Problem 8.1
[1]
Given:
Air entering duct
Find:
Flow rate for turbulence; Entrance length
Solution:
The governing equations are
Re
V D
⋅
ν
=
Re
crit
2300
=
Q
π
4
D
2
⋅
V
⋅
=
The given data is
D
6 in
⋅
=
From Table A.9
ν
1.62
10
4
−
×
ft
2
s
⋅
=
L
laminar
0.06 Re
crit
⋅
D
⋅
=
or, for turbulent, L
turb
= 25D to 40D
Hence
Re
crit
Q
π
4
D
2
⋅
D
⋅
ν
=
or
Q
Re
crit
π
⋅
ν
⋅
D
⋅
4
=
Q
2300
π
4
×
1.62
×
10
4
−
×
ft
2
s
⋅
1
2
×
ft
⋅
=
Q
0.146
ft
3
s
⋅
=
For laminar flow
L
laminar
0.06 Re
crit
⋅
D
⋅
=
L
laminar
0.06
2300
×
6
×
in
⋅
=
L
laminar
69.0 ft
⋅
=
For turbulent flow
L
min
25 D
⋅
=
L
min
12.5 ft
⋅
=
L
max
40 D
⋅
=
L
max
20 ft
⋅
=

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Problem 8.2
[2]