Problem 2.86[Difficulty: 3]
Given:Geometry of and flow rate through tapered nozzle
Find:At which point becomes turbulent
Solution:
Basic equationFor pipe flow (Section 2-6)ReρV⋅D⋅
μ
=2300=for transition to turbulence
Also flow rate Q is given byQπD2
⋅
4V⋅=
We can combine these equations and eliminate V to obtain an expression for Re in terms of D and Q
Re ρV⋅D⋅
μ
=ρD⋅
μ
4Q⋅
πD2
⋅
⋅=4Q⋅ρ⋅
πμ⋅D⋅
=Re 4Q⋅ρ⋅
πμ⋅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 DD4Q⋅ρ⋅
2300π⋅μ⋅
=
The nozzle is tapered:Din50mm⋅=Dout
Din
5
=Dout22.4mm⋅=
Carbon tetrachloride:μCT103−Ns⋅
m2
⋅=(Fig A.2)For waterρ1000 kg
m3
⋅=
SG1.595=(Table A.2)ρCTSGρ⋅=ρCT1595kg
m3
=
For the given flow rateQ2
L
min
⋅=
4Q⋅ρCT
⋅
πμ
CT
⋅Din
⋅1354=LAMINAR
4Q⋅ρCT
⋅
πμ
CT
⋅Dout
⋅3027=TURBULENT
For the diameter at which we reach turbulenceD
4Q⋅ρCT
⋅
2300π⋅μCT
⋅
=D29.4mm⋅=
ButL250 mm⋅=and linear ratios leads to the distance from Din at which D29.4 mm⋅=
Lturb
L
DD
in
−
DoutDin
−
=
LturbL
DD
in
−
DoutDin
−
⋅=Lturb186 mm⋅=