Fluid Mechanics-Frank M White Solution Ch5

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Fluid Mechanics-Frank M White Solution Ch5에 대한 보고서 자료입니다.

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5.1 For axial flow through a circular tube, the Reynolds number for transition to turbulence is approximately 2300 [see Eq. (6.2)], based upon the diameter and average velocity. If d = 5 cm and the fluid is kerosene at 20°C, find the volume flow rate in m3/h which causes transition.
Solution: For kerosene at 20°C, take ρ = 804 kg/m3 and µ = 0.00192 kg/m⋅s. The only unknown in the transition Reynolds number is the fluid velocity:

Retr ≈ 2300=
ρ Vd µ
=
(804)V(0.05) 0.00192
, solve for Vtr =0.11m/s
Then Q=VA=(0.11)
π 4(0.05)2 =2.16E−4 m3 s ×3600≈0.78 m3 hr Ans.

P5.2 A prototype automobile is designed for cold weather in Denver, CO (-10°C, 83 kPa). Its drag force is to be tested in on a one-seventh-scale model in a wind tunnel at 150 mi/h and at 20°C and 1 atm. If model and prototype satisfy dynamic similarity, what prototype velocity, in mi/h, is matched? Comment on your result. Solution: First assemble the necessary air density and viscosity data: Denver: T =263K ; ρ p= p RT = 83000 287(263)=1.10 kg m3 ; µ p = 1.75E−5 kg m−s Windtunnel: T =293K ; ρ m= p RT = 101350 287(293)=1.205 kg m3 ; µ m = 1.80E−5 kg m−s
Convert 150 mi/h = 67.1 m/s. For dynamic similarity, equate the Reynolds numbers: Rep = ρ VL µ |p = (1.10)Vp(7Lm) 1.75E−5 = Rem = ρ VL µ |m = (1.205)(67.1)(Lm) 1.80E−5 Solvefor Vprototype = 10.2m/s = 22.8mi/h Ans.
This is too slow, hardly fast enough to turn into a driveway. Since the tunnel can go no faster, the model drag must be corrected for Reynolds number effects. Note that we did not need to know the actual length of the prototype auto, only that it is 7 times larger than the model length.


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P5.3 The transfer of energy by viscous dissipation is dependent upon viscosity
µ , thermal conductivity k, stream velocity U, and stream temperature To. Group these quantities, if possible, into the dimensionless Brinkman number, which is proportional to µ .

Solution: Here we have only a single dimensionless group. List the dimensions, from Table 5.1:

µ k U To {ML-1T-1} {MLT-3Θ-1} {LT-1} {Θ}
Four dimensions, four variables (MLTΘ) – perfect for making a pi group. Put µ in the numerator:
Brinkman number = kaUbToc µ 1 yields Br = µ U2 /(kTo) Ans.

5.4 When tested in water at 20°C flowing at 2 m/s, an 8-cm-diameter sphere has a measured drag of 5 N. What will be the velocity and drag force on a 1.5-m-diameter weather balloon moored in sea-level standard air under dynamically similar conditions?
Solution: For water at 20°C take ρ ≈ 998 kg/m3 and µ ≈ 0.001 kg/m⋅s. For sea-level standard air take ρ ≈ 1.2255 kg/m3 and µ ≈ 1.78E-5 kg/m⋅s. The balloon velocity follows from dynamic similarity, which requires identical Reynolds numbers: Remodel = ρ VD µ |model= 998(2.0)(0.08) 0.001 =1.6E5=Reproto =1.2255Vballoon(1.5) 1.78E−5 or Vballoon ≈ 1.55 m/s. Ans. Then the two spheres will have identical drag coefficients: CD,model = F ρ V2D2 = 5 N 998(2.0)2(0.08)2 =0.196=CD,proto = Fballoon 1.2255(1.55)2(1.5)2 Solve for Fballoon ≈1.3 N Ans.

5.5 An automobile has a characteristic length and area of 2.45 m and 5.57 m2, respectively. When tested in sea-level standard air, it has the following measured drag force versus speed: V, km/h: 32 64 96 Drag, N: 137.9 511.5 1107.6 The same car travels in Colorado at 104 km/h at an altitude of 3500 m. Using dimensional analysis, estimate (a) its drag force and (b) the horsepower required to overcome air drag. Solution: For sea-level air in SI units, take ρ ≈1.23 kg/m3 and µ ≈1.79×10−5N⋅s/m2. Convert the raw drag and velocity data into dimensionless form: V (km/h): 32 64 96
CD = F/( ρ V2L2): 0.236 0.219 0.211 ReL = ρ VL/ µ : 1.50E6 3.00E6 4.50E6
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Drag coefficient plots versus Reynolds number in a very smooth fashion and is well fit (to ±1%) by the Power-law formula CD ≈ 1.07ReL−0.106. (a) The new velocity is V = 29.06 m/s, and for air at 3500-m Standard Altitude (Table A-6) take ρ = 0.8633 kg/m3 and calculate

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