Chapter 9
Natural Convection
9103E
The components of an electronic device located in a horizontal duct of rectangular cross section
is cooled by forced air. The heat transfer from the outer surfaces of the duct by natural convection and
the average temperature of the duct are to be determined.
Assumptions
1
Steady operating conditions exist.
2
Air is an ideal gas with constant properties.
3
The
local atmospheric pressure is 1 atm.
4
Radiation effects are negligible.
5
The thermal resistance of the
duct is negligible.
Properties
The properties of air at 1 atm and the anticipated film
temperature of (
T
s
+
T
∞
)/2 = (120+80)/2 = 100
°
F are (Table A15E)
1

2
3
R
001786
.
0
R
)
460
100
/(
1
/
1
726
.
0
Pr
/s
ft
10
1808
.
0
F
Btu/h.ft.
01529
.
0
=
+
=
=
=
×
=
°
=

f
T
k
β
υ
Analysis
(
a
) Using air properties at the average temperature of (85+100))/2 = 92.5
°
F and 1 atm for the
forced air, the mass flow rate of air and the heat transfer rate by forced convection are determined to be
lbm/min
581
.
1
)
/min
ft
22
)(
lbm/ft
07186
.
0
(
3
3
=
=
=
V
m
ρ
Btu/h
1
.
342
F
)
85
100
)(
F
Btu/lbm.
2405
.
0
)(
lbm/h
60
581
.
1
(
)
(
=
°

°
×
=

=
in
out
p
forced
T
T
C
m
Q
Noting that radiation heat transfer is negligible, the rest of the 180 W heat generated must be dissipated
by natural convection,
Btu/h
272
=

×
=

=
1
.
342
)
412
.
3
180
(
forced
total
natural
Q
Q
Q
(
b
) We start the calculations by “guessing” the surface temperature to be 120
°
F for the evaluation of the
properties and
h
. We will check the accuracy of this guess later and repeat the calculations if necessary.
Horizontal top surface:
The characteristic length is
ft
2222
.
0
ft)
6/12
+
ft
2(4
ft)
ft)(6/12
4
(
=
=
=
P
A
L
s
c
. Then,
5
2
2
3
3
1
2
2
3
10
604
.
5
)
726
.
0
(
)
/s
ft
10
1808
.
0
(
)
ft
2222
.
0
(
R)
80
120
)(
R
001786
.
0
)(
ft/s
2
.
32
(
Pr
)
(
×
=
×

=

=

∞
υ
β
c
s
L
T
T
g
Ra
77
.
14
)
10
604
.
5
(
54
.
0
54
.
0
4
/
1
5
4
/
1
=
×
=
=
Ra
Nu
bottom
top
c
top
A
A
Nu
L
k
h
=
=
=
°
=
°
=
=
2
2
ft
2
)
ft
12
/
6
)(
ft
4
(
F
.
Btu/h.ft
016
.
1
)
77
.
14
(
ft
2222
.
0
F
Btu/h.ft.
01529
.
0
Horizontal bottom surface:
The Nusselt number for this geometry and orientation can be determined
from
387
.
7
)
10
604
.
5
(
27
.
0
27
.
0
4
/
1
5
4
/
1
=
×
=
=
Ra
Nu
F
.
Btu/h.ft
5082
.
0
)
387
.
7
(
ft
2222
.
0
F
Btu/h.ft.
01529
.
0
2
°
=
°
=
=
Nu
L
k
h
c
bottom
Vertical side surfaces:
The characteristic length in this case is the height of the duct,
L
c
= L
= 6 in.
Then,
6
2
2
3
3
1
2
2
3
10
383
.
6
)
726
.
0
(
)
/s
ft
10
1808
.
0
(
)
ft
5
.
0
(
R)
80
120
)(
R
001786
.
0
)(
ft/s
2
.
32
(
Pr
)
(
×
=
×

=

=

∞
υ
β
L
T
T
g
Ra
s
997
180 W
L
= 4 ft
Air duct
6 in
×
6 in
Air
85
°
F
22 cfm
100
°
F
Air
80
°
F
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Chapter 9
Natural Convection
57
.
27
726
.
0
492
.
0
1
)
10
383
.
6
(
387
.
0
825
.
0
Pr
492
.
0
1
Ra
387
.
0
825
.
0
2
27
/
8
16
/
9
6
/
1
6
2
27
/
8
16
/
9
6
/
1
=
+
×
+
=
+
+
=
Nu
2
2
ft
4
)
ft
5
.
0
)(
ft
4
(
2
F
.
Btu/h.ft
843
.
0
)
57
.
27
(
ft
5
.
0
F
Btu/h.ft.
01529
.
0
=
=
°
=
°
=
=
side
side
A
Nu
L
k
h
Then the total heat loss from the duct can be expressed as
[(
)
(
)
(
)
](
)
Q
Q
Q
Q
hA
hA
hA
T
T
total
top
bottom
side
top
bottom
side
s
=
+
+
=
+
+

∞
Substituting and solving for the surface temperature,
F
)
80
](
F
Btu/h.
)
4
843
.
0
2
5082
.
0
2
016
.
1
[(
Btu/h
272
°

°
×
+
×
+
×
=
s
T
T
s
=
122.4
°
F
which is sufficiently close to the assumed value of 120
°
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 Spring '08
 BENARD
 Convection, Heat Transfer, natural convection, υ

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