Consider the microchannel cooling arrangement ofProblem 8.107. However, instead of assuming theentire chip and cap to be at a uniform temperature.adopt a more conservative (and realistic) approach thatprescribes a temperature of T s = 350 K at the base ofthe channels ( x = 0 ) and allows for a decrease in temperature with increasing x along the side walls of eachchannel. (a) For the operating conditions prescribed in Problem 8.107 and a chip thermal conductivity of k c h = 140 W/m ⋅ K , determine the water outlet temperature and the chip power dissipation. Heat transfer from the sides of the chip to the surroundings and from the side walls of a channel to the cap may be neglected. Note that the spacing between channels. δ = S − W , is twice the spacing between the side wall of an outer channel and the outer surface of the chip. The channel pitch is S = L / N , where L = 1 0 mm is the chip width and N = 5 0 is the number of channels (b) The channel geometry prescribed in Problem 8.107 and considered in part (a) is not optimized, and larger heat rates may be dissipated by adjusting related dimensions. Consider the effect of reducing the pitch to a value of S = 100 μ m . while retaining a width of W = 50 μ m and a flow rate per channel of m 1 = 10 − 4 kg/s.
Consider the microchannel cooling arrangement ofProblem 8.107. However, instead of assuming theentire chip and cap to be at a uniform temperature.adopt a more conservative (and realistic) approach thatprescribes a temperature of T s = 350 K at the base ofthe channels ( x = 0 ) and allows for a decrease in temperature with increasing x along the side walls of eachchannel. (a) For the operating conditions prescribed in Problem 8.107 and a chip thermal conductivity of k c h = 140 W/m ⋅ K , determine the water outlet temperature and the chip power dissipation. Heat transfer from the sides of the chip to the surroundings and from the side walls of a channel to the cap may be neglected. Note that the spacing between channels. δ = S − W , is twice the spacing between the side wall of an outer channel and the outer surface of the chip. The channel pitch is S = L / N , where L = 1 0 mm is the chip width and N = 5 0 is the number of channels (b) The channel geometry prescribed in Problem 8.107 and considered in part (a) is not optimized, and larger heat rates may be dissipated by adjusting related dimensions. Consider the effect of reducing the pitch to a value of S = 100 μ m . while retaining a width of W = 50 μ m and a flow rate per channel of m 1 = 10 − 4 kg/s.
Solution Summary: The author describes the water outlet temperature, chip power dissipation, and thermal conductivity of the chip. The expression for the hydraulic diameter is given as, mathrmRe_D
Consider the microchannel cooling arrangement ofProblem 8.107. However, instead of assuming theentire chip and cap to be at a uniform temperature.adopt a more conservative (and realistic) approach thatprescribes a temperature of
T
s
=
350
K at the base ofthe channels (
x
=
0
) and allows for a decrease in temperature with increasing x along the side walls of eachchannel.
(a) For the operating conditions prescribed in Problem 8.107 and a chip thermal conductivity of
k
c
h
=
140
W/m
⋅
K
, determine the water outlet temperature and the chip power dissipation. Heat transfer from the sides of the chip to the surroundings and from the side walls of a channel to the cap may be neglected. Note that the spacing between channels.
δ
=
S
−
W
, is twice the spacing between the side wall of an outer channel and the outer surface of the chip. The channel pitch is
S
=
L
/
N
, where
L
=
1
0
mm
is the chip width and
N
=
5
0
is the number of channels
(b) The channel geometry prescribed in Problem 8.107 and considered in part (a) is not optimized, and larger heat rates may be dissipated by adjusting related dimensions. Consider the effect of reducing the pitch to a value of
S
=
100
μ
m
. while retaining a width of
W
=
50
μ
m
and a flow rate per channel of
m
1
=
10
−
4
kg/s.
6. Make a diagram and show the step-by-step process. Do not use shortcut methods. Make it as detailed as it can be.
Encode (not hand-written)! DO NOT COPY CHEGG'S ANSWER
a buildiug has the following calculated cooling loads:
RSH gain 300 kW
RLH gain = 110 kW
The space is maintained at the following conditions
Room DBT 25°C, Room RH = 55%
Outdoor is at 40°C and RH 50%. And 10% by mass of air supplied to the buildings is
out door. If the air supplied to the space is not to be at a temperature lower than
18°C. Find
(i) Miu amount of air supplied to space in m/sec.
(ii) Mass flow rates of return air, exhaust air and out door air
(i) Capacity, BPFand sensible heat factor of cooling coil.
Given:
Condition
DBT
RH
Sp. Humidity
Enthalpy
°C
g.w.v./kg d.a.
kJ/kg d.a.
outside
40
50
23.6.
100
inside
25
55
10.6
52.5
supply
Specitic valume of air supply point 0.836 m³/kg, coil ADP=9°C.
18
41.2
EXL/ Calculate Cooling load temperature di fference (CLTD) for
Sonth wall from grouPDat 32° North latitude
Assume darK Colour,design outoor temperature 35 ,desion
Îndoor temperature TR=25.5 è, mean outdoor temperature
Tm= 29.4 c and daily range DR=.6¢ at hour 14.
on october21.
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