three modes of heat transfer
DESCRIPTION
Three Modes of Heat Transfer. “radiation”. “conduction”. “convection”. Conduction. Convection. = Cooling by mass motion (diffusion + advection) in a fluid. Radiation. - PowerPoint PPT PresentationTRANSCRIPT
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 1
Three Modes of Heat Transfer
“conduction”
“radiation”
“convection”
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 2
Conduction
TH
dTQ kAdx
LRkA
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 3
Convection
= Cooling by mass motion (diffusion + advection) in a fluid
TH
Q
R
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 4
Radiation
Note: Usually nothing is a perfect “black body” and parts of the emissive spectrum may be missing (ex: photonic band gap crystals).
4 41 1 2 1 1 2( )Q F A T T
20
50
2( )exp( / ) 1bb
B
hcehc k T
Linearize (when and why?):
8 3 24(5.67 10 )(300) 6 W/m Kradh
For black body (Ɛ=1) at 300 K:
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 5
• 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:
Boundaries and Lumped Elements
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 6
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 7
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 8
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/dxscale qualitatively within each layer?)
2) Assume UNIFORM internal heat generation
0ln( / )2
icyl
r rR
lk
slabLRkA
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 9
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 10
Notes on Finite-Element Heat Diffusion
RC
T0T0
TN
TiTi-1
Ti+1
Δx
L
0)(')( 0 TTgpTkA
1 1
2i i idu u u
dx x
21 1
22
2i i i id u u u udx x
T1
Boundary conditions:(heat flux conservation)
1 011
C
T TdTk Adx R
1 02 11
C
T TT Tk Ax R
=
M T b
T1
TN
Matlab:T = M\b
b1
bN
… …
M11 M12 0 …
MNN
………M21 M22 M23 0 …
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 11
More Comments on “Fin Equation”
• 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
2
02 ( ) 0d T hp T Tdx kA
Si
toxSiO2
d
T0
L
Wx x+dx
generalsolution
/ /0 1 2
H Hx L x LT T C e C e sinh, cosh, tanh … etc.
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 12
Fin Efficiency (how long is too long?)
• Fin efficiency η = actual heat loss by fin / heat loss if entire fin was at base temperature TB
• Actual heat loss:
• Here
-2 -1 0 1 2-4
-2
0
2
4
6
8
x/LH
sinh,
cos
h, ta
nh, e
xp
sinh
coshtanhT=TB
dT/dx ≈ 0
LW d
tanh( / )/
H
H
L LL L
Not worth making cooling fins much >> LH !
exp
© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 13
Poisson Equation Analogy
• 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
with solution
nt nt
ox
dC
and electrostatic screening lengthLiu (1993)Knoch (2006)