ch08
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ch08


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Chip and cooling channel dimensions. Channel flowrate and inlet temperature. Chip
temperature.
FIND: Water outlet temperature and chip power.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible kinetic and potential energy changes for water, (2) Uniform channel
surface temperature, (3) mT = 300 K, (4) Fully developed flow.
PROPERTIES: Table A-6, Water ( mT = 300 K): cp = 4179 J/kg×K, m = 855 ´ 10
-6
 kg/s×m, k =
0.613 W/m×K, Pr = 583.
ANALYSIS: Using the hydraulic diameter, find the Reynolds number,
( )
( )
( ) 2 6 5
h
4 H W 2 50 200 m
D 10 m/ m 8 10 m
2 H W 250 m
m
m
m
- -´ ´= = = ´
+
( )
( ) ( )
4 5
m h 1 h
D 12 2 6c
10 kg/s 8 10 m u D m D
Re 936.
A 50 200 10 m 855 10 kg/s m
r
m m
- -
- -
´
= = = =
´ ´ ×
&
Hence, the flow is laminar and, from Table 8.1, NuD = 4.44, so that
( ) 2
D 5h
4.44 0.613 W/m Kk
h Nu 34,022 W/m K.
D 8 10 m-
×
= = = ×
´
With P = 2(H + W) = 2(250 mm) 10
-6
 m/mm = 5 ´10
-4
 m, Eq. 8.42b yields
6 2 2
s m,o m,o
4s m,i 1 p
T T 350K T PL 5 10 m 34,022 W/m K
exp h exp
T T 60 K m c 10 kg/s 4179 J/kg K
-
-
æ öæ ö- - ´ ´ ×ç ÷= = - = -ç ÷ç ÷ ç ÷- ´ ×è ø è ø&
( )m,oT 350K 60 K exp 0.407 310 K.= - - = <
Hence, from Eq. 8.37,
( ) ( ) ( ) ( )4p m,o m,i 1 p m,o m,iq m c T T Nm c T T 50 10 kg/s 4179 J/kg K 20 K 418 W.-= - = - = ´ × =& & <
COMMENTS: (1) The chip heat flux of 418 W/cm
2
 is extremely large and the method provides a
very efficient means of heat removal from high power chips. However, clogging of the microchannels
is a potential problem which could seriously compromise reliability. (2) L/Dh = 125 and 0.05 ReDPr =
272. Hence, fully developed conditions are not realized and h > 34,022. The actual power dissipation
is therefore greater than 418 W.
PROBLEM 8.89
KNOWN: Chip and cooling channel dimensions. Channel flow rate and inlet temperature.
Temperature of chip at base of channel.
FIND: (a) Water outlet temperature and chip power, (b) Effect of channel width and pitch on power
dissipation.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible flow work and kinetic and potential energy changes for water, (2)
Flow may be approximated as fully developed and channel walls as isothermal for purposes of
estimating the convection coefficient, (3) One-dimensional conduction along channel side walls, (4)
Adiabatic condition at end of side walls, (5) Heat dissipation is exclusively through fluid flow in
channels, (6) Constant properties.
PROPERTIES: Table A-6, Water ( mT = 300K): cp = 4179 J/kg\u22c5K, µ = 855 × 10
-6
 kg/s\u22c5m, k = 0.613
W/m\u22c5K, Pr = 5.83.
ANALYSIS: (a) The channel sidewalls act as fins, and a unit channel/sidewall combination is shown
in schematic (a), where the total number of unit cells corresponds to N = L/S. With N = 50 and L =
10 mm, S = 200 µm and \u3b4 = S \u2013 W = 150 µm. Alternatively, the unit cell may be represented in terms
of a single fin of thickness \u3b4, as shown in schematic (b). The thermal resistance of the unit cell may
be obtained from the expression for a fin array, Eq. (3.103), Rt,o = (\u3b7ohAt)-1, where At = Af + Ab = L
(2 H + W) = 0.01m (4 × 10-4 + 0.5 × 10-4) m = 4.5 × 10-6 m2. With Dh = 4 (H × W)/2 (H + W) = 4 (2
× 10-4m × 0.5 × 10-4m)/2 (2.5 × 10-4m) = 8 × 10-5m, the Reynolds number is ReD = \u3c1um Dh/µ = 1m\ufffd
Dh/Acµ = 10
-4
 kg/s × 8 × 10-5m/(2 × 10-4m × 0.5 × 10-4m) 855 × 10-6 kg/s\u22c5m = 936. Hence, the flow
is laminar, and assuming fully developed conditions throughout a channel with uniform surface
temperature, Table 8.1 yields NuD = 4.44. Hence,
2
D 5h
k 0.613W / m K 4.44h Nu 34,022 W / m K
D 8 10 m\u2212
\u22c5 ×
= = = \u22c5
×
With m = (2h/kch\u3b4)1/2 = (68,044 W/m2\u22c5K/140 W/m\u22c5K × 1.5 × 10-4m)1/2 = 1800 m-1 and mH = 0.36,
the fin efficiency is
f
tanh mH 0.345 0.958
mH 0.36
\u3b7 = = =
and the overall surface efficiency is
( ) ( )
6
f
o f 6t
A 4.0 10l l l l 0.958 0.963
A 4.5 10
\u3b7 \u3b7
\u2212
\u2212
×
= \u2212 \u2212 = \u2212 \u2212 =
×
The thermal resistance of the unit cell is then
Continued \u2026..
PROBLEM 8.89 (Cont.)
( ) ( ) 11 2 6 2t,o o tR h A 0.963 34,022 W / m K 4.5 10 m 6.78K / W\u3b7 \u2212\u2212 \u2212= = × \u22c5 × × =
The outlet temperature follows from Eq. (8.46b),
( ) ( )
p t,o
l
exp 350K 60Km,o s s m,i
m c R
T T T T
\uf8eb \uf8f6
\u2212 = \u2212 ×\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8
= \u2212 \u2212
\ufffd
4
1
exp 307.8K
10 kg / s 4179J / kg K 6.78K / W\u2212
\uf8eb \uf8f6\uf8ec \uf8f7\u2212 =\uf8ec \uf8f7× \u22c5 ×\uf8ed \uf8f8
<
The heat rate per channel is then
( ) ( )41 1 p m,o m,iq m c T T 10 kg / s 4179 J / kg K 17.8K 7.46 W\u2212= \u2212 = × \u22c5 =\ufffd
and the chip power dissipation is
lq Nq 50 7.46 W 373W= = × = <
(b) The foregoing result indicates significant heat transfer from the channel side walls due to the large
value of \u3b7f. If the pitch is reduced by a factor of 2 (S = 100 µm), we obtain
1S 100 m, W 50 m, 50 m, N 100 : q 7.04 W, q 704 Wµ µ \u3b4 µ= = = = = = <
Hence, although there is a reduction in \u3b7f due to the reduction in \u3b4 (\u3b7f = 0.89) and therefore a slight
reduction in the value of ql, the effect is more than compensated by the increase in the number of
channels. Additional benefit may be derived by further reducing the pitch to whatever minimum
value of \u3b4 is imposed by manufacturing or structural limitations. There would also be an advantage to
increasing the channel hydraulic diameter and or flowrate, such that turbulent flow is achieved with a
correspondingly larger value of h.
COMMENTS: (1) Because electronic devices fail by contact with a polar fluid such as water, great
care would have to be taken to hermetically seal the devices from the coolant channels. In lieu of
water, a dielectric fluid could be used, thereby permitting contact between the fluid and the
electronics. However, all such fluids, such as air, are less effective as coolants. (2) With L/Dh = 125
and L/Dh)fd \u2248 0.05 ReD Pr = 273, fully developed flow is not achieved and the value of h = hfd
underestimates the actual value of h in the channel. The coefficient is also underestimated by using a
Nusselt number that presumes heat transfer from all four (rather than three) surfaces of a channel.
PROBLEM 8.90
KNOWN: Chip and cooling channel dimensions. Channel flow rate and inlet temperature.
Temperature of chip at base of channel.
FIND: (a) Outlet temperature and chip power dissipation for dielectric liquid, (b) Outlet temperature
and chip power dissipation for air.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible flow work and kinetic and potential energy changes, (2) Flow may
be approximated as fully developed and channel walls as isothermal for purposes of estimating the
convection coefficient, (3) One-dimensional conduction along the channel side walls, (4) Adiabatic
condition at end of side walls, (5) Heat dissipation is exclusively through fluid flow in channels, (6)
Constant properties.
PROPERTIES: Prescribed. Dielectric liquid: cp = 1050 J/kg\u22c5K, k = 0.065 W/m\u22c5K, µ = 0.0012
N\u22c5s/m2, Pr = 15. Air: cp = 1007 J/kg\u22c5K, k = 0.0263 W/m\u22c5K, µ = 185 × 10
-7
 N\u22c5s/m2, Pr = 0.707.
ANALYSIS: (a) The channel side walls act as fins, and a unit channel/sidewall combination is
shown in schematic (a), where \u3b4 = S \u2013 W = 150 µm. Alternatively, the unit cell may be represented in
terms of a single fin of thickness \u3b4, as shown in schematic (b). The thermal resistance of the unit cell
may be obtained from the expression for a fin array, Eq. (3.103), Rt,o = (\u3b7o h At)-1, where At = Af +
Ab = L (2 H + W) = 4.5 × 10-6 m2. With Ac = H × W = 10-8 m2 and Dh = 4 Ac/2(H + W) = 8 × 10-
5
m, the Reynolds number is ReD = \u3c1umDh/µ = 1m\ufffd Dh/Acµ = 667. Hence, the flow is laminar, and
assuming fully developed conditions throughout a channel with uniform surface temperature, Table
8.1 yields NuD = 4.44. Hence, 2D 5h
k 0.065 W / m K 4.44h Nu 3608 W / m K
D 8 10 m\u2212
\u22c5 ×
= = = \u22c5
×
With m = (2 h/kch \u3b4)1/2 = 586 m-1 and mH = 0.117, the fin efficiency is
f
tanh mH 0.1167 0.995
mH 0.117
\u3b7 =