8.1 Potential Flow
BASICS
When a body is moving in an otherwise stationary ﬂuid, there is no vorticity present in the undisturbed ﬂuid. To create a steady ﬂow, a uniform ﬂow with the body’s velocity is superimposed on the ﬂow ﬁeld so that the vorticitfree ﬂow moves by the stationary body, as in a wind tunnel. For high Reynoldsnumber ﬂows, the viscous effects are concentrated in the boundary layer and the wake (the wake includes the separated region). For a streamlined body and over the front part of a blunt body, the ﬂow outside the boundary layer is free of viscous effects so it is an inviscid ﬂow. The solution of the inviscid ﬂow problem provides the velocity ﬁeld and the pressure ﬁeld. The pressure is not signiﬁcantly inﬂuenced by the boundary layer so the pressure integrated over the body’s surface will provide the lift. The velocity at the surface of the body2 from the inviscid ﬂow solution will be the velocity at the outer edge of the thin boundary layer needed in the boundarylayer solution (to be presented in Sec. 8.5). So, before the boundary layer can be analyzed on a body, the inviscid ﬂow must be known.
A potential ﬂow (or irrotational ﬂow), is one in which the velocity ﬁeld can be expressed as the gradient of a scalar function, that is,
where f is the velocity potential. For a potential ﬂow, the vorticity is zero:
This can be shown to be true by expanding in rectangular coordinates and using Eq. (8.7).
To understand why vorticity cannot exist in regions of an irrotational ﬂow, con sider the effect of the three types of forces that can act on a cubic ﬂuid element: the pressure and body forces act through the center of the element, and consequently, cannot impart a rotary motion to the element. It is only the viscous shear forces that are able to give rotary motion to ﬂuid particles. Hence, if the viscous effects are nonexistent, vorticity cannot be introduced into an otherwise potential ﬂow.
If the velocity is given by Eq. (8.7), the continuity equation (5.8) for an incompressible ﬂow provides
which is the famous Laplace equation. In rectangular coordinates it is written as
With the required boundary conditions, this equation could be solved. But, rather than attempting to solve the resulting boundaryvalue problem directly, we will restrict our interest to plane ﬂows, identify several simple ﬂows that satisfy Laplace equation, and then superimpose those simple ﬂows to form more complex ﬂows of interest. Since Laplace equation is linear, the superimposed ﬂows will also satisfy Laplace equation.
First, however, we will deﬁne another scalar function that will be quite useful in our study. For the plane ﬂows of interest, the stream function y, is deﬁned by
so that the continuity equation (5.8) with ∂w/∂ z = 0 (for a plane ﬂow) is satisﬁed for all plane ﬂows. The vorticity [see Eqs. (8.8) and (3.14)] then provides
The stream function also satisﬁes the Laplace equation. So, from the above equations we have.
The equations between f and y in Eq. (8.14) form the CauchyRiemann equations and f and y are referred to as harmonic functions. The function f + iy is the complex velocity potential. The mathematical theory of complex variables is thus applicable to this subset of ﬂuid ﬂows: steady, incompressible, inviscid plane ﬂows.
Three items of interest contained in the above equations are:
• The stream function is constant along a streamline.
• The streamlines and lines of constant potential intersect at right angles.
• The difference of the stream functions between two streamlines is the ﬂow rate q per unit depth between the two streamlines, i.e., q = ψ 2 − ψ1.
EXAMPLE 8.5
Show that y is constant along a streamline.
Solution
A streamline is a line to which the velocity vector is tangent. This is expressed in vector form as V × dr = 0, which, for a plane ﬂow (no z variation), using dr = dx i + dy j takes the form udy − vdx = 0. Using Eq. (8.11), this becomes
This is the deﬁnition of dy from calculus, thus dy = 0 along a streamline, or, in other words, y is constant along a streamline.
SEVERAL SIMPLE FLOWS
Several of the simple ﬂows to be presented are much easier understood using polar (cylindrical) coordinates. The Laplace equation, the continuity equation, and the expressions for the velocity components for a plane ﬂow (see Table 5.1) are
where the expressions relating the velocity components to the stream function are selected so that the continuity equation is always satisﬁed. We now deﬁne four simple ﬂows that satisfy the Laplace equation:
These simple plane ﬂows are shown in Fig. 8.8. If a ycomponent is desired for the uniform ﬂow, an appropriate term is added. The source strength q in the line source is the ﬂow rate per unit depth; adding a minus sign creates a sink. The vortex strength
Γ is the circulation about the origin, deﬁned as
Figure 8.8 Four simple plane potential ﬂows.
where L is a closed curve, usually a circle, about the origin with clockwise being positive. The heavy arrow in the negative xdirection represents the doublet strength m in Fig. 8.8d. (A doublet can be thought of as a source and a sink of equal strengths separated by a very small distance.)
The velocity components are used quite often for the four simple ﬂows presented. They follow for both polar and rectangular coordinates:
These four simple ﬂows can be superimposed to create more complicated ﬂows of interest. This will be done in the following section.
EXAMPLE 8.6
If the stream function of a ﬂow is given as ψ = Aθ, determine the potential function φ .
Solution
We use Eq. (8.17) to relate the stream function to the potential function assuming polar coordinates because of the presence of θ :
Since we are only interested in the derivatives of the potential functions needed to provide the velocity and pressure ﬁelds, we simply let the constant be zero and thus
So, we see that the potential function can be found if the stream function is known. Conversely, the stream function can be found if the potential function is known.
SUPERIMPOSED FLOWS
The simple ﬂows just deﬁned can be superimposed to create complicated plane ﬂows. Divide a surface, such as an airfoil, into a large number of segments and position sources or sinks (or doublets) at the center of each segment; in addition, add a uniform ﬂow and a vortex. Then, adjust the various strengths so that the normal velocity component at each segment is zero and the rear stagnation point is located at the trailing edge. Obviously, a computer program would be used to create such a ﬂow. We will not attempt it in this book but will demonstrate how ﬂow around a circular cylinder can be created.
Superimpose the stream functions of a uniform ﬂow and a doublet:
Thus, two stagnation points occur at q = 0° and 180°. The streamline pattern would appear as in the sketch of Fig. 8.9. The circular streamline represents the cylinder, which is typically a solid, and hence our interest is in the ﬂow outside the circle. For a real ﬂow, there would be a separated region on the rear of the cylinder but the ﬂow over the front part (perhaps over the whole front half, depending on the Reynolds number) could be approximated by the potential ﬂow shown in the ﬁgure. The velocity that exists outside the thin boundary layer that would be present on a real
Figure 8.9 Potential ﬂow around a circular cylinder. (The dashed lines are lines of constant f.)
cylinder would be approximated as the velocity on the cylinder of the potential ﬂow, that is, it would be given by
The pressure that would exist on the cylinder’s surface would be found by applying Bernoulli’s equation between the stagnation point (V = 0) where the pressure is p0 and some general point at rc and θ:
This pressure would approximate the actual pressure for high Reynoldsnumber ﬂows up to separation. For low Reynoldsnumber ﬂows, say below Re ≈ 50, viscous effects are not conﬁned to a thin boundary layer so potential ﬂow does not approximate the real ﬂow.
To create ﬂow around a rotating cylinder, as in Fig. 8.10, add a vortex to the stream function of Eq. (8.27) [use the cylinder’s radius of Eq. (8.29)]:
Figure 8.10 Flow around a rotating cylinder.
The cylinder’s radius remains unchanged since a vortex does not affect vr . The stagnation points change, however, and are located by letting vθ = 0 on r = rc:
(8.35) is not valid (this would give sinθ > 1 ), so the stagnation point exists off the cylinder as shown in Fig. 8.10b. The radius is found by setting the velocity components equal to zero and the angle q = 270°. The pressure on the surface of the rotating cylinder is found using Bernoulli’s equation to be
If p dA is integrated around the surface of the cylinder, the component in the ﬂow direction, the drag, would be zero and the component normal to the ﬂow direction, the lift, would be
It turns out that this expression for the lift is applicable for all cylinders including the airfoil. It is known as the KuttaJoukowski theorem; it is exact for potential ﬂows and is an approximation for real ﬂows.
EXAMPLE 8.7
A 20cmdiameter cylinder rotates clockwise at 200 rpm in an atmospheric air stream ﬂowing at 10 m/s. Locate any stagnation points and ﬁnd the minimum pressure.
Solution
First, ﬁnd the circulation. It is Γ= 0∫L V ⋅ ds which is the velocity rc Ω multiplied by 2πrc , since V is in the direction of ds on the cylinder’s surface:
using P= 1.2 kg/m^{3} for atmospheric air. (If the temperature is not given, assume
standard conditions.)
8.2 BoundaryLayer Flow
GENERAL INFORMATION
The observation that for a high Reynoldsnumber ﬂow all the viscous effects can be conﬁned to a thin layer of ﬂuid near the surface gives rise to boundarylayer theory. Outside the boundary layer the ﬂuid acts as an inviscid ﬂuid. So, the potential ﬂow theory of the previous section provides both the velocity at the
outer edge of the boundary layer and the pressure at the surface. In this section we will provide the integral and differential equations needed to solve for the velocity distribution in the boundary layer. But, those equations are quite difﬁcult to solve for curved surfaces, so we will restrict our study to ﬂow on a ﬂat plate with zero pressure gradient.
The outer edge of a boundary layer cannot be observed, so we arbitrarily assign its thickness d (x), as shown in Fig. 8.11, to be the locus of points where the velocity is 99 percent of the freestream velocity U(x) (the velocity at the surface from the inviscid ﬂow solution). The pressure at the surface is not inﬂuenced by the presence of the thin boundary layer, so it is the pressure on the surface from the inviscid ﬂow. Note that the xycoordinate system is oriented so that the xcoordinate is along the surface; this is done for the boundarylayer equations and is possible because the boundary layer is so thin that curvature terms do not appear in the describing equations.
A boundary layer is laminar near the leading edge or near a stagnation point. It undergoes transition at xT to a turbulent ﬂow if there is sufﬁcient length, as shown in Fig. 8.12. This transition occurs when the critical Reynolds number U x /v = 5 × 103
on smooth, rigid ﬂat plates in a zero pressuregradient ﬂow with low freestream
Figure 8.13 Laminar and turbulent boundarylayer proﬁles.
ﬂuctuation intensity3 and U x /n = 3 × 105 for ﬂow on rough ﬂat plates or with high freestream ﬂuctuation intensity (intensity of at least 0.1). The transition region from laminar to turbulent ﬂow is relatively short and is typically ignored so a turbulent ﬂow is assumed to exist at the location of the ﬁrst burst.
The turbulent boundary layer thickens more rapidly than a laminar boundary layer and contains signiﬁcantly more momentum (if it has the same thickness), as observed from a sketch of the velocity proﬁles in Fig. 8.13. It also has a much greater slope at the wall resulting in a much larger wall shear stress. The instantaneous turbulent boundary layer varies randomly with time and position and can be 20 percent thicker or 60 percent thinner at any position at an instant in time or at any time at a given position. So, we usually sketch a timeaverage boundarylayer thickness. The viscous wall layer with thickness dv, in which turbulent bursts are
thought to originate, is quite thin compared to the boundarylayer thickness, as
shown.
It should be kept in mind that a turbulent boundary layer is very thin for most

applications. On a ﬂat plate with U
= 5 m/s, the boundary layer would be about 7 cm
thick after 4 m. If this were drawn to scale, the fact that the boundary layer is very thin would be quite apparent. Because the boundary layer is so thin and the velocity varies from zero at the wall to U(x) at the edge of the boundary layer, it is possible to approximate the velocity proﬁle in the boundary layer by assuming a parabolic or cubic proﬁle for a laminar layer and a powerlaw proﬁle for a turbulent layer. With the velocity proﬁle assumed, the integral equations, which follow, give the quantities of interest.
THE INTEGRAL EQUATIONS
An inﬁnitesimal control volume of thickness dx is shown in Fig. 8.14 with mass ﬂuxes in (b) and momentum ﬂuxes in (d). The continuity equation provides the mass ﬂux m top that crosses into the control volume through the top; it is
Figure 8.14 The inﬁnitesimal control volume for a boundary layer.
where we have neglected4 pdd and dpdd since they are of smaller order than the remaining terms; we also used mo·mtop = U (x)m· top . Divide by (–dx) and obtain the von Karman integral equation:
Ordinary derivatives have been used since after the integration only a function of x remains (d is a function of x). Also, the density r is assumed constant over the boundary layer.
For ﬂow on a ﬂat plate with zero pressure gradient, i.e., U(x) = U∞ and ∂ p/∂ x = 0, Eq. (8.41) can be put in the simpliﬁed form
If a velocity proﬁle u(x, y) is assumed for a particular ﬂow, Eq. (8.42) along withallows both d (x) and t0(x) to be determined. Two additional lengths are used in the study of boundary layers. They are the displacement thickness dd and the momentum thickness q deﬁned by
The displacement thickness is the distance the streamline outside the boundary layer is displaced because of the slower moving ﬂuid inside the boundary layer. The momentum thickness is the thickness of a ﬂuid layer with velocity U that possesses the momentum lost due to viscous effects; it is often used as the characteristic length for turbulent boundarylayer studies. Note that Eq. (8.42) can be written as
LAMINAR AND TURBULENT BOUNDARY LAYERS
The boundary conditions that must be met for the velocity proﬁle in a boundary layer on a ﬂat plate with a zero pressure gradient are
For a laminar boundary layer, we can either solve the xcomponent NavierStokes equation or we can assume a proﬁle such as a parabola. Since the boundary layer is so thin, an assumed proﬁle gives rather good results. Assume the parabolic proﬁle
The wall shear stress is also given by
The more accurate coefﬁcients for t0, cf , and Cf are 0.332, 0.664, and 1.33, so the assumption of a parabolic velocity proﬁle for laminar boundarylayer ﬂow has an error of about 10 percent.
Turbulent Boundary Layers
For a turbulent boundary layer we often assume a powerlaw velocity proﬁle5 as we did for ﬂow in a pipe. It is
Substitute this into the Blasius formula and ﬁnd the local skin friction coefﬁcient to be
The above formulas can actually be used up to Re ≅ 108 without substantial error.
If there is a signiﬁcant laminar part of the boundary layer, it should be included. If transition occurs at Re = 5 × 105, then the skin friction coefﬁcient should be modiﬁed as
For a rough plate, recall that Re = 3 × 105; the constant of 1700 in Eq. (8.66) should be replaced with 1060.
The displacement and momentum thicknesses can be evaluated using the power law velocity proﬁle to be
There are additional quantities often used in the study of turbulent boundary layers. We will introduce two such quantities here. One is the shear velocity u_{τ} deﬁned to be
It is a ﬁctitious velocity and often appears in turbulent boundarylayer relation ships. The other is the thickness dv of the highly ﬂuctuating viscous wall layer, displayed in Figs. 8.12 and 8.13. It is in this very thin layer that the turbulent bursts
are thought to originate. It has been related to the shear velocity through experimental observations by
Atmospheric air at 20°C ﬂows at 10 m/s over a smooth, rigid 2mwide, 4mlong ﬂat plate aligned with the ﬂow. How long is the laminar portion of the boundary layer? Predict the drag force on the laminar portion on one side of the plate.
Solution
Assuming the air to free of highintensity disturbances, use the critical Reynolds number to be 5 × 105, i.e.,
Water at 20°C ﬂows over a 2mlong, 3mwide ﬂat plate at 12 m/s. Estimate the shear velocity, the viscous walllayer thickness, and the boundarylayer thick ness at the end of the plate (assume a turbulent layer from the leading edge). Also, predict the drag force on one side of the plate.
Solution
The Reynolds number is Re = U_{∞} x/ν = 12 × 2 /10^{6} = 2.4 × 10^{7}.So, with n = 7 Eq. (8.64) provides
LAMINAR BOUNDARYLAYER DIFFERENTIAL EQUATIONS
The laminar ﬂow solution given in the preceding section was an approximate solution. In this section we will present a more accurate solution using the xcomponent Navier Stokes equation. It is, for horizontal plane ﬂow (no zvariation),
We can simplify this equation and actually obtain a solution. First, recall that the boundary layer is very thin so that there is no pressure variation normal to the
boundary layer, i.e., the pressure depends on x only and it is the pressure at the wall from the potential ﬂow solution. Since the pressure is considered known, the unknowns in Eq. (8.70) are u and v. The continuity equation
also relates u and v. So, we have two equations and two unknowns. Consider Figs. 8.12 and 8.13; u changes from zero to U over the very small distance d, resulting in very large gradients in the ydirection, whereas u changes quite slowly in the xdirection (holding y ﬁxed). Consequently, we conclude that
The two acceleration terms on the left side of the equation are retained, since v may be quite small but the gradient ∂u/∂y is quite large. Equation (8.73) is the Prandtl boundarylayer equation.
For ﬂow on a ﬂat plate with dp/dx = 0, and in terms of the stream function y (recall that u = ∂ψ /∂ y and v = −∂ψ /∂ x), Eq. (8.73) takes the form
This equation appears more formidable than Eq. (8.74), but if we let
The numerical solution to the boundaryvalue problem is presented in Table 8.5. The last two columns allow the calculation of v and t0, respectively. We deﬁned the
boundarylayer thickness to be that thickness where u = 0.99U∞ and we observe that this occurs at h = 5. So, from this numerical solution,
Air at 30°C ﬂows over a 2mwide, 4mlong ﬂat plate with a velocity of 2 m/s and dp/dx = 0. At the end of the plate, estimate (a) the wall shear stress, (b) the maximum value of v in the boundary layer, and (c) the ﬂow rate through the boundary layer. Assume laminar ﬂow over the entire length.
Solution
The Reynolds number is so laminar ﬂow is reasonable.
(a) The wall shear stress (this requires F ′′ at the wall) at x = 4 m is
(b) The maximum value of v requires the use of (h F ′ – F). Its maximum value occurs at the outer edge of the boundary layer and is 0.860. The maximum value of v is
1. Estimate the drag coefﬁcient for air at 10°C ﬂowing around a 4.1cmdiameter golf ball traveling at 35 m/s.
(A) 0.25
(B) 0.86
(C) 0.94
(D) 1.2
2. A ﬂuid ﬂows by a ﬂat circular disk with velocity V normal to the disk with Re > 103. Estimate the drag coefﬁcient if the pressure is assumed constant over the face of the disk. Assume the pressure is zero on the backside.
(A) 1.1
(B) 1.0
(C) 0.9
(D) 0.8
3. Atmospheric air at 20°C is ﬂowing at 10 m/s normal to a 10cmwide, 20cmlong rectangular plate. The drag force is nearest
A) 2.95 N
(B) 2.41 N
(C) 1.76 N
(D) 1.32 N
4. A 20cmdiameter smooth sphere is rigged with a strain gage calibrated to measure the force on the sphere. Estimate the wind speed in 20°C air if the gage measures 0.5 N.
(A) 21 m/s
(B) 25 m/s
(C) 29 m/s
(D) 33 m/s
5. A 2.2mlong hydrofoil with chord length of 50 cm operates 40 cm below the water’s surface with an angle of attack of 4°. For a speed of 15 m/s, the lift is nearest
(A) 75 kN
(B) 81 kN
(C) 99 kN
(D) 132 kN
6. Estimate the takeoff speed for an aircraft with conventional airfoils if the aircraft with payload weighs 120 000 N and the effective wing area is
20 m2 assuming a temperature of 30°C. An angle of attack at takeoff of
8° is desired.
(A) 100 m/s
(B) 95 m/s
(C) 90 m/s
(D) 85 m/s
7. Find the associated stream function if the potential function is φ = 10 y in a potential ﬂow.
(A) 10x
(B) 10y
(C) −10x
(D) −10y
8. A ﬂow is represented by ψ = 10 ln(x2 + y2 ) . Find the pressure along the negative xaxis if atmospheric air is ﬂowing and p = 0 at x = −∞.
(Α) −120/x^{2 }
(B) −160/x^{2}
(C) −200/x^{2}
(D) −240/x^{2}
9. A uniform ﬂow V = 10i m/s is superimposed on a doublet with strength 40 m3/s. The velocity distribution v (θ) on the cylinder is
(A) 10 sinθ
(B) 20 sinθ
(C) 40 sinθ
(D) 80 sinθ
10. A turbulent boundary layer is studied in a zero pressuregradient ﬂow on a ﬂat plate. Atmospheric air at 20°C ﬂows over the plate at 10 m/s. How far from the leading edge can turbulence be expected if the freestream ﬂuctuation intensity is low?
(A) 60 cm
(B) 70 cm
(C) 80 cm
(D) 90 cm
11. Given where and U_{∞} = 1 m/s. The drag force on one side of a 2mwide, 4mlong plate, over which 20°C atmospheric air ﬂows, is nearest (A) 0.013 N
(B) 0.024 N
(C) 0.046 N
(D) 0.067 N
12. A laminar boundary layer of 20°C water moving at 0.8 m/s exists on one side of a 2mwide, 3mlong ﬂat plate. At x = 3 m, v_{max} is nearest
(A) 0.008 m/s
(B) 0.006 m/s
(C) 0.0004 m/s
(D) 0.0002 m/s
Quiz No. 2
1. Estimate the drag coefﬁcient for air at 0°C moving past a 10cmdiameter, 4mhigh pole at 2 m/s.
2. Atmospheric air at 20°C is ﬂowing at 10 m/s normal to a 10cmdiameter, 80cmlong smooth cylinder with free ends. Calculate the drag force.
3. A 220cmsquare sign is impacted straight on by a 50 m/s 10°C wind. If the sign is held by a single 3mhigh post imbedded in concrete, what moment would exist at the base of the post?
4. A sensor is positioned downstream a short distance from a 4cmdiameter cylinder in a 20°C atmospheric airﬂow. It senses vortex shedding at a frequency of 0.16 Hz. Estimate the airspeed.
5. A 2000kg airplane is designed to carry a 4000N payload when cruising near sea level. For a conventional airfoil with an effective wing area of 25 m2, estimate the stall speed at elevation of 2000 m.
6. Find the associated potential function if the stream function is ψ = 20xy in a potential ﬂow.
7. A ﬂow is represented by ψ = 10 ln(x^{2} + y^{2} ) m2/s. Find the xcomponent of the acceleration at (−4, 0).
8. Locate any stagnation points in the ﬂow represented by
9. A body is formed by the streamline that separates the source ﬂow of strength q = 10π m2/s from a uniform ﬂow parallel to the xaxis of 10 m/s. Locate the positive yintercept of the body formed by the streamline that separates the source ﬂow from the uniform ﬂow.
10. Assume a linear velocity proﬁle in a laminar boundary layer on a ﬂat plate with a zero pressure gradient. Find d (x).
11. If the walls in a wind tunnel are parallel, the ﬂow will accelerate due to the boundary layers on each of the walls. If a wind tunnel using 20°C atmospheric air is square, how should one of the walls be displaced outward for a zero pressure gradient to exist if U = 10 m/s? Assume turbulent ﬂow.
12. Atmospheric air at 20°C ﬂows over a 3mlong and 2mwide ﬂat plate at 16 m/s. Assume a turbulent ﬂow from the leading edge and calculate drag force on one side of the plate.
13. A long cigarshaped dirigible is proposed to take rich people on cruises.
It is proposed to be 1000 m long and 150 m in diameter. How much horsepower is needed to move the dirigible through sealevel air at 12 m/s if the drag on the front and rear is neglected. Use the drag coefﬁcient of Eq. (8.67).