Download L18L19

Three Modes of Heat Transfer
“radiation”
“conduction”
“convection”
© 2010 Eric Pop, UIUC
ECE 598EP: Hot Chips
1
Conduction
dT
Q  kA
dx
L
RTH 
kA
© 2010 Eric Pop, UIUC
ECE 598EP: Hot Chips
2
Convection
Q
RTH 
= Cooling by mass motion (diffusion + advection) in a fluid
© 2010 Eric Pop, UIUC
ECE 598EP: Hot Chips
3
Radiation
2 hc02
ebb ( )  5
  exp(hc0 / k BT  )  1
Q  1 F1 2 A1 (T14  T24 )
Linearize (when and why?):
For black body (Ɛ=1) at 300 K:
hrad  4(5.67 108 )(300)3  6 W/m 2 K
Note: Usually nothing is a perfect “black body” and parts of the emissive
spectrum may be missing (ex: photonic band gap crystals).
© 2010 Eric Pop, UIUC
ECE 598EP: Hot Chips
4
Boundaries and Lumped Elements
• All problems have boundaries!
• Heat diffusion equation needs boundary conditions
• Dirichlet (fixed T):
• Neumann (fixed flux ~ dT/dx):
• When is it OK to “lump” a body as a single R or C?
• Biot number:
© 2010 Eric Pop, UIUC
ECE 598EP: Hot Chips
5
Transient Cooling of Lumped Body
Source: Lienhard book, http://web.mit.edu/lienhard/www/ahtt.html (2008)
© 2010 Eric Pop, UIUC
ECE 598EP: Hot Chips
6
What if Biot Number is Large
• Bi = hL/kb << 1 implies Tb(x) ~ Tsurf (lumped OK)
• Bi = hL/kb >> 1 implies significant internal Tb(x) gradient
© 2010 Eric Pop, UIUC
ECE 598EP: Hot Chips
7
Lumped Body Examples (Steady State)
Boundary conditions:
TL = 400 oC, TR = 100 oC
1) Assume NO internal heat generation
(how does the temperature slope dT/dx
scale qualitatively within each layer?)
2) Assume UNIFORM internal heat generation
ln(r0 / ri )
Rcyl 
2 lk
Rslab 
© 2010 Eric Pop, UIUC
L
kA
ECE 598EP: Hot Chips
8
Contact Resistance
• RC = 1/hCA
• BUT, also remember the fundamental solid-solid contact
resistance given by density of states, acoustic/diffuse
phonon mismatch ~Cv/4! (prof. Cahill’s lecture)
© 2010 Eric Pop, UIUC
ECE 598EP: Hot Chips
9
Notes on Finite-Element Heat Diffusion
Ti-1
Δx
T1
Ti
A(kT )  p' g (T  T0 )  0
Ti+1
RC
dui ui 1  ui 1

dx
2x
T0
d 2ui ui 1  2ui  ui 1

2
2
dx

x
 
TN
L
T0
Boundary conditions:
(heat flux conservation)
© 2010 Eric Pop, UIUC
b1
M21 M22 M23 0 …
…

T2  T1 T1  T0 
k
A

 1


x
RC 

T1
…
dT1 T1  T0
k1 A

dx
RC
M11 M12 0 …
TN
bN
T
b
=
MNN
M
ECE 598EP: Hot Chips
Matlab:
T = M\b
10
More Comments on “Fin Equation”
L
x
x+dx
d
W
tox
SiO2
Si
T0
• Same as Poisson equation with various BC’s
• BC’s can be given flux (dT/dx) or given temperature (T0)
• Very useful to know:
– Thermal healing length LH (Poisson: screening length)
– General, qualitative shape or solution
d 2T hp
 (T  T0 )  0
2
dx
kA
© 2010 Eric Pop, UIUC
general
solution
T  T0  C1e  x / LH  C2 e x / LH
sinh, cosh, tanh … etc.
ECE 598EP: Hot Chips
11
Fin Efficiency (how long is too long?)
L
d
W
T=TB
dT/dx ≈ 0
sinh, cosh, tanh, exp
8
6
exp
4
2
cosh
tanh
0
-2
-4
-2
sinh
-1
0
x/LH
1
2
• Fin efficiency η = actual heat loss by fin / heat loss if
entire fin was at base temperature TB
• Actual heat loss:
tanh( L / LH )
• Here  
L / LH
© 2010 Eric Pop, UIUC
Not worth making cooling fins much >> LH !
ECE 598EP: Hot Chips
12
Poisson Equation Analogy
with solution
Liu (1993)
Knoch (2006)
and electrostatic screening length

 nt dnt
Cox
• Thermal fin is ~ mathematically same problem as 1-D
transistor electrostatics, e.g. nanowire or SOI transistor
• L < λ  short fin, or “short channel” FET
• L >> λ  long fin (too long?!), or “long channel” FET
© 2010 Eric Pop, UIUC
ECE 598EP: Hot Chips
13