Problem 2.86 - Problem 2.86 [Difficulty: 3] Given: Geometry...

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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 2-6) 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.

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