, R =
resistance, L = length, σ = conductivity, A = cross-sectional area
Fourier’s Law
ΔT
ΔT
Q′ = A κ
=
L
(L / κ A) = θ
Q′ = rate of heat flow or the heat current, A = cross-sectional area, κ
= thermal conductivity (material-dependent constant of
proportionality), ΔT = temperature difference between ends of
component, L = length of component
Ohm’s Law
ΔV
ΔV
I=
=
R
(L / σ A)
=R
I = electric current, ΔV = voltage difference across the conductor, R =
resistance, L = length, σ = conductivity, A = cross-sectional area
Definition of Thermal Resistance
Q′ =
ΔT
θ
Q′ = rate of heat flow, ΔT = temperature difference, θ = thermal
resistance
Thermal Resistance
L
θ=
Aκ
θ = thermal resistance, L = length, A = cross-sectional area, κ =
thermal conductivity
Analogy between thermal and electrical phenomena
THERMAL PHENOMENA ELECTRICAL PHENOMENA
Q = rate of heat flow
I = Current
ΔT = temperature difference
ΔV = bias (voltage)
Θ = thermal resistance
R = resistance
Q′ = ΔT/θ Force)
EMF (Electromotive
Heat reservoir ΔT
Absolute
Hot zero
Q′ =
ΔT
θ
Cold
Heat generator
Q′
A
L
(a)
Q′
ΔT
Current supply
Ground
Q′
θ
(b)
Fig. 2.23: Conduction of heat through a component in (a) can be
modeled as a thermal resistance θ shown in (b) where Q′ = ΔT/θ.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
ΔV
I=
R
Analogy between thermal and electrical phenomena
THERMAL PHENOMENA ELECTRICAL PHENOMENA
Q = rate of heat flow
I = Current
ΔT = temperature difference
ΔV = bias (voltage)
Θ = thermal resistance
R = resistance
Heat reservoir
EMF (Electromotive Force)
Absolute zero
Ground
Heat generator
Current supply
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Analogy between thermal and electrical phenomena
THERMAL PHENOMENA ELECTRICAL PHENOMENA
Q = rate of heat flow
I = Current
ΔT = temperature difference
ΔV = bias (voltage)
Θ = thermal resistance
R = resistance
Heat reservoir
EMF (Electromotive Force)
Absolute zero
Ground
Heat generator
Current supply
Ti
T0
Q’=(Ti-T0)/Θ
IR2
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Analogy between thermal and electrical phenomena
THERMAL PHENOMENA ELECTRICAL PHENOMENA
Q = rate of heat flow
I = Current
ΔT = temperature difference
ΔV = bias (voltage)
Θ = thermal resistance
R = resistance
Heat reservoir
EMF (Electromotive Force)
Absolute zero
Ground
Heat generator
Current supply
C = thermal capacitance
C = capacitance
δQ = C δT
Q′ = C
δT
δt
δQ = C δV
I =C
δV
δt
Analogy between thermal and electrical phenomena.
Equivalent circuit of transistor
THERMAL PHENOMENA ELECTRICAL PHENOMENA
Q = rate of heat flow
I = Current
ΔT = temperature difference
ΔV = bias (voltage)
Θ = thermal resistance
R = resistance
C = thermal capacitance
C = capacitance
Heat reservoir
EMF (Electromotive Force)
Absolute zero
Ground
Heat generator
Current supply
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Transistor specifications: estimation of required heat sink
BJT(2N3716)
Pd = 15 W
Tj= 110 0C
θjc= 1.17 0C/W
θcs= 0.5 0C/W
θsa= ??
T0= 25 0C
θja = θjc +θcs + θsa
θ ja =
T j − T0
Pd
1100 C − 250 C
=
= 5.67 0 C / W
15W
P d = Q' =
T j − T0
θ ja
=
T j − T0
θ jc + θ cs + θ sa
θca = θja -θjc - θcs = 5.67- 1.17 – 0.5 = 4 0C/W
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Q ' = −κ (2πrL)
Q'
dT
dr
dr
= −κ (2πL)dT
r
T
b
0
' dr
Q ∫ = −κ (2πL) ∫ dT
r
a
Ti
b
Q ' ln( ) = −κ 2πL(Ti − T0 )
a
Q' =
a = 5 mm
b = 3 mm
ρ = 27 nΩ m – aluminum
κ = 0.3 W m-1 K-1 – polyethylene
I = 500 A
L=1m
Q' = I 2
κ 2πL(Ti − T0 )
b
ln( )
a
ρL
= 85.9W
2
πa
ΔT = Q 'θ = 21.50 C
⇒
b
ln( )
T −T
θ= i ' 0= a
Q
2πκL
b
ln( )
θ = a = 0.250 C / W
2πκL
Ti = 41.50 C
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Transistor specifications:
derated power
Tj= 150 0C
θjc= ?
Pmax=?
θcs= 0
θsa= 5 0C/W
T0= 25 0C
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Transistor specifications: non-steady-state regime
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Stefan’s law
Pradiated = εσ s S (T − T )
4
4
0
σs = 5.67×10-8 Wm-2K-4 – Stefan’s constant,
ε = emissivity of the surface,
S = surface area emitting the radiation,
T = temperature of the surface,
T0 = ambient temperature
Effective thermal resistance
≅
T>>T0
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Emissivities of different materials
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
What is the temperature of the filament?
40 W
0.333 A
120 V
Power radiated from a light bulb at 2408 °C is equal to the electrical
power dissipated in the filament.
P = 40 W
V = 120 V
L = 38.1 cm
D = 33 μm
ρ (273K) = 5.51×10-8 Ωm
ρ (T)~ T1.2
What is the temperature of filament in electric bulb ?
40 W
T = ??
0.333 A
P = 40 W
V = 120 V
L = 38.1 cm
D = 33 μm
___________________
σs = 5.67×10-8 Wm-2K-4
120 V
ε = 0.35 - 0.39
Power radiated from a light bulb at 2408 °C is equal to the electrical
power dissipated in the filament.
P = Pradiated = εσ s S (T 4 − T04 )
S = πDL = 3.14 × 33 ×10 −6 × 0.381m 2 = 3.95 ×10 −5 m 2
40W = (0.35)(5.67 × 10 −8 )(3.95 ×10 −5 )(T 4 − (293K ) 4 )
⇒ T = 2673K
TW = 3680 K
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Emission spectra of heated bodies
Sun spectrum
-- in outer space
-- on equator
-- slanting sunlight
T=2673 K
T=6050 K
250
500
750
1000
Wavelength, nm
1250
1500
How long does it take to light an electric bulb ?
tf = 0.042 s = 42 ms
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Convection
SOLID
GAS
M
V
v
m
Gas Atom
Fig. 1.23: Solid in equilibrium in air. During collisions between the
gas and solid atoms, kinetic energy is exchanged.
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Convection : Newton’s law of cooling
Newton’s law of cooling
Q’ = heat flow,
h = coefficient of convective heat transfer,
S = surface area ,
Q = hS (Ts − T0 )
'
Ts = temperature of the surface,
T0 = ambient temperature
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
1000 -
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
P = 750 W
S = 1m × 0.75m
h ≅ 6 W m-2 0C-1
____________________
T = ??
Solution
ΔT
P = Q' =
= hS × ΔT
Θ convection
ΔT =
750W
P
=
= 83.30
− 2 0 −1
hS (6Wm C ) × 2 × (1m × 0.75m)
=>
T = 83.3 + 25 = 108.30 C
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
κ = 0.76 Wm-1 0C-1
S = 1m × 0.75m
l = 10 mm
hi = 15 W m-2 0C-1
ho = 25 W m-2 0C-1
Ti = 250C
T1
T2
To = -400C
________________________________
T1 = ?? T2 = ?? Q’ = ??
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
κ = 0.76 Wm-1 0C-1
Ti = 250C
T1
hi = 15 W m-2 0C-1
ho = 25 W m-2 0C-1
T2
S = 1m × 0.75m
l = 10 mm
hi = 15 W m-2 0C-1
ho = 25 W m-2 0C-1
________________________________
T1 = ?? T2 = ?? Q’ = ??
To = -400C
T2 = -18.5 0C
T1 = -11.2 0C
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
ρ = 18 nΩ m
I = 700 A
a = 5 mm
b-a = 1.5 mm
c-b = 2 mm
κ1 = 0.3 W m-1 0C-1
κ2 = 0.25 W m-1 0C-1
T0 = 200C
h = 25 W m-2 K-1
T∞ = ?? Th = ??
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
ρ = 18 nΩ m
I = 700 A
a = 5 mm
b-a = 1.5 mm
c-b = 2 mm
κ1 = 0.3 W m-1 0C-1
κ2 = 0.25 W m-1 0C-1
T0 = 200C
h = 25 W m-2 K-1
T∞ = ?? Th = ??
ASSUMPTIONS
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
Q ' = −κ (2πrL)
Q'
dT
dr
dr
= −κ (2πL)dT
r
T
b
0
' dr
Q∫
= −κ (2πL) ∫ dT
r
a
Ti
b
Q ' ln( ) = −κ 2πL(Ti − T0 )
a
Q' =
a = 5 mm
b = 3 mm
ρ = 27 nΩ m – aluminum
κ = 0.3 W m-1 K-1 – polyethylene
I = 500 A
L=1m
Q' = I 2
κ 2πL(Ti − T0 )
b
ln( )
a
ρL
= 85.9W
2
πa
ΔT = Q 'Θ = 21.50 C
⇒
b
ln( )
T −T
a
Θ= i ' 0 =
Q
2πκL
b
ln( )
a = 0.250 C / W
Θ=
2πκL
Ti = 41.50 C
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
ρ = 18 nΩ m
I = 700 A
a = 5 mm
b-a = 1.5 mm
c-b = 2 mm
b
ln( )
a
Θ=
2πκL
κ1 = 0.3 W m-1 0C-1
κ2 = 0.25 W m-1 0C-1
T0 = 200C
h =∞
T∞ = ?? Th = ??
Thermal Resistance
Ti = 58.9 0C
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
ρ = 18 nΩ m
I = 700 A
a = 5 mm
b-a = 1.5 mm
c-b = 2 mm
κ1 = 0.3 W m-1 0C-1
κ2 = 0.25 W m-1 0C-1
T0 = 200C
h = 25 W m-2 K-1
T∞ = 58.90C Th = ??
h =∞
h =25 W m-2 K-1
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
ρ = 18 nΩ m
I = 700 A
a = 5 mm
b-a = 1.5 mm
c-b = 2 mm
κ1 = 0.3 W m-1 0C-1
κ2 = 0.25 W m-1 0C-1
TC
T0 = 200C
h = 25 W m-2 K-1
T∞ = 58.90C Th = ??
T0
!!!
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet
T0 = 250C
Tj = 1000C
S = 100 cm2 =0.01 m2
ε = 0.75
h = 10 W m-2 K-1
Θjc = 15 0C/W
Θcs = 1 0C/W
___________________________________________________________________________________________________________________
Pd = ??
Θradiation = 15.4
0C/W
>
Θconvection = 10
0C/W
convective transfer is more important
Θsink ≅ 6.0 0C/W
= 5W
Essential Heat Transfer for Electrical Engineers (© S.O. Kasap, 2003: v.2.02) An e-Booklet