December 2009
Intermediate to advanced
523 pages
18h 30m
English
Content preview from Heat Transfer
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152 Heat Transfer: Thermal Management of Electronics
R
NhAANA hNAA
hs cr
ff fhsb fb fffhsb
,
()[(
=
+−
=
+
11
ηη−− NA h
fb
)]
.
(7.63)
The total n area for this heat sink is N
f
A
f
and the total heat sink heat transfer area
is A
t
= N
f
A
f
+ A
unnned
= N
f
A
f
+ A
hsb
– N
f
A
b
. Substituting A
hsb
– N
f
A
b
= A
t
– N
f
A
f
in the
above equation gives
R
NhAANA hANA
hs cr
ff ftff tfff
,
()[()]
=
+−
=
−−
11
1ηηhh
NA
A
Ah
ff
t
ft
=
−−
1
11()
.
η
(7.64)
The total heat transfer rate from heat sink is
Q
TT
R
NA
A
Ah T
hs
b
hs cr
ff
t
ftb
=
−
=− −
∞
,
()(11η−− T
∞
)
. (7.65)
Fin effectiveness was dened as the ratio of heat transfer rate from a n to the
heat transfer rate from the base of the n if there ...
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