At what distance does a seaplane's boundary layer transition to turbulence?

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SUMMARY

The boundary layer transition to turbulence for a seaplane flying at 100 mph through air at 45°F occurs at a characteristic length of 0.511 feet based on the Reynolds number calculation. The Reynolds number (Re) is set to 500,000 for turbulent flow, leading to the formula L = (Re)(u)/(p)(v). Using the viscosity of air at 45°F (u = 3.66 x 10^-7 lbf·s/ft²) and the density (p = 0.00245 slug/ft³), the calculated distance is 0.511 feet, which contradicts the book's answer of 0.295 feet. This discrepancy suggests a potential oversight in assumptions or calculations.

PREREQUISITES
  • Understanding of Reynolds number and its significance in fluid dynamics
  • Familiarity with the properties of air, including viscosity and density
  • Basic knowledge of fluid flow concepts, particularly laminar and turbulent flow
  • Proficiency in unit conversions, especially from mph to ft/s
NEXT STEPS
  • Review the derivation of the Reynolds number and its application in fluid mechanics
  • Investigate the effects of temperature on air viscosity and density
  • Explore the differences between laminar and turbulent flow characteristics
  • Learn about boundary layer theory and its implications in aerodynamics
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Aerospace engineers, fluid dynamics students, and anyone involved in the design and analysis of aircraft performance will benefit from this discussion.

Brian T
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Homework Statement


A seaplane flies at 100 mph through air at 45 *F. At what distance from the leading edge of the underside of the fuselage does the boundary layer transition to turbulence?

Homework Equations


Re = pvL/u
Re is Reynolds number
p (rho) is the density
V is the characteristic velocity
L is the characteristic length
u (mu) is the viscosity

The Attempt at a Solution


For turbulent flow (as said by my prof and the book), Re > 500000. So to find out when it transitions to turbulence, set Re to 500000 and solve for the characteric length L.

Solving for L, L = (Re)(u)/(p)(v)

From the appendix, the viscosity of air at 45*F is u = 3.66 x 10^-7 (lbf) (s) /ft^2 and p =. 00245 slug/ft^3. V is 100 mph which becomes 146.666 ft/s. Plugging these in, and Re = 500000, I get am answer of .511 ft.

The book, however, says the answer is. 295 ft. Can anyone help me out? I believe the calculation is correct as I ran it several times, but is there something I'm doing wrong or an assumption I'm missing? Thanks
 
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Please show us the calculation, including the manipulation of the units.

Chet
 

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