Unformatted text preview: Cylindrical wave pattern produced in a ripple tank. When not modified by the noslip condition at solid surfaces, waves are nearly inviscid and well represented by the potential theory of
this chapter. (Courtesy of Dr. E. R. Degginger/ColorPic Inc.)  v v 494  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 8
Potential Flow
and Computational
Fluid Dynamics Motivation. The basic partial differential equations of mass, momentum, and energy
were discussed in Chap. 4. A few solutions were then given for incompressible potential flow in Sec. 4.10 and for incompressible viscous flow in Sec. 4.11. The viscous solutions were limited to simple geometries and unidirectional flows, where the difficult
nonlinear convective terms were neglected. The potential flows were not limited by such
nonlinear terms. Then, in Chap. 7, we found an approximation: patching boundarylayer
flows onto an outer inviscid flow pattern. For more complicated viscous flows, we found
no theory or solutions, just experimental data.
The purposes of the present chapter are (1) to explore more examples of potential
theory and (2) to indicate some flows which can be approximated by computational
fluid dynamics (CFD). The combination of these two gives us a good picture of incompressibleflow theory and its relation to experiment. One of the most important applications of potentialflow theory is to aerodynamics and marine hydrodynamics. First,
however, we will review and extend the concepts of Sec. 4.10. 8.1 Introduction and Review Figure 8.1 reminds us of the problems to be faced. A free stream approaches two closely
spaced bodies, creating an “internal’’ flow between them and “external’’ flows above
and below them. The fronts of the bodies are regions of favorable gradient (decreasing pressure along the surface), and the boundary layers will be attached and thin: Inviscid theory will give excellent results for the outer flow if Re 104. For the internal flow between bodies, the boundary layers will grow and eventually meet, and the
inviscid core vanishes. Inviscid theory works well in a “short’’ duct L/D 10, such as
the nozzle of a wind tunnel. For longer ducts we must estimate boundarylayer growth
and be cautious about using inviscid theory.  v v 495  eText Main Menu  Textbook Table of Contents  Study Guide 496 Chapter 8 Potential Flow and Computational Fluid Dynamics
Inviscid external flow Separation
Boundary layer Boundary layer
Freestream Fully
viscous
flow Inviscid internal core
Boundary layer Fig. 8.1 Patching viscous and inviscidflow regions. Potential theory in this chapter does not apply
to the boundarylayer regions Boundary layer
Inviscid external flow Separation For the external flows above and below the bodies in Fig. 8.1, inviscid theory should
work well for the outer flows, until the surface pressure gradient becomes adverse (increasing pressure) and the boundary layer separates or stalls. After the separation point,
boundarylayer theory becomes inaccurate, and the outer flow streamlines are deflected
and have a strong interaction with the viscous nearwall regions. The theoretical analysis of separatedflow regions is an active research area at present. Review of VelocityPotential
Concepts Recall from Sec. 4.9 that if viscous effects are neglected, lowspeed flows are irrotational,
V 0, and the velocity potential exists, such that
or u ∂
∂x The continuity equation (4.73), V 0, reduces to Laplace’s equation for : V ∂2
∂x2 2 ∂2
∂y2 ∂
∂y
∂2
∂z2 w ∂
∂z 0 (8.1) (8.2) and the momentum equation (4.74) reduces to Bernoulli’s equation:
∂
∂t p 12
V
2 gz const where V (8.3) Typical boundary conditions are known freestream conditions
Outer boundaries: Known ∂
∂
∂
,
,
∂x ∂y ∂z (8.4) and no velocity normal to the boundary at the body surface:
Solid surfaces: ∂
∂n 0 where n is perpendicular to body (8.5)  v v Unlike the noslip condition in viscous flow, here there is no condition on the tangential surface velocity Vs ∂ /∂s, where s is the coordinate along the surface. This velocity is determined as part of the solution to the problem.  eText Main Menu  Textbook Table of Contents  Study Guide 8.1 Introduction and Review 497 Occasionally the problem involves a free surface, for which the boundary pressure
is known and equal to pa, usually a constant. The Bernoulli equation (8.3) then supplies a relation at the surface between V and the elevation z of the surface. For steady
flow, e.g.,
V2 Free surface: 2 const 2gzsurf (8.6) It should be clear to the reader that this use of Laplace’s equation, with known values
of the derivative of along the boundaries, is much easier than a direct attack using
the fully viscous NavierStokes equations. The analysis of Laplace’s equation is very
well developed and is termed potential theory, with whole books written about the general theory [1] and its application to fluid mechanics [2 to 4]. There are many analytical techniques, including superposition of elementary functions, conformal mapping,
numerical finite differences [5], numerical finite elements [6], numerical boundary elements [7], and electric or mechanical analogs [8] now outdated. Having found
(x, y, z, t) from such an analysis, we then compute V by direct differentiation in Eq.
(8.1), after which we compute p from Eq. (8.3). The procedure is quite straightforward,
and many interesting albeit idealized results can be obtained. Review of Stream Function
Concepts Recall from Sec. 4.7 that if a flow is described by only two coordinates, the stream
function also exists as an alternate approach. For plane incompressible flow in xy
coordinates, the correct form is
u ∂
∂y ∂
∂x (8.7) The condition of irrotationality reduces to Laplace’s equation for
2 z 0 ∂
∂x ∂u
∂y
∂2
∂x2 or ∂
∂x ∂
∂x ∂2
∂y2 also: ∂∂
∂y ∂y 0 (8.8) The boundary conditions again are known velocity in the stream and no flow through
any solid surface:
Free stream:
Solid surface: Known
body ∂∂
,
∂x ∂y (8.9a) const (8.9b)  v v Equation (8.9b) is particularly interesting because any line of constant in a flow
can therefore be interpreted as a body shape and may lead to interesting applications.
For the applications in this chapter, we may compute either or or both, and the
solution will be an orthogonal flow net as in Fig. 8.2. Once found, either set of lines
may be considered the lines, and the other set will be the lines. Both sets of lines
are laplacian and could be useful.  eText Main Menu  Textbook Table of Contents  Study Guide 498 Chapter 8 Potential Flow and Computational Fluid Dynamics
φ3 ψ1 φ2 ψ3
ψ2 φ1 φ2 ψ1 ψ3 Fig. 8.2 Streamlines and potential
lines are orthogonal and may reverse roles if results are useful: (a)
typical inviscidflow pattern; (b)
same as (a) with roles reversed. φ3 (a) Plane Polar Coordinates φ1 ψ2 (b) Many solutions in this chapter are conveniently expressed in polar coordinates (r, ).
Both the velocity components and the differential relations for and are then changed,
as follows:
r ∂
∂r 1∂
r∂ ∂
∂r 1∂
r∂ (8.10) Laplace’s equation takes the form
∂
1∂
r
∂r
r ∂r 1 ∂2
2
r2 ∂ 0 (8.11) Exactly the same equation holds for the polarcoordinate form of (r, ).
An intriguing facet of potential flow with no free surface is that the governing equations (8.2) and (8.8) contain no parameters, nor do the boundary conditions. Therefore
the solutions are purely geometric, depending only upon the body shape, the freestream
orientation, and — surprisingly — the position of the rear stagnation point.1 There is no
Reynolds, Froude, or Mach number to complicate the dynamic similarity. Inviscid flows
are kinematically similar without additional parameters — recall Fig. 5.6a. 8.2 Elementary PlaneFlow
Solutions Recall from Sec. 4.10 that we defined three elementary potential flows which are quite
useful: (1) uniform stream in the x direction, (2) line source or sink at the origin, and
(3) line vortex at the origin. (Recall Fig. 4.12 for these geometries.) Let us review these
special cases here:
Uniform stream iU: Uy Ux (8.12a) Line source or sink: m m ln r (8.12b) K (8.12c) Line vortex: K ln r  v v 1
The rear stagnation condition establishes the net amount of “circulation’’ about the body, giving rise
to a lift force. Otherwise the solution could not be unique. See Sec. 8.4.  eText Main Menu  Textbook Table of Contents  Study Guide 8.2 Elementary PlaneFlow Solutions 499 The source “strength’’ m and the vortex “strength’’ K have the same dimensions, namely,
velocity times length, or {L2/T}.
If the uniform stream is written in plane polar coordinates, it becomes
Uniform stream iU: Ur sin Ur cos (8.13) This makes it easier to superimpose, say, a stream and a source or vortex by using the
same coordinates. If the uniform stream is moving at angle with respect to the xaxis, i.e.,
u U cos ∂
∂y ∂
∂x U sin ∂
∂x ∂
∂y then by integration we obtain the correct functions for flow at an angle:
U(y cos x sin ) U(x cos y sin ) (8.14) These expressions are useful in airfoil angleofattack problems (Sec. 8.7). Circulation The linevortex flow is irrotational everywhere except at the origin, where the vorticity
V is infinite. This means that a certain line integral called the fluid circulation
does not vanish when taken around a vortex center.
With reference to Fig. 8.3, the circulation is defined as the counterclockwise line
integral, around a closed curve C, of arc length ds times the velocity component tangent to the curve
0
C V cos ds 0 V ds (u dx C dy w dz) (8.15) C From the definition of , V ds
ds d for an irrotational flow; hence normally in an irrotational flow would equal the final value of minus the initial value
of . Since we start and end at the same point, we compute
0, but not for vortex
K from Eq. (8.12c) there is a change in of amount 2 K as we make
flow: With
one complete circle:
Path enclosing a vortex: 2K (8.16) Γ = οC V cos α d s
Closed
curve C : dS α Streamline V  v v Fig. 8.3 Definition of the fluid circulation .  eText Main Menu  Textbook Table of Contents  Study Guide 500 Chapter 8 Potential Flow and Computational Fluid Dynamics Alternately the calculation can be made by defining a circular path of radius r around
the vortex center, from Eq. (8.15)
0 2 ds
C 0 K
rd
r 2K (8.17) In general, denotes the net algebraic strength of all the vortex filaments contained
within the closed curve. In the next section we shall see that a region of finite circulation within a flowing stream will be subjected to a lift force proportional to both U
and .
It is easy to show, by using Eq. (8.15), that a source or sink creates no circulation.
If there are no vortices present, the circulation will be zero for any path enclosing any
number of sources and sinks. 8.3 Superposition of PlaneFlow Solutions We can now form a variety of interesting potential flows by summing the velocitypotential and stream functions of a uniform stream, source or sink, and vortex. Most
of the results are classic, of course, needing only a brief treatment here. Graphical Method of
Superposition A simple means of accomplishing tot
i graphically is to plot the individual
stream functions separately and then look at their intersections. The value of tot at
each intersection is the sum of the individual values i which cross there. Connecting
intersections with the same value of tot creates the desired superimposed flow streamlines.
A simple example is shown in Fig. 8.4, summing two families of streamlines a and
b. The individual components are plotted separately, and four typical intersections are
shown. Dashed lines are then drawn through intersections representing the same sum
of a
b. These dashed lines are the desired solution. Often this graphical method
is a quick means of evaluating the proposed superposition before a fullblown numerical plot routine is executed.
ψ = ψ2
Family (a) ψ = ψ1 ψ = 2 ψ2
Combined
streamline ψ = ψ 1 + ψ2
ψ = 2 ψ1
ψ = ψ2
Family (b)  v v Fig. 8.4 Intersections of elementary
streamlines can be joined to form a
combined streamline. ψ = ψ1  eText Main Menu  Textbook Table of Contents  Study Guide 8.3 Superposition of PlaneFlow Solutions Some Examples from Chap. 4 501 In Sec. 4.10 we discussed a number of superposition examples.
1. Source m at ( a, 0) plus an equal sink at ( a, 0), Eq. (4.133), and Fig. 4.13:
m tan 1
2 x a)2
a)2 (x
1
m ln
(x
2 2ay
y2 a2 y2
y2 (4.133) The streamlines and potential lines are two families of orthogonal circles as
plotted in Fig. 4.13. They resemble a magnet with poles at (x, y) ( a, 0).
2. Sink m plus a vortex K, both at the origin, Eq. (4.134), and Fig. 4.14:
m K ln r m ln r K (4.134) The streamlines are logarithmic spirals swirling into the origin, as in Fig. 4.14.
They resemble a tornado or a bathtub vortex.
3. Uniform stream iU plus a source m at the origin, Eq. (4.135) and Fig. 4.15, the
Rankine halfbody:
U r sin m U r cos m ln r (4.135) If the origin contains a source, a plane halfbody is formed with its nose to
the left, as in Fig. 8.5a. If the origin is a sink, m 0, the halfbody nose is to
the right, as in Fig. 8.5c. In either case the stagnation point is at a position
a m/U away from the origin. Us (max) = 1.26 U∞ Laminar separation ψ = +π m
y
U∞  v v Fig. 8.5 The Rankine halfbody;
pattern (c) is not found in a real
fluid because of boundarylayer
separation: (a) uniform stream plus
a source equals a halfbody; stagnation point at x
a
m/U ; (b)
slight adverse gradient for s/a
greater than 3.0: no separation; (c)
uniform stream plus a sink equals
the rear of a halfbody; stagnation
point at x a m/U ; (d ) strong
adverse gradient for s/a
3.0:
separation.  y
x ψ=0 x a ψ = –π m
(a) (c) Us 1.0
U∞
0.5
0 Us 1.0
U∞
0.5
0 eText Main Menu 2 4 s
a 6 8 Separation 0
–8 –6 –4 (d) (b)  Textbook Table of Contents s
a  Study Guide –2 0 502 Chapter 8 Potential Flow and Computational Fluid Dynamics BoundaryLayer Separation
on a HalfBody Although the inviscidflow patterns, Fig. 8.5a and c, are mirror images, their viscous
(boundarylayer) behavior is different. The body shape and the velocity along the surface are repeated here from Sec. 4.10:
V2 U2 1 a2
r2 2a
cos
r along r m(
)
U sin (8.18) The computed surface velocities are plotted along the halfbody contours in Fig. 8.5b
and d as a function of arc length s/a measured from the stagnation point. These plots
are also mirror images. However, if the nose is in front, Fig. 8.5b, the pressure gradient there is favorable (decreasing pressure along the surface). In contrast, the pressure
gradient is adverse (increasing pressure along the surface) when the nose is in the rear,
Fig. 8.5d, and boundarylayer separation may occur.
Application to Fig. 8.5b of Thwaites’ laminarboundary method from Eqs. (7.54)
and (7.56) reveals that separation does not occur on the front nose of the halfbody.
Therefore Fig. 8.5a is a very realistic picture of streamlines past a halfbody nose. In
contrast, when applied to the tail, Fig. 8.5c, Thwaites’ method predicts separation at
about s/a
2.2, or
110°. Thus, if a halfbody is a solid surface, Fig. 8.5c is not
realistic and a broad separated wake will form. However, if the halfbody tail is a fluid
line separating the sinkdirected flow from the outer stream, as in Example 8.1, then
Fig. 8.5c is quite realistic and useful. Computations for turbulent boundarylayer theory would be similar: separation on the tail, no separation on the nose. EXAMPLE 8.1
An offshore power plant coolingwater intake sucks in 1500 ft3/s in water 30 ft deep, as in Fig.
E8.1. If the tidal velocity approaching the intake is 0.7 ft/s, (a) how far downstream does the intake effect extend and (b) how much width L of tidal flow is entrained into the intake? Halfbody
shape
Intake a? L? Solution
1500 ft3/s 0.7 ft /s Recall from Eq. (4.131) that the sink strength m is related to the volume flow Q and the depth
b into the paper Top view Q
2b m
E8.1 1500 ft3/s
2 (30 ft) 7.96 ft2/s Therefore from Fig. 8.5 the desired lengths a and L are
a 7.96 ft2/s
0.7 ft/s L Flow Past a Vortex m
U
2a 2 (11.4 ft) 11.4 ft Ans. (a) 71 ft Ans. (b) Consider a uniform stream U in the x direction flowing past a vortex of strength K
with center at the origin. By superposition the combined stream function is  v v stream  eText Main Menu  vortex U r sin Textbook Table of Contents K ln r  Study Guide (8.19) 8.3 Superposition of PlaneFlow Solutions 503 The velocity components are given by
r 1∂
r∂ ∂
∂r U cos U sin K
r (8.20) The streamlines are plotted in Fig. 8.6 by the graphical method, intersecting the circular streamlines of the vortex with the horizontal lines of the uniform stream.
By setting r
0 from (8.20) we find a stagnation point at
90°, r a
K/U , or (x, y) (0, a). This is where the counterclockwise vortex velocity K/r exactly cancels the stream velocity U .
Probably the most interesting thing about this example is that there is a nonzero lift
force normal to the stream on the surface of any region enclosing the vortex, but we
postpone this discussion until the next section. An Infinite Row of Vortices Consider an infinite row of vortices of equal strength K and equal spacing a, as in Fig.
8.7a. This case is included here to illustrate the interesting concept of a vortex sheet.
From Eq. (8.12c), the ith vortex in Fig. 8.7a has a stream function i
K ln ri,
so that the total infinite row has a combined stream function
K ln ri (8.21) i1 It can be shown [2, sec. 4.51] that this infinite sum of logarithms is equivalent to a
closedform function
1
2 K ln 2y
1
cosh
a
2 Since the proof uses the complex variable z
show the details here. x cos
iy, i 2x
a ( 1)1/2, we are not going to y x v v Fig. 8.6 Flow of a uniform stream
past a vortex constructed by the
graphical method.   eText Main Menu  Textbook Table of Contents  (8.22) Study Guide 504 Chapter 8 Potential Flow and Computational Fluid Dynamics
y ( x, y)
i th
vor tex ri
K K K K K K K K
x (a)
a a a a a a a y x (b) y u = – π K/a Fig. 8.7 Superposition of vortices:
(a) an infinite row of equal
strength; (b) streamline pattern for
part (a); (c) vortex sheet: part (b)
viewed from afar. (c) •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• x u = + π K/a The streamlines from Eq. (8.22) are plotted in Fig. 8.7b, showing what is called a
cat’seye pattern of enclosed flow cells surrounding the individual vortices. Above the
cat’s eyes the flow is entirely to the left, and below the cat’s eyes the flow is to the
right. Moreover, these left and right flows are uniform if y
a, which follows by differentiating Eq. (8.22)
u y y a K
a (8.23) where the plus sign applies below the row and the minus sign above the row. This uniform left and right streaming is sketched in Fig. 8.7c. We stress that this effect is induced by the row of vortices: There is no uniform stream approaching the row in this
example. The Vortex Sheet  v v When Fig. 8.7b is viewed from afar, the streaming motion is uniform left above and
uniform right below, as in Fig. 8.7c, and the vortices are packed so closely together  eText Main Menu  Textbook Table of Contents  Study Guide 8.3 Superposition of PlaneFlow Solutions 505 that they are smudged into a continuous vortex sheet. The strength of the sheet is defined as
2K
a (8.24) and in the general case can vary with x. The circulation about any closed curve which
encloses a short length dx of the sheet would be, from Eqs. (8.15) and (8.23),
d ul dx uu dx (ul uu) dx 2K
dx
a dx (8.25) where the subscripts l and u stand for lower and upper, respectively. Thus the sheet
strength
d /dx is the circulation per unit length of the sheet. Thus when a vortex
sheet is immersed in a uniform stream, is proportional to the lift per unit length of
any surface enclosing the sheet.
Note that there is no velocity normal to the sheet at the sheet surface. Therefore a
vortex sheet can simulate a thinbody shape, e.g., plate or thin airfoil. This is the basis of the thinairfoil theory mentioned in Sec. 8.7. The Doublet As we move far away from the sourcesink pair of Fig. 4.13, the flow pattern begins
to resemble a family of circles tangent to the origin, as in Fig. 8.8. This limit of vanishingly small distance a is called a doublet. To keep the flow strength large enough
to exhibit decent velocities as a becomes small, we specify that the product 2am remain constant. Let us call this constant . Then the stream function of a doublet is
lim a→0
2am m tan 1
2 x 2ay
y2 a2 2amy
x2 y2  v v Fig. 8.8 A doublet, or sourcesink
pair, is the limiting case of Fig.
4.13 viewed from afar. Streamlines
are circles tangent to the xaxis at
the origin. This figure was prepared
using the contour feature of MATLAB [34, 35].  eText Main Menu  Textbook Table of Contents  Study Guide y
2 x y2 (8.26) Chapter 8 Potential Flow and Computational Fluid Dynamics We have used the fact that tan 1
as becomes small. The quantity
the strength of the doublet.
Equation (8.26) can be rearranged to yield
x2 2 y 2 is called 2 2 so that, as advertised, the streamlines are circles tangent to the origin with centers on
the yaxis. This pattern is sketched in Fig. 8.8.
Although the author has in the past laboriously sketched streamlines by hand,
this is no longer necessary. Figure 8.8 was computerdrawn, using the contour
feature of the student version of MATLAB [35]. Simply set up a grid of points, spell
out the stream function, and call for a contour. For Fig. 8.8, the actual statements
were
[X,Y] meshgrid( 1:.02:1);
PSI
Y./(X.^2 Y.^2);
contour(X,Y,PSI,100)
This would produce 100 contour lines of from Eq. (8.26), with
1 for convenience. The plot would include grid lines, scale markings, and a surrounding box, and
the circles might look a bit elliptical. These blemishes can be eliminated with three
statements of cosmetic improvement:
axis square
grid off
axis off
The final plot, Fig. 8.8, has no markings but the streamlines themselves. MATLAB is
thus a recommended tool and, in addition, has scores of other uses. All this chapter’s
problem assignments which call for “sketch the streamlines/potential lines” can be completed using this contour feature. For further details, consult Ref. 34.
In a similar manner the velocity potential of a doublet is found by taking the limit
of Eq. (4.133) as a → 0 and 2am
x
doublet
2
x
y2
or
2 x 2 y2 2 (8.27) 2 The potential lines are circles tangent to the origin with centers on the xaxis. Simply
turn Fig. 8.8 clockwise 90° to visualize the lines, which are everywhere normal to
the streamlines.
The doublet functions can also be written in polar coordinates
sin
r cos
r (8.28) These forms are convenient for the cylinder flows of the next section.  v v 506  eText Main Menu  Textbook Table of Contents  Study Guide 8.4 Plane Flow Past ClosedBody Shapes 507 8.4 Plane Flow Past ClosedBody Shapes A variety of closedbody external flows can be constructed by superimposing a uniform stream with sources, sinks, and vortices. The body shape will be closed only if
the net source outflow equals the net sink inflow. The Rankine Oval A cylindrical shape called a Rankine oval, which is long compared with its height, is
formed by a sourcesink pair aligned parallel to a uniform stream, as in Fig. 8.9a.
From Eqs. (8.12a) and (4.133) the combined stream function is
Uy m tan 1 x2 2ay
y2 a2 U r sin m( 1 2) (8.29) When streamlines of constant are plotted from Eq. (8.29), an oval body shape appears, as in Fig. (8.9b). The halflength L and halfheight h of the oval depend upon
the relative strength of source and stream, i.e., the ratio m/(U a), which equals 1.0 in
Fig. 8.9b. The circulating streamlines inside the oval are uninteresting and not usually
shown. The oval is the line
0. y (x, y) U∞ r1
r2 –m +m θ2 θ1
a x a Source Sink
(a)
Shoulder
u max = 1.74 U∞ h
–m +m
a  v v Fig. 8.9 Flow past a Rankine oval:
(a) uniform stream plus a sourcesink pair; (b) oval shape and
streamlines for m/(U a) 1.0.  eText Main Menu L (b)  Textbook Table of Contents  Study Guide 508 Chapter 8 Potential Flow and Computational Fluid Dynamics There are stagnation points at the front and rear, x
L, and points of maximum
velocity and minimum pressure at the shoulders, y
h, of the oval. All these parameters are a function of the basic dimensionless parameter m/(U a), which we can
determine from Eq. (8.29):
h
a cot h/a
2m/(U a)
umax
U L
a 1 2m
Ua 1/2 (8.30) 2m/(U a)
1 h2/a2 1 As we increase m/(U a) from zero to large values, the oval shape increases in size and
thickness from a flat plate of length 2a to a huge, nearly circular cylinder. This is shown
in Table 8.1. In the limit as m/(U a) → , L/h → 1.0 and umax/U → 2.0, which is
equivalent to flow past a circular cylinder.
All the Rankine ovals except very thin ones have a large adverse pressure gradient
on their leeward surface. Thus boundarylayer separation will occur in the rear with a
broad wake flow, and the inviscid pattern is unrealistic in that region. Flow Past a Circular Cylinder
with Circulation From Table 8.1 at large source strength the Rankine oval becomes a large circle, much
greater in diameter than the sourcesink spacing 2a. Viewed on the scale of the cylinder, this is equivalent to a uniform stream plus a doublet. We also throw in a vortex at
the doublet center, which does not change the shape of the cylinder.
Thus the stream function for flow past a circular cylinder with circulation, centered
at the origin, is a uniform stream plus a doublet plus a vortex
U r sin sin
r K ln r const (8.31) The doublet strength has units of velocity times length squared. For convenience, let
U a2, where a is a length, and let the arbitrary constant in Eq. (8.31) equal
K ln a. Then the stream function becomes
U sin r a2
r K ln r
a (8.32) The streamlines are plotted in Fig. 8.10 for four different values of the dimensionless vortex strength K/(U a). For all cases the line
0 corresponds to the circle r
Table 8.1 RankineOval Parameters
from Eq. (8.30) v v  h/a L/a 0.0
0.01
0.1
1.0
10.0
100.0  m/(U a) 0.0
0.031
0.263
1.307
4.435
14.130 1.0
1.010
1.095
1.732
4.583
14.177 eText Main Menu  L/h umax/U 32.79
4.169
1.326
1.033
1.003
1.000 1.0
1.020
1.187
1.739
1.968
1.997
2.000 Textbook Table of Contents  Study Guide 8.4 Plane Flow Past ClosedBody Shapes a θ K (a) Fig. 8.10 Flow past a circular
cylinder with circulation for values
of K/(U a) of (a) 0, (b) 1.0, (c)
2.0, and (d) 3.0. 509 (b) (c) (d) a, that is, the shape of the cylindrical body. As circulation
2 K increases, the velocity becomes faster and faster below the cylinder and slower and slower above it.
The velocity components in the flow are given by
r 1∂
r∂ U cos ∂
∂r U sin The velocity at the cylinder surface r
r(r a) 0 a2
r2 1
1 a2
r2 (8.33) K
r a is purely tangential, as expected
(r a) 2U sin K
a For small K, two stagnation points appear on the surface at angles
from Eq. (8.34),
sin s K
2U a (8.34)
s where 0, or, (8.35)  v v Figure 8.10a is for K 0, s 0 and 180°, or doubly symmetric inviscid flow past a
cylinder with no circulation. Figure 8.10b is for K/(U a) 1, s 30 and 150°, and
Fig. 8.10c is the limiting case where the two stagnation points meet at the top,
K/(U a) 2, s 90°.  eText Main Menu  Textbook Table of Contents  Study Guide 510 Chapter 8 Potential Flow and Computational Fluid Dynamics For K 2U a, Eq. (8.35) is invalid, and the single stagnation point is above the
cylinder, as in Fig. 8.10d, at a point y h given by
h
a 1
[
2 In Fig. 8.10d, K/(U a) The KuttaJoukowski Lift
Theorem ( 2 K
Ua 4)1/2] 3.0, and h/a 2 (8.36) 2.6. For the cylinder flows of Fig. 8.10b to d there is a downward force, or negative lift,
called the Magnus effect, which is proportional to stream velocity and vortex strength.
We can see from the streamline pattern that the velocity on the top of the cylinder is
less and therefore the pressure higher from Bernoulli’s equation; this explains the force.
There is no viscous force, of course, because our theory is inviscid.
The surface velocity is given by Eq. (8.34). From Bernoulli’s equation (8.4), neglecting gravity, the surface pressure ps is given by
1
U2
2 p
or ps 1
2 p 1
2 ps
U2 (1 K
a 2U sin 4 sin2 4 sin 2 2 ) (8.37) K/(U a) and p is the freestream pressure. If b is the cylinder depth into
where
the paper, the drag D is the integral over the surface of the horizontal component of
pressure force
2 D
0 (ps p ) cos ba d where ps p is substituted from Eq. (8.37). But the integral of cos times any power
of sin over a full cycle 2 is identically zero. Thus we obtain the (perhaps surprising) result
D(cylinder with circulation) 0
(8.38)
This is a special case of d’Alembert’s paradox, mentioned in Sec. 1.10:
According to inviscid theory, the drag of any body of any shape immersed in a uniform stream is identically zero.
D’Alembert published this result in 1752 and pointed out himself that it did not square
with the facts for real fluid flows. This unfortunate paradox caused everyone to overreact and reject all inviscid theory until 1904, when Prandtl first pointed out the profound effect of the thin viscous boundary layer on the flow pattern in the rear, as in
Fig. 7.2b, for example.
The lift force L normal to the stream, taken positive upward, is given by summation of vertical pressure forces
2 L
0 (ps p ) sin Since the integral over 2 of any odd power of sin
parentheses in Eq. (8.37) contributes to the lift:  v v L  eText Main Menu 4K
1
U2
ba
aU
2  2 sin2 ba d
is zero, only the third term in the d U (2 K)b 0 Textbook Table of Contents  Study Guide 8.4 Plane Flow Past ClosedBody Shapes L
b or U 511 (8.39) Notice that the lift is independent of the radius a of the cylinder. Actually, though, as
we shall see in Sec. 8.7, the circulation depends upon the body size and orientation
through a physical requirement.
Equation (8.39) was generalized by W. M. Kutta in 1902 and independently by N.
Joukowski in 1906 as follows:
According to inviscid theory, the lift per unit depth of any cylinder of any shape immersed in a uniform stream equals u , where is the total net circulation contained within the body shape. The direction of the lift is 90° from the stream direction, rotating opposite to the circulation.
The problem in airfoil analysis, Sec. 8.7, is thus to determine the circulation
function of airfoil shape and orientation. Experimental Lift and Drag of
Rotating Cylinders as a It is nearly impossible to test Fig. 8.10 by constructing a doublet and vortex with the
same center and then letting a stream flow past them. But one physical realization
would be a rotating cylinder in a stream. The viscous noslip condition would cause
the fluid in contact with the cylinder to move tangentially at the cylinder peripheral
speed
a . A net circulation would be set up by this noslip mechanism, but it
turns out to be less than 50 percent of the value expected from inviscid theory, primarily because of flow separation behind the cylinder.
Figure 8.11 shows experimental lift and drag coefficients, based on planform area
2ba, of rotating cylinders. From Eq. (8.38) the theoretical drag is zero, but the actual
CD is quite large, more even than the stationary cylinder of Fig. 5.3. The theoretical
lift follows from Eq. (8.39)
CL 1
2 L
U2 (2ba) 2 U Kb
U2 ba 2
U s (8.40) where s K/a is the peripheral speed of the cylinder.
Figure 8.11 shows that the theoretical lift from Eq. (8.40) is much too high, but the
measured lift is quite respectable, much larger in fact than a typical airfoil of the same
chord length, e.g., Fig. 7.25. Thus rotating cylinders have practical possibilities. The
Flettner rotor ship built in Germany in 1924 employed rotating vertical cylinders which
developed a thrust due to any winds blowing past the ship. The Flettner design did not
gain any popularity, but such inventions may be more attractive in this era of high energy costs. EXAMPLE 8.2  v v The experimental Flettner rotor sailboat at the University of Rhode Island is shown in Fig. E8.2.
The rotor is 2.5 ft in diameter and 10 ft long and rotates at 220 r/min. It is driven by a small
lawnmower engine. If the wind is a steady 10 kn and boat relative motion is neglected, what is
the maximum thrust expected for the rotor? Assume standard air and water density.  eText Main Menu  Textbook Table of Contents  Study Guide 512 ;;;
;;;
;;; Chapter 8 Potential Flow and Computational Fluid Dynamics 10 Theory CL = 2 π aω
U∞ 8 ;;
;; Experiment
(Ref. 5) CL , CD 6 CL , CD 4 CL 2 ω
a U∞ Theory
CD = 0 0 Fig. 8.11 Theoretical and experimental lift and drag of a rotating
cylinder. (From Ref. 22.) 0 2 4 6 CD 8 Velocity ratio aω
U∞ (Courtesy of R. C. Lessmann, University of Rhode Island.)  v v E8.2  eText Main Menu  Textbook Table of Contents  Study Guide 8.4 Plane Flow Past ClosedBody Shapes 513 Solution
Convert the rotation rate to
ft/s, so the velocity ratio is 2 (220)/60
a
U 23.04 rad/s. The wind velocity is 10 kn (1.25 ft)(23.04 rad/s)
16.88 ft/s 16.88 1.71 Entering Fig. 8.11, we read CL 3.3 and CD 1.2. From Table A.6, standard air density is
0.002377 slug/ft3. Then the estimated rotor lift and drag are
L CL 1 U2 2ba
2 3.3( 1 )(0.002377)(16.88)2(2)(10)(1.25)
2 27.9 lbf
D CD 1 U2 2ba
2 L CD
CL 27.9 1.2
3.3 10.2 lbf The maximum thrust available is the resultant of these two
F [(27.9)2 (10.2)2]1/2 29 lbf Ans. If aligned along the boat’s keel, this thrust will drive the boat at a speed of about 5 kn through
the water. The Kelvin Oval A family of body shapes taller than they are wide can be formed by letting a uniform
stream flow normal to a vortex pair. If U is to the right, the negative vortex K is
placed at y
a and the counterclockwise vortex K placed at y
a, as in Fig.
8.12. The combined stream function is
Uy x2
1
K ln 2
x
2 (y
(y a)2
a)2 (8.41) The body shape is the line
0, and some of these shapes are shown in Fig. 8.12.
For K/(U a) 10 the shape is within 1 percent of a Rankine oval (Fig. 8.9) turned
90°, but for small K/(U a) the waist becomes pinched in, and a figureeight shape occurs at 0.5. For K/(U a) 0.5 the stream blasts right between the vortices and isolates two more or less circular body shapes, one surrounding each vortex.
A closed body of practically any shape can be constructed by proper superposition
of sources, sinks, and vortices. See the advanced work in Refs. 2 to 4 for further details. A summary of elementary potential flows is given in Table 8.2.  v v PotentialFlow Analogs For complicated potentialflow geometries, one can resort to other methods than superposition of sources, sinks, and vortices. There are a variety of devices which simulate solutions to Laplace’s equation.
From 1897 to 1900 HeleShaw [9] developed a technique whereby laminar flow between very closely spaced parallel plates simulated potential flow when viewed from
above the plates. Obstructions simulate body shapes, and dye streaks represent the
streamlines. The HeleShaw apparatus makes an excellent laboratory demonstration of
potential flow [10, pp. 8 – 10]. Figure 8.13a illustrates HeleShaw (potential) flow  eText Main Menu  Textbook Table of Contents  Study Guide 514 Chapter 8 Potential Flow and Computational Fluid Dynamics
y
K
= 1.5
U∞ a
1.0
0.75
0.55
0.5 –K
a
U∞ x +K Fig. 8.12 Kelvinoval body shapes
as a function of the vortexstrength
parameter K/(U a); outer streamlines not shown. through an array of cylinders, a flow pattern that would be difficult to analyze just using Laplace’s equation. However beautiful this array pattern may be, it is not a good
approximation to real (laminar viscous) array flow. Figure 8.13b shows experimental
streakline patterns for a similar staggeredarray flow at Re 6400. We see that the interacting wakes of the real flow (Fig. 8.13b) cause intensive mixing and transverse motion, not the smooth streaming passage of the potentialflow model (Fig. 8.13a). The
moral is that this is an internal flow with multiple bodies and, therefore, not a good
candidate for a realistic potentialflow model.
Other flowmapping techniques are discussed in Ref. 8. Electromagnetic fields also
satisfy Laplace’s equation, with voltage analogous to velocity potential and current
lines analogous to streamlines. At one time commercial analog field plotters were available, using thin conducting paper cut to the shape of the flow geometry. Potential lines
(voltage contours) were plotted by probing the paper with a potentiometer pointer. Table 8.2 Summary of Plane
Incompressible Potential Flows Type of flow
Stream iU
Line source (m
Line vortex
Halfbody Potential functions 0) or sink (m Rankine oval Ur sin v v Cylinder with circulation  eText Main Menu Uy
m
K ln r
Ur sin
Ur cos Ux
m ln r
K
m
m ln r sin
r Doublet  0) U sin  Remarks
See Fig. 4.12a
See Fig. 4.12b
See Fig. 4.12c
See Fig. 8.5 cos
r See Fig. 8.8 2) See Fig. 8.9 m( 1
a2
r
r Textbook Table of Contents r
K ln
a  See Fig. 8.10 Study Guide 8.4 Plane Flow Past ClosedBody Shapes 515 (a ) Fig. 8.13 Flow past a staggered array of cylinders: (a) potentialflow
model using the HeleShaw apparatus (Tecquipment Ltd., Nottingham,
England); (b) experimental streaklines for actual staggeredarray
flow at ReD 6400. (From Ref. 36,
courtesy of Jack Hoyt, with the permission of the American Society of
Mechanical Engineers.)
(b) Handsketching “curvilinear square’’ techniques were also popular. The availability and
the simplicity of digitalcomputer potentialflow methods [5 to 7] have made analog
models obsolete. EXAMPLE 8.3  v v A Kelvin oval from Fig. 8.12 has K/(U a)
the oval in terms of U .  eText Main Menu  1.0. Compute the velocity at the top shoulder of Textbook Table of Contents  Study Guide 516 Chapter 8 Potential Flow and Computational Fluid Dynamics Solution
We must locate the shoulder y
differentiation. At
0 and y h from Eq. (8.41) for
0 and then compute the velocity by
h and x 0, Eq. (8.41) becomes h
a h/a
K
ln
h/a
Ua 1
1 With K/(U a) 1.0 and the initial guess h/a 1.5 from Fig. 8.12, we iterate and find the location h/a 1.5434.
By inspection
0 at the shoulder because the streamline is horizontal. Therefore the shoulder velocity is, from Eq. (8.41),
u
Introducing K ∂
∂y yh U a and h yh K K U h a h a 1.5434a, we obtain ushoulder U (1.0 1.84 0.39) 2.45U Ans. Because they are shortwaisted compared with a circular cylinder, all the Kelvin ovals have shoulder velocity greater than the cylinder result 2.0U from Eq. (8.34). 8.5 Other Plane Potential
Flows2 References 2 to 4 treat many other potential flows of interest in addition to the cases
presented in Secs. 8.3 and 8.4. In principle, any plane potential flow can be solved by
the method of conformal mapping, by using the complex variable
z x iy ( 1)1/2 i It turns out that any arbitrary analytic function of this complex variable z has the remarkable property that both its real and its imaginary parts are solutions of Laplace’s
equation. If
f(z) f(x ∂2 f1
∂x2 then iy)
∂2 f1
∂y2 f1(x, y)
0 ∂2 f2
∂x2 i f2(x, y)
∂2 f2
∂y2 (8.42) We shall assign the proof of this as a problem. Even more remarkable if you have never
seen it before is that lines of constant f1 will be everywhere perpendicular to lines of
constant f2:
dy
dx f1 C 1
(dy/dx)f2 (8.43)
C We also leave this proof as a problem exercise. This is true for totally arbitrary f (z) as
long as this function is analytic; i.e., it must have a unique derivative df/dz at every
point in the region.
The net result of Eqs. (8.42) and (8.43) is that the functions f1 and f2 can be interpreted to be the potential lines and streamlines of an inviscid flow. By long custom we  v v 2  This section may be omitted without loss of continuity. eText Main Menu  Textbook Table of Contents  Study Guide 8.5 Other Plane Potential Flows 517 let the real part of f (z) be the velocity potential and the imaginary part be the stream
function
f (z) (x, y) i (x, y) (8.44) We try various functions f (z) and see whether any interesting flow pattern results. Of
course, most of them have already been found, and we simply report on them here.
We shall not go into the details, but there are excellent treatments of this complexvariable technique on both an introductory [4, chap. 5; 10, chap. 5] and a more advanced [2, 3,] level. The method is less important now because of the popularity of
digitalcomputer techniques.
As a simple example, consider the linear function
f (z) Uz Ux iU y It follows from Eq. (8.44) that
U x and
U y, which, we recall from Eq.
(8.12a), represents a uniform stream in the x direction. Once you get used to the complex variable, the solution practically falls in your lap.
To find the velocities, you may either separate and from f(z) and differentiate
or differentiate f directly
∂
∂x df
dz i ∂
∂x i ∂
∂y ∂
∂y u (8.45) i Thus the real part of df/dz equals u(x, y), and the imaginary part equals
(x, y). To
get a practical result, the derivative df/dz must exist and be unique, hence the requirement that f be an analytic function. For Eq. (8.45), df/dz U
u, since it is real, and
0, as expected.
Sometimes it is convenient to use the polarcoordinate form of the complex variable
z
where x
r rei iy
(x2 r cos y2)1/2 ir sin
tan 1 y
x This form is especially convenient when powers of z occur. Uniform Stream at an Angle
of Attack All the elementary plane flows of Sec. 8.2 have a complexvariable formulation. The
uniform stream U at an angle of attack has the complex potential
f(z) i U ze (8.46) Compare this form with Eq. (8.14). Line Source at a Point z0 Consider a line source of strength m placed off the origin at a point z0
complex potential is
f(z) m ln (z z0) x0 iy0. Its
(8.47)  v v This can be compared with Eq. (8.12b), which is valid only for the source at the origin. For a line sink, the strength m is negative.  eText Main Menu  Textbook Table of Contents  Study Guide 518 Chapter 8 Potential Flow and Computational Fluid Dynamics Line Vortex at a Point z0 If a line vortex of strength K is placed at point z0, its complex potential is
f (z) iK ln (z z0) (8.48) to be compared with Eq. (8.12c). Also compare to Eq. (8.47) to see that we reverse the
meaning of and simply by multiplying the complex potential by i. Flow around a Corner of
Arbitrary Angle Corner flow is an example of a pattern that cannot be conveniently produced by superimposing sources, sinks, and vortices. It has a strikingly simple complex representation
Azn f (z) Arnein Arn cos n iArn sin n where A and n are constants.
It follows from Eq. (8.44) that for this pattern
Arn cos n Arn sin n (8.49) Streamlines from Eq. (8.49) are plotted in Fig. 8.14 for five different values of n.
/n. Patterns in
The flow is seen to represent a stream turning through an angle
Fig. 8.14d and e are not realistic on the downstream side of the corner, where separation will occur due to the adverse pressure gradient and sudden change of direction. In
general, separation always occurs downstream of salient, or protruding corners, except
in creeping flows at low Reynolds number Re 1.
Since 360° 2 is the largest possible corner, the patterns for n 1 do not repre2
sent corner flow. They are peculiarlooking, and we ask you to plot one as a problem.
If we expand the plot of Fig. 8.14a to c to double size, we can represent stagnation
flow toward a corner of angle 2
2 /n. This is done in Fig. 8.15 for n 3, 2, and n=3 n=2
1 n=2 (a) (b) 3 n=2 2 n=3  v v Fig. 8.14 Streamlines for corner
flow, Eq. (8.49) for corner angle
of (a) 60°, (b) 90°, (c) 120°, (d)
270°, and (e) 360°. (c)  eText Main Menu (e) (d)  Textbook Table of Contents  Study Guide 8.5 Other Plane Potential Flows 519 n=3 3 n=2 ( a) n=2 ( b) ( c) Fig. 8.15 Streamlines for stagnation flow from Eq. (8.49) for corner angle 2 of (a) 120°, (b) 180°, and (c) 240°. 1.5. These are very realistic flows; although they slip at the wall, they can be patched
to boundarylayer theories very successfully. We took a brief look at corner flows before, in Examples 4.5 and 4.9 and in Probs. 4.49 to 4.51. Flow Normal to a Flat
Plate We treat this case separately because the Kelvin ovals of Fig. 8.12 failed to degenerate into a flat plate as K became small. The flat plate normal to a uniform stream is an
extreme case worthy of our attention.
Although the result is quite simple, the derivation is very complicated and is given,
e.g., in Ref. 2, sec. 9.3. There are three changes of complex variable, or mappings, beginning with the basic cylinderflow solution of Fig. 8.10a. First the uniform stream is
rotated to be vertical upward, then the cylinder is squeezed down into a plate shape,
and finally the free stream is rotated back to the horizontal. The final result for complex potential is
U (z2 i f(z) where 2a is the height of the plate. To isolate
real and imaginary parts  v v 2  eText Main Menu  2 U2 (x2 y2 Textbook Table of Contents a2)1/2 (8.50) or , square both sides and separate
a2)  U2 xy Study Guide 520 Chapter 8 Potential Flow and Computational Fluid Dynamics We can solve for to determine the streamlines
4 2 U2 (x2 y2 a2) U 4 x 2y2 (8.51) Equation (8.51) is plotted in Fig. 8.16a, revealing a doubly symmetric pattern of streamlines which approach very closely to the plate and then deflect up and over, with very
high velocities and low pressures near the plate tips.
The velocity s along the plate surface is found by computing df/dz from Eq. (8.50)
and isolating the imaginary part
s U plate surface (1 y/a
y2/a2)1/2 (8.52) Some values of surface velocity can be tabulated as follows: y/a 0.0 0.2 0.4 0.6 0.71 0.8 0.9 s/U 0.0 0.204 0.436 0.750 1.00 1.33 2.07 y U∞ a
C
L x (a) U∞ a Broad, low pressure
region of
separated flow
C
L x (b)  v v Fig. 8.16 Streamlines in upper halfplane for flow normal to a flat plate
of height 2a: (a) continuous potentialflow theory, Eq. (8.51); (b) actual measured flow pattern; (c) discontinuous potential theory with
k 1.5. U∞
Constantpressure
a
region
C
L Free streamline
discontinuity
at V = k U∞ x (c)  eText Main Menu  Textbook Table of Contents  Study Guide 1.0 8.6 Images 521 The origin is a stagnation point; then the velocity grows linearly at first and very rapidly
near the tip, with both velocity and acceleration being infinite at the tip.
As you might guess, Fig. 8.16a is not realistic. In a real flow the sharp salient edge
causes separation, and a broad, lowpressure wake forms in the lee, as in Fig. 8.16b.
Instead of being zero, the drag coefficient is very large, CD 2.0 from Table 7.2.
A discontinuouspotentialflow theory which accounts for flow separation was devised by Helmholtz in 1868 and Kirchhoff in 1869. This freestreamline solution is
shown in Fig. 8.16c, with the streamline which breaks away from the tip having a constant velocity V kU . From Bernoulli’s equation the pressure in the deadwater re1
2
gion behind the plate will equal pr p
k 2) to match the pressure along
2 U (1
the free streamline. For k 1.5 this HelmholtzKirchoff theory predicts pr p
0.625 U2 and an average pressure on the front pf p
0.375 U2 , giving an overall drag coefficient of 2.0, in agreement with experiment. However, the coefficient k is
a priori unknown and must be tuned to experimental data, so freestreamline theory
can be considered only a qualified success. For further details see Ref. 2, sec. 11.2. 8.6 Images3 The previous solutions have all been for unbounded flows, such as a circular cylinder
immersed in a broad expanse of uniformly streaming fluid, Fig. 8.10a. However, many
practical problems involve a nearby rigid boundary constraining the flow, e.g., (1)
groundwater flow near the bottom of a dam, (2) an airfoil near the ground, simulating
landing or takeoff, or (3) a cylinder mounted in a wind tunnel with narrow walls. In
such cases the basic unboundedpotentialflow solutions can be modified for wall effects by the method of images.
Consider a line source placed a distance a from a wall, as in Fig. 8.17a. To create
the desired wall, an image source of identical strength is placed the same distance below the wall. By symmetry the two sources create a planesurface streamline between
them, which is taken to be the wall.
In Fig. 8.17b a vortex near a wall requires an image vortex the same distance below but of opposite rotation. We have shaded in the wall, but of course the pattern could
also be interpreted as the flow near a vortex pair in an unbounded fluid.
In Fig. 8.17c an airfoil in a uniform stream near the ground is created by an image
airfoil below the ground of opposite circulation and lift. This looks easy, but actually
it is not because the airfoils are so close together that they interact and distort each
other’s shapes. A rule of thumb is that nonnegligible shape distortion occurs if the body
shape is within two chord lengths of the wall. To eliminate distortion, a whole series
of “corrective’’ images must be added to the flow to recapture the shape of the original isolated airfoil. Reference 2, sec. 7.75, has a good discussion of this procedure,
which usually requires digitalcomputer summation of the multiple images needed.
Figure 8.17d shows a source constrained between two walls. One wall required only
one image in Fig. 8.17a, but two walls require an infinite array of image sources above
and below the desired pattern, as shown. Usually computer summation is necessary,
but sometimes a closedform summation can be achieved, as in the infinite vortex row
of Eq. (8.22).  v v 3  This section may be omitted without loss of continuity. eText Main Menu  Textbook Table of Contents  Study Guide 522 Chapter 8 Potential Flow and Computational Fluid Dynamics a a (a) (c) Fig. 8.17 Constraining walls can be
created by image flows: (a) source
near a wall with identical image
source; (b) vortex near a wall with
image vortex of opposite sense; (c)
airfoil in ground effect with image
airfoil of opposite circulation; (d)
source between two walls requiring
an infinite row of images. (b) (d ) EXAMPLE 8.4
For the source near a wall as in Fig. 8.17a, the wall velocity is zero between the sources, rises
to a maximum moving out along the wall, and then drops to zero far from the sources. If the
source strength is 8 m2/s, how far from the wall should the source be to ensure that the maximum velocity along the wall will be 5 m/s? Solution  v v At any point x along the wall, as in Fig. E8.4, each source induces a radial outward velocity
m/r, which has a component r cos along the wall. The total wall velocity is thus
r  eText Main Menu  Textbook Table of Contents  Study Guide 8.7 Airfoil Theory 523
Source m = 8 m2/s
r a =m
r r θ x Wall θ
a E8.4 r r Source m 2 uwall
From the geometry of Fig. E8.4, r
can be expressed as (x2 0 and at x →
du
dx 0 cos a2)1/2 and cos x/r. Then the total wall velocity 2mx
x2 a2 u
This is zero at x
to zero r . To find the maximum velocity, differentiate and set equal at x a and umax m
a We have omitted a bit of algebra in giving these results. For the given source strength and maximum velocity, the proper distance a is
a m
umax 8 m2/s
5 m/s 1.625 m Ans. For x a, there is an adverse pressure gradient along the wall, and boundarylayer theory should
be used to predict separation. 8.7 Airfoil Theory4 As mentioned in conjunction with the KuttaJoukowski lift theorem, Eq. (8.39), the
problem in airfoil theory is to determine the net circulation as a function of airfoil
shape and freestream angle of attack . The Kutta Condition Even if the airfoil shape and freestream angle of attack are specified, the potentialflowtheory solution is nonunique: An infinite family of solutions can be found corresponding to different values of circulation . Four examples of this nonuniqueness were
shown for the cylinder flows in Fig. 8.10. The same is true of the airfoil, and Fig. 8.18
shows three mathematically acceptable “solutions’’ to a given airfoil flow for small
(Fig. 8.18a), large (Fig. 8.18b), and medium (Fig. 8.18c) net circulation. You can guess  v v 4  This section may be omitted without loss of continuity. eText Main Menu  Textbook Table of Contents  Study Guide 524 Chapter 8 Potential Flow and Computational Fluid Dynamics
Γ < ΓKutta (a) Γ > ΓKutta (b) Fig. 8.18 The Kutta condition properly simulates the flow about an
airfoil; (a) too little circulation,
stagnation point on rear upper surface; (b) too much, stagnation point
on rear lower surface; (c) just right,
Kutta condition requires smooth
flow at trailing edge. Γ = ΓKutta (c) which case best simulates a real airfoil from the earlier discussion of transientlift development in Fig. 7.23. It is the case (Fig. 8.18c) where the upper and lower flows
meet and leave the trailing edge smoothly. If the trailing edge is rounded slightly, there
will be a stagnation point there. If the trailing edge is sharp, approximating most airfoil designs, the upper and lowersurface flow velocities will be equal as they meet
and leave the airfoil.
This statement of the physically proper value of is generally attributed to W. M.
Kutta, hence the name Kutta condition, although some texts give credit to Joukowski
and/or Chaplygin. All airfoil theories use the Kutta condition, which is in good agreement with experiment. It turns out that the correct circulation Kutta depends upon flow
velocity, angle of attack, and airfoil shape. FlatPlate Airfoil VortexSheet
Theory The flat plate is the simplest airfoil, having no thickness or “shape,’’ but even its theory is not so simple. The problem can be solved by a complexvariable mapping [2,
p. 480], but here we shall use a vortexsheet approach. Figure 8.19a shows a flat plate
of length C simulated by a vortex sheet of variable strength (x). The free stream U
is at an angle of attack with respect to the plate chord line.
To make the lift “up’’ with flow from left to right as shown, we specify here that
the circulation is positive clockwise. Recall from Fig. 8.7c that there is a jump in tangential velocity across a sheet equal to the local strength  v v uu  eText Main Menu  ul (x) Textbook Table of Contents (8.53)  Study Guide 8.7 Airfoil Theory 525 1
If we omit the free stream, the sheet should cause a rightward flow u
on the
2
upper surface and an equal and opposite leftward flow on the lower surface, as shown
in Fig. 8.19a. The Kutta condition for this sharp trailing edge requires that this velocity difference vanish at the trailing edge to keep the exit flow smooth and parallel (C) 0 (8.54) The proper solution must satisfy this condition, after which the total lift can be computed by summing the sheet strength over the whole airfoil. From Eq. (8.39) for a foil
of depth b
C L Ub (x) dx (8.55) 0 y
γ ( x) δu ≈ α x=C 1
δu ≈ γ
2 0 U∞ 1γ
2 (a) x 8
CL ≈ area between curves 6
4 Cpl = 2 1
2 ( x C 1)
– 2
Cp
sin α 0
–2
Cpu = – Cpl –4
–6
–8 (b) 1.2
Uu
U∞  v v Fig. 8.19 Vortexsheet solution for
the flatplate airfoil; (a) sheet
geometry; (b) theoretical pressure
coefficient on upper and lower surfaces; (c) uppersurface velocity
with laminar separation points S.  eText Main Menu Separation:
S (6°)
S (5°)
1.1 1.0 α = 3°
4°
0 5° 0.2 S (3°) 6°
0.4 0.6
(c)  S (4°) x
C Textbook Table of Contents 0.8  1.0 Study Guide Chapter 8 Potential Flow and Computational Fluid Dynamics An alternate way to compute lift is from the dimensionless pressure coefficient Cp
on the upper and lower surfaces
pu,l Cpu,l 1
2 p 2
Uu,l
U2 1 2 U (8.56) where the last expression follows from Bernoulli’s equation. The surface velocity
squared is given by combining the uniform stream and the vortexsheet velocity components from Fig. 8.19a:
2
Uu,l u)2 (U cos (U sin )2 2u
(8.57)
U
where we have made the approximations u
U and cos in the last expression, assuming a small angle of attack. Equations (8.56) and (8.57) combine to the firstorder approximation
2u
(8.58)
Cpu,l
U
U
The lift force is the integral of the pressure difference over the length of the airfoil, assuming depth b
2
Uu,l U2 2U u2 u cos U2 1 C L
0 or CL 1
2 (pl pu)b dx 1 L
U2 bC 0 (Cpl Cpu) 1 dx
C 2
0 U d x
C (8.59) Equations (8.55) and (8.59) are entirely equivalent within the smallangle approximations.
The sheet strength (x) is computed from the requirement that the net normal velocity (x) be zero at the sheet (y 0), since the sheet represents a solid plate or
stream surface. Consider a small piece of sheet dx located at position x0. The velocity at point x on the sheet is that of an infinitesimal line vortex of strength d
dx
d
dx
d
2 r x0→x
2 (x0 x)
x
The total normal velocity induced by the entire sheet at point x is thus
C
sheet
0 dx
2 (x0 (8.60) x) Meanwhile, from Fig. 8.19a, the uniform stream induces a constant normal velocity at every point on the sheet given by
stream Setting the sum of sheet and stream
C
0 U sin equal to zero gives the integral equation dx
x0 x 2 U sin (8.61) to be solved for (x) subject to the Kutta condition (C)  v v 526  eText Main Menu  Textbook Table of Contents 0 from Eq. (8.54).  Study Guide 8.7 Airfoil Theory 527 Although Eq. (8.61) is quite formidable (and not only for beginners), in fact it was
solved long ago by using integral formulas developed by Poisson in the nineteenth century. The sheet strength which satisfies Eq. (8.61) is
(x) 1/2 C
x 2U sin 1 (8.62) From Eq. (8.58) the surfacepressure coefficients are thus
Cpu,l C
x 2 sin 1/2 1 (8.63) Details of the calculations are given in advanced texts [for example, 11, chap. 5].
The pressure coefficients from Eq. (8.63) are plotted in Fig. 8.19b, showing that the
upper surface has pressure continually increasing with x, that is, an adverse gradient.
1
The uppersurface velocity Uu U
uU
is plotted in Fig. 8.19c for var2
ious angles of attack. Above
5° the sheet contribution u is about 20 percent of
U so that the smalldisturbance assumption is violated. Figure 8.19c also shows separation points computed by Thwaites’ laminarboundarylayer method, Eqs. (7.54) and
(7.55). The prediction is that a flat plate would be extensively stalled on the upper surface for
6°, which is approximately correct.
The lift coefficient of the airfoil is proportional to the area between cpl and cpu in
Fig. 8.19b, from Eq. (8.59):
1 CL 2
0 U d x
C 1 C
x 4 sin
0 1/2 1 d x
C 2 sin 2 (8.64) This is a classic result which was alluded to earlier in Eq. (7.70) without proof.
Also of interest is the moment coefficient about the leading edge (LE) of the airfoil, taken as positive counterclockwise
CMLE 1
2 MLE
U 2 bC2 1
0 (Cpl Cpu) x
x
d
C
C 2 sin 1
CL
4 (8.65) Thus the center of pressure (CP), or position of the resultant lift force, is at the onequarterchord point
x
C CP 1
4 (8.66)  v v This theoretical result is independent of the angle of attack.
These results can be compared with experimental results for NACA airfoils in Fig.
8.20). The thinnest NACA airfoil is t/C 0.06, and the thickest is 24 percent, or
t/C 0.24. The liftcurve slope dCL /d is within 9 percent of the theoretical value
of 2 for all the various airfoil families at all thicknesses. Increasing thickness tends
to increase both CL,max and the stall angle. The stall angle at t/C 0.06 is about 8°
and would be even less for a flat plate, verifying the boundarylayer separation estimates in Fig. 8.19c. Best performance is usually at about the 12 percent thickness
point for any airfoil.  eText Main Menu  Textbook Table of Contents  Study Guide 528 Chapter 8 Potential Flow and Computational Fluid Dynamics 7
2 π (1 + dCL
dα 0.77 t /C ) 2π 2π 6
65series
63, 64series
4digit, 5digit
5 6% 9% 12% 15% 18% t
C 2.0 CL max Series: 1.0 0 00
24632306% 9% 12% 15% 18% t
C
20° α stall Fig. 8.20 Lift characteristics of
smooth NACA airfoils as a function of thickness ratio, for infinite
aspect ratio. (From Ref. 12.) 0° 6% 9% 12% 15% 18% The theory of thick cambered airfoils is covered in advanced texts [for example, 2 to
4]; Ref. 13 has a thorough and comprehensive review of both inviscid and viscous aspects of airfoil behavior.
Basically the theory uses a complexvariable mapping which transforms the flow
about a cylinder with circulation in Fig. 8.10 into flow about a foil shape with circulation. The circulation is then adjusted to match the Kutta condition of smooth exit
flow from the trailing edge.
Regardless of the exact airfoil shape, the inviscid mapping theory predicts that the
correct circulation for any thick cambered airfoil is v v 00
2463230 t
C Potential Theory for Thick
Cambered Airfoils  Series: 10°  eText Main Menu  Textbook Table of Contents  Study Guide 8.7 Airfoil Theory 529 bCU Kutta 1 t
sin (
C 0.77 ) (8.67) where
tan 1 (2h/C) and h is the maximum camber, or maximum deviation of the
airfoil midline from its chord line, as in Fig. 8.21a.
The lift coefficient of the infinitespan airfoil is thus
CL U
U2 bC 1
2 2 1 0.77 t
sin (
C ) Midline
Chordline
t h C
(a) 0.28
63X X X 0.27 64X X X
xcp
C 0.26 65X X X
00X X, 14X X 0.25 24X X, 44X X
0.24
230X X
0.23 6% 9% 12% 15% t
C 18% (b) 0.015
CD min Rough (all) 0.010 4digit, 5digit 0.005 636465
66 (Smooth)  v v Fig. 8.21 Characteristics of NACA
airfoils: (a) typical thick cambered
airfoil; (b) centerofpressure data;
and (c) minimum drag coefficient.  eText Main Menu 0 6% 9% 12% 15% t
C 18% (c)  Textbook Table of Contents  Study Guide (8.68) 530 Chapter 8 Potential Flow and Computational Fluid Dynamics This reduces to Eq. (8.64) when the thickness and camber are zero. Figure 8.20 shows
that the thickness effect 1 0.77t/C is not verified by experiment. Some airfoils increase lift with thickness, others decrease, and none approach the theory very closely,
the primary reason being the boundarylayer growth on the upper surface affecting the
airfoil “shape.’’ Thus it is customary to drop the thickness effect from the theory
CL 2 sin ( ) (8.69) The theory correctly predicts that a cambered airfoil will have finite lift at zero angle
of attack and zero lift (ZL) at an angle
1 tan ZL 2h
C (8.70) Equation (8.70) overpredicts the measured zerolift angle by 1° or so, as shown in Table
8.3. The measured values are essentially independent of thickness. The designation XX
in the NACA series indicates the thickness in percent, and the other digits refer to camber and other details. For example, the 2415 airfoil has 2 percent maximum camber
(the first digit) occurring at 40 percent chord (the second digit) with 15 percent maximum thickness (the last two digits). The maximum thickness need not occur at the
same position as the maximum camber.
Figure 8.21b shows the measured position of the center of pressure of the various
NACA airfoils, both symmetric and cambered. In all cases xCP is within 0.02 chord
length of the theoretical quarterchord point predicted by Eq. (8.66). The standard cambered airfoils (24, 44, and 230 series) lie slightly forward of x/C 0.25 and the lowdrag (60 series) foils slightly aft. The symmetric airfoils are at 0.25.
Figure 8.21c shows the minimum drag coefficient of NACA airfoils as a function
of thickness. As mentioned earlier in conjunction with Fig. 7.25, these foils when
smooth actually have less drag than turbulent flow parallel to a flat plate, especially
the lowdrag 60 series. However, for standard surface roughness all foils have about
the same minimum drag, roughly 30 percent greater than that for a smooth flat plate. Wings of Finite Span The results of airfoil theory and experiment in the previous subsection were for twodimensional, or infinitespan, wings. But all real wings have tips and are therefore of
finite span or finite aspect ratio AR, defined by
AR Table 8.3 ZeroLift Angle of
NACA Airfoils v v  Camber h/C, % 24XX
44XX
230XX
632XX
634XX
641XX  Airfoil series 2.0
4.0
1.8
2.2
4.4
1.1 eText Main Menu  b2
Ap b
C Measured (8.71) ZL, deg 2.1
4.0
1.3
1.8
3.1
0.8 Textbook Table of Contents Theory , deg
2.3
4.6
2.1
2.5
5.0
1.2  Study Guide 8.7 Airfoil Theory 531 where b is the span length from tip to tip and Ap is the planform area of the wing as
seen from above. The lift and drag coefficients of a finiteaspectratio wing depend
strongly upon the aspect ratio and slightly upon the planform shape of the wing.
Vortices cannot end in a fluid; they must either extend to the boundary or form a
closed loop. Figure 8.22a shows how the vortices which provide the wing circulation
bend downstream at finite wing tips and extend far behind the wing to join the starting vortex (Fig. 7.23) downstream. The strongest vortices are shed from the tips, but U• (a)
y = 1b
2 y Circulation
Γ( y) y=0
y = –1b
2
x (b) y, η
Wing
replaced by
“ lifting line”
x  v v Fig. 8.22 Liftingline theory for a
finite wing: (a) actual trailingvortex system behind a wing; (b) simulation by vortex system “bound’’
to the wing; (c) downwash on the
wing due to an element of the trailingvortex system.  eText Main Menu γ (η ) dη = vor tex sheet element
d w = downwash due to γ d η (c)  Textbook Table of Contents  Study Guide Chapter 8 Potential Flow and Computational Fluid Dynamics some are shed from the body of the wing, as sketched schematically in Fig. 8.22b. The
effective circulation (y) of these trailing shed vortices is zero at the tips and usually
a maximum at the center plane, or root, of the wing. In 1918 Prandtl successfully modeled this flow by replacing the wing by a single lifting line and a continuous sheet of
semiinfinite trailing vortices of strength (y) d /dy, as in Fig. 8.22c. Each elemental piece of trailing sheet ( ) d induces a downwash, or downward velocity, dw(y),
given by
( )d
4 (y
) dw(y) at position y on the lifting line. Note the denominator term 4 rather than 2 because
the trailing vortex extends only from 0 to rather than from
to
.
The total downwash w(y) induced by the entire trailing vortex system is thus
1
4 w(y) (1/2)b ( )d
y (1/2)b (8.72) When the downwash is vectorially added to the approaching free stream U , the effective angle of attack at this section of the wing is reduced to
eff i i 1 tan w
U w
U (8.73) where we have used a smallamplitude approximation w U .
The final step is to assume that the local circulation (y) is equal to that of a twodimensional wing of the same shape and same effective angle of attack. From thinairfoil theory, Eqs. (8.55) and (8.64), we have the estimate
CL 1
2 or Ub
U2 bC
CU 2 eff (8.74) eff Combining Eqs. (8.72) and (8.74), we obtain Prandtl’s liftingline theory for a finitespan wing
C(y)U (y) (y) 1
4U (1/2)b
(1/2)b (d /d ) d
y (8.75) This is an integrodifferential equation to be solved for (y) subject to the conditions
( 1 b)
( 1 b) 0. It is similar to the thinairfoil integral equation (8.61) and
2
2
even more formidable. Once it is solved, the total wing lift and induced drag are
given by
(1/2)b L U (1/2)b (y) dy
(1/2)b Di U
(1/2)b (y) i(y) dy (8.76) Here is a case where the drag is not zero in a frictionless theory because the downwash causes the lift to slant backward by angle i so that it has a drag component parallel to the freestream direction, dDi dL sin i dL i.
The complete solution to Eq. (8.75) for arbitrary wing planform C(y) and arbitrary  v v 532  eText Main Menu  Textbook Table of Contents  Study Guide 8.7 Airfoil Theory 533 twist (y) is treated in advanced texts [for example, 11]. It turns out that there is a simple representative solution for an untwisted wing of elliptical planform
2 1/2 2y
b C0 1 C(y) (8.77) The area and aspect ratio of this wing are
(1/2)b Ap 1
4 C dy
(1/2)b bC0 AR 4b
C0 (8.78) The solution to Eq. (8.75) for this C(y) is an elliptical circulation distribution of exactly similar shape
(y) 0 2 1/2 2y
b 1 (8.79) Substituting into Eq. (8.75) and integrating give a relation between
0 1 C0U
2/AR 0 and C0
(8.80) where is assumed constant across the untwisted wing.
Substitution into Eq. (8.76) gives the ellipticalwing lift
L 2 bC0 U2 /(1 1
4 or 2/AR) 2 CL 1 2/AR (8.81) If we generalize this to a thick cambered finite wing of approximately elliptical planform, we obtain
CL 2 sin (
)
1 2/AR (8.82) This result was given without proof as Eq. (7.70). From Eq. (8.72) the computed downwash for the elliptical wing is constant
w(y) 2U
2 AR const (8.83) Finally, the induced drag coefficient from Eq. (8.76) is
CDi CL 2
CL
AR w
U (8.84)  v v This was given without proof as Eq. (7.71).
Figure 8.23 shows the effectiveness of this theory when tested against a nonelliptical cambered wing by Prandtl in 1921 [14]. Figure 8.23a and b shows the measured
lift curves and drag polars for five different aspect ratios. Note the increase in stall angle and drag and the decrease in lift slope as the aspect ratio decreases.
Figure 8.23c shows the lift data replotted against effective angle of attack eff
(
)/(1 2/AR), as predicted by Eq. (8.82). These curves should be equivalent to  eText Main Menu  Textbook Table of Contents  Study Guide 534 Chapter 8 Potential Flow and Computational Fluid Dynamics 1.5 1.5
AR = 7 5 AR = 7 3
2 1.0 1 CL 1.0 1
0.5
CDo ≈ 0.01 β = – 5°
0° 10° α 0 20° 0 0.1
CD 0.2 (b) (a) 1.5 3
2 CL 0.5 0
– 5° 5 1.5 2 π (α + β )
17 AR = 2 3 5 AR = 5 1.0 7 1 3,2
1.0 CL CL
0.5 Fig. 8.23 Comparison of theory and
experiment for a finite wing: (a)
measured lift [14]; (b) measured
drag polar [14]; (c) lift reduced to
infinite aspect ratio; (d) drag polar
reduced to infinite aspect ratio. 0
– 5° 0.5 0° 10°
α +β
1 + 2 /AR 20° 0 0 0.05
C2
CD – L
πAR (c) 0.1 (d ) an infiniteaspectratio wing, and they do collapse together except near stall. Their common slope dCL/d is about 10 percent less than the theoretical value 2 , but this is
consistent with the thickness and shape effects noted in Fig. 8.20.
Figure 8.23d shows the drag data replotted with the theoretical induced drag CDi
C2 /( AR) subtracted out. Again, except near stall, the data collapse onto a single line
L
of nearly constant infiniteaspectratio drag CD0 0.01. We conclude that the finitewing theory is very effective and may be used for design calculations. 8.8 Axisymmetric Potential
Flow5 The same superposition technique which worked so well for plane flow in Sec. 8.3 is
also successful for axisymmetric potential flow. We give some brief examples here.
Most of the basic results carry over from plane to axisymmetric flow with only slight
changes owing to the geometric differences. Consider the following related flows:  v v 5  This section may be omitted without loss of continuity. eText Main Menu  Textbook Table of Contents  Study Guide 8.8 Axisymmetric Potential Flow 535
Basic plane flow Counterpart axisymmetric flow Uniform stream
Line source or sink
Line doublet
Line vortex
Rankine halfbody cylinder
Rankineoval cylinder
Circular cylinder
Symmetric airfoil Uniform stream
Point source or sink
Point doublet
No counterpart
Rankine halfbody of revolution
Rankine oval of revolution
Sphere
Tearshaped body Since there is no such thing as a point vortex, we must forgo the pleasure of studying
circulation effects in axisymmetric flow. However, as any smoker knows, there is an
axisymmetric ring vortex, and there are also ring sources and ring sinks, which we
leave to advanced texts [for example, 3]. Spherical Polar Coordinates Axisymmetric potential flows are conveniently treated in the spherical polar coordinates of Fig. 8.24. There are only two coordinates (r, ), and flow properties are constant on a circle of radius r sin about the xaxis.
The equation of continuity for incompressible flow in these coordinates is
∂2
(r
∂r r sin ) ∂
(r
∂ sin ) 0 (8.85) where r and
are radial and tangential velocity as shown. Thus a spherical polar
stream function6 exists such that
∂
∂ 1
r2 sin r 1
r sin ∂
∂r (8.86) In like manner a velocity potential (r, ) exists such that
r y 1∂
r∂ ∂
∂r (8.87) Properties vary with θ on a
circle about z axis νθ
νr
r
θ Fig. 8.24 Spherical polar coordinates for axisymmetric flow. Axis of
symmetry x Properties do not vary on a
circle about x axis z  v v 6
It is often called Stokes’ stream function, having been used in a paper Stokes wrote in 1851 on viscous sphere flow.  eText Main Menu  Textbook Table of Contents  Study Guide 536 Chapter 8 Potential Flow and Computational Fluid Dynamics These formulas serve to deduce the
axisymmetric potential flows. Uniform Stream in the x Direction and functions for various elementary A stream U in the x direction has components
U cos r U sin Substitution into Eqs. (8.86) and (8.87) and integrating give
1
2 Uniform stream: U r2 sin2 U r cos (8.88) As usual, arbitrary constants of integration have been neglected. Point Source or Sink Consider a volume flux Q issuing from a point source. The flow will spread out radially and at radius r will equal Q divided by the area 4 r2 of a sphere. Thus
Q
4 r2 r with m 0 (8.89) Q/(4 ) for convenience. Integrating (8.86) and (8.87) gives Point source m
r m cos For a point sink, change m to Point Doublet m
r2 (8.90) m in Eq. (8.90). Exactly as in Fig. 8.8, place a source at (x, y) ( a, 0) and an equal sink at (
0), taking the limit as a becomes small with the product 2am
held constant
lim (m cos doublet a→0
2am source m cos sin2
r sink) a, (8.91) We leave the proof of this limit as a problem. The pointdoublet velocity potential doublet m
rsource lim a→0
2am cos
r2 m
rsink (8.92) The streamlines and potential lines are shown in Fig. 8.25. Unlike the plane doublet
flow of Fig. 8.8, neither set of lines represents perfect circles. Uniform Stream plus a
Point Source By combining Eqs. (8.88) and (8.90) we obtain the stream function for a uniform stream
plus a point source at the origin
1
2 U r2 sin2 m cos (8.93) From Eq. (8.86) the velocity components are, by differentiation,  v v r  eText Main Menu  U cos m
r2 U sin Textbook Table of Contents  Study Guide (8.94) 8.8 Axisymmetric Potential Flow 537
y Potential
lines x Fig. 8.25 Streamlines and potential
lines due to a point doublet at the
origin, from Eqs. (8.91) and (8.92). Setting these equal to zero reveals a stagnation point at
180° and r a
(m/U )1/2, as shown in Fig. 8.26. If we let m U a2, the stream function can be rewritten as
cos 2 Ua r
a 1
2 2 sin2 (8.95) The stream surface which passes through the stagnation point (r, ) (a, ) has the
value
U a2 and forms a halfbody of revolution enclosing the point source, as
shown in Fig. 8.26. This halfbody can be used to simulate a pitot tube. Far downstream the halfbody approaches the constant radius R 2a about the xaxis. The maximum velocity and minimum pressure along the halfbody surface occur at
70.5°, Vs max = 1.155 U∞ y
Stagnation
point θ a U∞ 2a r
x Source
Halfbody
θ
r
a = csc 2 2a  v v Fig. 8.26 Streamlines for a Rankine
halfbody of revolution.  eText Main Menu  Textbook Table of Contents  Study Guide 538 Chapter 8 Potential Flow and Computational Fluid Dynamics r a 3, Vs 1.155U . Downstream of this point there is an adverse gradient as Vs
slowly decelerates to U , but boundarylayer theory indicates no flow separation. Thus
Eq. (8.95) is a very realistic simulation of a real halfbody flow. But when the uniform
stream is added to a sink to form a halfbody rear surface, e.g., similar to Fig. 8.5c,
separation is predicted and the inviscid pattern is not realistic. Uniform Stream plus a
Point Doublet From Eqs. (8.88) and (8.91), combination of a uniform stream and a point doublet at
the origin gives
1
U r2 sin2
2 r sin2 (8.96) Examination of this relation reveals that the stream surface
sphere of radius
r a 2
U 0 corresponds to the 1/3 (8.97) This is exactly analogous to the cylinder flow of Fig. 8.10a formed by combining a
uniform stream and a line doublet.
1
3
Letting
2 U a for convenience, we rewrite Eq. (8.96) as
1
2 2 Ua sin2 r2
a2 a
r (8.98) The streamlines for this sphere flow are plotted in Fig. 8.27. By differentiation from
Eq. (8.86) the velocity components are
r U cos a3
r3 1 1
U sin
2 2 a3
r3 (8.99) We see that the radial velocity vanishes at the sphere surface r a, as expected. There
is a stagnation point at the front (a, ) and the rear (a, 0) of the sphere. The maximum Vmax = 1.5 U∞
Potential lines a θ U∞ S  v v Fig. 8.27 Streamlines and potential
lines for inviscid flow past a
sphere. Laminar
separation
at 76°  eText Main Menu  Textbook Table of Contents  Study Guide 8.8 Axisymmetric Potential Flow 539
1
2 velocity occurs at the shoulder (a,
surfacevelocity distribution is
Vs ), where
3
2 ra r 0 and U sin 1.5U . The
(8.100) Note the similarity to the cylinder surface velocity equal to 2U sin from Eq. (8.34)
with zero circulation.
Equation (8.100) predicts, as expected, an adverse pressure gradient on the rear (
90°) of the sphere. If we use this distribution with laminarboundarylayer theory [for
example, 15, p. 298], separation is computed to occur at about
76°, so that in the
actual flow pattern of Fig. 7.14 a broad wake forms in the rear. This wake interacts
with the free stream and causes Eq. (8.100) to be inaccurate even in the front of the
sphere. The measured maximum surface velocity is equal only to about 1.3U and occurs at about
107° (see Ref. 15, sec. 4.10.4, for further details). The Concept of Hydrodynamic
Mass When a body moves through a fluid, it must push a finite mass of fluid out of the way.
If the body is accelerated, the surrounding fluid must also be accelerated. The body
behaves as if it were heavier by an amount called the hydrodynamic mass (also called
the added or virtual mass) of the fluid. If the instantaneous body velocity is U(t), the
summation of forces must include this effect
F mh) (m dU
dt (8.101) where mh, the hydrodynamic mass, is a function of body shape, the direction of motion, and (to a lesser extent) flow parameters such as the Reynolds number.
According to potential theory [2, sec. 6.4; 3, sec. 9.22], mh depends only on the
shape and direction of motion and can be computed by summing the total kinetic energy of the fluid relative to the body and setting this equal to an equivalent body energy
1
2 KEfluid 2
dm Vrel 1
2 mhU2 (8.102) The integration of fluid kinetic energy can also be accomplished by a bodysurface integral involving the velocity potential [16, sec. 11].
Consider the previous example of a sphere immersed in a uniform stream. By subtracting out the stream velocity we can replot the flow as in Fig. 8.28, showing the
streamlines relative to the moving sphere. Note the similarity to the doublet flow in
Fig. 8.25. The relativevelocity components are found by subtracting U from Eqs. (8.99)
Ua3 cos
r3 r Ua3 sin
2r3 The element of fluid mass, in spherical polar coordinates, is
dm
2
When dm and Vrel
uated 2
r 2 (2 r sin )r dr d are substituted into Eq. (8.102), the integral can be eval  v v KEfluid  eText Main Menu  1
3 Textbook Table of Contents a3U2  Study Guide 540 Chapter 8 Potential Flow and Computational Fluid Dynamics
Fluid
particle:
dm d ( KE ) = 1 dm V 2
2 V U Fig. 8.28 Potentialflow streamlines
relative to a moving sphere. Compare with Figs. 8.25 and 8.27. or mh(sphere) 2
3 a3 (8.103) Thus, according to potential theory, the hydrodynamic mass of a sphere equals onehalf of its displaced mass, independent of the direction of motion.
A similar result for a cylinder moving normal to its axis can be computed from Eqs.
(8.33) after subtracting out the stream velocity. The result is
mh(cylinder) a2L (8.104) for a cylinder of length L, assuming twodimensional motion. The cylinder’s hydrodynamic mass equals its displaced mass.
Tables of hydrodynamic mass for various body shapes and directions of motion are
given by Patton [17]. See also Ref. 21. 8.9 Numerical Analysis When potential flow involves complicated geometries or unusual stream conditions,
the classical superposition scheme of Secs. 8.3 and 8.4 becomes less attractive. Conformal mapping of body shapes, by using the complexvariable technique of Sec. 8.5,
is no longer popular. Numerical analysis is the appropriate modern approach, and at
least three different approaches are in use:  v v 1. The finiteelement method (FEM) [6, 19]
2. The finitedifference method (FDM) [5, 20]
3. a. Integral methods with distributed singularities [18]
b. The boundaryelement method [7]  eText Main Menu  Textbook Table of Contents  Study Guide 8.9 Numerical Analysis 541 Methods 3a and 3b are closely related, having first been developed on an ad hoc basis by aerodynamicists in the 1960s [18] and then generalized into a multipurpose appliedmechanics technique in the 1970s [7].
Methods 1 (or FEM) and 2 (or FDM), though strikingly different in concept, are
comparable in scope, mesh size, and general accuracy. We concentrate here on the latter method for illustration purposes. The FiniteElement Method The finiteelement method [19] is applicable to all types of linear and nonlinear partial differential equations in physics and engineering. The computational domain is divided into small regions, usually triangular or quadrilateral. These regions are delineated with a finite number of nodes where the field variables — temperature, velocity,
pressure, stream function, etc. — are to be calculated. The solution in each region is approximated by an algebraic combination of local nodal values. Then the approximate
functions are integrated over the region, and their error is minimized, often by using a
weighting function. This process yields a set of N algebraic equations for the N unknown nodal values. The nodal equations are solved simultaneously, by matrix inversion or iteration. For further details see Ref. 6 or 19. The FiniteDifference Method Although textbooks on numerical analysis [5, 20] apply finitedifference techniques to
many different problems, here we concentrate on potential flow. The idea of FDM is
to approximate the partial derivatives in a physical equation by “differences’’ between
nodel values spaced a finite distance apart — a sort of numerical calculus. The basic
partial differential equation is thus replaced by a set of algebraic equations for the nodal
values. For potential (inviscid) flow, these algebraic equations are linear, but they are
generally nonlinear for viscous flows. The solution for nodal values is obtained by iteration or matrix inversion. Nodal spacings need not be equal.
Here we illustrate the twodimensional Laplace equation, choosing for convenience
the streamfunction form
∂2
∂x2 ∂2
∂y2 0 (8.105) subject to known values of along any body surface and known values of ∂ /∂x and
∂ /∂y in the free stream.
Our finitedifference technique divides the flow field into equally spaced nodes, as
shown in Fig. 8.29. To economize on the use of parentheses or functional notation, subscripts i and j denote the position of an arbitrary, equally spaced node, and i,j denotes
the value of the stream function at that node
i,j (x0 i x, y0 j y) Thus, i 1,j is just to the right of i,j, and i,j 1 is just above.
An algebraic approximation for the derivative ∂ /∂x is  v v ∂
∂x  eText Main Menu  (x x, y)
x Textbook Table of Contents  (x, y) Study Guide 542 Chapter 8 Potential Flow and Computational Fluid Dynamics
ψi, j + 1 ∆y ψi –1, j ψi +1, j ψi, j
∆x ∆x ∆y Fig. 8.29 Definition sketch for a
twodimensional rectangular finitedifference grid. ψi, j – 1 A similar approximation for the second derivative is
∂2
∂x2 (x 1
x x, y)
x (x, y) (x, y) (x
x x, y) The subscript notation makes these expressions more compact
∂
∂x 1
(
x ∂2
∂x2 1
(
x2 i,j) i 1,j (8.106)
i 1, j 2 i 1, j) i, j These formulas are exact in the calculus limit as x → 0, but in numerical analysis we
keep x and y finite, hence the term finite differences.
In an exactly similar manner we can derive the equivalent difference expressions for
the y direction
∂
∂y
∂2
∂y2 1
(
y
1
(
y2 i, j) i, j 1 i, j 1 2 i, j 1) i, j (8.107) The use of subscript notation allows these expressions to be programmed directly into
a scientific computer language such as BASIC or FORTRAN.
When (8.106) and (8.107) are substituted into Laplace’s equation (8.105), the result
is the algebraic formula  v v 2(1  eText Main Menu )  i, j i 1, j i 1, j ( i, j 1 Textbook Table of Contents  i, j 1) Study Guide (8.108) 8.9 Numerical Analysis 543 where
( x/ y)2 depends upon the mesh size selected. This finitedifference model
of Laplace’s equation states that every nodal streamfunction value i, j is a linear combination of its four nearest neighbors.
The most commonly programmed case is a square mesh (
1), for which Eq.
(8.108) reduces to
1
4 i, j ( i, j 1 i, j 1 i 1, j) i 1, j (8.109) Thus, for a square mesh, each nodal value equals the arithmetic average of the four
neighbors shown in Fig. 8.29. The formula is easily remembered and easily programmed. If P(I, J) is a subscripted variable stream function, the BASIC or FORTRAN
statement of (8.109) is
P(I, J) 0.25 * (P(I, J 1) P(I, J 1) P(I 1, J) P(I 1, J)) (8.110) This is applied in iterative fashion sweeping over each of the internal nodes (I, J), with
known values of P specified at each of the surrounding boundary nodes. Any initial
guesses can be specified for the internal nodes P(I, J), and the iteration process will
converge to the final algebraic solution in a finite number of sweeps. The numerical
error, compared with the exact solution of Laplace’s equation, is proportional to the
square of the mesh size.
Convergence can be speeded up by the successive overrelaxation (SOR) method,
discussed by Patankar [5]. The modified SOR form of the iteration is
P(I, J) P(I, J)
P(I 0.25 * A * (P(I, J
1, J) P(I 1, J) 1) P(I, J 4 * P(I, J)) 1)
(8.111) The recommended value of the SOR convergence factor A is about 1.7. Note that the
value A 1.0 reduces Eq. (8.111) to (8.110).
Let us illustrate the finitedifference method with an example. EXAMPLE 8.5
Make a numerical analysis, using x
y 0.2 m, of potential flow in the duct expansion
shown in Fig. 8.30. The flow enters at a uniform 10 m/s, where the duct width is 1 m, and is
assumed to leave at a uniform velocity of 5 m/s, where the duct width is 2 m. There is a straight
section 1 m long, a 45° expansion section, and a final straight section 1 m long. Solution
Using the mesh shown in Fig. 8.30 results in 45 boundary nodes and 91 internal nodes, with i
varying from 1 to 16 and j varying from 1 to 11. The internal points are modeled by Eq. (8.110).
For convenience, let the stream function be zero along the lower wall. Then since the volume
flow is (10 m/s)(1 m) 10 m2/s per unit depth, the stream function must equal 10 m2/s along
the upper wall. Over the entrance and exit planes, the stream function must vary linearly to give
uniform velocities: v   (1, J) Exit: v Inlet: (16, J) eText Main Menu  2 * (J
J 6) 1 Textbook Table of Contents for J 7 to 10 for J 2 to 10  Study Guide 544 Chapter 8 Potential Flow and Computational Fluid Dynamics
(1, 11) y=2m (16, 11) 10
m /s
(i, j)
5
m /s
(1, 6) y=1m (6, 6) 45° j Fig. 8.30 Numerical model of potential flow through a twodimensional 45° expansion. The nodal
points shown are 20 cm apart.
There are 45 boundary nodes and
91 internal nodes. i (11, 1) 1m y=0m 1m (16, 1) 1m All these boundary values must be input to the program and are shown printed in Fig. 8.31.
Initial guesses are stored for the internal points, say, zero or an average value of 5.0 m2/s.
The program then starts at any convenient point, such as the upper left (2, 10), and evaluates Eq.
(8.110) at every internal point, repeating this sweep iteratively until there are no further changes
(within some selected maximum change) in the nodal values. The results are the finitedifference simulation of this potential flow for this mesh size; they are shown printed in Fig. 8.31 to
threedigit accuracy. The reader should test a few nodes in Fig. 8.31 to verify that Eq. (8.110)
is satisfied everywhere. The numerical accuracy of these printed values is difficult to estimate,
since there is no known exact solution to this problem. In practice, one would keep decreasing
the mesh size to see whether there were any significant changes in nodal values.
This problem is well within the capability of a small personal computer. The values shown
in Fig. 8.31 were obtained after 100 iterations, or 6 min of execution time, on a Macintosh SE
personal computer, using BASIC. ψ = 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 8.00 8.02 8.04 8.07 8.12 8.20 8.30 8.41 8.52 8.62 8.71 8.79 8.85 8.91 8.95 9.00 6.00 6.03 6.06 6.12 6.22 6.37 6.58 6.82 7.05 7.26 7.44 7.59 7.71 7.82 7.91 8.00 4.00 4.03 4.07 4.13 4.26 4.48 4.84 5.24 5.61 5.93 6.19 6.41 6.59 6.74 6.88 7.00 2.00 2.02 2.05 2.09 2.20 2.44 3.08 3.69 4.22 4.65 5.00 5.28 5.50 5.69 5.85 6.00 ψ = 0.00 0.00 0.00 0.00 0.00 0.00 1.33 2.22 2.92 3.45 3.87 4.19 4.45 4.66 4.84 5.00 0.00 1.00 1.77 2.37 2.83 3.18 3.45 3.66 3.84 4.00 0.00 v v   0.80 1.42 1.90 2.24 2.50 2.70 2.86 3.00 0.00 Fig. 8.31 Streamfunction nodal values for the potential flow of Fig.
8.30. Boundary values are known inputs. Internal nodes are solutions to
Eq. (8.110). 0.63 1.09 1.40 1.61 1.77 1.89 2.00 0.00 eText Main Menu  0.44 0.66 0.79 0.87 0.94 1.00 0.00 0.00 0.00 0.00 0.00 0.00 Textbook Table of Contents  Study Guide 8.9 Numerical Analysis 545 Although Fig. 8.31 is the computer solution to the problem, these numbers must be manipulated to yield practical engineering results. For example, one can interpolate these numbers to
sketch various streamlines of the flow. This is done in Fig. 8.32a. We see that the streamlines
are curved both upstream and downstream of the corner regions, especially near the lower wall.
This indicates that the flow is not onedimensional.
The velocities at any point in the flow can be computed from finitedifference formulas such
as Eqs. (8.106) and (8.107). For example, at the point (I, J) (3, 6), from Eq. (8.107), the horizontal velocity is approximately
u(3, 6) (3, 7) (3, 6) 2.09 y 0.00 10.45 m/s 0.2 and the vertical velocity is zero from Eq. (8.106). Directly above this on the upper wall, we estimate
u(3, 11) (3, 11) (3,10)
y 10.00 8.07
0.2 9.65 m/s ψ = 10
8 p1 6 V1 4
(a)
2 0 1.0
0.8 Cp = p – p1 ρ V12 / 2 0.75 0.6
0.4 Upper surface 0.2
0.0 Onedimensional
approximation
Eq. (1) – 0.2
– 0.4  v v Fig. 8.32 Useful results computed
from Fig. 8.31: (a) streamlines of
the flow; (b) pressurecoefficient
distribution along each wall.  – 0.6 Lower surface – 0.8
(b) eText Main Menu  Textbook Table of Contents  Study Guide 546 Chapter 8 Potential Flow and Computational Fluid Dynamics
The flow is not truly onedimensional in the entrance duct. The lower wall, which contains
the diverging section, accelerates the fluid, while the flat upper wall is actually decelerating the fluid.
Another output function, useful in making boundarylayer analyses of the wall regions, is the
pressure distribution along the walls. If p1 and V1 are the pressure and velocity at the entrance (I
1), conditions at any other point are computed from Bernoulli’s equation (8.3), neglecting gravity
p 1
2 V2 p1 1
2 2
V1 which can be rewritten as a dimensionless pressure coefficient
Cp p
1
2 p1
2
V1 1 V
V1 2 This determines p after V is computed from the streamfunction differences in Fig. 8.31.
Figure 8.32b shows the computed wallpressure distributions as compared with the onedimensional continuity approximation V1A1 V(x)A(x), or
Cp(onedim) 1 A1
A 2 (1) The onedimensional approximation, which is rather crude for this large (45°) expansion, lies
between the upper and lower wall pressures. Onedimensional theory would be much more accurate for a 10° expansion.
Analyzing Fig. 8.32b, we predict that boundarylayer separation will probably occur on the
lower wall between the corners, where pressure is strongly rising (highly adverse gradient). Therefore potential theory is probably not too realistic for this flow, where viscous effects are strong.
(Recall Figs. 6.27 and 7.8.)
Potential theory is reversible; i.e., when we reverse the flow arrows in Fig. 8.32a, then Fig.
8.32b is still valid and would represent a 45° contraction flow. The pressure would fall on both
walls (no separation) from x 3 m to x 1 m. Between x 1 m and x 0, the pressure rises
on the lower surface, indicating possible separation, probably just downstream of the corner.
This example should give the reader an idea of the usefulness and generality of numerical
analysis of fluid flows. The BoundaryElement Method A relatively new technique for numerical solution of partial differential equations is
the boundaryelement method (BEM). Reference 7 is an introductory textbook outlining the concepts of BEM, including FORTRAN programs for potential theory and elastostatics. There are no interior elements. Rather, all nodes are placed on the boundary
of the domain, as in Fig. 8.33. The “element’’ is a small piece of the boundary surrounding the node. The “strength’’ of the element can be either constant or variable.
For plane potential flow, the method takes advantage of the particular solution
* 1
1
ln
2
r (8.112)  v v which satisfies Laplace’s equation, 2
0. Each element i is assumed to have a different strength i. Then r represents the distance from that element to any other point
in the flow field. Summing all these elemental effects, with proper boundary conditions, will give the total solution to the potentialflow problem.
At each element of the boundary, we typically know either the value of or the
value of ∂ /∂n, where n is normal to the boundary. (Mixed combinations of and  eText Main Menu  Textbook Table of Contents  Study Guide 8.9 Numerical Analysis
n 547 Element j
Node j Element i
rj Node i Domain:
2ψ = 0
∆ ds Fig. 8.33 Boundary elements of
constant strength in plane potential
flow. ∂ /∂n are also possible but are not discussed here.) The correct strengths i are such
that these boundary conditions are satisfied at every element. Summing these effects
over N elements requires integration by parts plus a careful evaluation of the (singular) effect of element i upon itself. The mathematical details are given in Ref. 7. The
result is a set of N algebraic equations for the unknown boundary values. In the case
of elements of constant strength, the final expression is
1
2 N
i 0
j j1 j ∂*
ds
∂n N
j1 ∂
∂n 0 * ds
j i 1 to N (8.113) j The integrals, which involve the logarithmic particular solution * from Eq. (8.112),
are evaluated numerically for each element. Reference 7 recommends and gives a
program for gaussian quadrature formulas.
Equations (8.113) contain 2N element values, i and (∂ /∂n)i, of which N are known
from the given boundary conditions. The remaining N are solved simultaneously from
Eqs. (8.113). Generally this completes the analysis only the boundary solution is
computed, and interior points are not studied. In most cases, the boundary velocity and
pressure are all that is needed.
We illustrated the method with stream function . Naturally the entire technique
also applies to velocity potential , if we are given proper conditions on or ∂ /∂n
at each boundary element. The method is readily extended to three dimensions [7].
Reference 7 gives a complete FORTRAN listing for solving Eqs. (8.113) numerically for constant, linear, and quadratic element strength variations. We now use their
constantelementstrength program, POCONBE [7], to take an alternate look at Example 8.5, which used the finitedifference method. EXAMPLE 8.6
Solve the duct expansion problem, Example 8.5, using boundary elements. Use the same grid
spacing x
y 0.2 m for the element sizes. Solution  v v The boundary nodes are equally spaced, as shown in Fig. 8.34. There are only 45 nodes, whereas
there were 91 interior points for the FDM solution of Example 8.5. We expect the same accuracy for 50 percent fewer nodes. (Had we reduced the grid size to 0.1 m, there would be 90  eText Main Menu  Textbook Table of Contents  Study Guide 548 Chapter 8 Potential Flow and Computational Fluid Dynamics
nodes as opposed to 406 interior points a savings of 78 percent.) The program POCONBE [7]
asks you to input the location of these 45 nodes. The streamfunction values are known all around
the boundary: equals 0 on the bottom and 10.0 on the top and is linearly increasing from 0 to
10.0 at entrance and exit. These values of , shown on the outside in Fig. 8.34, are inputted into
the program.
Once the input of nodes and element values is complete, the program immediately computes
and displays or stores the 45 unknowns, which in this case are the values of ∂ / ∂n all around
the boundary. These values are shown on the inside of the top and bottom surfaces in Fig. 8.34
and represent the local surface velocity near each element, in m/s. The values of ∂ / ∂n at entrance and exit, which are small fractions representing vertical velocity components, are not
shown here. 10 10 10 10 10 Stream function values listed outside
10
10
10
10
10
10 10 10 10 10 10 10.8 9.73 9.68 9.46 9.12 8.68 8.16 7.62 7.10 6.63 6.23 5.88 5.59 5.28 5.54
8 9 Boundary velocities listed inside 8 6
U = 10.0 m/s 7 [In BEM there are
no interior nodes.] 4 6
U = 5.0 2 5 10.9 10.1 10.4 10.8 14.1
0 Fig. 8.34 Boundary elements corresponding to the same grid size as
Fig. 8.31. Nodal values of stream
function and computed surface velocity are shown. 0 0 0 0 0 4 11.2
0 3 7.14
0 5.69
0 2
4.49
0 2.73 2.23 3.61 4.15 4.48 5.31
0 0 0 0 0 1 0 The reader may verify that use of the surface velocities in Fig. 8.34 to compute surface pressure coefficients, as in Example 8.5, leads to curves very similar to those shown in Fig. 8.32.
The BEM approach, using the same boundary nodes, has accuracy comparable to that for an
FDM computation. For further details see Ref. 7. Our previous finitedifference model of Laplace’s equation, e.g., Eq. (8.109), was very
well behaved and converged nicely with or without overrelaxation. Much more care is
needed to model the full NavierStokes equations. The challenges are quite different,
and they have been met to a large extent, so there are now many textbooks [20, 23 to
27] on (fully viscous) computational fluid dynamics (CFD). This is not a textbook on
CFD, but we will address some of the issues in this section. OneDimensional Unsteady Flow We begin with a simplified problem, showing that even a single viscous term introduces new effects and possible instabilities. Recall (or review) Prob. 4.85, where a wall
moves and drives a viscous fluid parallel to itself. Gravity is neglected. Let the wall be
the plane y 0, moving at a speed U0(t), as in Fig. 8.35. A uniform vertical grid, of
spacing y, has nodes n at which the local velocity unj is to be calculated, where su  v v ViscousFlow Computer Models  eText Main Menu  Textbook Table of Contents  Study Guide 8.9 Numerical Analysis 549 perscript j denotes the timestep j t. The wall is n
1. If u u(y, t) only and
w 0, continuity,
V 0, is satisfied and we need only solve the xmomentum
NavierStokes equation:
∂u
∂t ∂2u
∂y2 v (8.114) where
/ . Utilizing the same finitedifference approximations as in Eq. (8.106),
we may model Eq. (8.114) algebraically as a forward time difference and a central spatial difference:
j
un 1 j
un j
un j
2u n
2
y 1 t j
un 1 Rearrange and find that we can solve explicitly for un at the next timestep j
j
un 1 (1 j
2 ) un j
(un j
un 1 1) t
y2 1:
(8.115) Thus u at node n at the next timestep j 1 is a weighted average of three previous
values, similar to the “fournearestneighbors” average in the laplacian model of Eq.
(8.109). Since the new velocity is calculated immediately, Eq. (8.115) is called an explicit model. It differs from the wellbehaved laplacian model, however, because it may
be unstable. The weighting coefficients in Eq. (8.115) must all be positive to avoid divergence. Now is positive, but (1 2 ) may not be. Therefore, our explicit viscous
flow model has a stability requirement:
t
y2 1
2 (8.116) Normally one would first set up the mesh size y in Fig. 8.35, after which Eq. (8.116)
would limit the timestep t. The solutions for nodal values would then be stable, but
not necessarily that accurate. The mesh sizes y and t could be reduced to increase
accuracy, similar to the case of the potentialflow laplacian model (8.109).
For example, to solve Prob. 4.85 numerically, one sets up a mesh with plenty of
nodes (30 or more y within the expected viscous layer); selects t according to Eq. n 1 y
n
y
n 1 y  v v Fig. 8.35 An equally spaced finitedifference mesh for onedimensional viscous flow [Eq. (8.114)].  y
u
Wall eText Main Menu n  U0 1 Textbook Table of Contents  Study Guide 550 Chapter 8 Potential Flow and Computational Fluid Dynamics (8.116); and sets two boundary conditions for all j: u1 U0 sin t† and uN 0, where
N is the outermost node. For initial conditions, perhaps assume the fluid initially at
1
rest: un 0 for 2 n N 1. Sweeping the nodes 2 n N 1 using Eq. (8.115)
j
(an Excel spreadsheet is excellent for this), one generates numerical values of un for
as long as one desires. After an initial transient, the final “steady” fluid oscillation will
approach the classical solution in viscousflow textbooks [15]. Try Prob. 8.115 to
demonstrate this. An Alternate Implicit Approach In many finitedifference problems, a stability limitation such as Eq. (8.116) requires
an extremely small timestep. To allow larger steps, one can recast the model in an implicit fashion by evaluating the secondderivative model in Eq. (8.114) at the next timestep:
j
un 1 j
un j
un 1
1 j
2un 1
y2 t This rearrangement is unconditionally stable for any
knowns:
j
un 1
1 (1 j
2 )un 1 j
un j
un 1
1 , but now we have three un1
1 j
un (8.117) This is an implicit model, meaning that one must solve a large system of algebraic
equations for the new nodal values at time j 1. Fortunately, the system is narrowly
banded, with the unknowns confined to the principal diagonal and its two nearest diagonals. In other words, the coefficient matrix of Eq. (8.117) is tridiagonal, a happy
event. A direct method, called the tridiagonal matrix algorithm (TDMA), is available
and explained in most CFD texts [20, 23 to 27]. Appendix A of Ref. 20 includes a
complete program for solving the TDMA. If you have not learned the TDMA yet, Eq.
(8.117) converges satisfactorily by rearrangement and iteration:
j
un 1 j
un j
(u n 1
1
12 j
u n 1)
1 (8.118) At each timestep j 1, sweep the nodes 2
n
N
1 over and over, using Eq.
(8.118), until the nodal values have converged. This implicit method is stable for any
, however large. To ensure accuracy, though, one should keep t and y small compared to the basic time and length scales of the problem. This author’s habit is to keep
t and y small enough that nodal values change no more than 10 percent from one
(n, j) to the next. EXAMPLE 8.7
SAE 30 oil at 20°C is at rest near a wall when the wall suddenly begins moving at a constant 1
m/s. Using the explicit model of Eq. (8.114), estimate the oil velocity at y 3 cm after 1 second of wall motion. †  v v Finite differences are not analytical; one must set U0 and  eText Main Menu  equal to numerical values. Textbook Table of Contents  Study Guide 8.9 Numerical Analysis 551 Solution
For SAE 30 oil, from Table A3,
0.29/891 3.25 E4 m2/s. For convenience in putting a
node exactly at y 3 cm, choose y 0.01 m. The stability limit (8.116) is t/ y2 0.5, or
t 0.154 s. Again for convenience, to hit t 1 s on the nose, choose t 0.1 s, or
0.3255
and (1 2 ) 0.3491. Then our explicit algebraic model (8.115) for this problem is
j
un 1 0.3491 unj j
0.3255(u n u nj 1 1) (1) We apply this relation from n 2 out to at least n N 15, to make sure that the desired value
j
of u at n 3 is accurate. The wall noslip boundary requires u1 1.0 m/s constant for all j.
The outer boundary condition is uN 0. The initial conditions are u1 0 for n 2. We then
n
apply Eq. (1) repeatedly for n 2 until we reach j 11, which corresponds to t 1 s. This is
easily programmed on a spreadsheet such as Excel. Here we print out only j 1, 6, and 11 as
follows: j
1
6
11 t u1 u2 u3 u4 u5 u6 u7 u8 u9 u10 u11 0.000
0.500
1.000 1.000
1.000
1.000 0.000
0.601
0.704 0.000
0.290
0.446 0.000
0.107
0.250 0.000
0.027
0.123 0.000
0.004
0.052 0.000
0.000
0.018 0.000
0.000
0.005 0.000
0.000
0.001 0.000
0.000
0.000 0.000
0.000
0.000 Note: Units for t and u’s are s and m/s, respectively. Our numerical estimate is u11 u(3 cm, 1 s) 0.250 m/s, which is about 4 percent high —
4
this problem has a known exact solution, u 0.241 m/s [15]. We could improve the accuracy
indefinitely by decreasing y and t. Steady TwoDimensional
Laminar Flow The previous example, unsteady onedimensional flow, had only one viscous term and
no convective accelerations. Let us look briefly at incompressible twodimensional
steady flow, which has four of each type of term, plus a nontrivial continuity equation:
∂u
∂x Continuity: ∂
∂y 0 (8.119a) x momentum: u ∂u
∂x ∂u
∂y 1 ∂p
∂x ∂2u
∂x2 ∂2u
∂y2 (8.119b) y momentum u ∂
∂x ∂
∂y 1 ∂p
∂y ∂2
∂x2 ∂2
∂y2 (8.119c)  v v These equations, to be solved for (u, , p) as functions of (x, y), are familiar to us from
analytical solutions in Chaps. 4 and 6. However, to a numerical analyst, they are odd,
because there is no pressure equation, that is, a differential equation for which the dominant derivatives involve p. This situation has led to several different “pressureadjustment” schemes in the literature [20, 23 to 27], most of which manipulate the continuity equation to insert a pressure correction.
A second difficulty in Eq. (8.119b and c) is the presence of nonlinear convective
accelerations such as u(∂u/∂x), which create asymmetry in viscous flows. Early at  eText Main Menu  Textbook Table of Contents  Study Guide 552 Chapter 8 Potential Flow and Computational Fluid Dynamics tempts, which modeled such terms with a central difference, led to numerical instability. The remedy is to relate convection finite differences solely to the upwind flow entering the cell, ignoring the downwind cell. For example, the derivative ∂u/∂x could be
modeled, for a given cell, as (uupwind ucell)/ x. Such improvements have made fully
viscous CFD an effective tool, with various commercial userfriendly codes available.
For details beyond our scope, see Refs. 20 and 23 to 27.
Mesh generation and gridding have also become quite refined in modern CFD. Figure 8.36 illustrates a CFD solution of twodimensional flow past an NACA 66(MOD)
hydrofoil [28]. The gridding in Fig. 8.36a is of the C type, which wraps around the
leading edge and trails off behind the foil, thus capturing the important nearwall and
wake details without wasting nodes in front or to the sides. The grid size is 262 by 91.
The CFD model for this hydrofoil flow is also quite sophisticated: a full NavierStokes solver with turbulence modeling [29] and allowance for cavitation bubble formation when surface pressures drop below the local vapor pressure. Figure 8.36b compares computed and experimental surface pressure coefficients for an angle of attack of
1°. The dimensionless pressure coefficient is defined as Cp (psurface p )/( V 2 /2).
The agreement is excellent, as indeed it is also for cases where the hydrofoil cavitates
[28]. Clearly, when properly implemented for the proper flow cases, CFD can be an extremely effective tool for engineers. The coming of the third millennium has seen an enormous emphasis on computer applications in nearly every field, fluid mechanics being a prime example. It is now possible, at least for moderately complex geometries and flow patterns, to model on a digital computer, approximately, the equations of motion of fluid flow, with dedicated CFD
textbooks available [20, 23 to 27]. The flow region is broken into a fine grid of elements
and nodes, which algebraically simulate the basic partial differential equations of flow.
While simple twodimensional flow simulations have long been reported and can be
programmed as student exercises, threedimensional flows, involving thousands or even
millions of grid points, are now solvable with the modern supercomputer.
Although elementary computer modeling was treated briefly here, the general topic
of CFD is essentially for advanced study or professional practice. The big change over
the past decade is that engineers, rather than laboriously programming CFD problems
themselves, can now take advantage of any of several commercial CFD codes. These
are extensive software packages which allow engineers to construct a geometry and
boundary conditions to simulate a given viscousflow problem. The software then grids
the flow region and attempts to compute flow properties at each grid element. The convenience is great; the danger is also great. That is, computations are not merely automatic, like when using a hand calculator, but rather require care and concern from the
user. Convergence and accuracy are real problems for the modeler. Use of the codes
requires some art and experience. In particular, when the flow Reynolds number, Re
VL/ , goes from moderate (laminar flow) to high (turbulent flow), the accuracy of
the simulation is no longer assured in any real sense. The reason is that turbulent flows
are not completely resolved by the full equations of motion, and one resorts to using
approximate turbulence models.
Turbulence models [29] are developed for particular geometries and flow conditions
and may be inaccurate or unrealistic for others. This is discussed by Freitas [30], who  v v Commercial CFD Codes  eText Main Menu  Textbook Table of Contents  Study Guide 8.9 Numerical Analysis 553 (a)
0.6
G 0.5
0.4 G G G G G G 0.3 G 0.2
0.1 G 0.0 Fig. 8.36 CFD results for water
flow past an NASA 66(MOD) hydrofoil [from Ref 28, with permission of the American Society of
Mechanical Engineers]: (a) C gridding, 262 by 91 nodes; (b) surface
pressures at
1°. G
G G Cp Expt.
Comp. G G 0.1
0.2
0.0 0.5
x/C 1.0 (b)  v v compared eight different commercialcode calculations (FLOW3D, FLOTRAN, STARCD, N3S, CFDACE, FLUENT, CFDSFLOW3D, and NISA/3DFLUID) with experimental results for five benchmark flow experiments. Calculations were made by the
vendors themselves. Freitas concludes that commercial codes, though promising in general, can be inaccurate for certain laminar and turbulentflow situations. Further research is recommended before engineers can truly rely upon such software to give generally accurate fluidflow predictions.
In spite of the above warning to treat CFD codes with care, one should also realize
that the results of a given CFD simulation can be spectacular. Figure 8.37 illustrates
turbulent flow past a cube mounted on the floor of a channel whose clearance is twice  eText Main Menu  Textbook Table of Contents  Study Guide 554 Chapter 8 Potential Flow and Computational Fluid Dynamics (a)  v v Fig. 8.37 Flow over a surfacemounted cube creates a complex
and perhaps unexpected pattern: (a)
experimental oilstreak visualization of surface flow at Re 40,000
(based on cube height) ( from Ref.
31, courtesy of Robert Martinuzzi,
with the permission of the American Society of Mechanical Engineers); (b) computational largeeddy simulation of the surface flow
in (a) ( from Ref. 32, courtesy of
Kishan Shah, Stanford University);
and (c) a side view of the flow in
(a) visualized by smoke generation
and a laser light sheet ( from Ref.
31, courtesy of Robert Martinuzzi,
with the permission of the American Society of Mechanical Engineers). (b) (c)  eText Main Menu  Textbook Table of Contents  Study Guide Problems 555 the cube height. Compare Fig. 8.37a, a top view of the experimental surface flow [31]
as visualized by oil streaks, with Fig. 8.37b, a CFD supercomputer result using the
method of largeeddy simulation [32, 33]. The agreement is remarkable. The Cshaped
flow pattern in front of the cube is caused by formation of a horseshoe vortex, as seen
in a side view of the experiment [31] in Fig. 8.37c. Horseshoe vortices commonly result when surface shear flows meet an obstacle. We conclude that CFD has a tremendous potential for flow prediction. Summary This chapter has analyzed a highly idealized but very useful type of flow: inviscid, incompressible, irrotational flow, for which Laplace’s equation holds for the velocity potential (8.1) and for the plane stream function (8.7). The mathematics is well developed, and solutions of potential flows can be obtained for practically any body shape.
Some solution techniques outlined here are (1) superposition of elementary line or
point solutions in both plane and axisymmetric flow, (2) the analytic functions of a
complex variable, (3) use of variablestrength vortex sheets, and (4) numerical analysis on a digital computer. Potential theory is especially useful and accurate for thin
bodies such as airfoils. The only requirement is that the boundary layer be thin, i.e.,
that the Reynolds number be large.
For blunt bodies or highly divergent flows, potential theory serves as a first approximation, to be used as input to a boundarylayer analysis. The reader should consult the
advanced texts [for example, 2 to 4, 10 to 13] for further applications of potential theory. Section 8.9 discusses computational methods for viscous (nonpotential) flows. Problems
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 8.1 to 8.115 (categorized in the problem list
below) are followed by word problems W8.1 to W8.7, comprehensive problems C8.1 to C8.3, and design projects D8.1 to D8.3. P8.1
P8.2 Prove that the streamlines (r, ) in polar coordinates from
Eqs. (8.10) are orthogonal to the potential lines (r, ).
The steady plane flow in Fig. P8.2 has the polar velocity
components
r and r 0. Determine the circulation around the path shown.
R2 R1 Problem Distribution
Section Topic
Introduction and review
Elementary planeflow solutions
Superposition of plane flows
Plane flow past closedbody shapes
The complex potential
Images
Airfoil theory: Twodimensional
Airfoil theory: Finitespan wings
Axisymmetric potential flow
Hydrodynamic mass
Numerical methods 8.1 – 8.7
8.8 – 8.17
8.18 – 8.34
8.35 – 8.59
8.60 – 8.71
8.72 – 8.79
8.80 – 8.84
8.85 – 8.90
8.91 – 8.103
8.104 – 8.105
8.106 – 8.115  v v 8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.7
8.8
8.8
8.9 Problems  eText Main Menu  P8.2
P8.3 P8.4 P8.5 Using cartesian coordinates, show that each velocity component (u, , w) of a potential flow satisfies Laplace’s equation separately.
Is the function 1/r a legitimate velocity potential in plane
polar coordinates? If so, what is the associated stream function (r, )?
Consider the twodimensional velocity distribution u
By,
Bx, where B is a constant. If this flow possesses a Textbook Table of Contents  Study Guide 556 Chapter 8 Potential Flow and Computational Fluid Dynamics stream function, find its form. If it has a velocity potential,
find that also. Compute the local angular velocity of the flow,
if any, and describe what the flow might represent.
P8.6 An incompressible flow has the velocity potential
2Bxy,
where B is a constant. Find the stream function of this flow,
sketch a few streamlines, and interpret the pattern.
P8.7 Consider a flow with constant density and viscosity. If the
flow possesses a velocity potential as defined by Eq. (8.1),
show that it exactly satisfies the full NavierStokes equations (4.38). If this is so, why for inviscid theory do we
back away from the full NavierStokes equations?
P8.8 For the velocity distribution of Prob. 8.5, evaluate the circulation around the rectangular closed curve defined by
(x, y) (1, 1), (3, 1), (3, 2), and (1, 2). Interpret your result, especially visàvis the velocity potential.
P8.9 Consider the twodimensional flow u
Ax,
Ay,
where A is a constant. Evaluate the circulation around
the rectangular closed curve defined by (x, y) (1, 1),
(4, 1), (4, 3), and (1, 3). Interpret your result, especially
visàvis the velocity potential.
P8.10 A mathematical relation sometimes used in fluid mechanics is the theorem of Stokes [1]
V ds ( C V) n dA A where A is any surface and C is the curve enclosing that surface. The vector ds is the differential arc length along C,
and n is the unit outward normal vector to A. How does this
relation simplify for irrotational flow, and how does the resulting line integral relate to velocity potential?
P8.11 A power plant discharges cooling water through the manifold in Fig. P8.11, which is 55 cm in diameter and 8 m
high and is perforated with 25,000 holes 1 cm in diameter. Does this manifold simulate a line source? If so, what
is the equivalent source strength m? P8.12 Consider the flow due to a vortex of strength K at the origin. Evaluate the circulation from Eq. (8.15) about the
clockwise path from (r, ) (a, 0) to (2a, 0) to (2a, 3 /2)
to (a, 3 /2) and back to (a, 0). Interpret the result.
P8.13 A wellknown exact solution to the NavierStokes equations (4.38) is the unsteady circulating motion [15]
K
1
2r r2
4t exp r z 0 where K is a constant and is the kinematic viscosity. Does
this flow have a polarcoordinate stream function and/or velocity potential? Explain. Evaluate the circulation for this
motion, plot it versus r for a given finite time, and interpret
compared to ordinary line vortex motion.
P8.14 A tornado may be modeled as the circulating flow shown
in Fig. P8.14, with r
0 and (r) such that
z v r 2
R r r R r R Determine whether this flow pattern is irrotational in either the inner or outer region. Using the rmomentum
equation (D.5) of App. D, determine the pressure distribution p(r) in the tornado, assuming p p as r → .
Find the location and magnitude of the lowest pressure.
θ (r) r
R P8.14
P8.15 Evaluate Prob. 8.14 for the particular case of a smallscale
tornado, R 100 m, ,max 65 m/s, with sealevel conditions at r
. Plot p(r) out to r 400 m.
P8.16 Consider inviscid stagnation flow,
Kxy (see Fig.
8.15b) superimposed with a source at the origin of strength
m. Plot the resulting streamlines in the upper half plane,
using the length scale a (m/K)1/2. Give a physical interpretation of the flow pattern.
P8.17 Examine the flow of Fig. 8.30 as an analytical (not a numerical) problem. Give the appropriate differential equation and the complete boundary conditions for both the
stream function and the velocity potential. Is a Fourierseries solution possible? Inlet  v v P8.11  eText Main Menu  Textbook Table of Contents  Study Guide Problems 557
P8.18 P8.19 P8.20 P8.21 P8.22 P8.23 Plot the streamlines and potential lines of the flow due
to a line source of strength m at (a, 0) plus a source
3m at ( a, 0). What is the flow pattern viewed from
afar?
Plot the streamlines and potential lines of the flow due to
a line source of strength 3m at (a, 0) plus a sink m at
( a, 0). What is the pattern viewed from afar?
Plot the streamlines of the flow due to a line vortex K at
(0, a) and a vortex K at (0, a). What is the pattern
viewed from afar?
Plot the streamlines of the flow due to a line vortex K at
( a, 0) and a vortex 2K at ( a, 0). What is the pattern
viewed from afar?
Plot the streamlines of a uniform stream V iU plus a
clockwise line vortex K located at the origin. Are there
any stagnation points?
Find the resultant velocity vector induced at point A
in Fig. P8.23 by the uniform stream, vortex, and line
source. β 40 m P8.25
P8.26 Find the resultant velocity vector induced at point A in Fig.
P8.26 by the uniform stream, line source, line sink, and
EES
vortex. m = 12 m2 /s
1m
2m 2m
20° K = 25 m2 /s A
1m U = 6 m /s
1.5 m m = – 10 m2 /s 2m
U = 8 m /s
m = 15 m2 /s 1m
A P8.23
P8.24 Line sources of equal strength m Ua, where U is a reference velocity, are placed at (x, y) (0, a) and (0, a).
Sketch the stream and potential lines in the upper half
plane. Is y 0 a “wall’’? If so, sketch the pressure coefficient
Cp p
1
2 p0
U2 v v along the wall, where p0 is the pressure at (0, 0). Find the
minimum pressure point and indicate where flow separation might occur in the boundary layer.
P8.25 Let the vortex/sink flow of Eq. (4.134) simulate a tornado as in Fig. P8.25. Suppose that the circulation
about the tornado is
8500 m2/s and that the pressure at r 40 m is 2200 Pa less than the farfield pressure. Assuming inviscid flow at sealevel density, estimate (a) the appropriate sink strength m, (b) the
pressure at r 15 m, and (c) the angle at which the
streamlines cross the circle at r 40 m (see Fig.
P8.25).   eText Main Menu K = 9 m2 /s  P8.26
P8.27 A counterclockwise line vortex of strength 3K at
(x, y) (0, a) is combined with a clockwise vortex K at
(0, a). Plot the streamline and potentialline pattern,
and find the point of minimum velocity between the two
vortices.
P8.28 Sources of equal strength m are placed at the four symmetric positions (x, y) (a, a), ( a, a), ( a, a), and
(a, a). Sketch the streamline and potentialline patterns.
Do any plane “walls’’ appear?
P8.29 A uniform water stream, U
20 m/s and
998 kg/m3,
combines with a source at the origin to form a halfbody.
At (x, y) (0, 1.2 m), the pressure is 12.5 kPa less than
p . (a) Is this point outside the body? Estimate (b) the appropriate source strength m and (c) the pressure at the nose
of the body.
P8.30 Suppose that the total discharge from the manifold in Fig.
P8.11 is 450 m3/s and that there is a uniform ocean current of 60 cm/s to the right. Sketch the flow pattern from
above, showing the dimensions and the region where the
coolingwater discharge is confined.
P8.31 A Rankine halfbody is formed as shown in Fig. P8.31.
For the stream velocity and body dimension shown, compute (a) the source strength m in m2/s, (b) the distance
a, (c) the distance h, and (d) the total velocity at point A. Textbook Table of Contents  Study Guide 558 Chapter 8 Potential Flow and Computational Fluid Dynamics A (0, 3 m) h y
x a
7 m /s (4m, 0) +m
Source P8.31
P8.32 Sketch the streamlines, especially the body shape, due to
equal line sources m at ( a, 0) and ( a, 0) plus a uniform stream U
ma.
P8.33 Sketch the streamlines, especially the body shape, due to
equal line sources m at (0, a) and (0, a) plus a uniform stream U
ma.
P8.34 Consider three equally spaced sources of strength m placed
at (x, y) (0, a), (0, 0), and (0, a). Sketch the resulting streamlines, noting the position of any stagnation
points. What would the pattern look like from afar?
P8.35 Consider three equal sources m in a triangular configuration: one at (a/2, 0), one at ( a/2, 0), and one at (0, a).
Plot the streamlines for this flow. Are there any stagnation
points? Hint: Try the MATLAB contour command [34].
P8.36 When a line sourcesink pair with m 2 m2/s is combined
with a uniform stream, it forms a Rankine oval whose minimum dimension is 40 cm. If a 15 cm, what are the
stream velocity and the velocity at the shoulder? What is
the maximum dimension?
P8.37 A Rankine oval 2 m long and 1 m high is immersed in a
stream U
10 m/s, as in Fig. P8.37. Estimate (a) the velocity at point A and (b) the location of point B where a
particle approaching the stagnation point achieves its maximum deceleration. P8.40 Consider a uniform stream U plus line sources m at (x,
y) ( a, 0) and ( a, 0) and a single line sink 2m at
the origin. Does a closedbody shape appear? If so, plot
its shape for m/(U a) equal to (a) 1.0 and (b) 5.0.
P8.41 A Kelvin oval is formed by a linevortex pair with K 9
m2/s, a 1 m, and U 10 m/s. What are the height,
width, and shoulder velocity of this oval?
P8.42 For what value of K/(U a) does the velocity at the shoulder of a Kelvin oval equal 4U ? What is the height h/a of
EES
this oval?
P8.43 Consider water at 20°C flowing at 6 m/s past a 1mdiameter circular cylinder. What doublet strength in m3/s
is required to simulate this flow? If the stream pressure is
200 kPa, use inviscid theory to estimate the surface pressure at equal to (a) 180°, (b) 135°, and (c) 90°.
P8.44 Suppose that circulation is added to the cylinder flow of
Prob. 8.43 sufficient to place the stagnation points at
equal to 50° and 130°. What is the required vortex strength
K in m2/s? Compute the resulting pressure and surface velocity at (a) the stagnation points and (b) the upper and
lower shoulders. What will the lift per meter of cylinder
width be?
P8.45 What circulation K must be added to the cylinder flow in
Prob. 8.43 to place the stagnation point exactly at the upper shoulder? What will the velocity and pressure at the
lower shoulder be then? What value of K causes the lower
shoulder pressure to be 10 kPa?
P8.46 A cylinder is formed by bolting two semicylindrical channels together on the inside, as shown in Fig. P8.46. There
are 10 bolts per meter of width on each side, and the inside pressure is 50 kPa (gage). Using potential theory for
the outside pressure, compute the tension force in each bolt
if the fluid outside is sealevel air.
D=2m A
U = 25 m /s B? 1m 10 m/s p=
50 k Pa
(gage) P8.46 2m P8.37  v v P8.47 A circular cylinder is fitted with two surfacemounted pres180° and pb at
105°.
sure sensors, to measure pa at
P8.38 A uniform stream U in the x direction combines with a
The intention is to use the cylinder as a stream velocimesource m at (a, 0) and a sink m at ( a, 0). Plot the reter. Using inviscid theory, derive a formula for estimating
sulting streamlines and note any stagnation points.
U in terms of pa, pb, , and the cylinder radius a.
P8.39 Sketch the streamlines of a uniform stream U past a line
sourcesink pair aligned vertically with the source at a *P8.48 Wind at U and p flows past a Quonset hut which is a
halfcylinder of radius a and length L (Fig. P8.48). The inand the sink at a on the yaxis. Does a closedbody shape
ternal pressure is pi. Using inviscid theory, derive an exform?  eText Main Menu  Textbook Table of Contents  Study Guide Problems 559
pression for the upward force on the hut due to the difference between pi and ps.
ps (θ ) pi A U∞ , p∞ a θ ft2. As sketched in Fig. P8.54, it had two rotors 50 ft high
and 9 ft in diameter rotating at 750 r/min, which is far outside the range of Fig. 8.11. The measured lift and drag coefficients for each rotor were about 10 and 4, respectively.
If the ship is moored and subjected to a crosswind of 25
ft/s, as in Fig. P8.54, what will the wind force parallel and
normal to the ship centerline be? Estimate the power required to drive the rotors. P8.48
P8.49 In strong winds the force in Prob. 8.48 can be quite large.
Suppose that a hole is introduced in the hut roof at point
A to make pi equal to the surface pressure there. At what
angle should hole A be placed to make the net wind force
zero?
P8.50 It is desired to simulate flow past a twodimensional ridge
or bump by using a streamline which passes above the flow
over a cylinder, as in Fig. P8.50. The bump is to be a/2
high, where a is the cylinder radius. What is the elevation
h of this streamline? What is Umax on the bump compared
with stream velocity U?
U a/2 h? Umax? Bump a
U P8.50  v v P8.51 Modify Prob. 8.50 as follows. Let the bump be such that
Umax 1.5U. Find (a) the upstream elevation h and (b)
the height of the bump.
P8.52 The Flettner rotor sailboat in Fig. E8.2 has a water
drag coefficient of 0.006 based on a wetted area of 45
ft2. If the rotor spins at 220 r/min, find the maximum
boat velocity that can be achieved in a 15mi/h wind.
What is the optimum angle between the boat and the
wind?
P8.53 Modify Prob. 8.52 as follows. For the same sailboat data,
find the wind velocity, in mi/h, which will drive the boat
at an optimum speed of 10 kn parallel to its keel.
P8.54 The original Flettner rotor ship was approximately 100 ft
long, displaced 800 tons, and had a wetted area of 3500  eText Main Menu  ω ω U∞ P8.54
P8.55 Assume that the Flettner rotorship of Fig. P8.54 has a water resistance coefficient of 0.005. How fast will the ship
sail in seawater at 20°C in a 20ft/s wind if the keel aligns
itself with the resultant force on the rotors? Hint: This is
a problem in relative velocities.
P8.56 The measured drag coefficient of a cylinder in crossflow,
based on frontal area DL, is approximately 1.0 for the laminarboundarylayer range (see Fig. 7.16a). Boundarylayer separation occurs near the shoulder (see Fig. 7.13a).
This suggests an analytical model: the standard inviscidflow solution on the front of the cylinder and constant pressure (equal to the shoulder value) on the rear. Use this
model to predict the drag coefficient and comment on the
results with reference to Fig. 7.13c.
P8.57 In principle, it is possible to use rotating cylinders as aircraft wings. Consider a cylinder 30 cm in diameter, rotating at 2400 r/min. It is to lift a 55kN airplane cruising at
100 m/s. What should the cylinder length be? How much
power is required to maintain this speed? Neglect end effects on the rotating wing.
P8.58 Plot the streamlines due to the combined flow of a line
sink m at the origin plus line sources m at (a, 0) and
(4a, 0). Hint: A cylinder of radius 2a will appear.
P8.59 By analogy with Prob. 8.58 plot the streamlines due to
counterclockwise line vortices K at (0, 0) and (4a, 0)
plus a clockwise vortex K at (a, 0). Again a cylinder appears. Textbook Table of Contents  Study Guide 560 Chapter 8 Potential Flow and Computational Fluid Dynamics P8.60 One of the cornerflow patterns of Fig. 8.15 is given by
the cartesian stream function
A(3yx 2
y3). Which
one? Can the correspondence be proved from Eq. (8.49)?
P8.61 Plot the streamlines of Eq. (8.49) in the upper right quadrant for n 4. How does the velocity increase with x outward along the xaxis from the origin? For what corner angle and value of n would this increase be linear in x? For
what corner angle and n would the increase be as x5?
P8.62 Combine stagnation flow, Fig. 8.14b, with a source at the
origin:
Az 2 f (z) m ln z Plot the streamlines for m AL2, where L is a length scale.
Interpret.
P8.63 The superposition in Prob. 8.62 leads to stagnation flow
near a curved bump, in contrast to the flat wall of Fig.
8.14b. Determine the maximum height H of the bump as
a function of the constants A and m.
P8.64 Determine qualitatively from boundarylayer theory (Chap.
7) whether any of the three stagnationflow patterns of Fig.
8.15 can suffer flow separation along the walls.
P8.65 Potential flow past a wedge of halfangle leads to an
important application of laminarboundarylayer theory
called the FalknerSkan flows [15, pp. 242 – 247]. Let x denote distance along the wedge wall, as in Fig. P8.65, and
let
10°. Use Eq. (8.49) to find the variation of surface
velocity U(x) along the wall. Is the pressure gradient adverse or favorable? P8.69. What hyphenated word (originally French) might
describe such a flow pattern?
(ψ = 0) y=a Plot the streamlines
inside this region y P8.69 x
1 P8.70 Show that the complex potential f U {z 4 a coth
[ (z/a)]} represents flow past an oval shape placed mid1
way between two parallel walls y
2 a. What is a practical application?
P8.71 Figure P8.71 shows the streamlines and potential lines of
flow over a thinplate weir as computed by the complex potential method. Compare qualitatively with Fig. 10.16a.
State the proper boundary conditions at all boundaries. The
velocity potential has equally spaced values. Why do the
flownet “squares’’ become smaller in the overflow jet? Weir U (x)
x θ
θ P8.71
P8.72 Use the method of images to construct the flow pattern for
a source m near two walls, as shown in Fig. P8.72.
Sketch the velocity distribution along the lower wall (y
0). Is there any danger of flow separation along this wall? P8.65  v v *P8.66 The inviscid velocity along the wedge in Prob. 8.65 has
the analytic form U(x) Cx m, where m n 1 and n is
the exponent in Eq. (8.49). Show that, for any C and n,
computation of the boundary layer by Thwaites’ method,
Eqs. (7.53) and (7.54), leads to a unique value of the
Thwaites parameter . Thus wedge flows are called similar [15, p. 244].
P8.67 Investigate the complex potential function f(z)
U (z a 2/z) and interpret the flow pattern.
P8.68 Investigate the complex potential function f(z) U z
m ln [(z a)/(z a)] and interpret the flow pattern.
P8.69 Investigate the complex potential f(z) A cosh [ (z/a)],
and plot the streamlines inside the region shown in Fig.  eText Main Menu  y +m a P8.72 a x 0 P8.73 Set up an image system to compute the flow of a source at
unequal distances from two walls, as in Fig. P8.73. Find the
point of maximum velocity on the yaxis. Textbook Table of Contents  Study Guide Problems 561
y D
4a +m
B
U∞ a
2a 2a
A x P8.73 4a
C P8.74 A positive line vortex K is trapped in a corner, as in Fig.
P8.74. Compute the total induced velocity vector at point
B, (x, y) (2a, a), and compare with the induced velocity when no walls are present.
y
K
2a V? P8.76
P8.77 Discuss how the flow pattern of Prob. 8.58 might be interpreted to be an imagesystem construction for circular
walls. Why are there two images instead of one?
*P8.78 Indicate the system of images needed to construct the flow
of a uniform stream past a Rankine halfbody constrained
between two parallel walls, as in Fig. P8.78. For the particular dimensions shown in this figure, estimate the position of the nose of the resulting halfbody. B a y a
0 a U∞ x 2a x P8.74
P8.75 The flow past a cylinder very near a wall might be simulated by doublet images, as in Fig. P8.75. Explain why the
result is not very successful and the cylinder shape becomes badly distorted. 2a
a P8.78
P8.79 Explain the system of images needed to simulate the flow
of a line source placed unsymmetrically between two parallel walls as in Fig. P8.79. Compute the velocity on the
lower wall at x a. How many images are needed to estimate this velocity within 1 percent?
y P8.75 2a +m  v v P8.76 Use the method of images to approximate the flow pattern
past a cylinder a distance 4a from a single wall, as in Fig.
P8.76. To illustrate the effect of the wall, compute the velocities at corresponding points A, B and C, D, comparing
with a cylinder flow in an infinite expanse of fluid.  eText Main Menu  a
0 P8.79 Textbook Table of Contents  Study Guide x 562 Chapter 8 Potential Flow and Computational Fluid Dynamics *P8.80 The beautiful expression for lift of a twodimensional airfoil, Eq. (8.69), arose from applying the Joukowski transformation,
z a2/z, where z x iy and
i.
The constant a is a length scale. The theory transforms a
certain circle in the z plane into an airfoil in the plane.
Taking a 1 unit for convenience, show that (a) a circle
with center at the origin and radius 1 will become an
ellipse in the plane and (b) a circle with center at x
1, y 0, and radius (1
) will become an airfoil shape in the plane. Hint: The Excel spreadsheet is
excellent for solving this problem.
P8.81 A twodimensional airfoil has 2 percent camber and 10 percent thickness. If C 1.75 m, estimate its lift per meter
when immersed in 20°C water at
6° and U 18 m/s.
P8.82 The ultralight plane Gossamer Condor in 1977 was the first
to complete the Kremer Prize figureeight course under human power. Its wingspan was 29 m, with Cav 2.3 m and
a total mass of 95 kg. The drag coefficient was approximately
1
0.05. The pilot was able to deliver 4 hp to propel the plane.
Assuming twodimensional flow at sea level, estimate (a) the
cruise speed attained, (b) the lift coefficient, and (c) the horsepower required to achieve a speed of 15 kn.
P8.83 Twodimensional liftdrag data for the NACA 2412 airfoil
with 2 percent camber (from Ref. 12) may be curvefitted
accurately as follows:
CL 0.178 CD 0.0089 0.109 0.00109 1.97 E4 1.35 E5 3 V/U (lower) 0.00
0.97
1.23
1.28
1.29
1.29
1.24
1.14
0.99
0.82 0.00
0.82
0.98
1.05
1.13
1.16
1.16
1.08
0.95
0.82 v v   P8.88
EES P8.89 4 V/U (upper) 0.0
0.025
0.05
0.1
0.2
0.3
0.4
0.6
0.8
1.0 P8.87 2 with in degrees in the range 4°
10°. Compare (a) the liftcurve slope and (b) the angle of zero lift
with theory, Eq. (8.69). (c) Prepare a polar liftdrag plot
and compare with Fig. 7.26.
P8.84 Reference 12 contains inviscidtheory calculations for the
upper and lower surface velocity distributions V(x) over an
airfoil, where x is the chordwise coordinate. A typical result for small angle of attack is as follows:
x/c P8.86 2 8.45 E5 9.92 E7 P8.85 eText Main Menu P8.90 P8.91 P8.92 P8.93  Use these data, plus Bernoulli’s equation, to estimate (a)
the lift coefficient and (b) the angle of attack if the airfoil
is symmetric.
A wing of 2 percent camber, 5in chord, and 30in span is
tested at a certain angle of attack in a wind tunnel with
sealevel standard air at 200 ft/s and is found to have lift
of 30 lbf and drag of 1.5 lbf. Estimate from wing theory
(a) the angle of attack, (b) the minimum drag of the wing
and the angle of attack at which it occurs, and (c) the maximum lifttodrag ratio.
An airplane has a mass of 20,000 kg and flies at 175 m/s
at 5000m standard altitude. Its rectangular wing has a 3m chord and a symmetric airfoil at 2.5° angle of attack.
Estimate (a) the wing span, (b) the aspect ratio, and (c)
the induced drag.
A freshwater boat of mass 400 kg is supported by a rectangular hydrofoil of aspect ratio 8, 2 percent camber, and
12 percent thickness. If the boat travels at 8 m/s and
3.5°, estimate (a) the chord length, (b) the power required
if CD
0.01, and (c) the top speed if the boat is refitted
with an engine which delivers 50 hp to the water.
The Boeing 727 airplane has a gross weight of 125,000
lbf, a wing area of 1200 ft2, and an aspect ratio of 6. It is
fitted with two turbofan engines and cruises at 532 mi/h
at 30,000ft standard altitude. Assume for this problem that
its airfoil is the NACA 2412 section described in Prob.
8.83. If we neglect all drag except the wing, what thrust
is required from each engine for these conditions?
The Beechcraft T34C aircraft has a gross weight of 5500
lbf and a wing area of 60 ft2 and flies at 322 mi/h at 10,000ft standard altitude. It is driven by a propeller which delivers 300 hp to the air. Assume for this problem that its
airfoil is the NACA 2412 section described in Prob. 8.83,
and neglect all drag except the wing. What is the appropriate aspect ratio for the wing?
When moving at 15 m/s in seawater at its maximum lifttodrag ratio of 18:1, a symmetric hydrofoil, of plan area
3 m2, develops a lift of 120 kN. Estimate from wing theory (a) the aspect ratio and (b) the angle of attack in degrees.
If (r, ) in axisymmetric flow is defined by Eq. (8.85)
and the coordinates are given in Fig. 8.24, determine what
partial differential equation is satisfied by .
A point source with volume flow Q 30 m3/s is immersed in a uniform stream of speed 4 m/s. A Rankine
halfbody of revolution results. Compute (a) the distance
from source to the stagnation point and (b) the two points
(r, ) on the body surface where the local velocity equals
4.5 m/s.
The Rankine halfbody of revolution (Fig. 8.26) could
simulate the shape of a pitotstatic tube (Fig. 6.30). Ac Textbook Table of Contents  Study Guide Problems 563
cording to inviscid theory, how far downstream from the
nose should the static pressure holes be placed so that the
local velocity is within 0.5 percent of U ? Compare
your answer with the recommendation x 8D in Fig.
6.30.
P8.94 Determine whether the Stokes streamlines from Eq. (8.86)
are everywhere orthogonal to the Stokes potential lines
from Eq. (8.87), as is the case for cartesian and plane polar coordinates.
P8.95 Show that theaxisymmetric potential flow formed by superposition of a point source m at (x, y) ( a, 0), a
point sink m at ( a, 0), and a stream U in the x direction forms a Rankine body of revolution as in Fig.
P8.95. Find analytic expressions for determining the length
2L and maximum diameter 2R of the body in terms of m,
U , and a. y
+m U∞ a r
θ –m
a x P8.95 P8.98 We have studied the point source (sink) and the line source
(sink) of infinite depth into the paper. Does it make any
sense to define a finitelength line sink (source) as in Fig.
P8.98? If so, how would you establish the mathematical
properties of such a finite line sink? When combined with
a uniform stream and a point source of equivalent strength
as in Fig. P8.98, should a closedbody shape be formed?
Make a guess and sketch some of these possible shapes for
various values of the dimensionless parameter m/(U L2).
Line sink of
total strength
–m +m
U∞ x 0 L P8.98
*P8.99 Consider air flowing past a hemisphere resting on a flat
surface, as in Fig. P8.99. If the internal pressure is pi, find
an expression for the pressure force on the hemisphere. By
analogy with Prob. 8.49, at what point A on the hemisphere
should a hole be cut so that the pressure force will be zero
according to inviscid theory? U∞ , p∞ P8.96 Suppose that a sphere with a single stagnation hole is to
be used as a velocimeter. The pressure at this hole is used
to compute the stream velocity, but there are errors if the
hole is not perfectly aligned with the oncoming stream.
Using inviscid incompressible theory, plot the percent error in stream velocity estimate as a function of misalignment angle . At what angle is the error 10 percent?
P8.97 The Rankine body of revolution in Fig. P8.97 is 60 cm
long and 30 cm in diameter. When it is immersed in the
lowpressure water tunnel as shown, cavitation may appear at point A. Compute the stream velocity U, neglecting surface wave formation, for which cavitation occurs. y Point
source pi 2a P8.99
P8.100 A 1mdiameter sphere is being towed at speed V in fresh
water at 20°C as shown in Fig. P8.100. Assuming inviscid theory with an undistorted free surface, estimate the
speed V in m/s at which cavitation will first appear on the
sphere surface. Where will cavitation appear? For this condition, what will be the pressure at point A on the sphere
which is 45° up from the direction of travel?
pa = 101.35 k Pa pa = 40 kPa
Water at 20˚ C 3m
A 80 cm A
V U D=1m P8.100
P8.101 Normally by its very nature inviscid theory is incapable of
predicting body drag, but by analogy with Fig. 8.16c we
can analyze flow approaching a hemisphere, as in Fig. Rankine ovoid  v v P8.97  eText Main Menu  Textbook Table of Contents  Study Guide 564 Chapter 8 Potential Flow and Computational Fluid Dynamics
P8.101. Assume that the flow on the front follows inviscid sphere theory, Eq. (8.96), and the pressure in the rear
equals the shoulder pressure. Compute the drag coefficient
and compare with experiment (Table 7.3). What are the
defects and limitations of this analysis?
Hemisphere
U∞ , p∞ a imum thickness. Use these results to derive a formula from
the time history U(t) of the cylinder if it is accelerated from
rest in a still fluid by the sudden application of a constant
force F.
P8.106 Laplace’s equation in plane polar coordinates, Eq. (8.11),
is complicated by the variable radius. Consider the finitedifference mesh in Fig. P8.106, with nodes (i, j) equally
spaced
and r apart. Derive a finitedifference model
for Eq. (8.11) similar to the cartesian expression (8.109). Rear pressure
assumed equal
to shoulder
pressure i, j + 1
rj + 1
∆r P8.101
P8.102 A golf ball weighs 0.102 lbf and has a diameter of 1.7 in.
A professional golfer strikes the ball at an initial velocity
of 250 ft/s, an upward angle of 20°, and a backspin (front
of the ball rotating upward). Assume that the lift coefficient on the ball (based on frontal area) follows Fig.
P7.108. If the ground is level and drag is neglected, make
a simple analysis to predict the impact point (a) without
spin and (b) with backspin of 7500 r/min.
P8.103 Modify Prob. 8.102 as follows. Golf balls are dimpled, not
smooth, and have higher lift and lower drag (CL 0.2 and
CD 0.3 for typical backspin). Using these values, make
a computer analysis of the ball trajectory for the initial conditions of Prob. 8.102. If time permits, investigate the effect of initial angle for the range 10°
50°.
0
P8.104 Consider a cylinder of radius a moving at speed U
through a still fluid, as in Fig. P8.104. Plot the streamlines
relative to the cylinder by modifying Eq. (8.32) to give the
relative flow with K 0. Integrate to find the total relative kinetic energy, and verify the hydrodynamic mass of
a cylinder from Eq. (8.104). νr νθ i + 1, j
rj i, j i – 1, j
∆r i, j – 1
∆θ rj – 1 ∆θ P8.106
P8.107 Set up the numerical problem of Fig. 8.30 for an expansion of 30°. A new grid system and a nonsquare mesh may
be needed. Give the proper nodal equation and boundary
conditions. If possible, program this 30° expansion and
solve on a digital computer.
P8.108 Consider twodimensional potential flow into a step contraction as in Fig. P8.108. The inlet velocity U1 7 m/s,
and the outlet velocity U2 is uniform. The nodes (i, j) are
labeled in the figure. Set up the complete finitedifference
algebraic relations for all nodes. Solve, if possible, on a
digital computer and plot the streamlines in the flow.
i=1
j=1 Still
fluid 2 3 4 5 6 7 8 9 2
a U∞ U2 3
U1 P8.104 4
5
6 v v *P8.105 In Table 7.2 the drag coefficient of a 4:1 elliptical cylinder in laminarboundarylayer flow is 0.35. According to
Patton [17], the hydrodynamic mass of this cylinder is
hb/4, where b is width into the paper and h is the max  10  eText Main Menu  7
8 P8.108 Textbook Table of Contents  Study Guide Problems 565
P8.109 Consider inviscid flow through a twodimensional 90°
bend with a contraction, as in Fig. P8.109. Assume uniform flow at the entrance and exit. Make a finitedifference computer analysis for small grid size (at least 150
nodes), determine the dimensionless pressure distribution
along the walls, and sketch the streamlines. (You may use
either square or rectangular grids.)
5m 6m V2 10 m 16 m
10 m P8.112 In his CFD textbook, Patankar [5] replaces the lefthand
sides of Eq. (8.119b and c) with the following two expressions, respectively:
∂2
(u )
∂x ∂
(u )
∂x and 1.4845 1/2 0.63 1.758 2 1.4215 15 m P8.109
P8.110 For fully developed laminar incompressible flow through
a straight noncircular duct, as in Sec. 6.6, the NavierStokes equations (4.38) reduce to
∂2u
∂y2 ∂2u
∂z2 1 dp
dx const Q 0.1143 dp
dx f ReDh 62.19 v v where Dh 4A/P 4b/3 for this case. Comment on the
possible truncation errors of your model.   eText Main Menu 3 0.5075 4 where
x/C and the maximum thickness tmax occurs at
0.3. Use this shape as part of the lower boundary for
zero angle of attack. Let the thickness be fairly large, say,
tmax 0.12, 0.15, or 0.18. Choose a generous number of
nodes ( 60), and calculate and plot the velocity distribution V/U along the airfoil surface. Compare with the theoretical results in Ref. 12 for NACA 0012, 0015, or 0018
airfoils. If time permits, investigate the effect of the boundary lengths L1, L2, and L3, which can initially be set equal
to the chord length C. 0 where (y, z) is the plane of the duct cross section and x is
along the ductaxis. Gravity is neglected. Using a nonsquare rectangular grid ( x, y), develop a finitedifference model for this equation, and indicate how it may be
applied to solve for flow in a rectangular duct of side
lengths a and b.
P8.111 Solve Prob. 8.110 numerically for a rectangular duct of
side length b by 2b, using at least 100 nodal points. Evaluate the volume flow rate and the friction factor, and compare with the results in Table 6.4:
b4 ∂2
()
∂y Are these equivalent expressions, or are they merely simplified approximations? Either way, why might these forms
be better for finitedifference purposes?
P8.113 Repeat Example 8.7 using the implicit method of Eq.
(8.118). Take t 0.2 s and y 0.01 m, which ensures
that an explicit model would diverge. Compare your accuracy with Example 8.7.
P8.114 If your institution has an online potentialflow boundaryelement computer code, consider flow past a symmetric
airfoil, as in Fig. P8.114. The basic shape of an NACA
symmetric airfoil is defined by the function [12]
2y
tmax V1 = 10 m/s ∂
( u)
∂y  L3 U∞
y L1 Airfoil halfcontour x=0 x x=C L2 P8.114
P8.115 Use the explicit method of Eq. (8.115) to solve Prob. 4.85
numerically for SAE 30 oil at 20°C with U0 1 m/s and
M rad/s, where M is the number of letters in your surname. (This author will solve the problem for M 5.)
When steady oscillation is reached, plot the oil velocity
versus time at y 2 cm. Textbook Table of Contents  Study Guide 566 Chapter 8 Potential Flow and Computational Fluid Dynamics Word Problems
W8.1 What simplifications have been made, in the potentialflow
theory of this chapter, which result in the elimination of the
Reynolds number, Froude number, and Mach number as important parameters?
W8.2 In this chapter we superimpose many basic solutions, a concept associated with linear equations. Yet Bernoulli’s equation (8.3) is nonlinear, being proportional to the square of
the velocity. How, then, do we justify the use of superposition in inviscidflow analysis?
W8.3 Give a physical explanation of circulation as it relates to
the lift force on an immersed body. If the line integral defined by Eq. (8.15) is zero, it means that the integrand is a
perfect differential — but of what variable? W8.4 Give a simple proof of Eq. (8.42), namely, that both the real
and imaginary parts of a function f(z) are laplacian if z
x iy. What is the secret of this remarkable behavior?
W8.5 Figure 8.14 contains five body corners. Without carrying out
any calculations, explain physically what the value of the
inviscid fluid velocity must be at each of these five corners.
Is any flow separation expected?
W8.6 Explain the Kutta condition physically. Why is it necessary?
W8.7 We have briefly outlined finitedifference and boundaryelement methods for potential flow but have neglected the
finiteelement technique. Do some reading and write a brief
essay on the use of the finiteelement method for potentialflow problems. Comprehensive Problems
C8.1 Did you know that you can solve simple fluid mechanics
problems with Microsoft Excel? The successive relaxation
technique for solving the Laplace equation for potentialflow problems is easily set up on a spreadsheet, since the
stream function at each interior cell is simply the average
of its four neighbors. As an example, solve for the irrotational potential flow through a contraction, as given in Fig.
C8.1. Note: To avoid the “circular reference” error, you must
turn on the iteration option. Use the help index for more information. For full credit, attach a printout of your spreadsheet, with stream function converged and the value of the
stream function at each node displayed to four digits of accuracy.
Wall, cm and (b) the instantaneous boundarylayer thickness
(where u 0.99 u ). Hint: There is a nonzero pressure
gradient in the outer (nearly shearfree) stream, n N,
which must be included in Eq. (8.114) and your explicit
model.
C8.3 Consider plane inviscid flow through a symmetric diffuser,
as in Fig. C8.3. Only the upper half is shown. The flow is to
expand from inlet halfwidth h to exit halfwidth 2h, as
shown. The expansion angle is 18.5° (L 3h). Set up a
nonsquare potentialflow mesh for this problem, and calculate and plot (a) the velocity distribution and (b) the pressure coefficient along the centerline. Assume uniform inlet
and exit flows. 5 5 3.333 4 2h
Outlet 3
Inlet 1.667 2 h V 1
Wall, 0 2h 0 L C8.3
C8.1 Wall, 0  v v C8.2 Use an explicit method, similar to but not identical to Eq.
(8.115), to solve the case of SAE 30 oil at 20°C starting
from rest near a fixed wall. Far from the wall, the oil accelerates linearly, that is, u
uN at, where a 9
m/s2. At t 1 s, determine (a) the oil velocity at y 1  eText Main Menu  C8.4 Use potential flow to approximate the flow of air being
sucked up into a vacuum cleaner through a twodimensional
slit attachment, as in Fig. C8.4. In the xy plane through the
centerline of the attachment, model the flow as a line sink of
strength ( m), with its axis in the zdirection at height a
above the floor. (a) Sketch the streamlines and locate any Textbook Table of Contents  Study Guide References 567
stagnation points in the flow. (b) Find the magnitude of velocity V(x) along the floor in terms of the parameters a and
m. (c) Let the pressure far away be p , where velocity is
zero. Define a velocity scale U m/a. Determine the variation of dimensionless pressure coefficient, Cp (p
p )/( U2/2), along the floor. (d) The vacuum cleaner is most
effective where Cp is a minimum, that is, where velocity is
maximum. Find the locations of minimum pressure coefficient along the xaxis. (e) At which points along the xaxis
do you expect the vacuum cleaner to work most effectively?
Is it best at x 0 directly beneath the slit, or at some other
x location along the floor? Conduct a scientific experiment
at home with a vacuum cleaner and some small pieces of
dust or dirt to test your prediction. Report your results and
discuss the agreement with prediction. Give reasons for any
disagreements. y a x C8.4
C8.5 Consider a threedimensional, incompressible, irrotational flow.
Use the following two methods to prove that the viscous term
in the NavierStokes equation is identically zero: (a) using vector notation; and (b) expanding out the scalar terms and substituting terms from the definition of irrotationality. Design Projects
D8.1 In 1927, Theodore von Kármán developed a scheme to use a
uniform stream, plus a row of sources and sinks, to generate an
arbitrary closedbody shape. A schematic of the idea is sketched
in Fig. D8.1. The body is symmetric and at zero angle of attack. A total of N sources and sinks are distributed along the
axis within the body, with strengths mi at positions xi, for i
1 to N. The object is to find the correct distribution of strengths
which approximates a given body shape y(x) at a finite number of surface locations and then to compute the approximate
surface velocity and pressure. The technique should work for
either twodimensional bodies (distributed line sources) or bodies of revolution (distributed point sources).
For our body shape let us select the NACA 0018 airfoil,
given by the formula in Prob. 8.114 with tmax 0.18. Develop the ideas stated above into N simultaneous algebraic
equations which can be used to solve for the N unknown line
source/sink strengths. Then program your equations for a
computer, with N 20; solve for mi; compute the surface
velocities; and compare with the theoretical velocities for this
shape in Ref. 12. Your goal should be to achieve accuracy
within 1 percent of the classical results. If necessary, you
should adjust N and the locations of the sources. Body shape Typical body point yj U∞
Axis Source m i
i=1 x
i=N D8.1 D8.2 Modify Prob. D8.1 to solve for the pointsource distribution
which approximates an “0018’’ bodyofrevolution shape.
Since no theoretical results are published, simply make sure
that your results converge to 1 percent.
D8.3 Consider water at 20°C flowing at 12 m/s in a water channel. A Rankine oval cylinder, 40 cm long, is to be placed
parallel to the flow, where the water static pressure is 120
kPa. The oval’s thickness is a design parameter. Prepare a
plot of the minimum pressure on the oval’s surface as a function of body thickness. Especially note the thicknesses where
(a) the local pressure is 50 kPa and (b) cavitation first occurs on the surface. References
O. D. Kellogg, Foundations of Potential Theory, Dover, New
York, 1969.
2. J. M. Robertson, Hydrodynamics in Theory and Application,
PrenticeHall, Englewood Cliffs, NJ, 1965.
3. L. M. MilneThomson, Theoretical Hydrodynamics, 4th ed.,
Macmillan, New York, 1960.  v v 1.  eText Main Menu  4. I. G. Currie, Fundamental Mechanics of Fluids, 2d ed.,
McGrawHill, New York, 1993.
5. S. V. Patankar, Numerical Heat Transfer and Fluid Flow,
McGrawHill, New York, 1980.
6. G. F. Carey and J. T. Oden, Finite Elements: Fluid Mechanics, vol. 6, PrenticeHall, Englewood Cliffs, NJ, 1986. Textbook Table of Contents  Study Guide 568 Chapter 8 Potential Flow and Computational Fluid Dynamics  v v 7. C. A. Brebbia and J. Dominquez, Boundary Elements — An Introductory Course, 2d ed., Computational Mechanics Publications, Southampton, England, and McGrawHill, New York,
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8. A. D. Moore, “Fields from Fluid Flow Mappers,’’ J. Appl.
Phys., vol. 20, 1949, pp. 790 – 804.
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40, 1898, p. 25.
10. R. H. F. Pao, Fluid Dynamics, Merrill, Columbus, OH,
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11. A. M. Kuethe and C.Y. Chow, Foundations of Aerodynamics, 5th ed., Wiley, New York, 1997.
12. I. H. Abbott and A. E. von Doenhoff, Theory of Wing Sections, Dover, New York, 1959.
13. B. Thwaites (ed.), Incompressible Aerodynamics, Clarendon
Press, Oxford, England, 1960.
14. L. Prandtl, “Applications of Modern Hydrodynamics to Aeronautics,’’ NACA Rep. 116, 1921.
15. F. M. White, Viscous Fluid Flow, 2d ed., McGrawHill, New
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Potential Flow about Arbitrary ThreeDimensional Bodies,’’
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19. K. H. Huebner, The Finite Element Method for Engineers, 3d
ed., Wiley, New York, 1994.
20. J. C. Tannehill, D. A. Anderson, and R. H. Pletcher, Computational Fluid Mechanics and Heat Transfer, 2d ed., Taylor
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21. J. N. Newman, Marine Hydrodynamics, M.I.T. Press, Cambridge, MA, 1977.  eText Main Menu  22.
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J. H. Ferziger and M. Peric, Computational Methods for Fluid
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C. Hirsch, Numerical Computation of Internal and External
Flows, 2 vols., Wiley, New York, 1990.
K. A. Hoffmann and S. T. Chiang, CFD for Engineers, 2 vols.,
Engineering Education System, New York, 1993.
J. D. Anderson, Computational Fluid Dynamics: The Basics
with Applications, McGrawHill, New York, 1995.
C. A. J. Fletcher, Computational Techniques for Fluid Dynamics, 2 vols., Springer Verlag, New York, 1997.
M. Deshpande, J. Feng, and C. L. Merkle, “Numerical Modeling of the Thermodynamic Effects of Cavitation,” J. Fluids
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P. A. Libby, An Introduction to Turbulence, Taylor and Francis, Bristol, PA, 1996.
C. J. Freitas, “Perspective: Selected Benchmarks from Commercial CFD Codes,” J. Fluids Eng., vol. 117, June 1995, pp.
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480 – 483. Textbook Table of Contents  Study Guide ...
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