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Céﬂ’i P4.35 From the NavierStokes 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 6momentum 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 (solidbody 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 NavierStokes 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 (noSlip)
dt dx y l i
l 5.4 When tested in water at 20°C ﬂowing at 2 m/s, an 8cmdiameter sphere has a measured drag
of 5 N. What will be the velocity and drag force on a 1.5mdiameter 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/ms. For sea—level standard
air take p z 1.2255 kg/m3 and ,a z 1.78E—5 kg/ms. 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 sealevel 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 Powerlaw formula CD ~ 1.07ReL—0106‘ 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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 Fall '08
 Schwartz,L
 Drag force, Fig, A. Shields, Shields Diagram

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