HW_10_solutions - «lib Hw/o fa/a 2; Céfl’i P4.35 From...

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Unformatted text preview: «lib Hw/o fa/a 2;» / Céfl’i P4.35 From the Navier-Stokes equations for incompressible flow in polar coordinates (App. E for cylindrical coordinates), find the most general case of purely circulating motion 09(r), Ur = ()2 = 0, for flow with no slip between two fixed concentric cylinders, as in Fig. P4.35. Fig. P4.35 problem is identical to Prob. 4.32 on an earlier page. That is, there are two possible solutions for purely circulating motion 09(r), hence Solution: The preliminary work for this C v9 = C1r+—l, r subject to v6(a) = 0 = Cla + CZ/a and v6,(b) = 0 = Clb + C2/b // Jo // I o This requires C1 = C2 = 0, or v.9: 0 (no steady motion possible between fixed walls) Ans. I' ,. I , r r 7., .7, Solution: Given V9 = f(r) and Vr 2 V2 = 0, we need only satisfy the 6-momentum relation: [.(Vrflflifle) 21%.)! 326%) +i20’2V9 “v; ’ (H/V 5r r 549 r66 rar 5r r 592 r >\ or: ,0(0+0)=—0+y li{r§£]+0—iz , or: f”+lf’—i2f=0 r dr dr r r r This is the ‘equidirnensional’ ODE and always has a solution in the form of a power—law, f = Cr“. The two relevant solutions for these particular coefficients are n = i1: f1 = Clr (solid-body rotation); f2 = C2/r (irrotational vortex) Ans. P4.37 A viscous liquid of constant density and viscosity falls due to gravity between two parallel plates a distance 2h h i h apart, as in the figure. The flow is fully a ' ' ' “u x developed, that is, w = w(x) only. There are no pressure gradients, only gravity. Set z W up and solve the Navier-Stokes equation for the velocity profile w(x). 4-— (N E The solution is very similar to Eqs. (4.142) to (4.143) of the text: w =$(h2 —X2) Ans. Solution: Only the z—component of Navier- Stokes is relevant: Fig. P4.37 dw d2 p— = 0 =pg+,u—12v-, or. w” = —$, w(—h) = w(+h)= 0 (no-Slip) dt dx y l i l 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 p z 998 kg/m3 and a = 0.001 kg/m-s. For sea—level standard air take p z 1.2255 kg/m3 and ,a z 1.78E—5 kg/m-s. The balloon velocity follows from dynamic similarity, which requires identical Reynolds numbers: 998(2.0)(0.08) 0.001 pVD model = _ 1.2255Vba1100n(1.5) R =1.6E5 = R _ e em" 1.78E—5 model = or Vballoon z 1.55 m/s. Ans. Then the two spheres will have identical drag coefficients: F _ 5 N pV2D2 998(2.0)2(0.08)2 Solve for Fballoon ~ 1.3 N Ans. 1:balloon 1.2255(1.55)2(1.5)2 = 0.196 = CD,proto = CD,model = V m , . ,. 5.5 An automobile has a characteristic length and area of 8 ft and 60 ftz, respectively. When tested in sea-level standard air, it has the following measured drag force versus speed: 40 60 V, mi/h: 20 Drag, lbf: 31 115 249 The same car travels in Colorado at 65 mi/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 BG units, take p :2 0.00238 slug/ft3 and a as 3.72E—7 slug/ft's. Convert the raw drag and velocity data into dimensionless form: V (mi/hr): 20 40 60 CD = F/(pVZLZ): 0.237 0.220 0.211 ReL = pVL/a: 1.50E6 3.00E6 4.50E6 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‘ 5.27 In studying sand transport by ocean waves, A. Shields in 1936 postulated that the bottom shear stress 1: required to move particles depends upon gravity g, particle size d and density pp, and water density p and viscosity ,a. Rewrite this in terms of dimensionless groups (which led to the Shields Diagram in 1936). Solution: There are six variables (17, g, d, pp, p, a) and three dimensions (M, L, T), hence we expect n —j = 6 — 3 = 3 Pi groups. The author used (,0, g, d) as repeating variables: 1/2 3/2 -1— = d ’ u ,0 pgd The shear parameter used by Shields himself was based on net weight: 17/[(pp —p)gd]. Ans. _ mm,” haw...”me ...
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