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Unformatted text preview: 3.8 THE JOUKOWSKI TRANSFORMATION 167 To use these results we ﬁrst combine (3.7.34) and (3.7.36) to obtain 1 ~_cos(6/2) _ _1"__
,3 —sin(5/2)’ ’3 ” 4hU' Thus, when the vortex strength ,6 and offset hr are prescribed, Equations (3.7.38) and (3.7.39) can be solved for the deflection angle 6 and the vortex image angle 11/. The shape of the jet can then be plotted by use of (3.7.37), the upper and lower streamlines in the ﬁgure corresponding respectively to the intervals (—00, 0), (0, 00) of the real I axis.
For the case shown in Figure 3.7.16, tan w = (3.7.39) 1“ E = ., = 3", =55.°.
4hU 05 6 51 w 7 hr=0, .3 3.8 The Joukowski transformation The Joukowski transformation §=§+ %—1, a>0, zl=x+iy,
(3.8.1)
.. z 1 1 . '
anditsmverse —=— {+— , §=§+lm
a 2 C have important applications to the theory of airfoils. It is conformal everywhere except
at z = 21:51, where dg/dz = 00 (i.e. at; = i1, where dz/d§ = O). In particular, this means
that smooth contours in the 4“ plane in § >‘ 1 map into smooth contours in the z plane.
Also, z ~ a; /2 when z and lgll are large, so that uniform ﬂow at large distances from
the origin in either plane maps into a uniform ﬂow in the same direction in the other plane. _ V
Consider steady flow in the g plane at speed V in a direction inclined at angle oz (0 < at < §) to the positive S axis, with complex potential Vge—i“. According to the circle
theorem (§3.2), when a circular cylinder I; l = ,8 with centre at the origin and radius )6
is inserted into the ﬂow the complex potential becomes Zia
w:V<;'e"“"+ﬁ§e The streamlines of this ﬂow are illustrated in Figure 3.8.1. We can plot them by ﬁrst
inverting (3.8.2) to obtain for g = E + in ‘ (3.8.2) wz
(2V5 )2 §+in = ﬁe”t + — 1), where w =' (p + iw; (3.8.3) the principal value of the square root is taken to ensure that to ~ V; e‘” as §l —+ 00.
This provides a parametric representation of the streamlines (on which 11/ = constant) in terms of<p (——oo < (0 < +00). 168 IDEAL FLOW IN TWO DIMENSIONS Figure 3.81 When the radius 5 > 1 the circular cylinder maps into a cylinder of elliptic cross
section in the 2 plane [Figure 3.8.1(b)]. If g“ = ﬁeie on the circular cylinder, the second of (3.8.1) gives
I Z _ l i@ 1 —ie
a _ 2 (ﬁe + ﬁe ) , so that the elliptic cylinder is given pararnetrically by x=§<ﬁ+%>cos®, y=g<ﬂ—%>sin®, x2 y2 V  a ,1 a 1 — ——= z ._ —_—_ _# 3.4
13, AZ+B2 1, where A 2<ﬁ+ﬁ>, B 2<ﬁ ﬂ), (8) a=\/A2—B2, ,3: A+B>1.' A—B 3.8 THE JOUKOWSKI TRANSFORMATION 169 Figure 3.8.2 Thus in the 2 plane the mean ﬂow in the at direction is around the elliptic cylinder.
Now as lg], z —> 00,
22V w ~ §Ve"i°‘ ~ —e‘
a ict because g ~ 2z/a. Hence, if the speed in the z plane at large distances from the elliptic
cylinder is U, we must set U = 2V/a, and the corresponding complex potential is then
given by (3.8.2) and the ﬁrst of (3.8.1) in the form U _ia '82 2 ia
11):? (Z‘l' zz—a2)e +'<z+;zf_azj
‘=%[<z+ aw We plot the streamline pattern around the elliptic cylinder [Figure 3.8.1(b)] by using
(3.8.3) to calculate g = S + in in terms of w 2 go +i1/r and then substituting into the
second of (3.8.1) to obtain the parametric representation of a point z = x + i y in terms
of go (—00 < go < 00) on each streamline 1/r = constant. I It is evident from the symmetry of both of the ideal ﬂows in Figure 3.8.1 that the
fore—aft surface pressure forces are in equilibrium, and that there is no net force on
either body (D’Alembert’s paradox). However, Bernoulli’s equation, 1 1 p = EpoUZ — EMWY; shows that the excess pressure p is a maximum on the surface at the stagnation points,
labelled C, C’ in the ﬁgure. For the elliptic cylinder this means that there must be a net
moment on the cylinder, tending to turn it in the clockwise direction, so that its major
axis is facing the oncoming mean stream. To verify this the moment M3 (per unit span)
about the origin is calculated by means of Blasius formula (3.3.3) (see Figure 3.8.2): loo dw 2
M3——R6[2 dZ], 170 IDEAL FLOW IN TWO DIMENSIONS where the integration can be taken around any contour enclosing the cylinder. In par
ticular, we can integrate around a circle of radius lzl —+ 00, where d_w_g —icv Z Zia _ Z
alz—2{e [1+(Z2_a2)%]+ﬂe [1 (Z2_42) NI— Then, by residues,
. 1 .
moment = Re [_§p0ﬂiU2a2 (6—21a7_ ’82)] 2 1 .
= —%p,,Uzsin2oz = 774.42 — BZMOUZsinZoz. (3.8.6) Thus the mean ﬂow exerts a clockwise couple on the cylinder, tending to turn it broadside
to the stream, a conclusion that is applicable for any elongated body (such as a boat in
a stream). The couple vanishes when at = 0, when the major axis of the cylinder is parallel to the
stream. However, this conﬁguration is clearly unstable. On the other hand, the other
equilibrium position oz 2 g is stable. ' 3.8.1 The ﬂat—plate airfoil When ,6 —> 1 elliptic image (3.8.4) in the z plane of the circular cylinder of Fig—
ure 3.8.1(a) collapses onto the strip [xl < a, y = O, which can be regarded as a ‘ﬂat—plate’
approximation to an airfoil (of inﬁnite span). The points g” = $1 on the cylinder cor—
respond respectively to the trailing edge 2 = —a and the leading edge z = +a of the
airfoil. These are singular points at which dt/dz = 00, where the Joukowski transfor
mation (3.8.1) ceases to be conformal. The generation of lift by an airfoil requires it to be inclined at a small angle 0 f attack or to the mean ﬂow direction. This is, of course, just the problem considered previously for the elliptic cylinder. We ﬁnd that the velocity potential of the mean ﬂow is given by
setting V = aU/2 and ﬂ = 1 in (3.8.2): .w = % (ge—za + . (3.8.7) In the 2 plane we have, setting ,8 = l in (3.8.5), warp—l—ir/r:U(zcosoz—isinon/z2—a2). (3.8.8) The streamline pattern (1/; = constant) is plotted in Figure 3.8.3. The flow exhibits
fore—aft asymmetry, with leading and trailing stagnation points A and B where the
excess pressure is a maximum. The complex velocity dw izsinoz
—=U cosoz———
dz Z2_a2 3.8 THE JOUKOWSKI TRANSFORMATION Figure 38.3 is singular at the edges z = in, where streamlines are required to turn through 180°.
There is no net force on the airfoil, but there is a couple of magnitude —%2m2 p0 U 2 sin 2a
tending to rotate it in the clockwise direction. The ﬂow past an airfoil in practice is subject to relatively strong viscous action in
the neighbourhood of the trailing edge. When the motion starts from rest the flow
everywhere is initially irrotational. The high—speed ﬂow around the trailing edge must
necessarily be rapidly slowed as it approaches the stagnation point B. In reality, however,
the ﬂow at the edge immediately separates resulting in the formation of a ‘Vstarting
vortex’ of positive circulation F that is convected away by the mean ﬂow. According to
Kelvin’s circulation theorem the progressive shedding of circulation from the trailing
edge must be compensated by the growth in an equal amount of negative circulation
around the airfoil, the effect of which is to progressively shift the stagnation point B to
the trailing edge where it ultimately cancels the irrotational ﬂow edge singularityThus
vorticity continues to be shed until the ﬂow at the trailing edge becomes continuous,
that is, until the edge ﬂow is smooth and leaves the edge tangentially. The process is very
rapid in practice, the shed vorticity coagulating into the starting vortex that is quickly carried away in the mean ﬂow, so that it eventually ceases to affect the motion at the I airfoil. However, the inﬂuence of the circulation induced around the airfoil by shedding
is to maintain the smooth ﬂow from the trailing edge. The requirement that the ﬂow be smooth at the trailing edge determines the magni
tude of P. In the r plane we obtain the velocity potential in the presence of the negative
circulation about the airfoil by adding (iF/Zyr) ln 9“ to the right—hand side of (3.8.7): aU . em ir
= — ‘1“ —— —— . 3.8.9
w 2<re +£>+2ﬂlnr ( ) 172 IDEAL FLOW IN TWO DIMENSIONS o) I
high speed, low pressure Figure 3.8.4 The complex velocity in the z plane then becomes dw izsinoz i1“ 
— = U 0080: — —— + ———. 3.8.10
dz ( 22 — a2) Zyrx/ z2 — a2 ( ) The value of F is found by application of the Kumz—Joukowski hypothesis that the
velocity should remain ﬁnite at z = a, which yields F = ZirUa sinoz, (3.8.11) in which case dw . . z—a
— = Ucosoz—zUsmoz . 3.8.12
dz z + a ( ) The ﬂow therefore leaves the trailing edge tangentially with velocity U cos at on both
sides of the airfoil. The situation is illustrated in Figure 3.8.4. This compares the stream
line patterns (a) without circulation, and (b) with circulation (3.8.11) around the airfoil,
when the airfoil and ﬂow are rotated clockwise through the angle of attack at, so that
the incident mean ﬂow is horizontal. '“T 3.8 THE JOUKOWSKI TRANSFORMATION 173 pOFUsina ‘\ pol‘Ucosa Figure 3.8.5 3.8.2 Calculation of the lift There is no net force between the airfoil and ﬂuid except in the presence of circulation.
In this case the force is determined in two dimensions by use of the timeindependent 
form of Blasius formula (3.3.2). For steady ﬂow dw /dz is given by (3.8.10) (with respect
to coordinate axes orientated as in Figure 3.8.3), so that ‘ ’ . 7 . . . . 7
_ 1,00 dw ‘ zpo zzU smoz ll" “
5—15: _ <_) dz: _— (mosa— _ +___ dz.
2 5 dz 2 S /z2_a2 .271 /Z2_a2 The integrand is regular throughout the region occupied by the fluid, and the value of
the integral is therefore the same as that evaluated on a large circle lzl = R —> 00. On
this circle only the term in the integrand that behaves like C / 2 (C = constant) can make
a nontrivial contribution to the integral, by an amount equal precisely to 217i C. Hence 1171—in = *pol‘Usinoz — ipol‘Ucosa, i.e., (F1, F2) 2 pOPU(— sinoz, cos oz). The net force F on the airfoil (orientated as in Figure 3.8.3) accordingly consists of a
component ,oOFUsinoc in the negative x direction and a component 'poFU cosoz in the
y direction. The overall force is therefore a lift force of magnitude lift = poFU a 27rp0U2a sina per unit span, (3.8.13) directed as indicated in Figure 3.8.4(b) at right angles to the impinging mean ﬂow. The / lift is the resultant of a component of magnitude poPU cos or normal to the plane of the
airfoil and the component poFU sina parallel to the airfoil (Figure 3.8.5); the latter is
produced by leading—edge suction. There is no suction at the trailing edge because the
flow velocity remains ﬁnite there. 3.8.3 Lift calculated from the Kirchhoff vector force formula The lift can also be calculated in a very convenient fashion by use of formula (4.5.15) of
Chapter 4..For steady motion in the x direction and when viscous forces are neglected? the lift (in the y direction) is given by lift: p; / VX2 . w /\ vreld3x, (3.8.14)
V 174 IDEAL FLOW IN TWO DIMENSIONS (6!) y (b) Circulation/1‘ Lift/poI'U 0 5 10 15 20 25
Ut/a Figure 3.8.6 where 2600 is the y component of the Kirchhoff vector (the velocity potential of ﬂow
past the airfoil having unit speed in the y direction at large distances from the airfoil)
and vrel is the convection velocity of the shed vorticity relative to the airfoil. When the motion is steady, all of the wake vorticity shed from the trailing edge has
been swept away towards x = +00. At such points X2 ~ y, where it may be assumed
that the convection velocity vrel = (U, 0): lift per unit span = pO/ w(x — Ur, y)dedy E pOFU,
V where a) is the shed vorticity distribution of total circulation 1". 3.8.4 Lift developed by a starting airfoil When the angle of attack or is small it is possible to obtain an analytical representation of
the growth of the lift for an airfoil that starts impulsively from rest at z‘ = O and proceeds
to translate at constant speed U (Wagner 1925). Let the motion be in the negative
x direction in an otherwise stationary ﬂuid [Figure 3.8.6(a)], and let the coordinate
origin translate with the airfoil at the midchord position. In a linearised approximation 3.9 THE JOUKOWSKI AIRFOIL 175 (to ﬁrst order in a) it may be assumed that the vorticity a) shed from the trailing edge is
conﬁned to a vortex sheet lying along the x axis between x = a and x = a + Ur [that is,
that vrel = (U, 0)], so that a)(x, t) = y0(x, t)6(y), ' (3.8.15) where y0(x, t) is the circulation per unit length of the sheet.
It may then be shown (by the method discussed below in §3.12) that
2a U 00 e—ik(Utx)dk VOL“: I) = — ——.—<1>_“Tr ‘1 < x < a + Ur, (3.8.16)
” °° (k+10)[H0 (ka) +iH1 {an} where H81) and H51) are Hankel functions, and the integration path passes just above
any singularities on the real k axis. This integral is easily evaluated numerically and can
be used with formula (3.8.14) to calculate the lift force as a function of time. The circulation in the wake at time t is a+Ut
/ yo(x, t)dx,
ll which is effectively equal to the total shed circulation l" N ZyromU when U r /a exceeds
about 10; it is plotted as the dashed curve in Figure 3.8.6(b). p Because to ~ 0(a), the Kirchhoff vector X’g in (3.8.14) can be approximated by its
value for an airfoilat zero angle of attack (Table 2.19.1), namely X2 = Re (4m). in terms of which y0(x, t)xdx When U r /a >> 1 the shed vorticity is far downstream of the airfoil, where 3X2 / 3 y —> 1,
and the lift tends to the Kutta—Joukowski value 271mm,, U2 E p01“ U. The approach to this limit is plotted as the solid curve in Figure 3.8.6(b), which also indicates that the lift
immediately jumps to half its ﬁnal value as soon as the airfoil begins to move. ajf2 a+Ut
lift per unit span = ,oOU/ y0(x, t)6(y)—6Fdxdy = ,0on (3.8.17)
V a 3.9 The Joukowski airfoil The generation and release of the starting vortex removes the singular velocity at the
trailing edge and eliminates the associated suction force. The corresponding pressure
distribution on the airfoil — a combination of leading—edge suction and a sideways pres
sure force — produces a lift force in a direction precisely at right angles to that of the
impinging mean ﬂow. For the thinplate airfoil, however, there remains a leading—edge
ﬂow with a nominally inﬁnite maximum velocity, which is not realisable in practice and
at which a real ﬂow would immediately separate; this would create a region above the
airfoil of relative low velocity and high pressure, ﬁlled with vorticity, that would ulti—
mately cause the airfoil to ‘stall’. This is avoided by proper airfoil design. The leading
edge must be sufﬁciently thickened and rounded and thereby furnished with a nonzero 176 IDEAL FLOW IN TWO DIMENSIONS high speed
low pressure . Figure 3.9.1 radius of curvature (see Figure 3.9.1). The effect is that separation does not occur pro
vided the angle of attack or is sufﬁciently small (say, less than about 8°—10°). Indeed, the
convergence of the streamlines in Figure 3.8.4(b) at the leading edge shows that ﬂow
over the airfoil forward of the stagnation point A is accelerated. This is also true for
a rounded edge: The ﬂow is accelerated into the lowpressure region above the airfoil
and therefore tends to remain attached to the surface provided the radius of curvature
at the leading edge is not too small. This would not happen at the trailing edge of Fig
ure 3.8.4(a), however, even when rounded, because the edge flow is directed towards
the high—pressure stagnation point 1B, which retards the ﬂuid and inevitably leads to
separation. / 3.9.1 Streamline ﬂow past an airfoil The Joukowski transformation (3.8.1) can be used to study airfoils with rounded nose
proﬁles. Recall that the edges of the thinplate airfoil 2': id correspond to the singular
points of the transformation where rig/dz : 00. These points are at g“ = :l:1 in the
5“ plane. They lie inside a cylinder 9* '= ,Bele (—7r < (9 5 7r) when )3 > 1, which must
therefore be mapped conformally into a smooth proﬁle (without edges) in the 2 plane.
According to the second of Equations (3.8.1) the proﬁle in the ,2 plane is the elliptic
cylinder (Figure 3.92) z x . y __ 1 1 i 1 . a a +10 _ 2 <ﬁ+ﬁ>cos®+§(ﬁ—E>sm®, —7r <®57r.
Suppose the circle § = B in Figure 3.9.2 is shifted a distance 6 to the left along the real axis until it just touches the circle 1;] :: 1 at § 2 1. Then [3 = 1 + 6, and the new circle (in Figure3.9.3) is §+ 5 = (1 + (3)610, —n < o 5 It, a > 0. (3.9.1) Mapping (3.8.1) of this circle is nonconforinal at g“ = 1, and its image in the z plane has
a sharp e'dge'at z = a. The leading edge remains smooth, however, because the other
singular point at g“ = —1 is still an interior point. The proﬁle in the z plane is the symmetric Joukowski airfoil illustrated in Figure 3.9.3
for 6 = 0.1. The airfoil extends along the x axis over the interval a
2 1.
<1+26+1+25> <x<a ,7 3.9 THE JOUKOWSKI AIRFOIL 177 5“ plane 2 plane Figure 3.9.2 with chord 2a(1 + 6 )2 / (1 + 28), and the radius of curvature of the nose is 8a62(1 + 6)3 = ~ W 7?.
In the 4“ plane the complex potential of ﬂow at speed U at angle at to the real axis is
given by the following modiﬁcation of (3.8.9): _Ua V 1 2 id
w _ 7 +5)e—“1 + Li (5+5) The trailing edge of the airfoil corresponds to the point 5 = l, and the velocity will
remain ﬁnite provided dw/d; = 0 there, which yields ' :1 + 1n(§ + 8). (392) l‘ = 271Ua(1 + 5) sina. When this condition is satisﬁed the velocity is ﬁnite everywhere on the airfoil. The airfoil
experiences a lift force equal to pal" U per unit span, as in the case of the thin—plate airfoil. 5“ plane z plane Figure 3.9.3 l
7....» i.
E
l
. l
} 178 IDEAL FLOW IN TWO DIMENSIONS T poFU
_,, U Figure 3.9.4 Figure 3.9.4 shows the typical streamline pattern when the angle of attack at = 10° (21
large angle for practical airfoils, chosen for the purpose of illustration), when the airfoil
and ﬂow are rotated through this angle so that the incident flow is horizontal.
Symmetric airfoils of this kind are used for tailplanes and rudders when there is
no preference in the desired direction of lift. For aircraft wings, however, improved
lift characteristics at smaller angles of attack are obtained by use of cambered proﬁles,
where the lower side of the airfoil is either ﬂat or concave. We can generate such proﬁles
by considering ﬂow around a circle in the 4" plane that again passes through the singular
point 4“ = 1, but with its centre shifted to g = —8 + i6’ (0’ in Figure 3.9.5), that is the circle
g: —5+is’+,/(1 +5)2 +5269, —7r < o 5 Jr. The leading edge of the airfoil is rounded because g = —1 lies within the circle. The case
shown in the ﬁgure corresponds to 6 = 0.1, 5’ = 0.2. The ﬂow pattern can be calculated
as before, but such details are left as an exercise for the reader. Z plane {—(—6+i6‘) = \/{(1+6)2+6'2}ei9 . Figure 3.95 ...
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 Fall '09
 RENWEIMEI

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