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Unformatted text preview: The turbulent wake behind a bluff body immersed in a stream flow is a subject of the present
chapter. This is a digitized video image showing the distribution of tracerdye concentration in
the wake of the body. Compare with Fig. 5.2a of the text, which is a laminar wake. (Courtesy
of R. Balachandar, by permission of the American Society of Mechanical Engineers.)  v v 426  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 7
Flow Past
Immersed Bodies Motivation. This chapter is devoted to “external” flows around bodies immersed in a
fluid stream. Such a flow will have viscous (shear and noslip) effects near the body
surfaces and in its wake, but will typically be nearly inviscid far from the body. These
are unconfined boundarylayer flows.
Chapter 6 considered “internal” flows confined by the walls of a duct. In that case
the viscous boundary layers grow from the sidewalls, meet downstream, and fill the
entire duct. Viscous shear is the dominant effect. For example, the Moody chart of Fig.
6.13 is essentially a correlation of wall shear stress for long ducts of constant cross
section.
External flows are unconfined, free to expand no matter how thick the viscous layers grow. Although boundarylayer theory (Sec. 7.3) is helpful in understanding external flows, complex body geometries usually require experimental data on the forces
and moments caused by the flow. Such immersedbody flows are commonly encountered in engineering studies: aerodynamics (airplanes, rockets, projectiles), hydrodynamics (ships, submarines, torpedos), transportation (automobiles, trucks, cycles), wind
engineering (buildings, bridges, water towers, wind turbines), and ocean engineering
(buoys, breakwaters, pilings, cables, moored instruments). This chapter provides data
and analysis to assist in such studies. 7.1 ReynoldsNumber
and Geometry Effects The technique of boundarylayer (BL) analysis can be used to compute viscous effects
near solid walls and to “patch” these onto the outer inviscid motion. This patching is
more successful as the body Reynolds number becomes larger, as shown in Fig. 7.1.
In Fig. 7.1 a uniform stream U moves parallel to a sharp flat plate of length L. If
the Reynolds number UL/ is low (Fig. 7.1a), the viscous region is very broad and extends far ahead and to the sides of the plate. The plate retards the oncoming stream
greatly, and small changes in flow parameters cause large changes in the pressure distribution along the plate. Thus, although in principle it should be possible to patch the  v v 427  eText Main Menu  Textbook Table of Contents  Study Guide 428 Chapter 7 Flow Past Immersed Bodies U δ≈L ;;
;; u = 0.99 U Large viscous
displacement
effect u<U L U ReL = 10 x Viscous
region Inviscid region U (a) Small
displacement
effect δ L U U u<U ReL = 10 7 x Laminar BL
Turbulent BL Fig. 7.1 Comparison of flow past a
sharp flat plate at low and high
Reynolds numbers: (a) laminar,
lowRe flow; (b) highRe flow. Viscous
Inviscid
region U (b) viscous and inviscid layers in a mathematical analysis, their interaction is strong and
nonlinear [1 to 3]. There is no existing simple theory for externalflow analysis at
Reynolds numbers from 1 to about 1000. Such thickshearlayer flows are typically
studied by experiment or by numerical modeling of the flow field on a digital computer [4].
A highReynoldsnumber flow (Fig. 7.1b) is much more amenable to boundarylayer
patching, as first pointed out by Prandtl in 1904. The viscous layers, either laminar or
turbulent, are very thin, thinner even than the drawing shows. We define the boundarylayer thickness as the locus of points where the velocity u parallel to the plate reaches
99 percent of the external velocity U. As we shall see in Sec. 7.4, the accepted formulas for flatplate flow are  v v x  eText Main Menu  5.0
Re1/2
x
0.16
Re1/7
x laminar (7.1a) turbulent (7.1b) Textbook Table of Contents  Study Guide 7.1 ReynoldsNumber and Geometry Effects 429 where Rex Ux/ is called the local Reynolds number of the flow along the plate surface. The turbulentflow formula applies for Rex greater than approximately 106.
Some computed values from Eq. (7.1) are
104 ( /x)lam 105 106 0.050 Rex 0.016 0.005 ( /x)turb 107 0.016 0.022 108 0.011 The blanks indicate that the formula is not applicable. In all cases these boundary layers are so thin that their displacement effect on the outer inviscid layer is negligible.
Thus the pressure distribution along the plate can be computed from inviscid theory as
if the boundary layer were not even there. This external pressure field then “drives”
the boundarylayer flow, acting as a forcing function in the momentum equation along
the surface. We shall explain this boundarylayer theory in Secs. 7.4 and 7.5.
For slender bodies, such as plates and airfoils parallel to the oncoming stream, we
conclude that this assumption of negligible interaction between the boundary layer and
the outer pressure distribution is an excellent approximation.
For a bluntbody flow, however, even at very high Reynolds numbers, there is a discrepancy in the viscousinviscid patching concept. Figure 7.2 shows two sketches of
flow past a two or threedimensional blunt body. In the idealized sketch (7.2a), there
is a thin film of boundary layer about the body and a narrow sheet of viscous wake in
the rear. The patching theory would be glorious for this picture, but it is false. In the
Beautifully behaved
but mythically thin
boundary layer
and wake Red = 10 5 Thin front
boundary layer (a) Outer stream grossly
perturbed by broad flow
separation and wake Red = 10 5  v v Fig. 7.2 Illustration of the strong
interaction between viscous and inviscid regions in the rear of bluntbody flow: (a) idealized and definitely false picture of bluntbody
flow; (b) actual picture of bluntbody flow.  eText Main Menu (b)  Textbook Table of Contents  Study Guide Chapter 7 Flow Past Immersed Bodies actual flow (Fig. 7.2b), the boundary layer is thin on the front, or windward, side of
the body, where the pressure decreases along the surface (favorable pressure gradient).
But in the rear the boundary layer encounters increasing pressure (adverse pressure
gradient) and breaks off, or separates, into a broad, pulsating wake. (See Fig. 5.2a for
a photograph of a specific example.) The mainstream is deflected by this wake, so that
the external flow is quite different from the prediction from inviscid theory with the
addition of a thin boundary layer.
The theory of strong interaction between bluntbody viscous and inviscid layers is
not well developed. Flows like that of Fig. 7.2b are normally studied experimentally.
Reference 5 is an example of efforts to improve the theory of separatedboundarylayer
flows. Reference 6 is a textbook devoted to separated flow.
EXAMPLE 7.1
A long, thin flat plate is placed parallel to a 20ft/s stream of water at 20°C. At what distance x
from the leading edge will the boundarylayer thickness be 1 in? Solution
Since we do not know the Reynolds number, we must guess which of Eqs. (7.1) applies. From
Table 1.4 for water,
1.09 10 5 ft2/s; hence
U
1.09
With 1 in 1
12 1.84 106 ft 1 5
(Ux/ )1/2 x
2 or 20 ft/s
10 5 ft2/s ft, try Eq. (7.1a): Laminar flow:
(U/v)
52 x ( 112 ft)2(1.84 106 ft 1)
511 ft 25 Now we can test the Reynolds number to see whether the formula applied:
Rex Ux (20 ft/s)(511 ft)
1.09 10 5 ft2/s 108 9.4 This is impossible since the maximum Rex for laminar flow past a flat plate is 3
try again with Eq. (7.1b):
Turbulent flow: or x (U/v)1/ 7
0.16 7/6 ( 112 ft)(1.84 106 ft 1)1/7 7/6 (4.09)7/6 0.16
Rex 106. So we 0.16
(Ux/ )1/7 x Test (20 ft/s)(5.17 ft)
1.09 10 5 ft2/s 9.5 5.17 ft Ans. 106 This is a perfectly proper turbulentflow condition; hence we have found the correct position x
on our second try.  v v 430  eText Main Menu  Textbook Table of Contents  Study Guide 7.2 MomentumIntegral Estimates 431 7.2 MomentumIntegral
Estimates When we derived the momentumintegral relation, Eq. (3.37), and applied it to a flatplate boundary layer in Example 3.11, we promised to consider it further in Chap. 7.
Well, here we are! Let us review the problem, using Fig. 7.3.
A shear layer of unknown thickness grows along the sharp flat plate in Fig. 7.3. The
noslip wall condition retards the flow, making it into a rounded profile u(y), which
merges into the external velocity U constant at a “thickness” y
(x). By utilizing
the control volume of Fig. 3.11, we found (without making any assumptions about laminar versus turbulent flow) in Example 3.11 that the drag force on the plate is given by
the following momentum integral across the exit plane
(x) D(x) b u(U u) dy (7.2) 0 where b is the plate width into the paper and the integration is carried out along a vertical plane x constant. You should review the momentumintegral relation (3.37) and
its use in Example 3.11. Kármán’s Analysis of the
Flat Plate Equation (7.2) was derived in 1921 by Kármán [7], who wrote it in the convenient form
of the momentum thickness
D(x) u
1
U bU2
0 u
dy
U (7.3) Momentum thickness is thus a measure of total plate drag. Kármán then noted that the
drag also equals the integrated wall shear stress along the plate
x D(x) w(x) b
0 dD
dx or b (7.4) w Meanwhile, the derivative of Eq. (7.3), with U
dD
dx dx bU2 constant, is
d
dx By comparing this with Eq. (7.4) Kármán arrived at what is now called the momentumintegral relation for flatplate boundarylayer flow
y
U U
p = pa δ (x) τ w (x) u (x, y)
x  v v Fig. 7.3 Growth of a boundary
layer on a flat plate.  x=0 eText Main Menu x=L  Textbook Table of Contents  Study Guide Chapter 7 Flow Past Immersed Bodies d
dx U2 w (7.5) It is valid for either laminar or turbulent flatplate flow.
To get a numerical result for laminar flow, Kármán assumed that the velocity profiles had an approximately parabolic shape
u(x, y) U y2 2y 0 2 y (x) (7.6) which makes it possible to estimate both momentum thickness and wall shear
y2 2y 2 0 y2 2y 1
u
y w 2 2
15 dy (7.7) 2U y0 By substituting (7.7) into (7.5) and rearranging we obtain
d
where
ing edge 15 U dx (7.8) / . We can integrate from 0 to x, assuming that
1
2
x 5.5 0 at x 0, the lead 15 x
U 2 1/2 or 5.5
Re1/2
x Ux (7.9) This is the desired thickness estimate. It is all approximate, of course, part of Kármán’s momentumintegral theory [7], but it is startlingly accurate, being only 10 percent higher than the known exact solution for laminar flatplate flow, which we gave
as Eq. (7.1a).
By combining Eqs. (7.9) and (7.7) we also obtain a shearstress estimate along the
plate
cf 2w
U2 8
15 Rex 1/2 0.73
Re1/2
x (7.10) Again this estimate, in spite of the crudeness of the profile assumption (7.6) is only 10
percent higher than the known exact laminarplateflow solution cf 0.664/Re1/2,
x
treated in Sec. 7.4. The dimensionless quantity cf, called the skinfriction coefficient,
is analogous to the friction factor f in ducts.
A boundary layer can be judged as “thin” if, say, the ratio /x is less than about 0.1.
This occurs at /x 0.1 5.0/Re1/2 or at Rex 2500. For Rex less than 2500 we can
x
estimate that boundarylayer theory fails because the thick layer has a significant effect
on the outer inviscid flow. The upper limit on Rex for laminar flow is about 3 106,
where measurements on a smooth flat plate [8] show that the flow undergoes transition
to a turbulent boundary layer. From 3 106 upward the turbulent Reynolds number may
be arbitrarily large, and a practical limit at present is 5 109 for oil supertankers.  v v 432  eText Main Menu  Textbook Table of Contents  Study Guide 7.2 MomentumIntegral Estimates 433
y = h +δ * y U U U
y=h h Outer streamline h u Fig. 7.4 Displacement effect of a
boundary layer. Displacement Thickness δ* 0 Simulated
effect x Another interesting effect of a boundary layer is its small but finite displacement of
the outer streamlines. As shown in Fig. 7.4, outer streamlines must deflect outward a
distance *(x) to satisfy conservation of mass between the inlet and outlet
h Ub dy ub dy h * (7.11) 0 0 The quantity * is called the displacement thickness of the boundary layer. To relate it
to u(y), cancel and b from Eq. (7.11), evaluate the left integral, and slyly add and
subtract U from the right integrand:
Uh (U u U) dy U(h *) 0 (u U) dy 0 or * u
dy
U 1
0 (7.12) Thus the ratio of */ varies only with the dimensionless velocityprofile shape u/U.
Introducing our profile approximation (7.6) into (7.12), we obtain by integration the
approximate result
* 1
3 1.83
Re1/2
x *
x (7.13) These estimates are only 6 percent away from the exact solutions for laminar flatplate
1.721x/Re1/2. Since * is much smaller than x
flow given in Sec. 7.4: * 0.344
x
for large Rex and the outer streamline slope V/U is proportional to *, we conclude
that the velocity normal to the wall is much smaller than the velocity parallel to the
wall. This is a key assumption in boundarylayer theory (Sec. 7.3).
We also conclude from the success of these simple parabolic estimates that Kármán’s momentumintegral theory is effective and useful. Many details of this theory
are given in Refs. 1 to 3. EXAMPLE 7.2  v v Are lowspeed, smallscale air and water boundary layers really thin? Consider flow at U 1
ft/s past a flat plate 1 ft long. Compute the boundarylayer thickness at the trailing edge for (a)
air and (b) water at 20°C.  eText Main Menu  Textbook Table of Contents  Study Guide 434 Chapter 7 Flow Past Immersed Bodies Solution
Part (a) From Table A.3, air 1.61 E4 ft2/s. The trailingedge Reynolds number thus is
UL ReL (1 ft/s)(1 ft)
1.61 E4 ft2/s 6200 Since this is less than 106, the flow is presumed laminar, and since it is greater than 2500, the
boundary layer is reasonably thin. From Eq. (7.1a), the predicted laminar thickness is
5.0
x
or, at x Part (b) 1 ft, From Table A.2 6200
0.0634 ft water 0.0634
0.76 in Ans. (a) 1.08 E5 ft2/s. The trailingedge Reynolds number is
(1 ft/s)(1 ft)
1.08 E5 ft2/s ReL 92,600 This again satisfies the laminar and thinness conditions. The boundarylayer thickness is
5.0
x
or, at x 1 ft, 92,600
0.0164 ft 0.0164
0.20 in Ans. (b) Thus, even at such low velocities and short lengths, both airflows and water flows satisfy the
boundarylayer approximations.  In Chaps. 4 and 6 we learned that there are several dozen known analytical laminarflow solutions [1 to 3]. None are for external flow around immersed bodies, although
this is one of the primary applications of fluid mechanics. No exact solutions are known
for turbulent flow, whose analysis typically uses empirical modeling laws to relate timemean variables.
There are presently three techniques used to study external flows: (1) numerical
(digitalcomputer) solutions, (2) experimentation, and (3) boundarylayer theory.
Computational fluid dynamics (CFD) is now well developed and described in advanced texts such as that by Anderson et al. [4]. Thousands of computer solutions
and models have been published; execution times, mesh sizes, and graphical presentations are improving each year. Both laminar and turbulentflow solutions have
been published, and turbulence modeling is a current research topic [9]. Except for
a brief discussion of computer analysis in Chap. 8, the topic of CFD is beyond our
scope here.
Experimentation is the most common method of studying external flows. Chapter
5 outlined the technique of dimensional analysis, and we shall give many nondimensional experimental data for external flows in Sec. 7.6.
The third tool is boundarylayer theory, first formulated by Ludwig Prandtl in 1904.
We shall follow Prandtl’s ideas here and make certain orderofmagnitude assumptions
to greatly simplify the NavierStokes equations (4.38) into boundarylayer equations
which are solved relatively easily and patched onto the outer inviscidflow field. v v 7.3 The BoundaryLayer
Equations  eText Main Menu  Textbook Table of Contents  Study Guide 7.3 The BoundaryLayer Equations 435 One of the great achievements of boundarylayer theory is its ability to predict the
flow separation illustrated in Fig. 7.2b. Before 1904 no one realized that such thin shear
layers could cause such a gross effect as flow separation. Unfortunately, even today theory cannot accurately predict the behavior of the separatedflow region and its interaction with the outer layer. This is the weakness of boundarylayer theory, which we hope
will be overcome by intensive research into the dynamics of separated flows [6]. Derivation for TwoDimensional
Flow We consider only steady twodimensional incompressible viscous flow with the x direction along the wall and y normal to the wall, as in Fig. 7.3.1 We neglect gravity,
which is important only in boundary layers where fluid buoyancy is dominant [2, sec.
4.13]. From Chap. 4, the complete equations of motion consist of continuity and the
x and ymomentum relations
u
x
u
u 0 y (7.14a) u
y x p
x 2 u
x2 2 p
y 2 2 y u
x 2 x u
y2 y2 (7.14b)
(7.14c) These should be solved for u, , and p subject to typical noslip, inlet, and exit boundary conditions, but in fact they are too difficult to handle for most external flows.
In 1904 Prandtl correctly deduced that a shear layer must be very thin if the Reynolds
number is large, so that the following approximations apply:
Velocities: u
u
x Rates of change: u
y (7.15a)
x y (7.15b) Our discussion of displacement thickness in the previous section was intended to justify these assumptions.
Applying these approximations to Eq. (7.14c) results in a powerful simplification
p
y 0 or p p(x) only (7.16) In other words, the ymomentum equation can be neglected entirely, and the pressure
varies only along the boundary layer, not through it. The pressuregradient term in Eq.
(7.14b) is assumed to be known in advance from Bernoulli’s equation applied to the
outer inviscid flow
p
x dp
dx U dU
dx 1 (7.17)  v v For a curved wall, x can represent the arc length along the wall and y can be everywhere normal to x
with negligible change in the boundarylayer equations as long as the radius of curvature of the wall is
large compared with the boundarylayer thickness [1 to 3].  eText Main Menu  Textbook Table of Contents  Study Guide 436 Chapter 7 Flow Past Immersed Bodies Presumably we have already made the inviscid analysis and know the distribution of
U(x) along the wall (Chap. 8).
Meanwhile, one term in Eq. (7.14b) is negligible due to Eqs. (7.15)
2 2 u
x2 u
y2 (7.18) However, neither term in the continuity relation (7.14a) can be neglected—another
warning that continuity is always a vital part of any fluidflow analysis.
The net result is that the three full equations of motion (7.14) are reduced to Prandtl’s
two boundarylayer equations
u
x Continuity:
Momentum along wall: u where u
x 0 y
u
y dU
dx U u
y
u
y (7.19a)
1 (7.19b) y laminar flow
u turbulent flow These are to be solved for u(x, y) and (x, y), with U(x) assumed to be a known function from the outer inviscidflow analysis. There are two boundary conditions on u and
one on :
At y
At y 0 (wall): u (x) (outer stream): u 0
U(x) (no slip) (7.20a) (patching) (7.20b) Unlike the NavierStokes equations (7.14), which are mathematically elliptic and must
be solved simultaneously over the entire flow field, the boundarylayer equations (7.19)
are mathematically parabolic and are solved by beginning at the leading edge and
marching downstream as far as you like, stopping at the separation point or earlier if
you prefer.2
The boundarylayer equations have been solved for scores of interesting cases of
internal and external flow for both laminar and turbulent flow, utilizing the inviscid
distribution U(x) appropriate to each flow. Full details of boundarylayer theory and
results and comparison with experiment are given in Refs. 1 to 3. Here we shall confine ourselves primarily to flatplate solutions (Sec. 7.4). 7.4 The FlatPlate Boundary
Layer The classic and most often used solution of boundarylayer theory is for flatplate flow,
as in Fig. 7.3, which can represent either laminar or turbulent flow. 2  v v For further mathematical details, see Ref. 2, sec. 2.8.  eText Main Menu  Textbook Table of Contents  Study Guide 7.4 The FlatPlate Boundary Layer Laminar Flow 437 For laminar flow past the plate, the boundarylayer equations (7.19) can be solved exactly for u and , assuming that the freestream velocity U is constant (dU/dx 0).
The solution was given by Prandtl’s student Blasius, in his 1908 dissertation from Göttingen. With an ingenious coordinate transformation, Blasius showed that the dimensionless velocity profile u/U is a function only of the single composite dimensionless
variable (y)[U/( x)]1/2:
u
U f( ) U
vx y 1/2 (7.21) where the prime denotes differentiation with respect to . Substitution of (7.21) into
the boundarylayer equations (7.19) reduces the problem, after much algebra, to a single thirdorder nonlinear ordinary differential equation for f
1
2 f ff 0 (7.22) The boundary conditions (7.20) become
At y 0: f (0) As y → : f (0) 0 (7.23a) f ( ) → 1.0 (7.23b) This is the Blasius equation, for which accurate solutions have been obtained only by
numerical integration. Some tabulated values of the velocityprofile shape f ( ) u/U
are given in Table 7.1.
Since u/U approaches 1.0 only as y → , it is customary to select the boundarylayer thickness as that point where u/U 0.99. From the table, this occurs at
5.0:
U 1/2
5.0
99%
x
or Table 7.1 The Blasius Velocity
Profile [1 to 3] x 5.0
Re1/2
x Blasius (1908) v v  u/U y[U/( x)]1/2 u/U 0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6  y[U/( x)]1/2 0.0
0.06641
0.13277
0.19894
0.26471
0.32979
0.39378
0.45627
0.51676
0.57477
0.62977
0.68132
0.72899
0.77246 2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0 0.81152
0.84605
0.87609
0.90177
0.92333
0.94112
0.95552
0.96696
0.97587
0.98269
0.98779
0.99155
1.00000 eText Main Menu  Textbook Table of Contents  Study Guide (7.24) Chapter 7 Flow Past Immersed Bodies With the profile known, Blasius, of course, could also compute the wall shear and displacement thickness
0.664
Re1/2
x cf 1.721
Re1/2
x *
x (7.25) Notice how close these are to our integral estimates, Eqs. (7.9), (7.10), and (7.13).
When cf is converted to dimensional form, we have
0.332 w(x) 1/2 1/2 U1.5 1/2 x 1/2 The wall shear drops off with x because of boundarylayer growth and varies as velocity to the 1.5 power. This is in contrast to laminar pipe flow, where w U and is
independent of x.
If w(x) is substituted into Eq. (7.4), we compute the total drag force
x D(x) b w(x) 0 dx 0.664b 1/2 1/2 U1.5x1/2 (7.26) The drag increases only as the square root of the plate length. The nondimensional
drag coefficient is defined as
2D(L)
U2bL CD 1.328
1/2
ReL 2cf (L) (7.27) Thus, for laminar plate flow, CD equals twice the value of the skinfriction coefficient
at the trailing edge. This is the drag on one side of the plate.
Kármán pointed out that the drag could also be computed from the momentum relation (7.2). In dimensionless form, Eq. (7.2) becomes
CD 2
L 0 u
1
U u
dy
U (7.28) This can be rewritten in terms of the momentum thickness at the trailing edge
CD
Computation of 2 (L)
L (7.29) from the profile u/U or from CD gives
x 0.664
Re1/2
x laminar flat plate (7.30) Since is so ill defined, the momentum thickness, being definite, is often used to correlate data taken for a variety of boundary layers under differing conditions. The ratio
of displacement to momentum thickness, called the dimensionlessprofile shape factor, is also useful in integral theories. For laminar flatplate flow
H * 1.721
0.664 2.59 (7.31) A large shape factor then implies that boundarylayer separation is about to
occur.  v v 438  eText Main Menu  Textbook Table of Contents  Study Guide 7.4 The FlatPlate Boundary Layer 439 1.0
Turbulent 0.8 Seventh
root profile,
Eq. (7.39) 10 5 = Rex
10 6
10 7
0.6
u
U
0.4 Exact Blasius profile
for all laminar Rex
( Table 7.1) 0.2
Parabolic
approximation,
Eq. ( 7.6)
0 0.2 Fig. 7.5 Comparison of dimensionless laminar and turbulent flatplate
velocity profiles. 0.4 0.6 0.8 1.0 y
δ If we plot the Blasius velocity profile from Table 7.1 in the form of u/U versus y/ ,
we can see why the simple integraltheory guess, Eq. (7.6), was such a great success.
This is done in Fig. 7.5. The simple parabolic approximation is not far from the true
Blasius profile; hence its momentum thickness is within 10 percent of the true value.
Also shown in Fig. 7.5 are three typical turbulent flatplate velocity profiles. Notice how
strikingly different in shape they are from the laminar profiles. Instead of decreasing
monotonically to zero, the turbulent profiles are very flat and then drop off sharply at
the wall. As you might guess, they follow the logarithmiclaw shape and thus can be
analyzed by momentumintegral theory if this shape is properly represented. Transition to Turbulence The laminar flatplate boundary layer eventually becomes turbulent, but there is no
unique value for this change to occur. With care in polishing the wall and keeping the
free stream quiet, one can delay the transition Reynolds number to Rex,tr 3 E6 [8].
However, for typical commercial surfaces and gusty free streams, a more realistic value
is Rex,tr 5 E5. EXAMPLE 7.3  v v A sharp flat plate with L 1 m and b 3 m is immersed parallel to a stream of velocity 2 m/s.
Find the drag on one side of the plate, and at the trailing edge find the thicknesses , *, and
for (a) air,
1.23 kg/m3 and
1.46 10 5 m2/s, and (b) water,
1000 kg/m3 and
6
2
1.02 10 m /s.  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 7 Flow Past Immersed Bodies Solution
Part (a) The airflow Reynolds number is
VL (2.0 m/s)(1.0 m)
1.46 10 5 m2/s 137,000 106, we assume that the boundary layer is laminar. From Eq. (7.27), Since this is less than 3
the drag coefficient is 1.328
(137,000)1/2 CD 0.00359 Thus the drag on one side in the airflow is
1
CD 2 U2bL D 0.00359( 1 )(1.23)(2.0)2(3.0)(1.0)
2 0.0265 N Ans. (a) The boundarylayer thickness at the end of the plate is
5.0
Re1/2
L L 5.0
(137,000)1/2 0.0135(1.0) or 0.0135 0.0135 m 13.5 mm Ans. (a) We find the other two thicknesses simply by ratios:
1.721
5.0 * *
2.59 4.65 mm 1.79 mm Ans. (a) Notice that no conversion factors are needed with SI units. Part (b) The water Reynolds number is
ReL 2.0(1.0)
1.02 10 6 1.96 106 This is rather close to the critical value of 3 106, so that a rough surface or noisy free stream
might trigger transition to turbulence; but let us assume that the flow is laminar. The water drag
coefficient is
CD
and D 1.328
(1.96 106)1/2 0.000949 0.000949( 1 )(1000)(2.0)2(3.0)(1.0)
2 5.70 N Ans. (b) The drag is 215 times more for water in spite of the higher Reynolds number and lower drag
coefficient because water is 57 times more viscous and 813 times denser than air. From Eq.
(7.26), in laminar flow, it should have (57)1/2(813)1/2 7.53(28.5) 215 times more drag.
The boundarylayer thickness is given by
L (1.96 5.0
106)1/2 0.00357(1000 mm) or 0.00357
3.57 mm Ans. (b) By scaling down we have  v v 440  eText Main Menu  Textbook Table of Contents  Study Guide 7.4 The FlatPlate Boundary Layer * 1.721
5.0 *
2.59 1.23 mm 0.48 mm 441 Ans. (b) The water layer is 3.8 times thinner than the air layer, which reflects the square root of the 14.3
ratio of air to water kinematic viscosity. Turbulent Flow There is no exact theory for turbulent flatplate flow, although there are many elegant
computer solutions of the boundarylayer equations using various empirical models for
the turbulent eddy viscosity [9]. The most widely accepted result is simply an integal
analysis similar to our study of the laminarprofile approximation (7.6).
We begin with Eq. (7.5), which is valid for laminar or turbulent flow. We write it
here for convenient reference:
U2 w(x) d
dx (7.32) From the definition of cf, Eq. (7.10), this can be rewritten as
cf 2 d
dx (7.33) Now recall from Fig. 7.5 that the turbulent profiles are nowhere near parabolic. Going
back to Fig. 6.9, we see that flatplate flow is very nearly logarithmic, with a slight
outer wake and a thin viscous sublayer. Therefore, just as in turbulent pipe flow, we
assume that the logarithmic law (6.21) holds all the way across the boundary layer
u
u* 1 ln yu* B w u* 1/2 (7.34) with, as usual,
0.41 and B 5.0. At the outer edge of the boundary layer, y
and u U, and Eq. (7.34) becomes
U
u* 1 u* ln B (7.35) But the definition of the skinfriction coefficient, Eq. (7.10), is such that the following
identities hold:
U
u* 2
cf 1/2 u* Re cf
2 1/2 (7.36) Therefore Eq. (7.35) is a skinfriction law for turbulent flatplate flow
2
cf cf
2 1/2 2.44 ln Re 1/2 5.0 (7.37) It is a complicated law, but we can at least solve for a few values and list them: v   104 105 106 107 cf v Re 0.00493 0.00315 0.00217 0.00158 eText Main Menu  Textbook Table of Contents  Study Guide Chapter 7 Flow Past Immersed Bodies Following a suggestion of Prandtl, we can forget the complex log friction law (7.37)
and simply fit the numbers in the table to a powerlaw approximation
cf 0.02 Re 1/6 (7.38) This we shall use as the lefthand side of Eq. (7.33). For the righthand side, we need
an estimate for (x) in terms of (x). If we use the logarithmiclaw profile (7.34), we
shall be up to our hips in logarithmic integrations for the momentum thickness. Instead
we follow another suggestion of Prandtl, who pointed out that the turbulent profiles in
Fig. 7.5 can be approximated by a oneseventhpower law
u
U 1/7 y (7.39) turb This is shown as a dashed line in Fig. 7.5. It is an excellent fit to the lowReynoldsnumber turbulent data, which were all that were available to Prandtl at the time. With
this simple approximation, the momentum thickness (7.28) can easily be evaluated:
y 1/7 1/7 y 1 dy 0 7
72 (7.40) We accept this result and substitute Eqs. (7.38) and (7.40) into Kármán’s momentum
law (7.33)
cf
or Re 0.02 Re
1/6 9.72 1/6 2 d
dx d
dx Re 0.16 Re6/7
x 7
72
d(Re )
d(Rex) 9.72 Separate the variables and integrate, assuming 0 at x or x (7.41)
0: 0.16
Re1/ 7
x (7.42) Thus the thickness of a turbulent boundary layer increases as x6/ 7, far more rapidly
than the laminar increase x1/2. Equation (7.42) is the solution to the problem, because
all other parameters are now available. For example, combining Eqs. (7.42) and (7.38),
we obtain the friction variation
cf 0.027
Re1/ 7
x (7.43) Writing this out in dimensional form, we have
0.0135
w,turb 1/ 7 6/ 7 U13/ 7 (7.44) 1/ 7 x Turbulent plate friction drops slowly with x, increases nearly as
insensitive to viscosity.
We can evaluate the drag coefficient from Eq. (7.29)
CD  v v 442  eText Main Menu  0.031
Re1/ 7
L and U2, and is rather 7
cf (L)
6 Textbook Table of Contents (7.45)  Study Guide 7.4 The FlatPlate Boundary Layer 443 Then CD is only 16 percent greater than the trailingedge skin friction [compare with
Eq. (7.27) for laminar flow].
The displacement thickness can be estimated from the oneseventhpower law and
Eq. (7.12):
* 1/ 7 y 1 dy 0 1
8 (7.46) The turbulent flatplate shape factor is approximately
H 1
8
7
72 * 1.3 (7.47) These are the basic results of turbulent flatplate theory.
Figure 7.6 shows flatplate drag coefficients for both laminarand turbulentflow
conditions. The smoothwall relations (7.27) and (7.45) are shown, along with the effect of wall roughness, which is quite strong. The proper roughness parameter here is
x/ or L/ , by analogy with the pipe parameter /d. In the fully rough regime, CD is independent of the Reynolds number, so that the drag varies exactly as U2 and is inde 0.014
200
Fully rough
Eq. ( 7.48b) L
ε = 300 0.012 500 0.010 1000
0.008
2000 CD
0.006 5000
10 4
2 × 10 4 0.004 v v   2 × 10 5
10 6 Transition 0.002 Fig. 7.6 Drag coefficient of laminar
and turbulent boundary layers on
smooth and rough flat plates. This
chart is the flatplate analog of the
Moody diagram of Fig. 6.13. 5 × 10 4 Turbulent
smooth
Eq. ( 7.45 ) Laminar:
Eq. ( 7.27 )
0
10 5 eText Main Menu Eq. ( 7.49 ) 10 6 10 7 10 8 ReL  Textbook Table of Contents  Study Guide 10 9 Chapter 7 Flow Past Immersed Bodies pendent of . Reference 2 presents a theory of rough flatplate flow, and Ref. 1 gives
a curve fit for skin friction and drag in the fully rough regime:
2.5 x cf 2.87 1.58 log CD 1.89 1.62 log (7.48a) L 2.5 (7.48b) Equation (7.48b) is plotted to the right of the dashed line in Fig. 7.6. The figure also
shows the behavior of the drag coefficient in the transition region 5 105 ReL
8 107, where the laminar drag at the leading edge is an appreciable fraction of the
total drag. Schlichting [1] suggests the following curve fits for these transition drag
curves depending upon the Reynolds number Retrans where transition begins: CD
CD 0.031
Re1/ 7
L 1440
ReL Retrans 5 105 (7.49a) 0.031
Re1/ 7
L 8700
ReL Retrans 3 106 (7.49b) EXAMPLE 7.4
A hydrofoil 1.2 ft long and 6 ft wide is placed in a water flow of 40 ft/s, with
1.99
slugs/ft3 and
0.000011 ft2/s. (a) Estimate the boundarylayer thickness at the end of the
plate. Estimate the friction drag for (b) turbulent smoothwall flow from the leading edge,
(c) laminar turbulent flow with Retrans 5 105, and (d) turbulent roughwall flow with
0.0004 ft. Solution
Part (a) The Reynolds number is
ReL UL (40 ft/s)(1.2 ft)
0.000011 ft2/s 106 4.36 Thus the trailingedge flow is certainly turbulent. The maximum boundarylayer thickness would
occur for turbulent flow starting at the leading edge. From Eq. (7.42),
(L)
L
or (4.36 0.16
106)1/7 0.018(1.2 ft) 0.018 0.0216 ft Ans. (a) This is 7.5 times thicker than a fully laminar boundary layer at the same Reynolds number. Part (b) For fully turbulent smoothwall flow, the drag coefficient on one side of the plate is, from Eq.
(7.45),
CD  v v 444  eText Main Menu  0.031
(4.36 106)1/7 0.00349 Textbook Table of Contents  Study Guide 7.5 Boundary Layers with Pressure Gradient 445 Then the drag on both sides of the foil is approximately
D Part (c) 2CD( 1 U2)bL
2 2(0.00349)( 1 )(1.99)(40)2(6.0)(1.2)
2 CD 1440
4.36 106 0.00349 Ans. (b) 105, Eq. (7.49a) applies: 5 With a laminar leading edge and Retrans 80 lb 0.00316 The drag can be recomputed for this lower drag coefficient:
2CD( 1 U2)bL
2 D Part (d) 72 lbf Ans. (c) Finally, for the rough wall, we calculate
L
From Fig. 7.6 at ReL
tion (7.48b) applies: 3000 106, this condition is just inside the fully rough regime. Equa 4.36
CD 1.2 ft
0.0004 ft (1.89 1.62 log 3000) 2.5 0.00644 and the drag estimate is
D 1
2CD( 2 U2)bL 148 lbf Ans. (d) This small roughness nearly doubles the drag. It is probable that the total hydrofoil drag is still
another factor of 2 larger because of trailingedge flowseparation effects. 7.5 Boundary Layers with
Pressure Gradient3 The flatplate analysis of the previous section should give us a good feeling for the behavior of both laminar and turbulent boundary layers, except for one important effect:
flow separation. Prandtl showed that separation like that in Fig. 7.2b is caused by excessive momentum loss near the wall in a boundary layer trying to move downstream
against increasing pressure, dp/dx 0, which is called an adverse pressure gradient.
The opposite case of decreasing pressure, dp/dx 0, is called a favorable gradient,
where flow separation can never occur. In a typical immersedbody flow, e.g., Fig. 7.2b,
the favorable gradient is on the front of the body and the adverse gradient is in the rear,
as discussed in detail in Chap. 8.
We can explain flow separation with a geometric argument about the second derivative of velocity u at the wall. From the momentum equation (7.19b) at the wall, where
0, we obtain
u
y wall u
y 2 u
y2 2 or 2 U wall wall dU
dx dp
dx 1 dp
dx 3  v v This section may be omitted without loss of continuity.  eText Main Menu  Textbook Table of Contents  Study Guide (7.50) 446 Chapter 7 Flow Past Immersed Bodies for either laminar or turbulent flow. Thus in an adverse gradient the second derivative of
velocity is positive at the wall; yet it must be negative at the outer layer (y
) to merge
smoothly with the mainstream flow U(x). It follows that the second derivative must pass
through zero somewhere in between, at a point of inflection, and any boundarylayer profile in an adverse gradient must exhibit a characteristic S shape.
Figure 7.7 illustrates the general case. In a favorable gradient (Fig. 7.7a) the profile U U u u PI
(a) Favorable
gradient:
dU
>0
dx
dp
<0
dx (b) Zero
gradient:
dU
=0
dx
dp
=0
dx No separation,
PI inside wall No separation,
PI at wall dp
>0
dx
U
U
U
u
u
u PI
PI
Backflow
PI (c) Weak adverse
gradient: τw = 0
(d ) Critical adverse
gradient: (e) Excessive adverse
gradient: dU
<0
dx v v  Backflow
at the wall: dp
>0
dx Fig. 7.7 Effect of pressure gradient
on boundarylayer profiles; PI
point of inflection. Zero slope
at the wall:
Separation Separated
flow region No separation,
PI in the flow  eText Main Menu  Textbook Table of Contents  Study Guide 7.5 Boundary Layers with Pressure Gradient 447 is very rounded, there is no point of inflection, there can be no separation, and laminar profiles of this type are very resistant to a transition to turbulence [1 to 3].
In a zero pressure gradient (Fig. 7.7b), e.g., flatplate flow, the point of inflection
is at the wall itself. There can be no separation, and the flow will undergo transition
at Rex no greater than about 3 106, as discussed earlier.
In an adverse gradient (Fig. 7.7c to e), a point of inflection (PI) occurs in the boundary layer, its distance from the wall increasing with the strength of the adverse gradient. For a weak gradient (Fig. 7.7c) the flow does not actually separate, but it is vulnerable to transition to turbulence at Rex as low as 105 [1, 2]. At a moderate gradient,
a critical condition (Fig. 7.7d) is reached where the wall shear is exactly zero ( u/ y
0). This is defined as the separation point ( w 0), because any stronger gradient will
actually cause backflow at the wall (Fig. 7.7e): the boundary layer thickens greatly,
and the main flow breaks away, or separates, from the wall (Fig. 7.2b).
The flow profiles of Fig. 7.7 usually occur in sequence as the boundary layer progresses along the wall of a body. For example, in Fig. 7.2a, a favorable gradient occurs on the front of the body, zero pressure gradient occurs just upstream of the shoulder, and an adverse gradient occurs successively as we move around the rear of the
body.
A second practical example is the flow in a duct consisting of a nozzle, throat, and
diffuser, as in Fig. 7.8. The nozzle flow is a favorable gradient and never separates, nor Separation
point
τw = 0 Boundary
layers Profile point
of inflection Backflow δ ( x)
Nearly
inviscid
core flow x
U(x) δ ( x) Separation Nozzle:
Decreasing
pressure
and area v v  eText Main Menu  Diffuser:
Increasing pressure
and area Velocity
constant Decreasing velocity Favorable
gradient Fig. 7.8 Boundarylayer growth and
separation in a nozzlediffuser configuration. Throat:
Constant
pressure
and area Increasing
velocity  U(x) Zero
gradient Adverse gradient
(boundary layer thickens) Textbook Table of Contents  Study Guide Dividing
streamline 448 Chapter 7 Flow Past Immersed Bodies does the throat flow where the pressure gradient is approximately zero. But the expandingarea diffuser produces low velocity and increasing pressure, an adverse gradient. If the diffuser angle is too large, the adverse gradient is excessive, and the boundary layer will separate at one or both walls, with backflow, increased losses, and poor
pressure recovery. In the diffuser literature [10] this condition is called diffuser stall, a
term used also in airfoil aerodynamics (Sec. 7.6) to denote airfoil boundarylayer separation. Thus the boundarylayer behavior explains why a largeangle diffuser has heavy
flow losses (Fig. 6.23) and poor performance (Fig. 6.28).
Presently boundarylayer theory can compute only up to the separation point, after
which it is invalid. New techniques are now developed for analyzing the strong interaction effects caused by separated flows [5, 6]. Laminar Integral Theory Both laminar and turbulent theories can be developed from Kármán’s general twodimensional boundarylayer integral relation [7], which extends Eq. (7.33) to variable
U(x)
1
cf
2 w U2 d
dx (2 H) dU
U dx (7.51) *(x)/ (x) is the shape factor. From
where (x) is the momentum thickness and H(x)
Eq. (7.17) negative dU/dx is equivalent to positive dp/dx, that is, an adverse gradient.
We can integrate Eq. (7.51) to determine (x) for a given U(x) if we correlate cf and
H with the momentum thickness. This has been done by examining typical velocity
profiles of laminar and turbulent boundarylayer flows for various pressure gradients.
Some examples are given in Fig. 7.9, showing that the shape factor H is a good indicator of the pressure gradient. The higher the H, the stronger the adverse gradient, and
separation occurs approximately at
3.5
2.4 H laminar flow
turbulent flow (7.52) The laminar profiles (Fig. 7.9a) clearly exhibit the S shape and a point of inflection
with an adverse gradient. But in the turbulent profiles (Fig. 7.9b) the points of inflection are typically buried deep within the thin viscous sublayer, which can hardly be
seen on the scale of the figure.
There are scores of turbulent theories in the literature, but they are all complicated algebraically and will be omitted here. The reader is referred to advanced texts [1, 2, 9].
For laminar flow, a simple and effective method was developed by Thwaites [11],
who found that Eq. (7.51) can be correlated by a single dimensionless momentumthickness variable , defined as
2 v dU
dx (7.53) Using a straightline fit to his correlation, Thwaites was able to integrate Eq. (7.51) in
closed form, with the result  v v 2  eText Main Menu  2
0 U0
U 6 0.45
U6 x U5 dx (7.54) 0 Textbook Table of Contents  Study Guide 7.5 Boundary Layers with Pressure Gradient
1.0 449 1.0
Flat plate Favorable
gradients: 0.9 0.8
2.2 = H = 0.8 δ*
θ 2.4
2.6 (Flat plate)
0.6 0.6
u
0.5
U 2.9
3.2 0.4 δ * = 1.3
θ
4
1. .5
1 .6
1 .7 8
1 1. 9
1. 1
4
.
.0 2 .3 2.
2 .2 2
2 0.7 2.7 u
U H= Separation 0.4 3.5 (Separation)
0.3
Points of
inflection
(adverse
gradients) 0.2 0 0.2 0.4 0.6 0.2
0.1 0.8 1.0 y
δ 0 0.1 0.5
y
δ (a) (b) 0.2 0.3 0.4 0.6 0.7 0.8 0.9 1.0 Fig. 7.9 Velocity profiles with pressure gradient: (a) laminar flow; (b) turbulent flow with adverse gradients. where 0 is the momentum thickness at x 0 (usually taken to be zero). Separation
(cf 0) was found to occur at a particular value of
Separation: 0.09 (7.55) Finally, Thwaites correlated values of the dimensionless shear stress S
, and his graphed result can be curvefitted as follows:
S( ) w 0.09)0.62 ( U w /( U) with (7.56) This parameter is related to the skin friction by the identity
S 1
2 cf Re (7.57) Equations (7.54) to (7.56) constitute a complete theory for the laminar boundary layer
with variable U(x), with an accuracy of 10 percent compared with exact digitalcomputer solutions of the laminarboundarylayer equations (7.19). Complete details of
Thwaites’ and other laminar theories are given in Refs. 2 and 3.
As a demonstration of Thwaites’ method, take a flat plate, where U constant,
0, and 0 0. Equation (7.54) integrates to  v v 2  eText Main Menu  0.45 x
U Textbook Table of Contents  Study Guide Chapter 7 Flow Past Immersed Bodies or 0.671
Re1/2
x x (7.58) This is within 1 percent of Blasius’ exact solution, Eq. (7.30).
With
0, Eq. (7.56) predicts the flatplate shear to be
w (0.09)0.62 U
or 0.671
Re1/2
x 2w
U2 cf 0.225
(7.59) This is also within 1 percent of the Blasius result, Eq. (7.25). However, the general accuracy of this method is poorer than 1 percent because Thwaites actually “tuned” his
correlation constants to make them agree with exact flatplate theory.
We shall not compute any more boundarylayer details here, but as we go along,
investigating various immersedbody flows, especially in Chap. 8, we shall
use Thwaites’ method to make qualitative assessments of the boundarylayer behavior. EXAMPLE 7.5
In 1938 Howarth proposed a linearly decelerating externalvelocity distribution
U(x) x
L U0 1 (1) as a theoretical model for laminarboundarylayer study. (a) Use Thwaites’ method to compute
the separation point xsep for 0 0, and compare with the exact digitalcomputer solution
xsep/L 0.119863 given by H. Wipperman in 1966. (b) Also compute the value of cf 2 w/( U2)
at x/L 0.1. Solution
Part (a) First note that dU/dx
U0/L constant: Velocity decreases, pressure increases, and the pressure gradient is adverse throughout. Now integrate Eq. (7.54)
2 0.45v
x/L)6 U 6(1
0 x
0 Then the dimensionless factor v   eText Main Menu L
U0 0.075 1 U0
L x
L 1 x
L 6 1 (2) 6 1 (3) 0.09 for separation 0.09
xsep
L  0.075 dx 2 From Eq. (7.55) we set this equal to or 5 is given by dU
v dx sep x
L U5 1
0 2 v 450 0.075 1
1 (2.2) 1/6 xsep
L 6 1 0.123 Textbook Table of Contents Ans. (a)  Study Guide 7.6 Experimental External Flows 451
This is less than 3 percent higher than Wipperman’s exact solution, and the computational effort is very modest. Part (b) To compute cf at x/L 0.1 (just before separation), we first compute
(x 0.1L) 0.075[(1 0.1) 6 1] at this point, using Eq. (3)
0.0661 Then from Eq. (7.56) the shear parameter is
S(x 0.1L) ( 0.0661 0.09)0.62 0.099 1
2f c Re (4) We can compute Re in terms of ReL from Eq. (2) or (3)
2 0.0661
UL/ L2
or 0.0661
ReL 0.257 Re1/2
L Re at x
L 0.1 Substitute into Eq. (4):
0.099
or cf 1
2 cf (0.257 Re1/2)
L 0.77
Re1/2
L ReL UL Ans. (b) We cannot actually compute cf without the value of, say, U0 L/ . 7.6 Experimental External
Flows Boundarylayer theory is very interesting and illuminating and gives us a great qualitative grasp of viscousflow behavior, but, because of flow separation, the theory does
not generally allow a quantitative computation of the complete flow field. In particular, there is at present no satisfactory theory for the forces on an arbitrary body immersed in a stream flowing at an arbitrary Reynolds number. Therefore experimentation is the key to treating external flows.
Literally thousands of papers in the literature report experimental data on specific
external viscous flows. This section gives a brief description of the following externalflow problems:
1. Drag of twoand threedimensional bodies
a. Blunt bodies
b. Streamlined shapes
2. Performance of lifting bodies
a. Airfoils and aircraft
b. Projectiles and finned bodies
c. Birds and insects  v v For further reading see the goldmine of data compiled in Hoerner [12]. In later chapters we shall study data on supersonic airfoils (Chap. 9), openchannel friction (Chap.
10), and turbomachinery performance (Chap. 11).  eText Main Menu  Textbook Table of Contents  Study Guide 452 Chapter 7 Flow Past Immersed Bodies Drag of Immersed Bodies Any body of any shape when immersed in a fluid stream will experience forces and
moments from the flow. If the body has arbitrary shape and orientation, the flow will
exert forces and moments about all three coordinate axes, as shown in Fig. 7.10. It is
customary to choose one axis parallel to the free stream and positive downstream. The
force on the body along this axis is called drag, and the moment about that axis the
rolling moment. The drag is essentially a flow loss and must be overcome if the body
is to move against the stream.
A second and very important force is perpendicular to the drag and usually performs
a useful job, such as bearing the weight of the body. It is called the lift. The moment
about the lift axis is called yaw.
The third component, neither a loss nor a gain, is the side force, and about this axis
is the pitching moment. To deal with this threedimensional forcemoment situation is
more properly the role of a textbook on aerodynamics [for example, 13]. We shall limit
the discussion here to lift and drag.
When the body has symmetry about the liftdrag axis, e.g., airplanes, ships, and cars
moving directly into a stream, the side force, yaw, and roll vanish, and the problem reduces to a twodimensional case: two forces, lift and drag, and one moment, pitch.
A final simplification often occurs when the body has two planes of symmetry, as
in Fig. 7.11. A wide variety of shapes such as cylinders, wings, and all bodies of revolution satisfy this requirement. If the free stream is parallel to the intersection of these
two planes, called the principal chord line of the body, the body experiences drag only,
with no lift, side force, or moments.4 This type of degenerate oneforce drag data is
what is most commonly reported in the literature, but if the free stream is not parallel
to the chord line, the body will have an unsymmetric orientation and all three forces
and three moments can arise in principle.
In lowspeed flow past geometrically similar bodies with identical orientation and
relative roughness, the drag coefficient should be a function of the body Reynolds number
CD f (Re) (7.60) Lift force
Yawing
moment
Arbitrary
body Drag force
Rolling moment Pitching moment Fig. 7.10 Definition of forces and
moments on a body immersed in a
uniform flow. V Freestream
velocity Side force  v v 4
In bodies with shed vortices, such as the cylinder in Fig. 5.2, there may be oscillating lift, side force,
and moments, but their mean value is zero.  eText Main Menu  Textbook Table of Contents  Study Guide 7.6 Experimental External Flows 453
Vertical plane of symmetry Horizontal plane
of symmetry Drag only if V
is parallel to
chord line Principal
chord line V Fig. 7.11 Only the drag force occurs if the flow is parallel to both
planes of symmetry. Doubly
symmetric
body The Reynolds number is based upon the freestream velocity V and a characteristic
length L of the body, usually the chord or body length parallel to the stream
VL Re (7.61) For cylinders, spheres, and disks, the characteristic length is the diameter D. Characteristic Area Drag coefficients are defined by using a characteristic area A which may differ depending upon the body shape:
CD 1
2 drag
V2A (7.62) The factor 1 is our traditional tribute to Euler and Bernoulli. The area A is usually one
2
of three types:
1. Frontal area, the body as seen from the stream; suitable for thick, stubby bodies,
such as spheres, cylinders, cars, missiles, projectiles, and torpedoes.
2. Planform area, the body area as seen from above; suitable for wide, flat bodies
such as wings and hydrofoils.
3. Wetted area, customary for surface ships and barges.
In using drag or other fluidforce data, it is important to note what length and area are
being used to scale the measured coefficients. Friction Drag and Pressure Drag As we have mentioned, the theory of drag is weak and inadequate, except for the flat
plate. This is because of flow separation. Boundarylayer theory can predict the separation point but cannot accurately estimate the (usually low) pressure distribution in
the separated region. The difference between the high pressure in the front stagnation
region and the low pressure in the rear separated region causes a large drag contribution called pressure drag. This is added to the integrated shear stress or friction drag
of the body, which it often exceeds:  v v CD  eText Main Menu  CD,press Textbook Table of Contents CD,fric  Study Guide (7.63) 454 Chapter 7 Flow Past Immersed Bodies The relative contribution of friction and pressure drag depends upon the body’s shape,
especially its thickness. Figure 7.12 shows drag data for a streamlined cylinder of very
large depth into the paper. At zero thickness the body is a flat plate and exhibits 100
percent friction drag. At thickness equal to the chord length, simulating a circular cylinder, the friction drag is only about 3 percent. Friction and pressure drag are about equal
at thickness t/c 0.25. Note that CD in Fig. 7.12b looks quite different when based
upon frontal area instead of planform area, planform being the usual choice for this
body shape. The two curves in Fig. 7.12b represent exactly the same drag data.
Figure 7.13 illustrates the dramatic effect of separated flow and the subsequent failure of boundarylayer theory. The theoretical inviscid pressure distribution on a circular cylinder (Chap. 8) is shown as the dashed line in Fig. 7.13c:
Cp p
1
2 p
V2 1 4 sin2 (7.64) where p and V are the pressure and velocity, respectively, in the free stream. The actual laminar and turbulent boundarylayer pressure distributions in Fig. 7.13c are startlingly different from those predicted by theory. Laminar flow is very vulnerable to the
adverse gradient on the rear of the cylinder, and separation occurs at
82°, which Friction drag percent 100 Data scatter
50 0 (a) Percentage of
pressure
drag 3
0 0.2 0.4 t
c 0.6 0.8 1.0
Circular cylinder 0.3
CD based upon frontal area (t b)
0.2 CD based upon planform area (c b) CD
Width b 0.1  v v Fig. 7.12 Drag of a streamlined
twodimensional cylinder at Rec
106: (a) effect of thickness ratio on
percentage of friction drag; (b) total
drag versus thickness when based
upon two different areas. t Flat
plate (b)  0 0 eText Main Menu V 0.2  0.4 0.6
t
Thickness ratio c Textbook Table of Contents c
0.8  1.0 Study Guide 7.6 Experimental External Flows 455
Separation
Separation θ θ
82° V
p∞ Broad
wake 120° V
p∞ CD = 1.2 CD = 0.3 (a) Narrow
wake (b) 1.0 0.0 p – p∞ Laminar
– 1.0 Cp = ρ V 2/ 2 Turbulent Inviscid
theory – 2.0 Cp = 1 – 4 sin2 θ Fig. 7.13 Flow past a circular
cylinder: (a) laminar separation; (b)
turbulent separation; (c) theoretical
and actual surfacepressure distributions. – 3.0
0° 45° 90° 135° 180° θ
(c)  v v certainly could not have been predicted from inviscid theory. The broad wake and very
low pressure in the separated laminar region cause the large drag CD 1.2.
The turbulent boundary layer in Fig. 7.13b is more resistant, and separation is delayed until
120°, with a resulting smaller wake, higher pressure on the rear, and
75 percent less drag, CD 0.3. This explains the sharp drop in drag at transition in
Fig. 5.3.
The same sharp difference between vulnerable laminar separation and resistant turbulent separation can be seen for a sphere in Fig. 7.14. The laminar flow (Fig. 7.14a)
separates at about 80°, CD 0.5, while the turbulent flow (Fig. 7.14b) separates at
120°, CD 0.2. Here the Reynolds numbers are exactly the same, and the turbulent
boundary layer is induced by a patch of sand roughness at the nose of the ball. Golf
balls fly in this range of Reynolds numbers, which is why they are deliberately dimpled to induce a turbulent boundary layer and lower drag. Again we would find the
actual pressure distribution on the sphere to be quite different from that predicted by
inviscid theory.  eText Main Menu  Textbook Table of Contents  Study Guide 456 Chapter 7 Flow Past Immersed Bodies Fig. 7.14 Strong differences in laminar and turbulent separation on an
8.5in bowling ball entering water
at 25 ft/s: (a) smooth ball, laminar
boundary layer; (b) same entry, turbulent flow induced by patch of
nosesand roughness. (U.S. Navy
photograph, Ordnance Test Station,
Pasadena Annex.) (a ) (b ) In general, we cannot overstress the importance of body streamlining to reduce
drag at Reynolds numbers above about 100. This is illustrated in Fig. 7.15. The rectangular cylinder (Fig. 7.15a) has rampant separation at all sharp corners and very
high drag. Rounding its nose (Fig. 7.15b) reduces drag by about 45 percent, but CD
is still high. Streamlining its rear to a sharp trailing edge (Fig. 7.15c) reduces its drag
another 85 percent to a practical minimum for the given thickness. As a dramatic contrast, the circular cylinder (Fig. 7.15d) has oneeighth the thickness and onethreehundredth the cross section (c) (Fig. 7.15c), yet it has the same drag. For highperformance vehicles and other moving bodies, the name of the game is drag reduction,
for which intense research continues for both aerodynamic and hydrodynamic applications [20, 39].
The drag of some representative widespan (nearly twodimensional) bodies is shown
versus the Reynolds number in Fig. 7.16a. All bodies have high CD at very low (creeping flow) Re 1.0, while they spread apart at high Reynolds numbers according to CD = 2.0 V  v v Fig. 7.15 The importance of
streamlining in reducing drag of a
body (CD based on frontal area):
(a) rectangular cylinder; (b)
rounded nose; (c) rounded nose and
streamlined sharp trailing edge; (d)
circular cylinder with the same
drag as case (c). CD = 1.1 V (b) (a) V CD = 0.15 V (c)  eText Main Menu (d )  Textbook Table of Contents  Study Guide 7.6 Experimental External Flows 457 their degree of streamlining. All values of CD are based on the planform area except
the plate normal to the flow. The birds and the sailplane are, of course, not very twodimensional, having only modest span length. Note that birds are not nearly as efficient as modern sailplanes or airfoils [14, 15].
Table 7.2 gives a few data on drag, based on frontal area, of twodimensional bodies of various cross section, at Re 104. The sharpedged bodies, which tend to cause
flow separation regardless of the character of the boundary layer, are insensitive to the
Reynolds number. The elliptic cylinders, being smoothly rounded, have the laminartoturbulent transition effect of Figs. 7.13 and 7.14 and are therefore quite sensitive to
whether the boundary layer is laminar or turbulent. 100
Smooth
circular
cylinder
L =∞
D
=5 10 Plate
normal
to stream Square
cylinder 1
CD
0.1 0.01 Pigeon Seagull Smooth
flat plate
parallel
to stream Vulture Sailplane Airfoil Transition
0.001
0.1 1 10 10 3
Re
(a) 100 10 4 10 5 10 6 10 7 100 10 CD 1 Stokes'
law:
24/Re Disk
Sphere
2:1
ellipsoid 0.1  v v Fig. 7.16 Drag coefficients of
smooth bodies at low Mach numbers: (a) twodimensional bodies;
(b) threedimensional bodies. Note
the Reynoldsnumber independence
of blunt bodies at high Re.  Airship hull
0.01
0.1 eText Main Menu 1  10 100 10 3
Re
(b) Textbook Table of Contents 10 4  10 5 10 6 Study Guide 10 7 458 Chapter 7 Flow Past Immersed Bodies Table 7.2 Drag of TwoDimensional Bodies at Re CD based
on frontal
area 104
Shape
Square cylinder: CD based
on frontal
area Shape CD based
on frontal
area Shape
Plate:
Plate: Halfcylinder:
2.1 1.6 2.0 1.2 1.7 Half tube: Thin plate
normal to
a wall:
1.4 Equilateral triangle:
1.6 1.2 Hexagon:
2.0 2.3 Shape 1.0 0.7 CD based on frontal area Rounded nose section:
H L/H::
L/H
CD: 0.5
1.16 1.0
0.90 0.4
2.3 0.7
2.7 2.0
0.70 1.2
2.1 4.0
0.68 6.0
0.64 L
Flat nose section
Rounded nose section:
L/H:
L/H:
CD:
D H 0.1
1.9 2.0
1.8 2.5
1.4 L Laminar Turbulent 1:1 1.2 0.3 2:1 0.6 0.2 4:1 0.35 0.15 8:1 0.25 0.1  v v Elliptical cylinder:  eText Main Menu  Textbook Table of Contents  Study Guide 3.0
1.3 6.0
0.9 7.6 Experimental External Flows 459 EXAMPLE 7.6
A square 6in piling is acted on by a water flow of 5 ft/s that is 20 ft deep, as shown in Fig.
E7.6. Estimate the maximum bending exerted by the flow on the bottom of the piling.
h = 6 in 5 ft/s L = 20 ft E7.6 Solution
Assume seawater with
1.99 slugs/ft3 and kinematic viscosity
ing width of 0.5 ft, we have
Reh (5 ft/s)(0.5 ft)
0.000011 ft2/s 2.3 0.000011 ft2/s. With a pil 105 This is the range where Table 7.2 applies. The worst case occurs when the flow strikes the flat
side of the piling, CD 2.1. The frontal area is A Lh (20 ft)(0.5 ft) 10 ft2. The drag is
estimated by
F CD( 1 V2A)
2 2.1( 1 )(1.99 slugs/ft3)(5 ft/s)2(10 ft2)
2 522 lbf If the flow is uniform, the center of this force should be at approximately middepth. Therefore
the bottom bending moment is
M0 FL
2 522(10) 5220 ft lbf Ans. According to the flexure formula from strength of materials, the bending stress at the bottom
would be
S M0y
I (5220 ft lb)(0.25 ft)
1
4
12 (0.5 ft) 251,000 lbf/ft2 1740 lbf/in2 to be multiplied, of course, by the stressconcentration factor due to the builtin end conditions.  v v Some drag coefficients of threedimensional bodies are listed in Table 7.3 and Fig.
7.16b. Again we can conclude that sharp edges always cause flow separation and high
drag which is insensitive to the Reynolds number. Rounded bodies like the ellipsoid
have drag which depends upon the point of separation, so that both the Reynolds num  eText Main Menu  Textbook Table of Contents  Study Guide 460 Chapter 7 Flow Past Immersed Bodies Table 7.3 Drag of ThreeDimensional Bodies at Re
CD based on
frontal area Body 104
Body CD based on frontal area Cone: Cube:
1.07 θ:
CD: θ 0.81 10˚
0.30 20˚
0.40 30˚
0.55 40˚
0.65 60˚
0.80 75˚
1.05 90˚
1.15 L /D:
CD: 1
0.64 2
0.68 3
0.72 5
0.74 10
0.82 20
0.91 40
0.98 Short cylinder,
laminar flow:
L
D Cup:
1.4 Porous parabolic
dish [23]: Porosity:
CD:
CD: 0.4 0
1.42
0.95 0.1
1.33
0.92 0.2
1.20
0.90 0.3
1.05
0.86 0.4
0.95
0.83 0.5
0.82
0.80 Average person:
Disk:
CD A ≈ 9 ft 2 1.17
Parachute
(Low porosity): Pine and spruce
trees [24]: 1.2 Body U, m/s:
CD: CD based on
frontal area Ratio C D A ≈ 1.2 ft 2 10
1.2 ± 0.2 30
0.7 ± 0.2 40
0.5 ± 0.2 Ratio Body Rectangular plate: 20
1.0 ± 0.2 CD based on
frontal area L /d 0.5
1
2
4
8 1.15
0.90
0.85
0.87
0.99 Flatfaced cylinder:
b/h 1
5
10
20
∞ h
b
h 1.18
1.2
1.3
1.5
2.0 Laminar Ellipsoid:
L /d 0.75
1
2
4
8 d  v v L  0.5
0.47
0.27
0.25
0.2 d
L Turbulent
0.2
0.2
0.13
0.1
0.08 eText Main Menu  Textbook Table of Contents  Study Guide ∞
1.20 7.6 Experimental External Flows 461 ber and the character of the boundary layer are important. Body length will generally decrease pressure drag by making the body relatively more slender, but sooner or later the
friction drag will catch up. For the flatfaced cylinder in Table 7.3, pressure drag decreases
with L/d but friction increases, so that minimum drag occurs at about L/d 2. Aerodynamic Forces on Road
Vehicles Automobiles and trucks are now the subject of much research on aerodynamic forces,
both lift and drag [21]. At least one textbook is devoted to the subject [22]. Consumer,
manufacturer, and government interest has cycled between high speed/high horsepower
and lower speed/lower drag. Better streamlining of car shapes has resulted over the
years in a large decrease in the automobile drag coefficient, as shown in Fig. 7.17a.
Modern cars have an average drag coefficient of about 0.35, based upon the frontal  v v Fig. 7.17 Aerodynamics of automobiles: (a) the historical trend for
drag coefficients [From Ref. 21];
(b) effect of bottom rear upsweep
angle on drag and downward lift
force [From Ref. 25].  eText Main Menu  Textbook Table of Contents  Study Guide 462 Chapter 7 Flow Past Immersed Bodies Horsepower required area. Since the frontal area has also decreased sharply, the actual raw drag force on
cars has dropped even more than indicated in Fig. 7.17a. The practical minimum, shown
tentatively for the year 2000, is CD 0.15 for a tearshaped vehicle, which can be
achieved any time the public is willing to purchase such a shape. Note that basing CD
on the frontal area is awkward, since one would need an accurate drawing of the automobile to estimate its frontal area. For this reason, some technical articles simply report the raw drag in newtons or poundforce, or the product CDA.
Many companies and laboratories have automotive wind tunnels, some fullscale
and/or with moving floors to approximate actual kinematic similarity. The blunt shapes
of most automobiles, together with their proximity to the ground, cause a wide variety of flow and geometric effects. Simple changes in part of the shape can have a large
influence on aerodynamic forces. Figure 7.17b shows force data by Bearman et al. [25]
for an idealized smooth automobile shape with upsweep in the rear of the bottom section. We see that by simply adding an upsweep angle of 25°, we can quadruple the
downward force, gaining tire traction at the expense of doubling the drag. For this
study, the effect of a moving floor was small — about a 10 percent increase in both drag
and lift compared to a fixed floor.
It is difficult to quantify the exact effect of geometric changes on automotive forces,
since, e.g., changes in a windshield shape might interact with downstream flow over
the roof and trunk. Nevertheless, based on correlation of many model and fullscale
tests, Ref. 26 proposes a formula for automobile drag which adds separate effects such
as front ends, cowls, fenders, windshield, roofs, and rear ends.
Figure 7.18 shows the horsepower required to drive a typical tractortrailer truck
at speeds up to 80 mi/h (117 ft/s or 36 m/s). The rolling resistance increases linearly
and the air drag quadratically with speed (CD 1.0). The two are about equally important at 55 mi/h, which is the nominal speed limit in the United States. As shown
in Fig. 7.18b, air drag can be reduced by attaching a deflector to the top of the tractor. If the angle of the deflector is adjusted to carry the flow smoothly over the top
and around the sides of the trailer, the reduction in CD is about 20 percent. Thus, at
55 mi/h the total resistance is reduced 10 percent, with a corresponding reduction in  v v Fig. 7.18 Drag reduction of a tractortrailer truck: (a) horsepower required to overcome resistance; (b)
deflector added to cab reduces air
drag by 20 percent. (Uniroyal Inc.) 550
500
450
400
350
300
250
200
150
100
50
0 Gross engine horsepower required Air
resistance Rolling
resistance, hp 0 10 20 30 40 50 60 70
Vehicle speed, mi/h 80 (a)  eText Main Menu  Textbook Table of Contents (b)  Study Guide 7.6 Experimental External Flows 463 fuel costs and/or trip time for the trucker. This type of applied fluids engineering can
be a large factor in many of the conservationoriented transportation problems of the
future. EXAMPLE 7.7
A highspeed car with m 2000 kg, CD 0.3, and A 1 m2 deploys a 2m parachute to slow
down from an initial velocity of 100 m/s (Fig. E7.7). Assuming constant CD, brakes free, and
no rolling resistance, calculate the distance and velocity of the car after 1, 10, 100, and 1000 s.
For air assume
1.2 kg/m3, and neglect interference between the wake of the car and the parachute. dp = 2 m V0 = 100 m/s x E7.7 Solution
Newton’s law applied in the direction of motion gives
m Fx dV
dt Fc 1
V 2(CDcAc
2 Fp CDp Ap) where subscript c denotes the car and subscript p the parachute. This is of the form
K2
V
m dV
dt K CD A 2 Separate the variables and integrate
V dV
V2 V0 or V0 1 K
m
V t dt
0 K
t
m 1 Rearrange and solve for the velocity V:
V V0
(K/m)V0t 1 (CDc Ac K CDp Ap)
2 (1) We can integrate this to find the distance traveled:
S V0 ln (1 K
V0
m t) 1.2; hence Now work out some numbers. From Table 7.3, CDp  v v CDc Ac  eText Main Menu  CDp Ap 0.3(1 m2) Textbook Table of Contents (2) 1.2  4 (2 m)2 4.07 m2 Study Guide 464 Chapter 7 Flow Past Immersed Bodies
1
2 K
V0
m Then (4.07 m2)(1.2 kg/m3)(100 m/s)
2000 kg 0.122 s 1 Now make a table of the results for V and S from Eqs. (1) and (2):
t, s 1 10 100 1000 V, m/s 89 45 7.6 0.8 S, m 94 654 2110 3940 Air resistance alone will not stop a body completely. If you don’t apply the brakes, you’ll be
halfway to the Yukon Territory and still going. Other Methods of Drag Reduction Sometimes drag is good, for example, when using a parachute. Do not jump out of an
airplane holding a flat plate parallel to your motion (see Prob. 7.81). Mostly, though,
drag is bad and should be reduced. The classical method of drag reduction is streamlining (Figs. 7.15 and 7.18). For example, nose fairings and body panels have produced
motorcycles which can travel over 200 mi/h. More recent research has uncovered other
methods which hold great promise, especially for turbulent flows.
1. Oil pipelines introduce an annular core of water to reduce the pumping power
[36]. The lowviscosity water rides the wall and reduces friction up to 60 percent.
2. Turbulent friction in liquid flows is reduced up to 60 percent by dissolving small
amounts of a highmolecularweight polymer additive [37]. Without changing
pumps, the TransAlaska Pipeline System (TAPS) increased oil flow 50 percent
by injecting small amounts of polymer dissolved in kerosene.
3. Streamoriented surface veegroove microriblets reduce turbulent friction up to 8
percent [38]. Riblet heights are of order 1 mm and were used on the Stars and
Stripes yacht hull in the Americas Cup races. Riblets are also effective on aircraft skins.
4. Small, nearwall largeeddy breakup devices (LEBUs) reduce local turbulent
friction up to 10 percent [39]. However, one must add these small structures to
the surface.
5. Air microbubbles injected at the wall of a water flow create a lowshear bubble
blanket [40]. At high void fractions, drag reduction can be 80 percent.
6. Spanwise (transverse) wall oscillation may reduce turbulent friction up to 30
percent [41].
Drag reduction is presently an area of intense and fruitful research and applies to many
types of airflows and water flows for both vehicles and conduits. The drag data above, such as Tables 7.2 and 7.3, are for bodies “fully immersed” in a
free stream, i.e., with no free surface. If, however, the body moves at or near a free liquid surface, wavemaking drag becomes important and is dependent upon both the
Reynolds number and the Froude number. To move through a water surface, a ship  v v Drag of Surface Ships  eText Main Menu  Textbook Table of Contents  Study Guide 7.6 Experimental External Flows 465 must create waves on both sides. This implies putting energy into the water surface
and requires a finite drag force to keep the ship moving, even in a frictionless fluid.
The total drag of a ship can then be approximated as the sum of friction drag and wavemaking drag:
F Ffric Fwave or CD CD,fric CD,wave The friction drag can be estimated by the (turbulent) flatplate formula, Eq. (7.45),
based on the belowwater or wetted area of the ship.
Reference 27 is an interesting review of both theory and experiment for wakemaking surface ship drag. Generally speaking, the bow of the ship creates a wave system whose wavelength is related to the ship speed but not necessarily to the ship length.
If the stern of the ship is a wave trough, the ship is essentially climbing uphill and has
high wave drag. If the stern is a wave crest, the ship is nearly level and has lower drag.
The criterion for these two conditions results in certain approximate Froude numbers
[27]:
Fr V
gL 0.53
N high drag if N
low drag if N 1, 3, 5, 7, . . . ; (7.65) 2, 4, 6, 8, . . . where V is the ship’s speed, L is the ship’s length along the centerline, and N is the
number of halflengths, from bow to stern, of the dragmaking wave system. The wave
drag will increase with the Froude number and oscillate between lower drag (Fr
0.38, 0.27, 0.22, . . .) and higher drag (Fr 0.53, 0.31, 0.24, . . .) with negligible variation for Fr 0.2. Thus it is best to design a ship to cruise at N 2, 4, 6, 8. Shaping
the bow and stern can further reduce wavemaking drag.
Figure 7.19 shows the data of Inui [27] for a model ship. The main hull, curve A,
shows peaks and valleys in wave drag at the appropriate Froude numbers 0.2. Introduction of a bulb protrusion on the bow, curve B, greatly reduces the drag. Adding
a second bulb to the stern, curve C, is still better, and Inui recommends that the design
speed of this twobulb ship be at N 4, Fr 0.27, which is a nearly “waveless” condition. In this figure CD,wave is defined as 2Fwave/( V 2L2) instead of using the wetted
area.
The solid curves in Fig. 7.19 are based on potentialflow theory for the belowwater hull shape. Chapter 8 is an introduction to potentialflow theory. Modern digital
computers can be programmed for numerical CFD solutions of potential flow over the
hulls of ships, submarines, yachts, and sailboats, including boundarylayer effects
driven by the potential flow [28]. Thus theoretical prediction of flow past surface ships
is now at a fairly high level. See also Ref. 15.  v v Body Drag at High Mach
Numbers  All the data presented above are for nearly incompressible flows, with Mach numbers
assumed less than about 0.5. Beyond this value compressibility can be very important,
with CD fcn(Re, Ma). As the stream Mach number increases, at some subsonic value
Mcrit 1 which depends upon the body’s bluntness and thickness, the local velocity at
some point near the body surface will become sonic. If Ma increases beyond Macrit,
shock waves form, intensify, and spread, raising surface pressures near the front of the
body and therefore increasing the pressure drag. The effect can be dramatic with CD eText Main Menu  Textbook Table of Contents  Study Guide 466 Chapter 7 Flow Past Immersed Bodies A Main hull (without bulb)
B With bowbulb
C With bow and sternbulbs
0.002
Potentialflow theory
CD, wave 0.001
A
B
C 0
0.10 Fig. 7.19 Wavemaking drag on a
ship model. (After Inui [27].) Note:
The drag coefficient is defined as
CDW 2F/( V2L2). 0.20 0.30 0.40 Design
speed 0.50 0.60 V
Fr =
√Lg increasing tenfold, and 70 years ago this sharp increase was called the sonic barrier,
implying that it could not be surmounted. Of course, it can be — the rise in CD is finite, as supersonic bullets have proved for centuries.
Figure 7.20 shows the effect of the Mach number on the drag coefficient of various
body shapes tested in air.5 We see that compressibility affects blunt bodies earlier, with
Macrit equal to 0.4 for cylinders, 0.6 for spheres, and 0.7 for airfoils and pointed projectiles. Also the Reynolds number (laminar versus turbulent boundarylayer flow) has
a large effect below Macrit for spheres and cylinders but becomes unimportant above
Ma 1. In contrast, the effect of the Reynolds number is small for airfoils and projectiles and is not shown in Fig. 7.20. A general statement might divide Reynolds and
Machnumber effects as follows:
Ma 0.4: Reynolds number important, Mach number unimportant
0.4 Ma 1: both Reynolds and Mach numbers important
Ma 1.0: Reynolds number unimportant, Mach number important
At supersonic speeds, a broad bow shock wave forms in front of the body (see Figs.
9.10b and 9.19), and the drag is mainly due to high shockinduced pressures on the
front. Making the bow a sharp point can sharply reduce the drag (Fig. 9.28) but does
not eliminate the bow shock. Chapter 9 gives a brief treatment of compressibility.
References 30 and 31 are more advanced textbooks devoted entirely to compressible
flow.
5  v v There is a slight effect of the specificheat ratio k which would appear if other gases were tested.  eText Main Menu  Textbook Table of Contents  Study Guide 7.6 Experimental External Flows 467 Biological Drag Reduction A great deal of engineering effort goes into designing immersed bodies to reduce their
drag. Most such effort concentrates on rigidbody shapes. A different process occurs
in nature, as organisms adapt to survive high winds or currents, as reported in a series
of papers by S. Vogel [33, 34]. A good example is a tree, whose flexible structure allows it to reconfigure in high winds and thus reduce drag and damage. Tree root systems have evolved in several ways to resist windinduced bending moments, and trunk
cross sections have become resistant to bending but relatively easy to twist and reconfigure. We saw this in Table 7.3, where tree drag coefficients [24] reduced by 60 percent as wind velocity increased. The shape of the tree changes to offer less resistance.
The individual branches and leaves of a tree also curl and cluster to reduce drag.
Figure 7.21 shows the results of wind tunnel experiments by Vogel [33]. A tulip tree
leaf, Fig. 7.21(a), broad and open in low wind, curls into a conical lowdrag shape as
wind increases. A compound black walnut leaf group, Fig. 7.21(b), clusters into a lowdrag shape at high wind speed. Although drag coefficients were reduced up to 50 percent by flexibility, Vogel points out that rigid structures are sometimes just as effective. An interesting recent symposium [35] was devoted entirely to the solid mechanics
and fluid mechanics of biological organisms. Forces on Lifting Bodies Lifting bodies (airfoils, hydrofoils, or vanes) are intended to provide a large force normal to the free stream and as little drag as possible. Conventional design practice has
evolved a shape not unlike a bird’s wing, i.e., relatively thin (t/c 0/18) with a rounded
leading edge and a sharp trailing edge. A typical shape is sketched in Fig. 7.22.
For our purposes we consider the body to be symmetric, as in Fig. 7.11, with the 2.0
1.8
Cylinder in cross flow:
1.6 Laminar, Re ≈ 1 E5
Turbulent, Re ≈ 1 E6 1.4
1.2
CD 1.0 Sphere
Laminar, Re ≈ 1 E5
Turbulent, Re ≈ 1 E6 0.8
0.6 Pointed body
of revolution 0.4  v v Fig. 7.20 Effect of the Mach number on the drag of various body
shapes. (Data from Refs. 23 and
29.)  Airfoil 0.2
0.0
0.0 eText Main Menu 1.0  2.0
Mach number 3.0 Textbook Table of Contents 4.0  Study Guide 468 Chapter 7 Flow Past Immersed Bodies 5 m/s
5 m/s 10 m/s 10 m/s 20 m/s Fig. 7.21 Biological adaptation to
wind forces: (a) a tulip tree leaf
curls into a conical shape at high
velocity; (b) black walnut leaves
cluster into a lowdrag shape as
wind increases. (From Vogel, Ref.
33.) 20 m/s (a) (b) freestream velocity in the vertical plane. If the chord line between the leading and
trailing edge is not a line of symmetry, the airfoil is said to be cambered. The camber
line is the line midway between the upper and lower surfaces of the vane.
The angle between the free stream and the chord line is called the angle of attack
. The lift L and the drag D vary with this angle. The dimensionless forces are defined
with respect to the planform area Ap bc:
Lift coefficient: CL Drag coefficient: CD L
1
2 (7.66a) V2Ap D
1
2 (7.66b) V2Ap Planform
area = bc
Lift
Drag
Angle of
attack t = thickness α b = span V
c = cho  v v Fig. 7.22 Definition sketch for a
lifting vane. rd  eText Main Menu  Textbook Table of Contents  Study Guide 7.6 Experimental External Flows 469 If the chord length is not constant, as in the tapered wings of modern aircraft,
Ap
c db.
For lowspeed flow with a given roughness ratio, CL and CD should vary with and
the chord Reynolds number
CL f ( , Rec) or CD f ( , Rec) where Rec Vc/ . The Reynolds numbers are commonly in the turbulentboundarylayer range and have a modest effect.
The rounded leading edge prevents flow separation there, but the sharp trailing edge
causes a separation which generates the lift. Figure 7.23 shows what happens when a
flow starts up past a lifting vane or an airfoil.
Just after startup in Fig. 7.23a the streamline motion is irrotational and inviscid.
The rear stagnation point, assuming a positive angle of attack, is on the upper surface,
and there is no lift; but the flow cannot long negotiate the sharp turn at the trailing
edge: it separates, and a starting vortex forms in Fig. 7.23b. This starting vortex is shed
downstream in Fig. 7.23c and d, and a smooth streamline flow develops over the wing,
leaving the foil in a direction approximately parallel to the chord line. Lift at this time
is fully developed, and the starting vortex is gone. Should the flow now cease, a stopping vortex of opposite (clockwise) sense will form and be shed. During flight, increases or decreases in lift will cause incremental starting or stopping vortices, always
with the effect of maintaining a smooth parallel flow at the trailing edge. We pursue
this idea mathematically in Chap. 8.
At a low angle of attack, the rear surfaces have an adverse pressure gradient but not
enough to cause significant boundarylayer separation. The flow pattern is smooth, as
in Fig. 7.23d, and drag is small and lift excellent. As the angle of attack is increased,
the uppersurface adverse gradient becomes stronger, and generally a separation bub (a)  v v Fig. 7.23 Transient stages in the
development of lift: (a) startup:
rear stagnation point on the upper
surface: no lift; (b) sharp trailing
edge induces separation, and a
starting vortex forms: slight lift; (c)
starting vortex is shed, and streamlines flow smoothly from trailing
edge: lift is now 80 percent developed; (d) starting vortex now shed
far behind, trailing edge now very
smooth: lift fully developed.  eText Main Menu (b) (c) (d )  Textbook Table of Contents  Study Guide 470 Chapter 7 Flow Past Immersed Bodies Fig. 7.24 At high angle of attack,
smokeflow visualization shows
stalled flow on the upper surface of
a lifting vane. [From Ref. 19, Illustrated Experiments in Fluid Mechanics (The NCFMF Book of Film
Notes), National Committee for
Fluid Mechanics Films, Education
Development Center, Inc., copyright 1972.] ble begins to creep forward on the upper surface.6 At a certain angle
15 to 20°,
the flow is separated completely from the upper surface, as in Fig. 7.24. The airfoil is
said to be stalled: Lift drops off markedly, drag increases markedly, and the foil is no
longer flyable.
Early airfoils were thin, modeled after birds’ wings. The German engineer Otto
Lilienthal (1848 – 1896) experimented with flat and cambered plates on a rotating arm.
He and his brother Gustav flew the world’s first glider in 1891. Horatio Frederick
Phillips (1845 – 1912) built the first wind tunnel in 1884 and measured the lift and drag
of cambered vanes. The first theory of lift was proposed by Frederick W. Lanchester
shortly afterward. Modern airfoil theory dates from 1905, when the Russian hydrodynamicist N. E. Joukowsky (1847 – 1921) developed a circulation theorem (Chap. 8) for
computing airfoil lift for arbitrary camber and thickness. With this basic theory, as extended and developed by Prandtl and Kármán and their students, it is now possible to
design a lowspeed airfoil to satisfy particular surfacepressure distributions and boundarylayer characteristics. There are whole families of airfoil designs, notably those developed in the United States under the sponsorship of the NACA (now NASA). Extensive theory and data on these airfoils are contained in Ref. 16. We shall discuss this
further in Chap. 8.
Figure 7.25 shows the lift and drag on a symmetric airfoil denoted as the NACA
0009 foil, the last digit indicating the thickness of 9 percent. With no flap extended,
6  v v For some airfoils the bubble leaps, not creeps, forward, and stall occurs rapidly and dangerously.  eText Main Menu  Textbook Table of Contents  Study Guide 7.6 Experimental External Flows 471 this airfoil, as expected, has zero lift at zero angle of attack. Up to about 12° the lift
coefficient increases linearly with a slope of 0.1 per degree, or 6.0 per radian. This is
in agreement with the theory outlined in Chap. 8:
CL,theory 2h
c 2 sin (7.67) where h/c is the maximum camber expressed as a fraction of the chord. The NACA
0009 has zero camber; hence CL 2 sin
0.11 , where is in degrees. This is
excellent agreement.
The drag coefficient of the smoothmodel airfoils in Fig. 7.25 is as low as 0.005,
which is actually lower than both sides of a flat plate in turbulent flow. This is misleading inasmuch as a commercial foil will have roughness effects; e.g., a paint job
will double the drag coefficient.
The effect of increasing Reynolds number in Fig. 7.25 is to increase the maximum
lift and stall angle (without changing the slope appreciably) and to reduce the drag coefficient. This is a salutary effect, since the prototype will probably be at a higher
Reynolds number than the model (107 or more).
For takeoff and landing, the lift is greatly increased by deflecting a split flap, as
shown in Fig. 7.25. This makes the airfoil unsymmetric (or effectively cambered) and
changes the zerolift point to
12°. The drag is also greatly increased by the flap,
but the reduction in takeoff and landing distance is worth the extra power needed.
A lifting craft cruises at low angle of attack, where the lift is much larger than the
drag. Maximum lifttodrag ratios for the common airfoils lie between 20 and 50.
Some airfoils, such as the NACA 6 series, are shaped to provide favorable gradients over much of the upper surface at low angles. Thus separation is small, and transition to turbulence is delayed; the airfoil retains a good length of laminar flow even
CL
Rec = 6 × 10 6
1.6 0.04 α CD Split
flap
1.2 With flap
at 60° 0.03
Rec = 9 × 10 6 With flap
at 60° 0.02 6 × 10 6
3 × 10 6 0.8 Rec = 3 × 10 6
6 × 10 6
9 × 10 6 No flap
0.4  v v Fig. 7.25 Lift and drag of a symmetric NACA 0009 airfoil of infinite span, including effect of a
splitflap deflection. Note that
roughness can increase CD from
100 to 300 percent.  0.01 No flap
–12 –8 eText Main Menu –4 0 4 8 12 16 –8 α , deg  Textbook Table of Contents –4 0 4 α , deg  Study Guide 8 12 16 472 Chapter 7 Flow Past Immersed Bodies Stall
1.2 NACA
0009 Stall
0009 with
split flap NACA
63 – 009
0.8
CL
Lowdrag
“bucket” 0.4 0 Fig. 7.26 Liftdrag polar plot for
standard (0009) and a laminarflow
(63009) NACA airfoil. 0 0.008 0.016 0.024 CD at high Reynolds numbers. The liftdrag polar plot in Fig. 7.26 shows the NACA 0009
data from Fig. 7.25 and a laminarflow airfoil, NACA 63 – 009, of the same thickness.
The laminarflow airfoil has a lowdrag bucket at small angles but also suffers lower
stall angle and lower maximum lift coefficient. The drag is 30 percent less in the bucket,
but the bucket disappears if there is significant surface roughness.
All the data in Figs. 7.25 and 7.26 are for infinite span, i.e., a twodimensional flow
pattern about wings without tips. The effect of finite span can be correlated with the
dimensionless slenderness, or aspect ratio, denoted (AR),
b2
Ap AR b
c (7.68) where c is the average chord length. Finitespan effects are shown in Fig. 7.27. The
lift slope decreases, but the zerolift angle is the same; and the drag increases, but the
zerolift drag is the same. The theory of finitespan airfoils [16] predicts that the effective angle of attack increases, as in Fig. 7.27, by the amount
CL
AR (7.69) When applied to Eq. (7.67), the finitespan lift becomes
CL
The associated drag increase is CD  v v CD  eText Main Menu  2 sin (
2h/c)
1 2/AR
CL sin
CD (7.70) CL , or 2
CL (7.71) AR Textbook Table of Contents  Study Guide 7.6 Experimental External Flows 473 where CD is the drag of the infinitespan airfoil, as sketched in Fig. 7.25. These correlations are in good agreement with experiments on finitespan wings [16].
The existence of a maximum lift coefficient implies the existence of a minimum
speed, or stall speed, for a craft whose lift supports its weight
L
or W 2
CL,max( 1 V s Ap)
2
1/2 2W Vs (7.72) CL,max Ap The stall speed of typical aircraft varies between 60 and 200 ft/s, depending upon the
weight and value of CL,max. The pilot must hold the speed greater than about 1.2Vs to
avoid the instability associated with complete stall.
The split flap in Fig. 7.25 is only one of many devices used to secure high lift at
low speeds. Figure 7.28a shows six such devices whose lift performance is given in
7.28b along with a standard (A) and laminarflow (B) airfoil. The doubleslotted flap
achieves CL,max 3.4, and a combination of this plus a leadingedge slat can achieve
CL,max 4.0. These are not scientific curiosities; e.g., the Boeing 727 commercial jet
aircraft uses a tripleslotted flap plus a leadingedge slat during landing.
Also shown as C in Fig. 7.28b is the KlineFogleman airfoil [17], not yet a reality.
The designers are amateur modelplane enthusiasts who did not know that conventional
aerodynamic wisdom forbids a sharp leading edge and a step cutout from the trailing
edge. The KlineFogleman airfoil has relatively high drag but shows an amazing continual increase in lift out to
45°. In fact, we may fairly say that this airfoil does
not stall and provides smooth performance over a tremendous range of flight conditions. No explanation for this behavior has yet been published by any aerodynamicist.
This airfoil is under study and may or may not have any commercial value.
Another violation of conventional aerodynamic wisdom is that military aircraft are
beginning to fly, briefly, above the stall point. Fighter pilots are learning to make quick
maneuvers in the stalled region as detailed in Ref. 32. Some planes can even fly continuously while stalled — the Grumman X29 experimental aircraft recently set a record
67°.
by flying at AR = ∞ CL AR = b2
Ap CD
AR =
∆ CD ≈ 2
CL
π AR b2
Ap AR = ∞
CL
∆α ≈
π AR  v v Fig. 7.27 Effect of finite aspect ratio on lift and drag of an airfoil: (a)
effective angle increase; (b) induced drag increase.  eText Main Menu CD ∞ α –β
(a)  Textbook Table of Contents (b)  Study Guide α 474 Chapter 7 Flow Past Immersed Bodies 4
D Optimum but
cumbersome
combination CL Plain flap or aileron H
3 E G Split flap F
KlineFogleman
airfoil E, D F
External airfoil flap 2 I G
Slotted flap A
1 H Fig. 7.28 Performance of airfoils
with and without highlift devices:
A NACA 0009; B NACA 63009; C KlineFogleman airfoil
(from Ref. 17); D to I shown in (a):
(a) types of highlift devices; (b)
lift coefficients for various devices. C B Doubleslotted flap
α, deg I
Leading edge slat
– 10 0 10 New Aircraft Designs 20 30 40 50° (b) (a) The KlineFogleman airfoil in Fig. 7.28 is a departure from conventional aerodynamics,
but there have been other striking departures, as detailed in a recent article [42]. These
new aircraft, conceived presently as small models, have a variety of configurations, as
shown in Fig. 7.29: ringwing, cruciform, flying saucer, and flapwing. A saucer configuration (Fig. 7.29c), with a diameter of 40 in, has been successfully flown by radio control, and its inventor, Jack M. Jones, plans for a 20ft twopassenger version. Another 18inspan microplane called the Bat (not shown), made by MLB Co., flies for 20 min at
40 mi/h and contains a video camera for surveillance. New engines have been reduced
to a 10 by 3mm size, producing 20 W of power. At the other end of the size spectrum,
Boeing and NASA engineers have proposed a jumbo flyingwing jetliner, similar in shape
to the stealth bomber, which would carry 800 passengers for a range of 7000 mi.
Further information on the performance of lifting craft can be found in Refs. 12,
13, and 16. We discuss this matter again briefly in Chap. 8. (a) Ringwing  v v Fig. 7.29 New aircraft designs do
not necessarily look like your typical jetliner. (From Ref. 42.) (b) Cruciform delta (c) Flying saucer  eText Main Menu (d) Flapwing dragonfly  Textbook Table of Contents  Study Guide 7.6 Experimental External Flows 475 EXAMPLE 7.8
An aircraft weighs 75,000 lb, has a planform area of 2500 ft2, and can deliver a constant
thrust of 12,000 lb. It has an aspect ratio of 7, and CD
0.02. Neglecting rolling resistance,
estimate the takeoff distance at sea level if takeoff speed equals 1.2 times stall speed. Take
CL,max 2.0. Solution
0.00237 slug/ft3, is The stall speed from Eq. (7.72), with sealevel density 2W Vs 1/2 2(75,000)
2.0(0.00237)(2500) CL,max Ap
1.2Vs Hence takeoff speed V0 1/2 112.5 ft/s 135 ft/s. The drag is estimated from Eq. (7.71) for AR CD 2
CL
7 0.02 0.02 7 as 2
0.0455C L A force balance in the direction of takeoff gives
Fs m dV
dt thrust drag kV2 T k Since we are looking for distance, not time, we introduce dV/dt
rate variables, and integrate
S0
0 or S0 V0 m
2 dS 0 d(V2)
T kV2 T
m
ln
2
T kV 0
2k k 1
2 CD Ap (1) V dV/ds into Eq. (1), sepa const T
m
ln
T D0
2k (2) where D0 kV 2 is the takeoff drag. Equation (2) is the desired theoretical relation for takeoff
0
distance. For the particular numerical values, take
75,000
32.2 m
CL0
CD0
k 1
2 1
2 W
V 2 Ap
0 75,000
(0.00237)(135)2(2500) 0.0455(CL0)2 0.02 CD0 Ap 1
2 2329 slugs 0.108 1
2 ( )(0.108)(0.00237)(2500)
D0 kV 2
0 1.39 0.319 slug/ft 5820 lb Then Eq. (2) predicts that
S0 2329 slugs
12,000
ln
2(0.319 slug/ft)
12,000 5820 3650 ln 1.94 2420 ft Ans.  v v A more exact analysis accounting for variable k [13] gives the same result to within 1 percent.  eText Main Menu  Textbook Table of Contents  Study Guide 476 Chapter 7 Flow Past Immersed Bodies Summary This chapter has dealt with viscous effects in external flow past bodies immersed in a
stream. When the Reynolds number is large, viscous forces are confined to a thin boundary layer and wake in the vicinity of the body. Flow outside these “shear layers” is essentially inviscid and can be predicted by potential theory and Bernoulli’s equation.
The chapter begins with a discussion of the flatplate boundary layer and the use of
momentumintegral estimates to predict the wall shear, friction drag, and thickness of
such layers. These approximations suggest how to eliminate certain small terms in the
NavierStokes equations, resulting in Prandtl’s boundarylayer equations for laminar
and turbulent flow. Section 7.4 then solves the boundarylayer equations to give very
accurate formulas for flatplate flow at high Reynolds numbers. Roughwall effects are
included, and Sec. 7.5 gives a brief introduction to pressuregradient effects. An adverse (decelerating) gradient is seen to cause flow separation, where the boundary layer
breaks away from the surface and forms a broad, lowpressure wake.
Boundarylayer theory fails in separated flows, which are commonly studied by
experiment. Section 7.6 gives data on drag coefficients of various two and threedimensional body shapes. The chapter ends with a brief discussion of lift forces generated by lifting bodies such as airfoils and hydrofoils. Airfoils also suffer flow separation or stall at high angles of incidence. Problems cm, a chord length of 1.5 m, and a wingspan of 12 m. What
is the appropriate value of the Reynolds number for correlating the lift and drag of this wing? Explain your selection.
Equation (7.1b) assumes that the boundary layer on the
plate is turbulent from the leading edge onward. Devise
a scheme for determining the boundarylayer thickness
more accurately when the flow is laminar up to a point
Rex,crit and turbulent thereafter. Apply this scheme to computation of the boundarylayer thickness at x 1.5 m in
40 m/s flow of air at 20°C and 1 atm past a flat plate.
Compare your result with Eq. (7.1b). Assume Rex,crit
1.2 E6.
Air at 20°C and 1 atm flows at 15 m/s past a flat plate
with Rex,crit 1 E6. At what point x will the boundarylayer thickness be 8 mm? Why do Eqs. (7.1) seem to fail?
Make a sketch illustrating the discrepancy; then use the
ideas in Prob. 7.3 to complete this problem correctly.
SAE 30 oil at 20°C flows at 1.8 ft3/s from a reservoir into
a 6indiameter pipe. Use flatplate theory to estimate the
position x where the pipewall boundary layers meet in the
center. Compare with Eq. (6.5), and give some explanations for the discrepancy.
For the laminar parabolic boundarylayer profile of Eq.
(7.6), compute the shape factor H and compare with the
exact Blasius result, Eq. (7.31).
Air at 20°C and 1 atm enters a 40cmsquare duct as
in Fig. P7.7. Using the “displacement thickness” concept of Fig. 7.4, estimate (a) the mean velocity and (b) the
mean pressure in the core of the flow at the position x Most of the problems herein are fairly straightforward. More difficult or openended assignments are labeled with an asterisk. Problems labeled with an EES icon will benefit from the use of the Engineering Equation Solver (EES), while problems labeled with a
computer disk may require the use of a computer. The standard
endofchapter problems 7.1 to 7.124 (categorized in the problem
list below) are followed by word problems W7.1 to W7.12, fundamentals of engineering exam problems FE7.1 to FE7.10, comprehensive problems C7.1 to C7.4, and design project D7.1. P7.3 Problem Distribution
Section
7.1
7.2
7.3
7.4
7.4
7.5
7.6
7.6
7.6 Topic Problems P7.4 Reynoldsnumber and geometry
Momentumintegral estimates
The boundarylayer equations
Laminar flatplate flow
Turbulent flatplate flow
Boundary layers with pressure gradient
Drag of twodimensional bodies
Drag of threedimensional bodies
Lifting bodies — airfoils 7.1 – 7.5
7.6 – 7.12
7.13 – 7.15
7.16 – 7.29
7.30 – 7.46
7.47 – 7.52
7.53 – 7.63
7.64 – 7.114
7.115 – 7.124 P7.5 P7.6 P7.2 For flow at 20 m/s past a thin flat plate, estimate the distances x from the leading edge at which the boundarylayer
thickness will be either 1 mm or 10 cm for (a) air and (b)
water at 20°C and 1 atm.
Air, equivalent to that at a standard altitude of 4000 m,
flows at 450 mi/h past a wing which has a thickness of 18  v v P7.1  eText Main Menu  P7.7 Textbook Table of Contents  Study Guide Problems 477
3 m. (c) What is the average gradient, in Pa/m, in this
section?
40 × 40 cm square duct
Boundary layers
2 m /s
Ucore 3m for the parabolic flatplate profile of Eq. (7.3). Yet when
this new profile is used in the integral analysis of Sec. 7.3,
we get the lousy result /x 9.2/Re1/2, which is 80 perx
cent high. What is the reason for the inaccuracy? [Hint:
The answer lies in evaluating the laminar boundarylayer
momentum equation (7.19b) at the wall, y 0.]
P7.13 Derive modified forms of the laminar boundarylayer
equations (7.19) for the case of axisymmetric flow along
the outside of a circular cylinder of constant radius
R, as in Fig. P7.13. Consider the two special cases (a)
R and (b)
R. What are the proper boundary
conditions? P7.7
y P7.8 1.2 kg/m3 and
1.8 E5 kg/(m s), flows at
Air,
10 m/s past a flat plate. At the trailing edge of the plate,
the following velocity profile data are measured: y, mm 0 0.5 1.0 2.0 3.0 4.0 5.0 0 1.75 3.47 6.58 8.70 9.68 10.0 δ (x)
u 6.0 u, m/s U r p ≈ constant 10.0
R P7.9 If the upper surface has an area of 0.6 m2, estimate, using
momentum concepts, the friction drag, in N, on the upper
surface.
Repeat the flatplate momentum analysis of Sec. 7.2 by
replacing the parabolic profile, Eq. (7.6), with a more accurate sinusoidal profile:
u
U P7.13
P7.14 Show that the twodimensional laminarflow pattern with
dp/dx 0 y sin x 2 u
Compute momentumintegral estimates of cf, /x, */x,
and H.
P7.10 Repeat Prob. 7.9, using the polynomial profile suggested
by K. Pohlhausen in 1921:
u
U 2 y 2 y3 y4 3 4 Does this profile satisfy the boundary conditions of laminar flatplate flow?
P7.11 Find the correct form for a cubic velocityprofile polynomial
u A By Cy2 Dy3  v v to replace Eq. (7.6) in a flatplate momentum analysis.
Find the value of / for this profile, but do not pursue the
analysis further.
P7.12 The velocity profile shape u/U 1 exp ( 4.605y/ ) is
a smooth curve with u 0 at y 0 and u 0.99U at
y
and thus would seem to be a reasonable substitute  eText Main Menu  U0(1 e Cy) 0 0 is an exact solution to the boundarylayer equations (7.19).
Find the value of the constant C in terms of the flow parameters. Are the boundary conditions satisfied? What
might this flow represent?
P7.15 Discuss whether fully developed laminar incompressible
flow between parallel plates, Eq. (4.143) and Fig. 4.16b,
represents an exact solution to the boundarylayer
equations (7.19) and the boundary conditions (7.20). In
what sense, if any, are duct flows also boundarylayer
flows?
P7.16 A thin flat plate 55 by 110 cm is immersed in a 6m/s
stream of SAE 10 oil at 20°C. Compute the total friction
drag if the stream is parallel to (a) the long side and (b)
the short side.
P7.17 Helium at 20°C and low pressure flows past a thin flat
plate 1 m long and 2 m wide. It is desired that the
total friction drag of the plate be 0.5 N. What is the
appropriate absolute pressure of the helium if U
35 m/s? Textbook Table of Contents  Study Guide 478 Chapter 7 Flow Past Immersed Bodies P7.18 In Prob. 7.7, when the duct is perfectly square, the core
velocity speeds up. Suppose we wish to hold the core velocity constant by slanting the upper and lower walls
while keeping the front and rear walls parallel. What angle of slant will be the best approximation? For this condition, what will be the total friction drag on the duct
walls?
P7.19 Program a method of numerical solution of the Blasius
flatplate relation, Eq. (7.22), subject to the conditions
in (7.23). You will find that you cannot get started without knowing the initial second derivative f (0), which
lies between 0.2 and 0.5. Devise an iteration scheme
which starts at f (0) 0.2 and converges to the correct
value. Print out u/U f ( ) and compare with Table
7.1.
P7.20 Air at 20°C and 1 atm flows at 20 m/s past the flat plate
in Fig. P7.20. A pitot stagnation tube, placed 2 mm from
the wall, develops a manometer head h 16 mm of
Meriam red oil, SG 0.827. Use this information to estimate the downstream position x of the pitot tube. Assume
laminar flow. Boundary layer
20 m/s
2 mm so, determine the correct dimensionless form for , assuming that
0 at the wall, y 0.
P7.23 Suppose you buy a 4 by 8ft sheet of plywood and put
it on your roof rack. (See Fig. P7.23.) You drive home
at 35 mi/h. (a) Assuming the board is perfectly aligned
with the airflow, how thick is the boundary layer at the
end of the board? (b) Estimate the drag on the sheet of
plywood if the boundary layer remains laminar. (c) Estimate the drag on the sheet of plywood if the boundarylayer is turbulent (assume the wood is smooth), and
compare the result to that of the laminar boundarylayer
case. P7.23
*P7.24 Air at 20°C and 1 atm flows past the flat plate in Fig. P7.24
under laminar conditions. There are two equally spaced
pitot stagnation tubes, each placed 2 mm from the wall.
The manometer fluid is water at 20°C. If U 15 m/s and
L 50 cm, determine the values of the manometer readings h1 and h2, in mm. Boundary layer
x h
U P7.20
L 2 mm L P7.21 For the experimental setup of Fig. P7.20, suppose the
stream velocity is unknown and the pitot stagnation tube
is traversed across the boundary layer of air at 1 atm and
20°C. The manometer fluid is Meriam red oil, and the following readings are made: 2 mm h1 h2 P7.24
y, mm 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 h, mm 1.2 4.6 9.8 15.8 21.2 25.3 27.8 29.0 29.7 29.7  v v Using these data only (not the Blasius theory) estimate
(a) the stream velocity, (b) the boundarylayer thickness,
(c) the wall shear stress, and (d) the total friction drag
between the leading edge and the position of the pitot
tube.
P7.22 For the Blasius flatplate problem, Eqs. (7.21) to (7.23),
does a twodimensional stream function (x, y) exist? If  eText Main Menu  P7.25 Modify Prob. 7.24 to the following somewhat more difficult scenario. Let the known data be U 15 m/s and h1
8 mm of water. Use this information to determine (a) L,
in cm, (b) h2, in mm.
P7.26 Consider laminar boundarylayer flow past the squareplate arrangements in Fig. P7.26. Compared to the friction
drag of a single plate 1, how much larger is the drag of
four plates together as in configurations (a) and (b)? Explain your results. Textbook Table of Contents  Study Guide Problems 479 1 3 2 4 1 P7.26 1 2 3 (a) 4 (b) *P7.27 A thin smooth disk of diameter D is immersed parallel to
a uniform stream of velocity U. Assuming laminar flow
and using flatplate theory as a guide, develop an approximate formula for the drag of the disk.
P7.28 Flow straighteners are arrays of narrow ducts placed in
wind tunnels to remove swirl and other inplane secondary
velocities. They can be idealized as square boxes constructed by vertical and horizontal plates, as in Fig. P7.28.
The cross section is a by a, and the box length is L. Assuming laminar flatplate flow and an array of N N
boxes, derive a formula for (a) the total drag on the bundle of boxes and (b) the effective pressure drop across the
bundle. Using strip theory, derive a formula for the drag coefficient of this plate. Compare this result with the drag of the
same plate immersed in a uniform stream U0.
y
U
L
y =δ δ u( y) x
a P7.32 U0
a P7.33 An alternate analysis of turbulent flatplate flow was given
by Prandtl in 1927, using a wall shearstress formula from
pipe flow L P7.28 w 0.0225 U2 1/4 U
Let the flow straighteners in Fig. P7.28 form an array
of 20 20 boxes of size a 4 cm and L 25 cm. If
Show that this formula can be combined with Eqs. (7.33)
the approach velocity is U0 12 m/s and the fluid is
and (7.40) to derive the following relations for turbulent
sealevel standard air, estimate (a) the total array drag
flatplate flow:
and (b) the pressure drop across the array. Compare with
0.072
0.0577
Sec. 6.6.
0.37
cf
CD
P7.30 Repeat Prob. 7.16 if the fluid is water at 20°C and the plate
Re1/5
Re1/5
Re1/5
x
x
L
x
surface is smooth.
These formulas are limited to Rex between 5 105 and
P7.31 Repeat Prob. 7.16 if the fluid is water at 20°C and the plate
107.
has an average roughness height of 1.5 mm.
P7.32 A flat plate of length L and height is placed at a wall *P7.34 A thin equilateraltriangle plate is immersed parallel to a
12 m/s stream of water at 20°C, as in Fig. P7.34. Assumand is parallel to an approaching boundary layer, as in Fig.
ing Retr 5 105, estimate the drag of this plate.
P7.32. Assume that the flow over the plate is fully turbuP7.35 The solutions to Prob. 7.26 are (a) F 2.83F1plate and (b)
lent and that the approaching flow is a oneseventhpower
F 2.0F1plate. Do not reveal these results to your friends.
law
Repeat Prob. 7.26 assuming the boundarylayer flow is tury 1/7
bulent, and comment on the striking increase in numerical
u(y) U0
values.  v v P7.29  eText Main Menu  Textbook Table of Contents  Study Guide P7.36
EES P7.37 P7.38 P7.39 P7.40 P7.41 Chapter 7 Flow Past Immersed Bodies
wall turbulentflow theory to estimate the position x of the
probe, in m.
2m
*P7.42 A fourbladed helicopter rotor rotates at n r/min in air with
properties ( , ). Each blade has chord length C and extends from the center of rotation out to radius R (the hub
2m
size is neglected). Assuming turbulent flow from the leading edge, develop an analytical estimate for the power P
required to drive this rotor.
2m
P7.43 In the flow of air at 20°C and 1 atm past a flat plate in Fig.
P7.43, the wall shear is to be determined at position x by a
12 m/s
floating element (a small area connected to a straingage
P7.34
force measurement). At x 2 m, the element indicates a
shear stress of 2.1 Pa. Assuming turbulent flow from the
leading edge, estimate (a) the stream velocity U, (b) the
boundarylayer thickness at the element, and (c) the boundA ship is 125 m long and has a wetted area of 3500 m2.
arylayer velocity u, in m/s, at 5 mm above the element.
Its propellers can deliver a maximum power of 1.1 MW
to seawater at 20°C. If all drag is due to friction, estimate
U
the maximum ship speed, in kn.
Floating element with
A wind tunnel has a test section 1 m square and 6 m long
x
negligible gap
with air at 20°C moving at an average velocity of 30 m/s.
It is planned to slant the walls outward slightly to account
for the growing boundarylayer displacement thickness on
the four walls, thus keeping the testsection velocity constant. At what angle should they be slanted to keep V constant between x 2 m and x 4 m?
Atmospheric boundary layers are very thick but follow forP7.43
mulas very similar to those of flatplate theory. Consider
wind blowing at 10 m/s at a height of 80 m above a smooth
beach. Estimate the wind shear stress, in Pa, on the beach P7.44 Extensive measurements of wall shear stress and local velocity for turbulent airflow on the flat surface of the Uniif the air is standard sealevel conditions. What will the
versity of Rhode Island wind tunnel have led to the folwind velocity striking your nose be if (a) you are standlowing proposed correlation:
ing up and your nose is 170 cm off the ground and (b) you
are lying on the beach and your nose is 17 cm off the
y2 w
uy 1.77
0.0207
ground?
2
A hydrofoil 50 cm long and 4 m wide moves at 28 kn in
seawater at 20°C. Using flatplate theory with Retr 5 E5,
Thus, if y and u(y) are known at a point in a flatplate boundestimate its drag, in N, for (a) a smooth wall and (b) a
ary layer, the wall shear may be computed directly. If the
rough wall,
0.3 mm.
answer to part (c) of Prob. 7.43 is u 27 m/s, determine
Hoerner [12, p. 3.25] states that the drag coefficient of a
whether the correlation is accurate for this case.
flag in winds, based on total wetted area 2bL, is approxi P7.45 A thin sheet of fiberboard weighs 90 N and lies on a
mated by CD 0.01 0.05L/b, where L is the flag length
rooftop, as shown in Fig. P7.45. Assume ambient air at
in the flow direction. Test Reynolds numbers ReL were 1
20°C and 1 atm. If the coefficient of solid friction between
E6 or greater. (a) Explain why, for L/b 1, these drag valboard and roof is
0.12, what wind velocity will genues are much higher than for a flat plate. Assuming seaerate enough fluid friction to dislodge the board?
level standard air at 50 mi/h, with area bL 4 m2, find (b) P7.46 A ship is 150 m long and has a wetted area of 5000 m2.
the proper flag dimensions for which the total drag is apIf it is encrusted with barnacles, the ship requires 7000 hp
proximately 400 N.
to overcome friction drag when moving in seawater at 15
Repeat Prob. 7.20 with the sole change that the pitot probe
kn and 20°C. What is the average roughness of the barnais now 10 mm from the wall (5 times higher). Show that
cles? How fast would the ship move with the same power
the flow there cannot possibly be laminar, and use smoothif the surface were smooth? Neglect wave drag.  v v 480  eText Main Menu  Textbook Table of Contents  Study Guide Problems 481 2m 3m xsep, θ sep 1m x 1.5 m θ U0
U Fiberboard R 2m P7.50
thin, show that the core velocity U(x) in the diffuser is
given approximately by Roof P7.45 U P7.47 As a case similar to Example 7.5, Howarth also proposed
the adversegradient velocity distribution U U0(1
x2/L2) and computed separation at xsep/L 0.271 by a seriesexpansion method. Compute separation by Thwaites’
method and compare.
P7.48 In 1957 H. Görtler proposed the adversegradient test cases
U (1 Constant width b θ U? U? x U0 U(x) θ
W
L P7.51
*P7.52 In Fig. P7.52 a slanted upper wall creates a favorable pressure gradient on the upper surface of the flat plate. Use
Thwaites’ theory to estimate P7.49 CD *P7.50 For flow past a cylinder of radius R as in Fig. P7.50, the
theoretical inviscid velocity distribution along the surface
is U 2U0 sin (x/R), where U0 is the oncoming stream
velocity and x is the arc length measured from the nose
(Chap. 8). Compute the laminar separation point xsep and
sep by Thwaites’ method, and compare with the digitalcomputer solution xsep/R 1.823 ( sep 104.5°) given by
R. M. Terrill in 1960.
P7.51 Consider the flatwalled diffuser in Fig. P7.51, which is
similar to that of Fig. 6.26a with constant width b. If x is
measured from the inlet and the wall boundary layers are v   eText Main Menu U0
(2x tan )/W where W is the inlet height. Use this velocity distribution
with Thwaites’ method to compute the wall angle for
which laminar separation will occur in the exit plane when
diffuser length L 2W. Note that the result is independent of the Reynolds number. U0
x/L)n and computed separation for laminar flow at n 1 to be
xsep/L 0.159. Compare with Thwaites’ method, assuming 0 0.
P7.49 Based strictly upon your understanding of flatplate theory plus adverse and favorable pressure gradients, explain
the direction (left or right) for which airflow past the slender airfoil shape in Fig. P7.49 will have lower total (friction pressure) drag. v 1  1
2 F
U 2 bL
0 Slanted
wall U0 h0 U(x)
x L, b
Flat plate P7.52 Textbook Table of Contents  Study Guide 1
2 h0 482 Chapter 7 Flow Past Immersed Bodies P7.53 P7.54 P7.55 P7.56 on the upper plate surface if U0 L/
105. Compare with
Eq. (7.27).
From Table 7.2, the drag coefficient of a wide plate normal to a stream is approximately 2.0. Let the stream conditions be U and p . If the average pressure on the front
of the plate is approximately equal to the freestream
stagnation pressure, what is the average pressure on the
rear?
A chimney at sea level is 2 m in diameter and 40 m high.
When it is subjected to 50 mi/h storm winds, what is the
estimated windinduced bending moment about the bottom of the chimney?
A ship tows a submerged cylinder, which is 1.5 m in diameter and 22 m long, at 5 m/s in fresh water at 20°C. Estimate the towing power, in kW, required if the cylinder is
(a) parallel and (b) normal to the tow direction.
A delivery vehicle carries a long sign on top, as in Fig.
P7.56. If the sign is very thin and the vehicle moves at 55
mi/h, estimate the force on the sign (a) with no crosswind
and (b) with a 10 mi/h crosswind. how fast can Joe ride into the head wind? (c) Why is the
result not simply 10 5.0 5.0 m/s, as one might first
suspect?
P7.60 A fishnet consists of 1mmdiameter strings overlapped
and knotted to form 1 by 1cm squares. Estimate the drag
of 1 m2 of such a net when towed normal to its plane at 3
m/s in 20°C seawater. What horsepower is required to tow
400 ft2 of this net?
P7.61 A filter may be idealized as an array of cylindrical fibers
normal to the flow, as in Fig. P7.61. Assuming that the
fibers are uniformly distributed and have drag coefficients given by Fig. 7.16a, derive an approximate expression for the pressure drop p through a filter of thickness L.
Filter section U
p+∆p 8m Phil's Pizza: 5555748 U
p 60 cm Array of
cylinders
(fibers) P7.61
P7.56  v v P7.57 The main crosscable between towers of a coastal suspension bridge is 60 cm in diameter and 90 m long. Estimate
the total drag force on this cable in crosswinds of 50 mi/h.
Are these laminarflow conditions?
P7.58 A long cylinder of rectangular cross section, 5 cm high
and 30 cm long, is immersed in water at 20°C flowing
at 12 m/s parallel to the long side of the rectangle. Estimate the drag force on the cylinder, per unit length, if
the rectangle (a) has a flat face or (b) has a rounded
nose.
P7.59 Joe can pedal his bike at 10 m/s on a straight level road
with no wind. The rolling resistance of his bike is 0.80
N s/m, i.e., 0.80 N of force per m/s of speed. The drag
area (CDA) of Joe and his bike is 0.422 m2. Joe’s mass
is 80 kg and that of the bike is 15 kg. He now encounters a head wind of 5.0 m/s. (a) Develop an equation for
the speed at which Joe can pedal into the wind. [Hint: A
cubic equation for V will result.] (b) Solve for V, i.e.,  eText Main Menu  P7.62 A sealevel smokestack is 52 m high and has a square cross
section. Its supports can withstand a maximum side force
of 90 kN. If the stack is to survive 90mi/h hurricane winds,
what is its maximum possible width?
P7.63 The cross section of a cylinder is shown in Fig. P7.63. Assume that on the front surface the velocity is given by potential theory (Sec. 8.4), V 2U sin , from which the
surface pressure is computed by Bernoulli’s equation. In
V = 2 U∞ sin θ
R
θ p∞ , U∞ Separatedflow wake P7.63 Textbook Table of Contents  Study Guide Problems 483
the separated flow on the rear, the pressure is assumed
equal to its value at
90°. Compute the theoretical drag
coefficient and compare with Table 7.2.
P7.64 A parachutist jumps from a plane, using an 8.5mdiameter chute in the standard atmosphere. The total mass
of the chutist and the chute is 90 kg. Assuming an open
chute and quasisteady motion, estimate the time to fall
from 2000 to 1000m altitude.
P7.65 The drag of a sphere at very low Reynolds numbers
ReD 1 was given analytically by G. G. Stokes in 1851:
F 3 VD [2, pp. 175 – 178]. This formula is an example of Stokes’ law of creeping motion, where inertia is negligible. Show that the drag coefficient in this region is
CD 24/ReD. A 1mmdiameter sphere falls in 20°C
glycerin at 2.5 mm/s. Is this a creeping motion? Compute
(a) the Reynolds number, (b) the drag, and (c) the specific
gravity of the sphere.
P7.66 A sphere of density s and diameter D is dropped from
rest in a fluid of density and viscosity . Assuming a
constant drag coefficient Cd 0, derive a differential equation for the fall velocity V(t) and show that the solution
is
V
C P7.68 P7.69 P7.70 3gCd 0(S 1)
4S2D 1/2 tanh Ct
1/2 where S
s / is the specific gravity of the sphere material.
Apply the theory of Prob. 7.66 to a steel sphere of diameter 2 cm, dropped from rest in water at 20°C. Estimate
the time required for the sphere to reach 99 percent of its
terminal (zeroacceleration) velocity.
A baseball weighs 145 g and is 7.35 cm in diameter. It is
dropped from rest from a 35mhigh tower at approximately sea level. Assuming a laminarflow drag coefficient, estimate (a) its terminal velocity and (b) whether it
will reach 99 percent of its terminal velocity before it hits
the ground.
Two baseballs from Prob. 7.68 are connected to a rod 7
mm in diameter and 56 cm long, as in Fig. P7.69. What
power, in W, is required to keep the system spinning at
400 r/min? Include the drag of the rod, and assume sealevel standard air.
A baseball from Prob. 7.68 is batted upward during a game
at an angle of 45° and an initial velocity of 98 mi/h. Neglect spin and lift. Estimate the horizontal distance traveled, (a) neglecting drag and (b) accounting for drag in a
numerical (computer) solution with a transition Reynolds
number ReD,crit 2.5 E5.  v v P7.67 4gD(S 1)
3Cd 0  eText Main Menu  28 cm
28 cm Baseball Ω Baseball P7.69
P7.71 A football weighs 0.91 lbf and approximates an ellipsoid
6 in in diameter and 12 in long (Table 7.3). It is thrown
upward at a 45° angle with an initial velocity of 80 ft/s.
Neglect spin and lift. Assuming turbulent flow, estimate
the horizontal distance traveled, (a) neglecting drag and
(b) accounting for drag with a numerical (computer)
model.
P7.72 A settling tank for a municipal water supply is 2.5 m deep,
and 20°C water flows through continuously at 35 cm/s.
Estimate the minimum length of the tank which will
ensure that all sediment (SG 2.55) will fall to the bottom for particle diameters greater than (a) 1 mm and
(b) 100 m.
P7.73 A balloon is 4 m in diameter and contains helium at 125
kPa and 15°C. Balloon material and payload weigh 200
N, not including the helium. Estimate (a) the terminal ascent velocity in sealevel standard air, (b) the final standard altitude (neglecting winds) at which the balloon will
come to rest, and (c) the minimum diameter ( 4 m) for
which the balloon will just barely begin to rise in sealevel
standard air.
P7.74 If D 4 m in Prob. 7.73 and the helium remains at
15°C and 125 kPa, estimate the time required for the
balloon to rise through the standard atmosphere from
sea level to its equilibrium altitude, which is approximately 4000 m.
P7.75 The heliumfilled balloon in Fig. P7.75 is tethered at 20°C
and 1 atm with a string of negligible weight and drag. The
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diameter is 50 cm, and the balloon material weighs 0.2 N, D = 50 cm
U θ P7.75 Textbook Table of Contents  Study Guide 484 P7.76 P7.77 P7.78 P7.79 P7.80 Chapter 7 Flow Past Immersed Bodies
not including the helium. The helium pressure is 120 kPa.
Estimate the tilt angle if the airstream velocity U is (a)
5 m/s or (b) 20 m/s.
P7.83
Extend Prob. 7.75 to make a smooth plot of tilt angle
versus stream velocity U in the range 1 U 12 mi/h.
(A spreadsheet is recommended for this task.) Comment
on the effectiveness of this system as an airvelocity instrument.
Modify Prob. 7.75 as follows: Let the fluid be water at P7.84
20°C, and let the 50cmdiameter sphere be solid cork
(SG 0.16). Estimate the tilt angle if the water velocity U is 3 m/s.
Apply Prob. 7.61 to a filter consisting of 300 mdiameter fibers packed 250 per square centimeter in the
plane of Fig. P7.61. For air at 20°C and 1 atm flowing at
1.5 m/s, estimate the pressure drop if the filter is 5 cm
thick.
Assume that a radioactive dust particle approximates a
sphere of density 2400 kg/m3. How long, in days, will
it take such a particle to settle to sea level from an altitude of 12 km if the particle diameter is (a) 1 m or
(b) 20 m?
A heavy sphere attached to a string should hang at an
angle when immersed in a stream of velocity U, as in
Fig. P7.80. Derive an expression for as a function of
the sphere and flow properties. What is if the sphere
is steel (SG 7.86) of diameter 3 cm and the flow is
sealevel standard air at U 40 m/s? Neglect the string *P7.85
drag. θ 1.2 ft2 falling feet first [12, p. 313]. What are the minimum and maximum terminal speeds that can be achieved
by a skydiver at 5000ft standard altitude?
A highspeed car has a drag coefficient of 0.3 and a frontal
area of 1 m2. A parachute is to be used to slow this car
from 80 to 40 m/s in 8 s. What should the chute diameter
be? What distance will be traveled during this deceleration? Take m 2000 kg.
A PingPong ball weighs 2.6 g and has a diameter of 3.8
cm. It can be supported by an air jet from a vacuum cleaner
outlet, as in Fig. P7.84. For sealevel standard air, what jet
velocity is required? P7.84
An aluminum cylinder (SG 2.7) slides concentrically
down a taut 1mmdiameter wire as shown in Fig. P7.85.
Its length L 8 cm, and its radius R 1 cm. A 2mmdiameter hole down the cylinder center is lubricated by SAE
30 oil at 20°C. Estimate the terminal fall velocity V of the
cylinder if ambient air drag is (a) neglected and (b) included. Assume air at 1 atm and 20°C.
Oil film U
D, ρs R P7.80
A typical U.S. Army parachute has a projected diameter of 28 ft. For a payload mass of 80 kg, (a) what terminal velocity will result at 1000m standard altitude?
For the same velocity and net payload, what size dragproducing “chute” is required if one uses a square flat
plate held (b) vertically and (c) horizontally? (Neglect
the fact that flat shapes are not dynamically stable in
free fall.)
P7.82 The average skydiver with parachute unopened weighs 175
lbf and has a drag area CD A 9 ft2 spreadeagled and  v v P7.81  eText Main Menu  L V P7.85 Textbook Table of Contents  Study Guide Problems 485
P7.86 Hoerner [Ref. 12, pp. 3 – 25] states that the drag coefficient
of a flag of 2 1 aspect ratio is 0.11 based on planform
area. The University of Rhode Island has an aluminum
flagpole 25 m high and 14 cm in diameter. It flies equalsized national and state flags together. If the fracture stress
of aluminum is 210 MPa, what is the maximum flag size
that can be used yet avoids breaking the flagpole in hurricane (75 mi/h) winds?
P7.87 A tractortrailer truck has a dragarea CD A 8 m2 bare
and 6.7 m2 with an aerodynamic deflector (Fig. 7.18b). Its
rolling resistance is 50 N for each mile per hour of speed.
Calculate the total horsepower required at sea level with
and without the deflector if the truck moves at (a) 55 mi/h
and (b) 75 mi/h.
P7.88 A pickup truck has a clean dragarea CD A of 35 ft2. Estimate the horsepower required to drive the truck at 55
mi/h (a) clean and (b) with the 3 by 6ft sign in Fig.
P7.88 installed if the rolling resistance is 150 lbf at sea
level. torque of 0.004 N m. Making simplifying assumptions
to average out the timevarying geometry, estimate and
plot the variation of anemometer rotation rate with wind
velocity U in the range 0 U 25 m/s for sealevel standard air.
D = 5 cm Ω
15 cm
D = 5 cm 15 cm U 6 ft P7.91 Eat at
Joe's 3 ft P7.92 A 1500kg automobile uses its drag area CD A 0.4 m2,
plus brakes and a parachute, to slow down from 50 m/s.
Its brakes apply 5000 N of resistance. Assume sealevel
standard air. If the automobile must stop in 8 s, what diameter parachute is appropriate?
P7.93 A hotfilm probe is mounted on a coneandrod system in
a sealevel airstream of 45 m/s, as in Fig. P7.93. Estimate
the maximum cone vertex angle allowable if the flowinduced bending moment at the root of the rod is not to exceed 30 N cm.
Ho t film
3 cm
45 m/s P7.88  v v P7.89 A water tower is approximated by a 15mdiameter sphere
mounted on a 1mdiameter rod 20 m long. Estimate the
bending moment at the root of the rod due to aerodynamic
forces during hurricane winds of 40 m/s.
P7.90 In the great hurricane of 1938, winds of 85 mi/h blew over
a boxcar in Providence, Rhode Island. The boxcar was 10
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ft high, 40 ft long, and 6 ft wide, with a 3ft clearance
above tracks 4.8 ft apart. What wind speed would topple
a boxcar weighing 40,000 lbf?
*P7.91 A cup anemometer uses two 5cmdiameter hollow hemispheres connected to 15cm rods, as in Fig. P7.91. Rod
drag is negligible, and the central bearing has a retarding  eText Main Menu  20 cm 5mm
dia. P7.93
P7.94 A rotary mixer consists of two 1mlong halftubes rotating around a central arm, as in Fig. P7.94. Using the drag
from Table 7.2, derive an expression for the torque T required to drive the mixer at angular velocity in a fluid Textbook Table of Contents  Study Guide 486 Chapter 7 Flow Past Immersed Bodies
mobile of mass 1500 kg and frontal area 2 m2, the following velocityversustime data are obtained during a
coastdown: of density . Suppose that the fluid is water at 20°C and
the maximum driving power available is 20 kW. What is
the maximum rotation speed r/min?
t, s
Ω V, m/s *P7.98
R=1m D = 5 cm P7.94
P7.95 An airplane weighing 12 kN, with a dragarea CD A
5 m2, lands at sea level at 55 m/s and deploys a drag parachute 3 m in diameter. No other brakes are applied.
(a) How long will it take the plane to slow down to
30 m/s? (b) How far will it have traveled in that time?
*P7.96 A Savonius rotor (Fig. 6.29b) can be approximated by
the two open halftubes in Fig. P7.96 mounted on a central axis. If the drag of each tube is similar to that in
Table 7.2, derive an approximate formula for the rotation rate as a function of U, D, L, and the fluid properties ( , ). 0 10 20 30 40 27.0 24.2 21.8 19.7 17.9 Estimate (a) the rolling resistance and (b) the drag coefficient. This problem is well suited for digitalcomputer
analysis but can be done by hand also.
A buoyant ball of specific gravity SG 1 dropped into
water at inlet velocity V0 will penetrate a distance h and
then pop out again, as in Fig. P7.98. Make a dynamic
analysis of this problem, assuming a constant drag coefficient, and derive an expression for h as a function of
the system properties. How far will a 5cmdiameter ball
with SG 0.5 and CD 0.47 penetrate if it enters at
10 m/s?
Diameter
D
(SG < 1) V0 h Ω P7.98
Axis D P7.99 Two steel balls (SG 7.86) are connected by a thin hinged
rod of negligible weight and drag, as in Fig. P7.99. A stop
prevents the rod from rotating counterclockwise. Estimate
the sealevel air velocity U for which the rod will first begin to rotate clockwise. D L U D = 2 cm L U 10 cm
Hinge
Stop
45˚ P7.96
10 cm A simple measurement of automobile drag can be found
by an unpowered coastdown on a level road with no
wind. Assume constant rolling resistance. For an auto  v v P7.97  eText Main Menu  D = 1 cm P7.99 Textbook Table of Contents  Study Guide Problems 487
P7.100 In creeping motion or Stokes’ flow at velocity U past a
sphere of diameter D, density (fluid inertia) is unimportant and the drag force is F 3 UD. This formula
is valid if ReD 1.0 (see Fig. 7.16b). (a) Verify that
Stokes’ formula is equivalent to CD 24/ReD. (b) Determine the largest diameter raindrop whose terminal fall
velocity follows Stokes’ formula in sealevel standard
air.
P7.101 Icebergs can be driven at substantial speeds by the wind.
Let the iceberg be idealized as a large, flat cylinder,
D
L, with oneeighth of its bulk exposed, as in Fig.
P7.101. Let the seawater be at rest. If the upper and
lower drag forces depend upon relative velocities between the iceberg and the fluid, derive an approximate
expression for the steady iceberg speed V when driven
by wind velocity U.
D> L
>
U
L/8 V
7L / 8 Iceberg P7.101  v v P7.102 Sand particles (SG 2.7), approximately spherical with
diameters from 100 to 250 m, are introduced into an upwardflowing stream of water at 20°C. What is the minimum water velocity which will carry all the sand particles
upward?
P7.103 When immersed in a uniform stream V, a heavy rod
hinged at A will hang at Pode’s angle , after an analysis
by L. Pode in 1951 (Fig. P7.103). Assume that the cylinder has normal drag coefficient CDN and tangential coefficient CDT which relate the drag forces to VN and VT, respectively. Derive an expression for Pode’s angle as a
function of the flow and rod parameters. Find for a steel
rod, L 40 cm, D 1 cm, hanging in sealevel air at
V 35 m/s.
P7.104 Suppose that the body in Fig. P7.103 is a thin plate
weighing 0.6 N, of length L 20 cm and width b 4
cm into the paper. For sealevel standard air, plot Pode’s
angle versus velocity V in the range 0 V 40 m/s.
Would this device make a good airvelocity meter
(anemometer)?  eText Main Menu  A θ
CD N , VN
CD T , VT V
L, D, ρs P7.103
P7.105 A ship 50 m long, with a wetted area of 800 m2, has the
hull shape tested in Fig. 7.19. There are no bow or stern
bulbs. The total propulsive power available is 1 MW. For
seawater at 20°C, plot the ship’s velocity V kn versus
power P for 0 P 1 MW. What is the most efficient
setting?
P7.106 A smooth steel 1cmdiameter sphere (W 0.04 N) is
fired vertically at sea level at the initial supersonic velocity V0 1000 m/s. Its drag coefficient is given by Fig.
7.20. Assuming that the speed of sound is constant at a
343 m/s, compute the maximum altitude of the projectile
(a) by a simple analytical estimate and (b) by a digitalcomputer program.
P7.107 Repeat Prob. 7.106 if the body is a 9mm steel bullet (W
0.07 N) which approximates the “pointed body of revolution” in Fig. 7.20.
P7.108 The data in Fig. P7.108 are for the lift and drag of a spinning sphere from Ref. 12, pp. 7 – 20. Suppose that a tennis
ball (W 0.56 N, D 6.35 cm) is struck at sea level with
initial velocity V0 30 m/s and “topspin” (front of the ball
rotating downward) of 120 r/s. If the initial height of the
ball is 1.5 m, estimate the horizontal distance traveled before it strikes the ground.
P7.109 Repeat Prob. 7.108 if the ball is struck with “underspin”
(front of the ball rotating upward).
P7.110 A baseball pitcher throws a curveball with an initial velocity of 65 mi/h and a spin of 6500 r/min about a vertical axis. A baseball weighs 0.32 lbf and has a diameter of
2.9 in. Using the data of Fig. P7.108 for turbulent flow,
estimate how far such a curveball will have deviated from
its straightline path when it reaches home plate 60.5 ft
away.
*P7.111 A table tennis ball has a mass of 2.6 g and a diameter
of 3.81 cm. It is struck horizontally at an initial velocity of 20 m/s while it is 50 cm above the table, as in
Fig. P7.111. For sealevel air, what spin, in r/min, will
cause the ball to strike the opposite edge of the table,
4 m away? Make an analytical estimate, using Fig.
P7.108, and account for the fact that the ball decelerates during flight. Textbook Table of Contents  Study Guide 488 Chapter 7 Flow Past Immersed Bodies Figu
re U
nava
ilab
le
P7.108
4m
20 m/s ω? 50 cm
? P7.111  v v P7.112 Repeat Prob. 7.111 by making a detailed digitalcomputer
solution for the flight path of the ball. Use Fig. P7.108 for
lift and drag.
P7.113 An automobile has a mass of 1000 kg and a dragarea
CD A 0.7 m2. The rolling resistance of 70 N is approximately constant. The car is coasting without brakes at 90
km/h as it begins to climb a hill of 10 percent grade
(slope tan 1 0.1 5.71°). How far up the hill will the
car come to a stop?
*P7.114 Suppose that the car in Prob. 7.113 is placed at the top of
the 10 percent grade hill and released from rest to coast
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down without brakes. What will be its speed, in km/h, after dropping a vertical distance of 20 m?  eText Main Menu  P7.115 An airplane weighs 180 kN and has a wing area of 160
m2 and a mean chord of 4 m. The airfoil properties are
given by Fig. 7.25. If the airplane cruises at 250 mi/h at
3000m standard altitude, what propulsive power is required to overcome wing drag?
P7.116 The airplane of Prob. 7.115 is designed to land at V0
1.2Vstall, using a split flap set at 60°. What is the proper
landing speed in mi/h? What power is required for takeoff
at the same speed?
P7.117 Suppose that the airplane of Prob. 7.115 takes off at sea
level without benefit of flaps, with CL constant so that the
takeoff speed is 100 mi/h. Estimate the takeoff distance if
the thrust is 10 kN. How much thrust is needed to make
the takeoff distance 1250 m?
*P7.118 Suppose that the airplane of Prob. 7.115 is fitted with all
the best highlift devices of Fig. 7.28. What is its minimum stall speed in mi/h? Estimate the stopping distance
if the plane lands at V0 1.25Vstall with constant CL 3.0
and CD 0.2 and the braking force is 20 percent of the
weight on the wheels.
P7.119 An airplane has a mass of 5000 kg, a maximum thrust of
7000 N, and a rectangular wing with aspect ratio 6.0. It
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takes off at sea level with a 60° split flap whose twodimensional properties are shown in Fig. 7.25. Assume
all lift and all drag are due to the wing. What is the
proper wing size if the takeoff distance is to be 1 km?
P7.120 Show that if Eqs. (7.70) and (7.71) are valid, the maximum lifttodrag ratio occurs when CD 2CD . What are Textbook Table of Contents  Study Guide Fundamentals of Engineering Exam Problems
(L/D)max and for a symmetric wing when AR 5 and
CD
0.009?
P7.121 In gliding (unpowered) flight, the lift and drag are in equilibrium with the weight. Show that if there is no wind, the
aircraft sinks at an angle
tan drag
lift For a sailplane of mass 200 kg, wing area 12 m2, and aspect ratio 11, with an NACA 0009 airfoil, estimate (a) the
stall speed, (b) the minimum gliding angle, and (c) the
maximum distance it can glide in still air when it is 1200
m above level ground.
P7.122 A boat of mass 2500 kg has two hydrofoils, each of chord
30 cm and span 1.5 m, with CL,max 1.2 and CD
0.08.
Its engine can deliver 130 kW to the water. For seawater
at 20°C, estimate (a) the minimum speed for which the 489 foils support the boat and (b) the maximum speed attainable.
P7.123 In prewar days there was a controversy, perhaps apocryphal,
about whether the bumblebee has a legitimate aerodynamic
right to fly. The average bumblebee (Bombus terrestris)
weighs 0.88 g, with a wing span of 1.73 cm and a wing area
of 1.26 cm2. It can indeed fly at 10 m/s. Using fixedwing
theory, what is the lift coefficient of the bee at this speed? Is
this reasonable for typical airfoils?
*P7.124 The bumblebee can hover at zero speed by flapping its
wings. Using the data of Prob. 7.123, devise a theory for
flapping wings where the downstroke approximates a short
flat plate normal to the flow (Table 7.3) and the upstroke
is feathered at nearly zero drag. How many flaps per second of such a model wing are needed to support the bee’s
weight? (Actual measurements of bees show a flapping
rate of 194 Hz.) Word Problems
W7.1 W7.2 W7.3
W7.4
W7.5
W7.6
W7.7 How do you recognize a boundary layer? Cite some physical properties and some measurements which reveal appropriate characteristics.
In Chap. 6 the Reynolds number for transition to turbulence in pipe flow was about Retr 2300, whereas in flatplate flow Retr 1 E6, nearly three orders of magnitude
higher. What accounts for the difference?
Without writing any equations, give a verbal description of
boundarylayer displacement thickness.
Describe, in words only, the basic ideas behind the “boundarylayer approximations.”
What is an adverse pressure gradient? Give three examples
of flow regimes where such gradients occur.
What is a favorable pressure gradient? Give three examples of flow regimes where such gradients occur.
The drag of an airfoil (Fig. 7.12) increases considerably if
you turn the sharp edge around 180° to face the stream.
Can you explain this? W7.8 In Table 7.3, the drag coefficient of a spruce tree decreases
sharply with wind velocity. Can you explain this?
W7.9 Thrust is required to propel an airplane at a finite forward
velocity. Does this imply an energy loss to the system? Explain the concepts of thrust and drag in terms of the first
law of thermodynamics.
W7.10 How does the concept of drafting, in automobile and bicycle racing, apply to the material studied in this chapter?
W7.11 The circular cylinder of Fig. 7.13 is doubly symmetric and
therefore should have no lift. Yet a lift sensor would definitely reveal a finite rootmeansquare value of lift. Can
you explain this behavior?
W7.12 Explain in words why a thrown spinning ball moves in a
curved trajectory. Give some physical reasons why a side
force is developed in addition to the drag. Fundamentals of Engineering Exam Problems  v v FE7.1 A smooth 12cmdiameter sphere is immersed in a stream
of 20°C water moving at 6 m/s. The appropriate Reynolds
number of this sphere is approximately
(a) 2.3 E5, (b) 7.2 E5, (c) 2.3 E6, (d) 7.2 E6, (e) 7.2 E7
FE7.2 If, in Prob. FE7.1, the drag coefficient based upon frontal
area is 0.5, what is the drag force on the sphere?
(a) 17 N, (b) 51 N, (c) 102 N, (d) 130 N, (e) 203 N
FE7.3 If, in Prob. FE7.1, the drag coefficient based upon frontal
area is 0.5, at what terminal velocity will an aluminum
sphere (SG 2.7) fall in still water?
(a) 2.3 m/s, (b) 2.9 m/s, (c) 4.6 m/s, (d) 6.5 m/s, (e) 8.2 m/s  eText Main Menu  FE7.4 For flow of sealevel standard air at 4 m/s parallel to a thin
flat plate, estimate the boundarylayer thickness at x 60
cm from the leading edge:
(a) 1.0 mm, (b) 2.6 mm, (c) 5.3 mm, (d) 7.5 mm,
(e) 20.2 mm
FE7.5 In Prob. FE7.4, for the same flow conditions, what is the
wall shear stress at x 60 cm from the leading edge?
(a) 0.053 Pa, (b) 0.11 Pa, (c) 0.16 Pa, (d) 0.32 Pa,
(e) 0.64 Pa
FE7.6 Wind at 20°C and 1 atm blows at 75 km/h past a flagpole
18 m high and 20 cm in diameter. The drag coefficient, Textbook Table of Contents  Study Guide 490 Chapter 7 Flow Past Immersed Bodies
FE7.9 An airplane has a mass of 19,550 kg, a wing span of 20
m, and an average wing chord of 3 m. When flying in air
of density 0.5 kg/m3, its engines provide a thrust of 12 kN
against an overall drag coefficient of 0.025. What is its
approximate velocity?
(a) 250 mi/h, (b) 300 mi/h, (c) 350 mi/h, (d) 400 mi/h,
(e) 450 mi/h
FE7.10 For the flight conditions of the airplane in Prob. FE7.9
above, what is its approximate lift coefficient?
(a) 0.1, (b) 0.2, (c) 0.3, (d) 0.4, (e) 0.5 based upon frontal area, is 1.15. Estimate the windinduced bending moment at the base of the pole.
(a) 9.7 kN m, (b) 15.2 kN m, (c) 19.4 kN m,
(d) 30.5 kN m, (e) 61.0 kN m
FE7.7 Consider wind at 20°C and 1 atm blowing past a chimney 30
m high and 80 cm in diameter. If the chimney may fracture
at a base bending moment of 486 kN m, and its drag coefficient based upon frontal area is 0.5, what is the approximate
maximum allowable wind velocity to avoid fracture?
(a) 50 mi/h, (b) 75 mi/h, (c) 100 mi/h, (d) 125 mi/h,
(e) 150 mi/h
FE7.8 A dust particle of density 2600 kg/m3, small enough to
satisfy Stokes’ drag law, settles at 1.5 mm/s in air at 20°C
and 1 atm. What is its approximate diameter?
(a) 1.8 m, (b) 2.9 m, (c) 4.4 m, (d) 16.8 m,
(e) 234 m Comprehensive Problems
C7.1 Jane wants to estimate the drag coefficient of herself on her
bicycle. She measures the projected frontal area to be 0.40
m2 and the rolling resistance to be 0.80 N s/m. The mass
of the bike is 15 kg, while the mass of Jane is 80 kg. Jane
coasts down a long hill which has a constant 4° slope. (See
Fig. C7.1.) She reaches a terminal (steady state) speed of
14 m/s down the hill. Estimate the aerodynamic drag coefficient CD of the rider and bicycle combination.
C7.2 Air at 20°C and 1 atm flows at Vavg 5 m/s between long,
smooth parallel heatexchanger plates 10 cm apart, as in Fig.
C7.2. It is proposed to add a number of widely spaced 1cmlong interrupter plates to increase the heat transfer, as shown.
Although the flow in the channel is turbulent, the boundary
layers over the interrupter plates are essentially laminar. Assume all plates are 1 m wide into the paper. Find (a) the pressure drop in Pa/m without the small plates present. Then find
(b) the number of small plates per meter of channel length
which will cause the pressure drop to rise to 10.0 Pa/m.
C7.3 A new pizza store is planning to open. They will, of course,
offer free delivery, and therefore need a small delivery car
with a large sign attached. The sign (a flat plate) is 1.5 ft U Interrupter plates
L 1 cm 5 m/s C7.2
high and 5 ft long. The boss (having no feel for fluid mechanics) mounts the sign bluntly facing the wind. One of his
drivers is taking fluid mechanics and tells his boss he can
save lots of money by mounting the sign parallel to the wind.
(See Fig. C7.3.) (a) Calculate the drag (in lbf) on the sign
alone at 40 mi/h (58.7 ft/s) in both orientations. (b) Suppose
the car without any sign has a drag coefficient of 0.4 and a
frontal area of 40 ft2. For V 40 mi/h, calculate the total
drag of the carsign combination for both orientations. (c) If
the car has a rolling resistance of 40 lbf at 40 mi/h, calculate the horsepower required by the engine to drive the car
at 40 mi/h in both orientations. (d) Finally, if the engine can
deliver 10 hp for 1 h on a gallon of gasoline, calculate the
fuel efficiency in mi/gal for both orientations at 40 mi/h. V  v v C7.1  eText Main Menu  Textbook Table of Contents  Study Guide References 491 C7.3
C7.4 Consider a pendulum with an unusual bob shape: a hemispherical cup of diameter D whose axis is in the plane of
oscillation, as in Fig. C7.4. Neglect the mass and drag of
the rod L. (a) Set up the differential equation for the oscillation (t), including different cup drag (air density )
in each direction, and (b) nondimensionalize this equation.
(c) Determine the natural frequency of oscillation for small
1 rad. (d) For the special case L 1 m, D 10 cm,
m 50 g, and air at 20°C and 1 atm, with (0) 30°, find
(numerically) the time required for the oscillation amplitude to drop to 1°. L Air
m Cup shape C7.4 Design Project
als—are to be selected through your analysis. Make suitable assumptions about the instantaneous drag of the cups
and rods at any given angle (t) of the system. Compute
the instantaneous torque T(t), and find and integrate the instantaneous angular acceleration of the device. Develop a
complete theory for rotation rate versus wind speed in the
range 0 U 50 mi/h. Try to include actual commercial
bearingfriction properties. D7.1 It is desired to design a cup anemometer for wind speed,
similar to Fig. P7.91, with a more sophisticated approach
than the “averagetorque” method of Prob. 7.91. The design
should achieve an approximately linear relation between
wind velocity and rotation rate in the range 20 U 40
mi/h, and the anemometer should rotate at about 6 r/s at
U 30 mi/h. All specifications—cup diameter D, rod
length L, rod diameter d, the bearing type, and all materi References
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