Page 45 - Design and Operation of Heat Exchangers and their Networks

32 Design and operation of heat exchangers and their networks

of 25mm in height, channel spacing of 3.5mm in width, and wall
thickness of 2.5mm. The thermal conductivity of the heat sink material
3
is 230W/mK. The air flow rate is 18m /h at the inlet air temperature
of 25°C. Assuming that the heat flux is uniform over the printed circuit
board, evaluate its highest temperature.
Solution
The properties of air are calculated with RefProp. With the air density
3
ρ in ¼1.169kg/m at 25°C, we have
_ m ¼ ρ V ¼ 1:169 18=3600 ¼ 0:005844 kg=s
in
_ m 0:005844 2
G ¼ ¼ ¼ 5:138 kg=m s
Nh fs s fs 13 0:025 0:0035
For the rectangular channel, the hydraulic diameter d h is given by

2h fs s fs 2 0:025 0:0035
d h ¼ ¼ ¼ 0:00614 m
h fs + s fs 0:025 + 0:0035

The aspect ratio is equal to
γ ¼ s fs =h fs ¼ 0:0035=0:025 ¼ 0:14
The wall thickness

δ f ¼ δ ¼ 0:0025 m
We assume at first the air mean temperature of 25°C and obtain the
specific isobaric thermal capacity c p ¼1007J/kgK. The outlet air
temperature is calculated as
Q 100
t out ¼ t in + ¼ 25 + ¼ 42°C
m_c p 1007
and the new mean temperature is

ð
t m ¼ t in + t out Þ=2 ¼ 25 + 42ð Þ=2 ¼ 33:5°C
Since the thermal capacity at the new mean temperature is almost the same,
3
no further iteration is needed. The air properties are given as ρ¼1.136kg/m ,
c p ¼1007J/kgK, μ¼1.886 10 5 sPa, λ¼0.02688W/mK, and Pr¼0.7062.
Thus, we obtain the Reynolds number as

5:138 0:00614
Gd h
Re ¼ ¼ ¼ 1673
μ 1:886 10 5
Because of the thick wall of the aluminum heat sink, the boundary
condition of constant heat flux in the flow direction and uniform