 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⋅= |
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