Category Fundamentals of Aerodynamics

Fluid Statics: Buoyancy Force

In aerodynamics, we are concerned about fluids in motion, and the resulting forces and moments on bodies due to such motion. However, in this section, we consider the special case of no fluid motion, i. e., fluid statics. A body immersed in a fluid will still experience a force even if there is no relative motion between the body and the fluid. Let us see why.

To begin, we must first consider the force on an element of fluid itself. Consider a stagnant fluid above the xz plane, as shown in Figure 1.31. The vertical direction is given by y. Consider an infinitesimally small fluid element with sides of length dx, dy, and dz. There are two types of forces acting on this fluid element: pressure forces from the surrounding fluid exerted on the surface of the element, and the gravity force due to the weight of the fluid inside the element. Consider forces in the у direction. The pressure on the bottom surface of the element is p, and hence the force on the bottom face is p(dx dz) in the upward direction, as shown in Figure

(p + ^dy)dx dz

image57

fluid.

 

1.31. The pressure on the top surface of the element will be slightly different from the pressure on the bottom because the top surface is at a different location in the fluid. Let dp/dy denote the rate of change of p with respect to y. Then the pressure exerted on the top surface will be p + (dp/dy) dy, and the pressure force on the top of the element will be [p + (dp/dy) dy](dx dz) in the downward direction, as shown in Figure 1.31. Hence, letting upward force be positive, we have

Net pressure force = p(dx dz) — ( p + —dy ) (dx dz)

dy J

= — — (dx dy dz) dy

Let p be the mean density of the fluid element. The total mass of the element is p(dx dy dz). Therefore,

Gravity force = —p(dx dy dz)g

where g is the acceleration of gravity. Since the fluid element is stationary (in equi­librium), the sum of the forces exerted on it must be zero:

Подпись: or Подпись: dp = ~gp dy Подпись: [1.52]

-—(dx dy dz) – gp(dxdydz) = 0 dy

Equation (1.52) is called the Hydrostatic equation; it is a differential equation which relates the change in pressure dp in a fluid with a change in vertical height dy.

The net force on the element acts only in the vertical direction. The pressure forces on the front and back faces are equal and opposite and hence cancel; the same is true for the left and right faces. Also, the pressure forces shown in Figure 1.31 act at the center of the top and bottom faces, and the center of gravity is at the center of the elemental volume (assuming the fluid is homogeneous); hence, the forces in Figure 1.31 are colinear, and as a result, there is no moment on the element.

Equation (1.52) governs the variation of atmospheric properties as a function of altitude in the air above us. It is also used to estimate the properties of other planetary atmospheres such as for Venus, Mars, and Jupiter. The use of Equation (1.52) in the analysis and calculation of the “standard atmosphere” is given in detail in Reference 2; hence, it will not be repeated here.

Let the fluid be a liquid, for which we can assume p is constant. Consider points 1 and 2 separated by the vertical distance Ah as sketched on the right side of Figure

1.31. The pressure and у locations at these points are pi, h, and /?2, hi. respectively. Integrating Equation (1.52) between points 1 and 2, we have

Г P2 rhl

dp = – pg dy J p J h

Подпись: [1.53]Подпись: orPi – Pi = ~Pg(hi — hj) — pg Ah

where Ah = h — hi. Equation (1.46) can be more conveniently expressed as

Подпись: or Подпись: p + pgh = constant Подпись: [1.54]

P2 + pgh2 = pi + pghi

Note that in Equations (1.53) and (1.54), increasing values of h are in the positive (upward) у direction.

A simple application of Equation (1.54) is the calculation of the pressure distri­bution on the walls of a container holding a liquid, and open to the atmosphere at the top. This is illustrated in Figure 1.32, where the top of the liquid is at a heght hi. The atmospheric pressure pa is impressed on the top of the liquid; hence, the pressure at h is simply pa. Applying Equation (1.54) between the top (where h = h) and an arbitrary height h, we have

P + Pgh = pі + pghi = pa+ Pgh і

or p = Pa + pg(h – h) [1.55]

Equation (1.55) gives the pressure distribution on the vertical sidewall of the container as a function of h. Note that the pressure is a linear function of h as sketched on the right of Figure 1.32, and that p increases with depth below the surface.

image58

Another simple and very common application of Equation (1.54) is the liquid – filled U-tube manometer used for measuring pressure differences, as sketched in Figure 1.33. The manometer is usually made from hollow glass tubing bent in the shape of the letter U. Imagine that we have an aerodynamic body immersed in an airflow (such as in a wind tunnel), and we wish to use a manometer to measure the surface pressure at point b on the body. A small pressure orifice (hole) at point b is connected to one side of the manometer via a long (usually flexible) pressure tube. The other side of the manometer is open to the atmosphere, where the pressure pa is a known value. The U tube is partially filled with a liquid of known density p. The tops of the liquid on the left and right sides of the U tube are at points 1 and 2, with heights h and h2, respectively. The body surface pressure pt, is transmitted through the pressure tube and impressed on the top of the liquid at point 1. The atmospheric pressure pa is impressed on the top of the liquid at point 2. Because in general рь ф pa, the tops of the liquid will be at different heights; i. e., the two sides of the manometer will show a displacement Ah = hi — h2 of the fluid. We wish to

The use of a U-tube manometer.

obtain the value of the surface pressure at point b on the body by reading the value of Ah from the manometer. From Equation (1.54) applied between points 1 and 2,

Pb + pghi = pa + pgh2

or Pb = Pa – pg{h – h2)

or pb = Pa – pg Ah [1.56]

In Equation (1.56), pa, p, and g are known, and Ah is read from the U tube, thus allowing рь to be measured.

At the beginning of this section, we stated that a solid body immersed in a fluid will experience a force even if there is no relative motion between the body and the fluid. We are now in a position to derive an expression for this force, henceforth called the buoyancy force. We will consider a body immersed in either a stagnant gas or liquid, hence p can be a variable. For simplicity, consider a rectangular body of unit width, length/, and height {h—h2), as shown in Figure 1.34. Examining Figure 1.34, we see that the vertical force F on the body due to the pressure distribution over the surface is

F = (p2~ pi)l(l) [1.57]

There is no horizontal force because the pressure distributions over the vertical faces of the rectangular body lead to equal and opposite forces which cancel each other. In Equation (1.57), an expression for p2 — p can be obtained by integrating the hydrostatic equation, Equation (1.52), between the top and bottom faces:

Подпись: Pg dy

image60

ГР2 phi ph

P2 ~ Pi = dp = — I pgdy = I

J pi J h і «/ hi

image61

Figure 1.34 Source of the buoyancy force on a body immersed in a fluid.

 

Substituting this result into Equation (1.57), we obtain for the buoyancy force

Подпись: F = K 1) / Jhh

pgdy [1.58]

2

Подпись: Pg dy
image62

Consider the physical meaning of the integral in Equation (1.58). The weight of a small element of fluid of height dy and width and length of unity as shown at the right of Figure 1.34 is pg dy (1)(1). In turn, the weight of a column of fluid with a base of unit area and a height (h — /12) is

which is precisely the integral in Equation (1.58). Moreover, if we place l of these fluid columns side by side, we would have a volume of fluid equal to the volume of the body on the left of Figure 1.34, and the weight of this total volume of fluid would be

l pgdy J /12

which is precisely the right-hand side of Equation (1.58). Therefore, Equation (1.58) states in words that

Buoyancy force

weight of fluid

on body

— displaced by body

We have just proved the well-known Archimedes principle, first advanced by the Greek scientist, Archimedes of Syracuse (287-212 B. C.). Although we have used a rectangular body to simplify our derivation, the Archimedes principle holds for bodies of any general shape. (See Problem 1.14 at the end of this chapter.) Also, note from our derivation that the Archimedes principle holds for both gases and liquids and does not require that the density be constant.

The density of liquids is usually several orders of magnitude larger than the density of gases; e. g., for water p = 103 kg/m3, whereas for air p = 1.23 kg/m3. Therefore, a given body will experience a buoyancy force a thousand times greater in water than in air. Obviously, for naval vehicles buoyancy force is all important, whereas for airplanes it is negligible. On the other hand, lighter-than-air vehicles, such as blimps and hot-air balloons, rely on buoyancy force for sustenation; they obtain sufficient buoyancy force simply by displacing huge volumes of air. For most problems in aerodynamics, however, buoyancy force is so small that it can be readily neglected.

A hot-air balloon with an inflated diameter of 30 ft is carrying a weight of 800 lb, which includes the weight of the hot air inside the balloon. Calculate (a) its upward acceleration at sea level the instant the restraining ropes are released and (b) the maximum altitude it can achieve. Assume that the variation of density in the standard atmosphere is given by p = 0.002377(1 — 7 x 10~6/t)4-21, where h is the altitude in feet and p is in slug/ft3.

Solution

(a) At sea level, where h = 0, p = 0.002377 slug/ft3. The volume of the inflated balloon is |я(15)3 = 14,137 ft3. Hence,

Buoyancy force = weight of displaced air

= gpV

where g is the acceleration of gravity and V is the volume.

Buoyancy force = В = (32.2)(0.002377)(14,137) = 1082 lb

The net upward force at sea level is F = В — W, where W is the weight. From Newton’s second law,

F = В — W = та

where m is the mass, m = Щ = 24.8 slug. Hence,

Подпись: 11.4 ft/s2Подпись: Example 1.8_ B-W 1082 – 800 a ~ m ~ 24Я

Подпись: P Подпись: В gv Подпись: 800 (32.2) (14,137) Подпись: 0.00176 slug/ft3

(b) The maximum altitude occurs when В = W = 800 lb. Since В = gpV, and assuming the balloon volume does not change,

From the given variation of p with altitude, h,

p = 0.002377(1 – 7 x 10~6/t)4-21 = 0.00176

Solving for h, we obtain

1

0.00176 1/4 21

h =————

1 – (————-

9842 ft

7 x 10~6

V 0.002377 j

Stream Function

In this section, we consider two-dimensional steady flow. Recall from Section 2.11 that the differential equation for a streamline in such a flow is given by Equation

 

image173

(2.118), repeated below

Подпись: [2.1 18] Equation (2.118) can be integrated Подпись:dy v dx и

If и and v are known functions of x and y, then to yield the algebraic equation for a streamline:

f(x, y) = c

where c is an arbitrary constant of integration, with different values for different streamlines. In Equation (2.139), denote the function of x and у by the symbol f. Hence, Equation (2.139) is written as

Подпись: [2.140]Ф(х, y)=C

The function jr(x, y) is called the stream function. From Equation (2.140) we see that the equation for a streamline is given by setting the stream function equal to a constant, i. e., сі, c2, сз, etc. Two different streamlines are illustrated in Figure 2.38; streamlines ab and cd are given by jr = a and iJr = C2, respectively.

There is a certain arbitrariness in Equations (2.139) and (2.140) via the arbitrary constant of integration c. Let us define the stream function more precisely in order to reduce this arbitrariness. Referring to Figure 2.38, let us define the numerical value of іJr such that the difference AiJr between f = ci for streamline cd and jr = c for streamline ab is equal to the mass flow between the two streamlines. Since Figure 2.38 is a two-dimensional flow, the mass flow between two streamlines is defined per unit depth perpendicular to the page. That is, in Figure 2.38 we are considering the mass flow inside a streamtube bounded by streamlines ab and cd, with a rectangular cross­sectional area equal to An times a unit depth perpendicular to the page. Here, An is the normal distance between ab and cd, as shown in Figure 2.38. Hence, mass flow between streamlines ab and cd per unit depth perpendicular to the page is

Ді/r = C2 — Ci [2.141]

Подпись: Figure 2.38 Different streamlines are given by different values of the stream function.

The above definition does not completely remove the arbitrariness of the constant of integration in Equations (2.139) and (2.140), but it does make things a bit more precise. For example, consider a given two-dimensional flow field. Choose one streamline of the flow, and give it an arbitrary value of the stream function, say, fr = a. Then, the value of the stream function for any other streamline in the flow, say, fr = сг,

is fixed by the definition given in Equation (2.141). Which streamline you choose to designate as ф = c and what numerical value you give c usually depend on the geometry of the given flow field, as we see in Chapter 3.

The equivalence between ф = constant designating a streamline, and Аф equal­ing mass flow (per unit depth) between streamlines, is natural. For a steady flow, the mass flow inside a given streamtube is constant along the tube; the mass flow across any cross section of the tube is the same. Since by definition Д ф is equal to this mass flow, then Аф itself is constant for a given streamtube. In Figure 2.38, if ф = C designates the streamline on the bottom of the streamtube, then ф2 = <"2 = c + Аф is also constant along the top of the streamtube. Since by definition of a streamtube (see Section 2.11) the upper boundary of the streamtube is a streamline itself, then ф2 = C2 = constant must designate this streamline.

We have yet to develop the most important property of the stream function, namely, derivatives of ф yield the flow-field velocities. To obtain this relationship, consider again the streamlines a b and cd in Figure 2.38. Assume that these streamlines are close together (i. e., assume An is small), such that the flow velocity V is a constant value across An. The mass flow through the streamtube per unit depth perpendicular to the page is

Подпись: [2.142]Подпись: [2.143]Подпись:Аф = pV An(l) Аф

~^=PV

An

Consider the limit of Equation (2.142) as An —» 0:

,, r АФ H

pV = lim ——— = —

An->0 An dn

Equation (2.143) states that if we know ф, then we can obtain the product (p V) by differentiating ф in the direction normal to V. To obtain a practical form of Equation (2.143) for cartesian coordinates, consider Figure 2.39. Notice that the directed normal distance An is equivalent first to moving upward in the у direction by the amount Ay and then to the left in the negative x direction by the amount — Ax. Due to conservation of mass, the mass flow through An (per unit depth) is equal to the sum of the mass flows through Ay and — Ax (per unit depth):

Mass flow = Аф = pV An = pu Ay + pv(-Ax) [2.144]

Letting cd approach ab, Equation (2.144) becomes in the limit

d’ijf = pudy — pvdx [2.145]

However, since ф = ф (x, y), the chain rule of calculus states

– 3 ф 3 ф

dф = —dx H——- dy [2.146]

dx dy

Comparing Equations (2.145) and (2.146), we have
[2.147a]

image175"

Подпись: Figure 2.39 Mass flow through An is the sum of the mass flows through Ay and - Ax.

[2.147b]

Equations (2.147a and b) are important. If ф(х, y) is known for a given flow field, then at any point in the flow the products pu and pv can be obtained by differentiating ф in the directions normal to и and v, respectively.

Stream Function Подпись: [2.148a] [2.148b]

If Figure 2.39 were to be redrawn in terms of polar coordinates, then a similar derivation yields

Such a derivation is left as a homework problem.

Note that the dimensions of ф are equal to mass flow per unit depth perpendicular to the page. That is, in SI units, ф is in terms of kilograms per second per meter perpendicular to the page, or simply kg/(s ■ m).

The stream function ф defined above applies to both compressible and incom­pressible flow. Now consider the case of incompressible flow only, where p = con-

stant. Equation (2.143) can be written as

Подпись:д(Ф/р)

dn

We define a new stream function, for incompressible flow only, as 1jr = i// /p. Then Equation (2.149) becomes

ЗіIr V = ~ dn

and Equations (2.147) and (2.148) become

image177[3.150a]

[3.150b]

and

image178[3.151a] [3.151b]

The incompressible stream function ф has characteristics analogous to its more gen­eral compressible counterpart iJr. For example, since ф(х, у) = c is the equation of a streamline, and since p is a constant for incompressible flow, then 1(r{x, y) = ф jp = constant is also the equation for a streamline (for incompressible flow only). In addition, since Д ф is mass flow between two streamlines (per unit depth perpendic­ular to the page), and since p is mass per unit volume, then physically Аф = Ді]r/p represents the volume flow (per unit depth) between two streamlines. In SI units, Аф is expressed as cubic meters per second per meter perpendicular to the page, or simply m2/s.

In summary, the concept of the stream function is a powerful tool in aerodynam­ics, for two primary reasons. Assuming that ф(х, y) [or ф(х, y)[ is known through the two-dimensional flow field, then:

1. ф = constant (or ф = constant) gives the equation of a streamline.

2. The flow velocity can be obtained by differentiating ф (or ф), as given by Equa­tions (2.147) and (2.148) for compressible flow and Equations (2.150) and (2.151) for incompressible flow. We have not yet discussed how ф(х, y) [or ф(х, у)] can be obtained in the first place; we are assuming that it is known. The ac­tual determination of the stream function for various problems is discussed in Chapter 3.

The Kutta Condition

The lifting flow over a circular cylinder was discussed in Section 3.15, where we ob­served that an infinite number of potential flow solutions were possible, corresponding to the infinite choice of Г. For example, Figure 3.28 illustrates three different flows over the cylinder, corresponding to three different values of Г. The same situation applies to the potential flow over an airfoil; for a given airfoil at a given angle of attack, there are an infinite number of valid theoretical solutions, corresponding to an

3 It is interesting to note that some recent research by NASA is hinting that even as complex a problem as flow separation, heretofore thought to be a completely viscous-dominated phenomenon, may in reality be an inviscid-dominated flow which requires only a rotational flow. For example, some inviscid flow-field numerical solutions for flow over a circular cylinder, when vorticity is introduced either by means of a nonuniform freestream or a curved shock wave, are accurately predicting the separated flow on the rearward side of the cylinder. However, as exciting as these results may be, they are too preliminary to be emphasized in this book. We continue to talk about flow separation in Chapters 15 to 20 as being a viscous-dominated effect, until definitely proved otherwise. This recent research is mentioned here only as another example of the physical connection between vorticity, vortex sheets, viscosity, and real life.

infinite choice of Г. For example, Figure 4.12 illustrates two different flows over the same airfoil at the same angle of attack but with different values of Г. At first, this may seem to pose a dilemma. We know from experience that a given airfoil at a given angle of attack produces a single value of lift (e. g., see Figure 4.5). So, although there is an infinite number of possible potential flow solutions, nature knows how to pick a particular solution. Clearly, the philosophy discussed in the previous section is not complete—we need an additional condition that fixes Г for a given airfoil at a given a.

To attempt to find this condition, let us examine some experimental results for the development of the flow field around an airfoil which is set into motion from an initial state of rest. Figure 4.13 shows a series of classic photographs of the flow over an airfoil, taken from Prandtl and Tietjens (Reference 8). In Figure 4.13a, the flow has just started, and the flow pattern is just beginning to develop around the airfoil. In these early moments of development, the flow tries to curl around the sharp trailing edge from the bottom surface to the top surface, similar to the sketch shown at the left of Figure 4.12. However, more advanced considerations of inviscid, incompressible flow (see, e. g., Reference 9) show the theoretical result that the velocity becomes infinitely large at a sharp comer. Hence, the type of flow sketched at the left of Figure 4.12, and shown in Figure 4.13a, is not tolerated very long by nature. Rather, as the real flow develops over the airfoil, the stagnation point on the upper surface (point 2 in Figure 4.12) moves toward the trailing edge. Figure 4.13b shows this intermediate stage. Finally, after the initial transient process dies out, the steady flow shown in Figure 4.13c is reached. This photograph demonstrates that the flow is smoothly leaving the top and the bottom surfaces of the airfoil at the trailing edge. This flow pattern is sketched at the right of Figure 4.12 and represents the type of pattern to be expected for the steady flow over an airfoil.

Reflecting on Figures 4.12 and 4.13, we emphasize again that in establishing the steady flow over a given airfoil at a given angle of attack, nature adopts that particular value of circulation (Г2 in Figure 4.12) which results in the flow leaving smoothly at the trailing edge. This observation was first made and used in a theoretical analysis by the German mathematician M. Wilhelm Kutta in 1902. Therefore, it has become known as the Kutta condition.

image314

Figure 4.12 Effect of different values of circulation on the potential flow over a given airfoil at a given angle of attack. Points 1 and 2 are stagnation points.

image315

(a)

image316

(b)

Figure 4.1 3 The development of steady flow over an airfoil; the airfoil is impulsively started from rest and attains a steady velocity through the fluid, (a) A moment just after starting, (b) An intermediate time. (Source: Prandtl and Tiejens, Reference 8.)

In order to apply the Kutta condition in a theoretical analysis, we need to be more precise about the nature of the flow at the trailing edge. The trailing edge can have a finite angle, as shown in Figures 4.12 and 4.13 and as sketched at the left of Figure 4.14, or it can be cusped, as shown at the right of Figure 4.14. First, consider the trailing edge with a finite angle, as shown at the left of Figure 4.14. Denote the velocities along the top surface and the bottom surface as V and to, respectively. Vt

image317

(f)

Figure 4.1 3 (continued) The development of steady flow over an airfoil; the airfoil is

impulsively started from rest and attains a steady velocity through the fluid, (c) The final steady flow. (Source: Prandtl and Tiejens, Reference 8.)

 

Finite angle Cusp

image318

At point a, Kj = V2 = 0 At point a: V1 = V2 Ф 0

Figure 4.1 4 Different possible shapes of the trailing edge and their relation to the Kutta condition.

 

is parallel to the top surface at point a, and V2 is parallel to the bottom surface at point a. For the finite-angle trailing edge, if these velocities were finite at point a, then we would have two velocities in two different directions at the same point, as shown at the left of Figure 4.14. Flowever, this is not physically possible, and the only recourse is for both Vj and V2 to be zero at point a. That is, for the finite trailing edge, point a is a stagnation point, where Vj = V2 = 0. In contrast, for the cusped trailing edge shown at the right of Figure 4.14, V and V2 are in the same direction at point a, and hence both Vj and V2 can be finite. Flowever, the pressure at point a, p2, is a single, unique value, and Bernoulli’s equation applied at both the top and bottom surfaces immediately adjacent to point a yields

Pa + pV i2 = Pa + pV2

Подпись:Vj = F2

Hence, for the cusped trailing edge, we see that the velocities leaving the top and bottom surfaces of the airfoil at the trailing edge are finite and equal in magnitude and direction.

We can summarize the statement of the Kutta condition as follows:

1. For a given airfoil at a given angle of attack, the value of Г around the airfoil is such that the flow leaves the trailing edge smoothly.

2. If the trailing-edge angle is finite, then the trailing edge is a stagnation point.

3. If the trailing edge is cusped, then the velocities leaving the top and bottom surfaces at the trailing edge are finite and equal in magnitude and direction.

Consider again the philosophy of simulating the airfoil with vortex sheets placed either on the surface or on the camber line, as discussed in Section 4.4. The strength of such a vortex sheet is variable along the sheet and is denoted by у (і). The statement of the Kutta condition in terms of the vortex sheet is as follows. At the trailing edge (ТЕ), from Equation (4.8), we have

У (ТЕ) = y(a) = V1- V2 [4.9]

However, for the finite-angle trailing edge, V = V2 = 0; hence, from Equation (4.9), у (ТЕ) = 0. For the cusped trailing edge, V = V2 Ф 0; hence, from Equation (4.9), we again obtain the result that у (ТЕ) = 0. Therefore, the Kutta condition expressed in terms of the strength of the vortex sheet is

[4.10]

First Law of Thermodynamics

Подпись: Figure 7.2 Thermodynamic system.

Consider a fixed mass of gas, which we define as the system. (For simplicity, assume a unit mass, for example, 1 kg or 1 slug.) The region outside the system is called the surroundings. The interface between the system and its surroundings is called the boundary, as shown in Figure 7.2. Assume that the system is stationary. Let Sq be an incremental amount of heat added to the system across the boundary, as sketched in Figure 7.2. Examples of the source of 8q are radiation from the surroundings which is absorbed by the mass in the system and thermal conduction due to temperature gradients across the boundary. Also, let 8 w denote the work done on the system by the surroundings (say, by a displacement of the boundary, squeezing the volume of the system to a smaller value). As discussed earlier, due to the molecular motion of the gas, the system has an internal energy e. The heat added and work done on the system cause a change in energy, and since the system is stationary, this change in

Sq + Sw = de

Подпись: [7.1 I]This is the first law of thermodynamics: It is an empirical result confirmed by experi­ence. In Equation (7.11), e is a state variable. Hence, de is an exact differential, and its value depends only on the initial and final states of the system. In contrast, Sq and Sw depend on the process in going from the initial to the final states.

For a given de, there are in general an infinite number of different ways (pro­cesses) by which heat can be added and work done on the system. We are primarily concerned with three types of processes:

1. Adiabatic process. One in which no heat is added to or taken away from the system

2. Reversible process. One in which no dissipative phenomena occur, that is, where the effects of viscosity, thermal conductivity, and mass diffusion are absent

3. Isentropic process. One that is both adiabatic and reversible

For a reversible process, it can be easily shown that Sw = —pdv, where dv is an incremental change in the volume due to a displacement of the boundary of the system. Thus, Equation (7.11) becomes

Подпись: [7.12]Sq — p dv = de

Typical Orthogonal Coordinate Systems

To describe mathematically the flow of fluid through three-dimensional space, we have to prescribe a three-dimensional coordinate system. The geometry of some aerody­namic problems best fits a rectangular space, whereas others are mainly cylindrical in nature, and yet others may have spherical properties. Therefore, we have interest in the three most common orthogonal coordinate systems: cartesian, cylindrical, and spherical. These systems are described below. (An orthogonal coordinate system is one where all three coordinate directions are mutually perpendicular. It is interesting to note that some modem numerical solutions of fluid flows utilize nonorthogonal coordinate spaces; moreover, for some numerical problems the coordinate system is allowed to evolve and change during the course of the solution. These so-called adaptive grid techniques are beyond the scope of this book. See Reference 7 for details.)

A cartesian coordinate system is shown in Figure 23a. The x, y, and z axes are mutually perpendicular, and i, j, and к are unit vectors in the x, y, and z direc­tions, respectively. An arbitrary point P in space is located by specifying the three coordinates (x, y, z). The point can also be located by the position vector r, where

г = XI + yj + zk

If A is a given vector in cartesian space, it can be expressed as

A = Axi + Ay j + Az к

where Ax, Ay, and Az are the scalar components of A along the x, y, and z directions, respectively, as shown in Figure 23b.

image88

Figure 2.3 Cartesian coordinates.

A cylindrical coordinate system is shown in Figure 2.4a. A “phantom” cartesian system is also shown with dashed lines to help visualize the figure. The location of point P in space is given by three coordinates (г, в, z), where r and в are measured in the xy plane shown in Figure 2.4a. The r coordinate direction is the direction of increasing r, holding в and z constant; er is the unit vector in the r direction. The в coordinate direction is the direction of increasing 0, holding r and z constant; eg is the unit vector in the 0 direction. The г coordinate direction is the direction of increasing z, holding r and 0 constant; e: is the unit vector in the z direction. If A is a given vector in cylindrical space, then

A = Arer + AgCt9 + Аге.

where Ar, Ag, and Az are the scalar components of A along the r, 0, and z directions, respectively, as shown in Figure 2Ab. The relationship, or transformation, between cartesian and cylindrical coordinates can be obtained from inspection of Figure 2.4a, namely,

x = r cos 0

у = r sin 0 [2.5]

г = z

or inversely,

r = y/x2 + y2

0 = arctan — [2.6]

JC

z = z

A spherical coordinate system is shown in Figure 2.5a. Once again, a phantom cartesian system is shown with dashed lines. (However, for clarity in the picture, the

image89

ia)

z axis is drawn vertically, in contrast to Figures 2.3 and 2.4.) The location of point P in space is given by the three coordinates (r, 0, Ф), where r is the distance of P from the origin, в is the angle measured from the z axis and is in the rz plane, and Ф is the angle measured from the x axis and is in the xy plane. The r coordinate direction is the direction of increasing r, holding в and Ф constant; e, is the unit vector in the r
direction. The в coordinate direction is the direction of increasing в, holding r and Ф constant; eg is the unit vector in the в direction. The Ф coordinate direction is the direction of increasing Ф, holding r and в constant; e<j> is the unit vector in the Ф direction. The unit vectors er, eg, and Єф are mutually perpendicular, if A is a given vector in spherical space, then

A = Are, + Agfig + АфЄф

where Ar, Ag, and Аф are the scalar components of A along the r,6, and Ф directions, respectively, as shown in Figure 2.5b. The transformation between cartesian and spherical coordinates is obtained from inspection of Figure 2.5a, namely,

x — r sin в cos Ф

у = r sin 9 sin Ф [2.7]

z = r cos в

or inversely,

r = /ї2 + y2 + z2

n z z r ,

0 = arccos – = arccos, [2.81

r 3/x2 + y2 + z2

X

Ф = arccos, :

2.2.2 Scalar and Vector Fields

A scalar quantity given as a function of coordinate space and time t is called a scalar field. For example, pressure, density, and temperature are scalar quantities, and

P – Pi(x, y, z, t) = p2(r, в, z, t) = p3(r, в, Ф, t) p = P(x, y, z, t) = p2(r, в, Z, t) = рз(r, в, Ф, t)

T = Ti(x, y, z, t) = T2(r, в, z, t) = Тз(г, в, Ф, t)

are scalar fields for pressure, density, and temperature, respectively. Similarly, a vector quantity given as a function of coordinate space and time is called a vector field. For example, velocity is a vector quantity, and

V = V, i + Vvj + V, k

where Vx = Vx(x, y, z, t)

V, = Vy(x, y,z, t)

Vz = Vz(x, y, z, t)

is the vector field for V in cartesian space. Analogous expressions can be written for vector fields in cylindrical and spherical space. In many theoretical aerodynamic problems, the above scalar and vector fields are the unknowns to be obtained in a solution for a flow with prescribed initial and boundary conditions.

2.2.3 Scalar and Vector Products

The scalar and vector products defined by Equations (2.3) and (2.4), respectively, can be written in terms of the components of each vector as follows.

Cartesian Coordinates Let

A = АдТ + Ay j + Azk

and

В — Bxi + By j + Bz к

Then

A

• В — AXBX Ay By AZBZ

[2.9]

and

,

j

к "

A x В =

A*

A,

A,

= і (AyBz — AzBy) +j(AzBx — AXBZ) + к (AxBy

– Ay Bx)

_ Bx

By

Bz

[2.10]

Cylindrical Coordinates Let

A = Arer + AgCg + Azez and В = Brtr + ВвЄд + B, ez

Then A • В = ArBr + AgBg + AZBZ [2.11]

er ee

ег

and

A x В =

Ar Ад

[2.12]

В, Вд

Bz

Spherical Coordinates Let

A = Arer

+ Адед + АфЄф

and

В = Brer

+ Bgtg + ВфЄф

Then

A • В = ArBr + AqBq

+ АфВф

[2.13]

Є, Єв

Сф

and

A x В =

Аг Ад

Аф

[2.14]

Вг Вд

Вф

2.2.4 Gradient of a Scalar Field

We now begin a review of some elements of vector calculus. Consider a scalar field
P = Pi(x, y, z) = рг(г, в, z) = рз(г, в, Ф)

The gradient of p, Vp, at a given point in space is defined as a vector such that:

0. Its magnitude is the maximum rate of change of p per unit length of the coordinate space at the given point.

1. Its direction is that of the maximum rate of change of p at the given point.

For example, consider a two-dimensional pressure field in cartesian space as sketched in Figure 2.6. The solid curves are lines of constant pressure; i. e., they connect points in the pressure field which have the same value of p. Such lines are called isolines. Consider an arbitrary point (jc, y) in Figure 2.6. If we move away from this point in an arbitrary direction, p will, in general, change because we are moving to another location in space. Moreover, there will be some direction from this point along which p changes the most over a unit length in that direction. This defines the direction of the gradient of p and is identified in Figure 2.6. The magnitude of Vp is the rate of change of p per unit length in that direction. Both the magnitude and direction of Vp will change from one point to another in the coordinate space. A line drawn in this space along which V/? is tangent at every point is defined as a gradient line, as sketched in Figure 2.6. The gradient line and isoline through any given point in the coordinate space are perpendicular.

Consider V/j at a given point (x, y) as shown in Figure 2.7. Choose some arbitrary direction s away from the point, as also shown in Figure 2.7. Let n be a unit vector in the s direction. The rate of change of p per unit length in the s direction is

image92"Подпись: Figure 2.6 Illustration of the gradient of a scalar field. [2.15]

у

image94

У

S

Figure 2.7 Sketch for the

directional derivative.

Pressure Coefficient

Подпись: P - Рос QOQ Подпись: [3.36]

Pressure, by itself, is a dimensional quantity (e. g., pounds per square foot, newtons per square meter). However, in Sections 1.7 and 1.8, we established the usefulness of certain dimensionless parameters such as M, Re, Cl – It makes sense, therefore, that a dimensionless pressure would also find use in aerodynamics. Such a quantity is the pressure coefficient Cp, first introduced in Section 1.5 and defined as

where qoo = pooVl-t

The definition given in Equation (3.36) is just that—a definition. It is used throughout aerodynamics, from incompressible to hypersonic flow. In the aerodynamic literature, it is very common to find pressures given in terms of Cp rather than the pressure itself. Indeed, the pressure coefficient is another similarity parameter that can be added to the list started in Sections 1.7 and 1.8.

For incompressible flow, Cp can be expressed in terms of velocity only. Consider the flow over an aerodynamic body immersed in a freestream with pressure Poo and velocity Toe. Pick an arbitrary point in the flow where the pressure and velocity are p and V, respectively. From Bernoulli’s equation,

Poo + ‘oPV^ = P + {pV2

or p – Poo = У (Vi – V2) [3.37]

Substituting Equation (3.37) into (3.36), we have

Подпись: or Подпись: [3.38]
image223

„ P – Poo p{V2oo – V2) p – „ ~ і „1/2

Equation (3.38) is a useful expression for the pressure coefficient; however, note that the form of Equation (3.38) holds for incompressible flow only.

Note from Equation (3.38) that the pressure coefficient at a stagnation point (where V = 0) in an incompressible flow is always equal to 1.0. This is the highest allowable value of Cp anywhere in the flow field. (For compressible flows, Cp at a stagnation point is greater than 1.0, as shown in Chapter 14.) Also, keep in mind that in regions of the flow where V > Vx or p < Poo, Cp will be a negative value.

Another interesting property of the pressure coefficient can be seen by rearranging the definition given by Equation (3.36), as follows:

P = Рос + ЧосСр


Clearly, the value of Cp tells us how much p differs from p^ in multiples of the dynamic pressure. That is, if Cp = 1 (the value at a stagnation point in an incom­pressible flow), then p = рос + q<oc, or the local pressure is “one times” the dynamic pressure above freestream static pressure. If Cp — —3, then p = p0Q — 3qoo, or the local pressure is three times the dynamic pressure below freestream static pressure.

Example 3. 7 I Consider an airfoil in a flow with a freestream velocity of 150 ft/s. The velocity at a given point
on the airfoil is 225 ft/s. Calculate the pressure coefficient at this point.

225 V[7] [8] [9] І50 /

Подпись: Cp = 1 — Подпись: -1.25
image224

Solution

speed of sound at standard sea level is 1117 ft/s; hence, the freestream Mach number is 300/1117 = 0.269. A flow where the local Mach number is less than 0.3 can be assumed to be essentially incompressible. Hence, the freestream Mach number satisfies this criterion. On the other hand, the flow rapidly expands over the top surface of the airfoil and accelerates to a velocity of 753 ft/s at the point of minimum pressure (the point of peak negative Cp). In the expansion, the speed of sound decreases. (We will find out why in Part 3.) Hence, at the point of minimum pressure, the local Mach number is greater than = 0.674. That is, the flow has expanded to such a high local Mach number that it is no longer incompressible. Therefore, the answer given in part (b) of Example 3.8 is not correct. (We will learn how to calculate the correct value in Part 3.) There is an interesting point to be made here. Just because a model is being tested in a low-speed, subsonic wind tunnel, it does not mean that the assumption of incompressible flow will hold for all aspects of the flow field. As we see here, in some regions of the flow field around a body, the flow can achieve such high local Mach numbers that it must be considered as compressible.

Historical Note: Kutta, Joukowski, and the Circulation Theory of Lift

Frederick W. Lanchester (1868-1946), an English engineer, automobile manufacturer, and self-styled aerodynamicist, was the first to connect the idea of circulation with lift. His thoughts were originally set forth in a presentation given before the Birmingham Natural History and Philosophical Society in 1894 and later contained in a paper

submitted to the Physical Society, which turned it down. Finally, in 1907 and 1908, he published two books, entitled Aerodynamics and Aerodonetics, where his thoughts on circulation and lift were described in detail. His books were later translated into German in 1909 and French in 1914. Unfortunately, Lanchester’s style of writing was difficult to read and understand; this is partly responsible for the general lack of interest shown by British scientists in Lanchester’s work. Consequently, little positive benefit was derived from Lanchester’s writings. (See Section 5.7 for a more detailed portrait of Lanchester and his work.)

Quite independently, and with total lack of knowledge of Lanchester’s thinking, M. Wilhelm Kutta (1867-1944) developed the idea that lift and circulation are related. Kutta was bom in Pitschen, Germany, in 1867 and obtained a Ph. D. in mathematics from the University of Munich in 1902. After serving as professor of mathematics at several German technical schools and universities, he finally settled at the Tech – nische Hochschule in Stuttgart in 1911 until his retirement in 1935. Kutta’s interest in aerodynamics was initiated by the successful glider flights of Otto Lilienthal in Berlin during the period 1890-1896 (see chapter 1 of Reference 2). Kutta attempted theoretically to calculate the lift on the curved wing surfaces used by Lilienthal. In the process, he surmised from experimental data that the flow left the trailing edge of a sharp-edged body smoothly and that this condition fixed the circulation around the body (the Kutta condition, described in Section 4.5). At the same time, he was con­vinced that circulation and lift were connected. Kutta was reluctant to publish these ideas, but after the strong insistence of his teacher, S. Finsterwalder, he wrote a paper entitled “Auftriebskrafte in Stromenden Flussigkecten” (Lift in Flowing Fluids). This was actually a short note abstracted from his longer graduation paper in 1902, but it represents the first time in history where the concepts of the Kutta condition as well as the connection of circulation with lift were officially published. Finsterwalder clearly repeated the ideas of his student in a lecture given on September 6, 1909, in which he stated:

On the upper surface the circulatory motion increases the translatory one, therefore

there is high velocity and consequently low pressure, while on the lower surface the

two movements are opposite, therefore there is low velocity with high pressure, with

the result of a thrust upward.

However, in his 1902 note, Kutta did not give the precise quantitative relation between circulation and lift. This was left to Nikolai Y. Joukowski (Zhukouski). Joukowski was bom in Orekhovo in central Russia on January 5, 1847. The son of an engineer, he became an excellent student of mathematics and physics, grad­uating with a Ph. D. in applied mathematics from Moscow University in 1882. He subsequently held a joint appointment as a professor of mechanics at Moscow Uni­versity and the Moscow Higher Technical School. It was at this latter institution that Joukowski built in 1902 the first wind tunnel in Russia. Joukowski was deeply interested in aeronautics, and he combined a rare gift for both experimental and theoretical work in the field. He expanded his wind tunnel into a major aerodynam­ics laboratory in Moscow. Indeed, during World War I, his laboratory was used as a school to train military pilots in the principles of aerodynamics and flight. When he died in 1921, Joukowski was by far the most noted aerodynamicist in Russia.

Much of Joukowski’s fame was derived from a paper published in 1906, wherein he gives, for the first time in history, the relation L’ = Vx Г—the Kutta-Joukowski

theorem. In Joukowski’s own words:

If an irrotational two-dimensional fluid current, having at infinity the velocity Vx surrounds any closed contour on which the circulation of velocity is Г, the force of the aerodynamic pressure acts on this contour in a direction perpendicular to the velocity and has the value

L’^p^V^Y

The direction of this force is found by causing to rotate through a right angle the vector Voc around its origin in an inverse direction to that of the circulation.

Joukowski was unaware of Kutta’s 1902 note and developed his ideas on circu­lation and lift independently. However, in recognition of Kutta’s contribution, the equation given above has propagated through the twentieth century as the “Kutta – Joukowski theorem.”

Hence, by 1906—just 3 years after the first successful flight of the Wright brothers—the circulation theory of lift was in place, ready to aid aerodynamics in the design and understanding of lifting surfaces. In particular, this principle formed the cornerstone of the thin airfoil theory described in Sections 4.7 and 4.8. Thin airfoil theory was developed by Max Munk, a colleague of Prandtl in Germany, during the first few years after World War I. However, the very existence of thin airfoil theory, as well as its amazingly good results, rests upon the foundation laid by Lanchester, Kutta, and Joukowski a decade earlier.

4.15 Summary

Return to the road map given in Figure 4.2. Make certain that you feel comfortable with the material represented by each box on the road map and that you understand the flow of ideas from one box to another. If you are uncertain about one or more aspects, review the pertinent sections before progressing further.

Some important results from this chapter are itemized below:

A vortex sheet can be used to synthesize the inviscid, incompressible flow over an airfoil. If the distance along the sheet is given by. v and the strength of the sheet per unit length is y(s), then the velocity potential induced at point (x, y) by a vortex sheet that extends from point a to point h is

ф(х, y) = j 9y(s) ds

[4.3]

The circulation associated with this vortex sheet is

r=f y(s)ds

[4.4]

Across the vortex sheet, there is a tangential velocity discontinuity, where

у = u, ~ ІІ2

[4.8]

The Kutta condition is an observation that for a lifting airfoil of given shape at a given angle of attack, nature adopts that particular value of circulation around the airfoil which results in the flow leaving smoothly at the trailing edge. If the trailing-edge angle is finite, then the trailing edge is a stagnation point. If the trailing edge is cusped, then the velocities leaving the top and bottom surfaces at the trailing edge are finite and equal in magnitude and direction. In either case,

у (ТЕ) = 0 [4.10]

 

Thin airfoil theory is predicated on the replacement of the airfoil by the mean camber line. A vortex sheet is placed along the chord line, and its strength adjusted such that, in conjunction with the uniform freestream, the camber line becomes a streamline of the flow while at the same time satisfying the Kutta condition. The strength of such a vortex sheet is obtained from the fundamental equation of thin airfoil theory:

 

1 fc Y(S)dS „ / dz

2л J0 x – f °° v dx

 

[4.18]

 

Results of thin airfoil theory:

Symmetric airfoil

1.

Сі = 2л a.

2.

Lift slope = dci/da = 2л.

3.

The center of pressure and the aerodynamic center are both at the quarter-chord point.

4.

Cm, с/4 — C/n, ac —

Cambered airfoil

1.

сі = 2л

1 г dz

a—– —— (cos вд – l)d90

[4.57]

_ л Jo dx _

2.

Lift slope = dci/da = 2л.

3.

The aerodynamic center is at the quarter-chord point.

4.

The center of pressure varies with the lift coefficient.

 

The vortex panel method is an important numerical technique for the solution of the inviscid, incompressible flow over bodies of arbitrary shape, thickness, and angle of attack. For panels of constant strength, the governing equations are

 

(i = 1, 2, …, n)

 

Lee cos Pi

 

and

 

Yi = ~Yi-1

 

which is one way of expressing the Kutta condition for the panels immediately above and below the trailing edge.

 

image395

Problems

Подпись: І.Подпись: 2. 3. 4. 5. 6. Consider the data for the NACA 2412 airfoil given in Figure 4.5. Calculate the lift and moment about the quarter chord (per unit span) for this airfoil when the angle of attack is 4° and the freestream is at standard sea level conditions with a velocity of 50 ft/s. The chord of the airfoil is 2 ft.

Consider an NACA 2412 airfoil with a 2-m chord in an airstream with a velocity of 50 m/s at standard sea level conditions. If the lift per unit span is 1353 N, what is the angle of attack?

Starting with the definition of circulation, derive Kelvin’s circulation theorem, Equation (4.11).

Starting with Equation (4.35), derive Equation (4.36).

Consider a thin, symmetric airfoil at 1.5° angle of attack. From the results of thin airfoil theory, calculate the lift coefficient and the moment coefficient about the leading edge.

Подпись: c image396,image397 Historical Note: Kutta, Joukowski, and the Circulation Theory of Lift
image398

The NACA 4412 airfoil has a mean camber line given by

Using thin airfoil theory, calculate (a) aL=o (b) ci when a = 3°

Подпись: 7. 8. 9. 10. For the airfoil given in Problem 4.6, calculate стхц and лср/г when a = 3°.

Compare the results of Problems 4.6 and 4.7 with experimental data for the NACA 4412 airfoil, and note the percentage difference between theory and experiment. (Hint: A good source of experimental airfoil data is Reference 11.)

Starting with Equations (4.35) and (4.43), derive Equation (4.62).

For the NACA 2412 airfoil, the lift coefficient and moment coefficient about the quarter-chord at —6° angle of attack are —0.39 and —0.045, respectively. At 4° angle of attack, these coefficients are 0.65 and —0.037, respectively. Calculate the location of the aerodynamic center.