Fluid Mechanics-Frank M White Solution Ch7

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Fluid Mechanics-Frank M White Solution Ch7에 대한 보고서 자료입니다.

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P7.1 An ideal gas, at 20°C and 1 atm, flows at 12 m/s past a thin flat plate. At a position 60 cm downstream of the leading edge, the boundary layer thickness is 5 mm. Which of the 13 gases in Table A.4 is this likely to be?
Solution: We are looking for the kinematic viscosity. For a gas at low velocity and a short distance, we can guess laminar flow. Then we can begin by trying Eq. (7.1a):
δ x =
0.005m 0.6m =
5.0 Rex =
5.0 Vx/
ν
= 5.0
ν (12m/s)(0.6m)
Solve for
ν
= 2.0E−5m2 /s
The only gas in Table A.4 which matches this viscosity is the last one, CH4. Ans. But wait! Is it laminar? Check Rex = (12)(0.6)/(2.0E-5) = 360,000. Yes, OK.


7.2 A gas at 20°C flows at 2.5 m/s past a smooth, sharp flat plate. At x = 206 cm, the boundary layer thickness is 5 cm. Which of the gases in Table A.4 is this most likely to be?
Solution: Distance x fairly long, but let’s begin by guessing a laminar boundary layer:




δ x
=
5cm 206cm
= 0.0243≈ 5 Rex
, solve Rex ≈ 42,400 (OK for laminar flow) If this iscorrect, Rex= 42,400 = Ux ν = (2.5)(2.06) ν , solve ν ≈ 1.21E-4 m2 s
This value of ν exactly matches helium in Table A.4. It is not far from the value for hydrogen, but the helium result is right on the money. Ans. Guessing turbulent flow, δ /x = 0.0243 = 0.16/Rex1/7, solve Rex ≈ 541,000 (too small for transition to turbulence on a smooth wall). This would give ν ≈ 9.3E-6, about 15% greater than the kinematic viscosity of CO2. But the Reynolds number is too low, so I reject this answer.


P7.3 Equation (7.1b) assumes that the boundary layer on the plate is turbulent from the leading edge onward. Devise a scheme for determining the boundary-layer thickness more accurately when the flow is laminar up to a point Rex,crit and turbulent thereafter. Apply this scheme to computation of the boundary-layer thickness at x = 1.5 m in 40 m/s flow of air at
2
20°C and 1 atm past a flat plate. Compare your result with Eq. (7.1b). Assume Rex,crit ≈ 1.2E6.

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