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Unformatted text preview: Homework Assignment #6 Read Chapter 2 in Wilcox (required) and Chapter 7 in Fox and
McDonald (optional) Do the following problems: 2.10 The classical StokesOseen equation for the drag, F, in low Reynolds number ﬂow past a sphere is F = 31erD + 1—961rp U2D2 where p, u, U and D are density, viscosity, velocity and sphere diameter, respectively. (a) Is this equation dimensionally homogeneous? (b) Rewrite this equation in terms of the Reynolds number, Re = p U D/ It and the drag coefﬁcient,
CD E F/(%p 02A), where A = go”. 2.30 For ﬂow through a smooth straight pipe, the pressure gradient, dp/dz, is a function of viscosity, [1,
velocity, U, pipe diameter, D, and density, p. How many independent dimensionless groupings are there?
What are the dimensionless groupings? (This is one of the original problems discussed by Buckingham.) 2.53 We have a 1:20 scale model of an aircraﬁ that will ﬂy at very low Mach number, so low that we
can consider the ﬂow to be incompressible. For windtunnel testing, we cool the air to the point where
its kinematic viscosity, :1, is half the value of normal atmospheric air at the intended altitude where the
aircraft will fly. To guarantee dynamic similitude, what is the required ratio of the velocity in the wind
tunnel to the velocity of the fullscale aircraﬁ? 2.54 A 1:16 scale model of a submarine is tested in a highly pressurized wind tunnel with p = 20 atm
and T = 100° F. The prototype submarine will move at Up = 12 mph in seawater, whose kinematic viscosity is up = 1.12105 ﬁz/sec.
(a) Assuming dynamic similitude can be achieved by matching just the Reynolds number. what must
the windtunnel ﬂow speed be?
(b) If the speed of sound of air at 100° F is 791 mph, what is the windtunnel Mach number? 2.58 A 1:10 scale model of a tractortrailer is tested in a pressurized wind tunnel with air at a density
pm = .00585 slug/ti3 and a temperature of 68° F. The height of the model is Hm = 1.2 ft and the flow speed is Um = 240 ft/sec. The measured drag is Fm = 72 lb. (21) Assuming that the Mach number is unimportant and that the prototype moves in air at 68° F and
1 atm, verify that to achieve dynamic similitude, the condition F = pU 2H 2 f (Ran) implies PPFP = PmFrn
(b) Determine the prototype velocity in mph and the prototype drag force in lb. (c) How much power, in horsepower, is required to move the tractor trailer at this speed? 54 CHAPTER 2. DIMENSIONAL ANALYSIS 2.10 Chapter 2, Problem 10 2.10(a): The dimensions of F, u, p, D and U are M L M M L [F] : "T—Q—i Lu] 2 E7 [pi = E, [D] = L, [U] = f Thus, the two terms in the StokesOseen formula have the following dimensions. M L ML
UD 2 —— = ——
U‘ ] LT T T2
M L2 ML
2 2 __ __ 2 =
[pUD]  L3T2L —T2 Since both terms have dimensions of force, the equation is dimensionally homogeneous. 2.10(b): By deﬁnition, the drag coefficient is F _ s F
§p U2A _ 7er2D2 CD: where we use the fact that A = {{Dz. Hence, from the StokesOseen equation, D 9— U21)2 9
CD: §W%i5:p___:24_ﬂ_.+_
7r pUD pUD 2 Finally, introducing the Reynolds number, Re = p U D/ M, and rearranging terms a bit, we arrive at 24 3
= — 1 —R
CD Re i + 16 6i 76 CHAPTER 2. DIMENSIONAL ANALYSIS 2.30 Chapter 2, Problem 30 The dimensional quantities and their dimensions are F 1_ML/T2 M [dp/dzizﬁ'f— L3 =L2T2
[u1=,;—“'§,, [U1=§:, [D1=L, [pH—1:53 There are 5 dimensional quantities and 3 independent dimensions (M, L, T), so that the
number of dimensionless groupings is 2. The appropriate dimensional equation is
[dp/dx] = [#l‘” [U l“? [0]” W4
Substituting the dimensions for each quantity yields ML‘2T‘2 = M“!L‘“1T'“1La2T‘“2L“3M“4L‘3a4
= 1Ua1+a4L—a1+02+aa—3a4Ta1a2 Thus, equating exponents, we arrive at the following three equations: 1 = (11 +0.4
—2 —a1+a2+a3—3a4
—2 = —a1 — a2 We can solve immediately for al and (12 from the ﬁrst and third equations, viz.,
a1=1——a4 and a2=1+a4
Substituting these values into the second equation yields
—2=—(1a4)+(1+a4)+a3—3a4 =2 a3=—2+a4
Therefore, the solution for (11, etc. is
a1=1—a4, a2=1+a4, a3=—2+a4 Substituting back into the dimensional equation, we have [3%] =[#]1—a4[U]1+a4[Bl—2+a4[p]a4 = [lg—02:] [£12] 4
wherefore the dimensionless groupings are
D2dp/dx # U M 2.53. CHAPTER 2, PROBLEM 53 105 2.53 Chapter 2, Problem 53 For dynamic similitude, because Mach number is unimportant for incompressible ﬂow,
we require the Reynolds numbers to be equal, i.e.,
UmLm _ (1pr Um Lp um => — = ——
Um up Up Lm up But, we are given Lp/Lm = 20 and I/p/l/m = 2, wherefore Um 1
—=20—=10
Up 2 2.54 Chapter 2, Problem 54 2.54(a): First, we must determine the kinematic viscosity of the air in the wind tunnel,
um. We can determine the dynamic viscosity, pm, from Sutherland’s law and the density,
pm, from the perfectgas law. Then, um = um/pm. So, with T = 55967" R, .  ‘8 3/2 In
um: ——————227 10 T =3.9610‘7 S g
T + 198.6 ft  sec
lb ft2
(20 atm) (2116.8 / >
Pm atm slug
pm=ﬁ_—ﬁ—.ﬁ3——='O44l —ﬁT
m 1716 559.670R
< slug ° R) ( )
Therefore, the kinematic viscosity is
2
um = ”ﬂ = 8.99  10—6 5—
pm sec
To achieve similitude based on Reynolds number, we must require
UmLm = UPLP => Um = Upﬁﬁﬂ
um up Lm up
Hence, the velocity in the wind tunnel, Um, is
8.99 106 ft2/sec
= __—————— = 154 m h 2.54(b): The Mach number in the wind tunnel, M m, is m 14 h
Alm=g— 5 mp __ —— = 0.195
am 791 mph 2.58. CHAPTER 2, PROBLEM 58 109 2.58 Chapter 2, Problem 58 The parameters for this problem are the same as those of the sphere example in the text.
Anticipating that the ﬂow will lie in the high Reynolds number range, we thus expect
that the force on the tractortrailer will vary according to UH
F = pU2H2f (ReH), Re” a 9— u 2.58(a): To achieve dynamic similitude, we must match Reynolds number. Since the temperature is the same for model and prototype and both ﬂuids are air, necessarily
pm 2 up. Thus, —memHm = ppUpHp => Um = 1,——
Mm Hp Pm Hm So, the force scales according to F_m _ p_m (L) (a) _ p_m (a) (53)? (it) = a
F? pp Up Hp pp pm Hm Hp pm Hence, we conclude that 9pr = mem 2.58(b): We are given Hp = 10Hm. Also, pp = .00234 slug/ft3 and pm = .00585 slug/ft3.
So, the velocity ratio is 92 _ £51 _ .00234 slug/ﬁs) (10) = 4
.00585 slug/ft3 Up — pm H m
Since the speed for ﬂow past the model is given to be 240 ft/sec, we ﬁnd U ft
z—n—lz60—=4lmh
Up 4 sec p Also, the force on the model is 72 lb, so that
.00585 slug/ft3
F = Fm— — 72 lb ——————
” pp ( ) (.00234 slug/ft3
2.58(c): The power, P, is simply the product of the force and the velocity. Thus, ft  lb
sec pm— )2180 lb P = FpUp = (180 lb)(60 ft/sec) = 10800
Finally, 1 horsepower is equal to 550 ft'lb/sec. Therefore,
P = 19.6 hp ...
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 '09
 BRUNGART,TIMOTHY

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