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Problem 2.86
[Difficulty: 3]
Given:
Geometry of and flow rate through tapered nozzle
Find:
At which point becomes turbulent
Solution:
Basic equation
For pipe flow (Section 26)
Re
ρ
V
⋅
D
⋅
μ
=
2300
=
for transition to turbulence
Also flow rate Q is given by
Q
π
D
2
⋅
4
V
⋅
=
We can combine these equations and eliminate V to obtain an expression for Re in terms of D and Q
Re
ρ
V
⋅
D
⋅
μ
=
ρ
D
⋅
μ
4Q
⋅
π
D
2
⋅
⋅
=
⋅
ρ
⋅
πμ
⋅
D
⋅
=
Re
⋅
ρ
⋅
⋅
D
⋅
=
For a given flow rate Q, as the diameter is reduced the Reynolds number increases (due to the velocity increasing with A
1
or D
2
).
Hence for turbulence (Re = 2300), solving for D
D
⋅
ρ
⋅
2300
π
⋅
μ
⋅
=
The nozzle is tapered:
D
in
50 mm
⋅
=
D
out
D
in
5
=
D
out
22.4 mm
⋅
=
Carbon tetrachloride:
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics

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