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Chapter 4
Potential flow
Potential or irrotational flow theory is a cornerstone of fluid dynamics, for two reasons. Historically, its importance grew from the developments made possible by the theory of harmonic functions, and the many fluids problems thus made accessible within the theory. But a second, more important point is that po- tential flow is actually realized in nature, or at least approximated, in many situations of practical importance. Water waves provide an example. Here fluid initially at rest is set in motion by the passage of a wave. Kelvin’s theorem insures that the resulting flow will be irrotational whenever the viscous stresses are negligible. We shall see in a later chapter that viscous stresses cannot in gen- eral be neglected near rigid boundaries. But often potential flow theory applies away from boundaries, as in effects on distant points of the rapid movements of a body through a fluid. An example of potential flow in a barotropic fluid is provided by the theory of sound. There the potential is not harmonic, but the irrotational property is acquired by the smallness of the nonlinear term u · ∇u in the momentum equation. The latter thus reduces to
∂u ∂t
ρ
∇p ≈ 0. (4.1)
Since sound produces very small changes of density, here we may take ρ to be will approximated by the constant ambient density. Thus u = ∇φ with ∂φ ∂t = −p/ρ.
4.1 Harmonic flows
In a potential flow we have u = ∇φ. (4.2)
We also have the Bernoulli relation (for body force f = −ρ∇Φ)
φt +
(∇φ)^2 +
dp ρ
48 CHAPTER 4. POTENTIAL FLOW
Figure 4.1: A domain V , bounded by surfaces Si,o where ∂φ ∂n is prescribed.
Finally, we have conservation of mass
ρt + ∇ · (ρ∇φ) = 0. (4.4)
The most extensive use of potential flow theory is to the case of constant density, where ∇ · u = ∇^2 φ = 0. These harmonic flows can thus make use of the highly developed mathematical theory of harmonic functions. in the problems we study here we shall usually consider explicit examples where existence is not an issue. On the other hand the question of uniqueness of harmonic flows is an important issue we discuss now. A typical problem is shown in figure 4.1. A harmonic function φ has prescribed normal derivatives on inner and outer boundaries Si, So of an annular region V. The difference ud = ∇φd of two solutions of this problem will have zero normal derivatives on these boundaries. That the difference must in fact be zero throughout V can be established by noting that ∇ · (φd∇φd) = (∇φd)^2 + φd∇^2 φd = (∇φd)^2. (4.5) The left-hand side of (4.5) integrates to zero over V to zero by Gauss’ theo- rem and the homogeneous boundary conditions of ∂φ ∂nd. Thus
V (∇φd)
(^2) dV = 0,
implying ud = 0. Implicit in this proof is the assumption that φd is a single-value function. A function φ is single-valued in V if and only if
C dφ^ = 0 on any closed contour C contained in V. In three dimensions this is insured by the fact that any such contour may be shrunk to a point in V. In two dimensions, the same conclusion applies to simply-connected domains. In non-simply connect domains uniqueness of harmonic flows in 2DS is not assured. Note for a harmonic flow ∮
C
dφ =
C
u · dx = ΓC , (4.6)
so that a potential which is not single valued is associated with a non-zero circulation on some contour. Since there is no vorticity within the domain of harmonicity, we must look outside of this domain to find the vorticity giving rise to the circulation.
50 CHAPTER 4. POTENTIAL FLOW
Figure 4.2: Flow onto a wedge of half-angle α.
the construction of potential flows. For this application we may start with the physical of z-plane, where the complex potential w(z) is desired. A conformal map z → Z transforms boundaries and boundary conditions and leads to a problem which can be solved to obtain a complex potential W (Z). Under the map values of ψ are preserved, so that streamlines map onto streamlines. Example 4.5: The flow onto a wedge-shaped body (see figure 4.2). Consider in the Z plane the complex potential of a uniform flow,−U Z, U > 0. The region above upper surface of the wedge to the left, and the and the positive x-axis to the right, is mapped onto the upper half-plane Y >) by the function Z = z
π π−α (^). Thus w(z) = −U z ππ−α .
Example 4.6: The map z(Z) = Z + b
2 Z maps the circle of radius^ a > b^ in the Z-plane onto the ellipse of semi-major axis a
(^2) +b 2 a and semi-minor (y)-axis^
a^2 −b^2 a in the z-plane. And the exterior is mapped onto the exterior. Uniform flow with velocity (U, 0)at infinity, past the circular cylinder |Z| = a, has complex potential W (Z) = U (Z + a^2 /Z). Inverting the map and requiring that Z ≈ z for large |z| gives Z = 12 (z +
z^2 − 4 b^2 ). Then w(z) = W (Z(z)) is the complex potential for uniform flow past the ellipse. Notice how the map satisfies (^) dZdz → 1 as z → ∞ This insures that that infinity maps by the identity and so the uniform flow imposed on the circular cylinder is also imposed on the ellipse.
4.1.2 The circle theorem
We now state a result which gives the mathematical realization of the physical act of “placing a rigid body in an ideal fluid flow”, at least in the two-dimensional case.
Theorem 3 Let a harmonic flow have complex potential f(z), analytic in the domain |z| ≤ a. If a circular cylinder of radius a is place at the origin, then the
new complex potential is w(z) = f(z) + f
a^2 ¯z
To show this we need to establish that the analytical properties of the new flow match those of the old, in particular that the analytic properties and the singularities in the flow are unchanged. Then we need to verify that the surface of the circle is a streamline. Taking the latter issue first, note that on the circle
4.1. HARMONIC FLOWS 51
a^2 z ¯ =^ z, so that there we have^ w^ =^ f(z) +^ f(z), implying^ ψ^ = 0 and so the circle is a streamline. Next, we note that the added term is an analytic function of z if it is not singular at z. (If f(z) is analytic at z, so is f(¯z). As for the location of
singularities of w, since f is analytic in |z| ≤ a it follows that f
a^2 z
is analytic
in |z| ≥ a, and the same is true of f
a^2 z ¯
. Thus the only singularities of w(z)
in |z| > a are those of f(z).
Example 4.7: If a cylinder of radius a is placed in a uniform flow, then f = U z and w = U z + U (a^2 /¯z) = U (z + a^2 /z) as we already know. If a cylinder is placed in the flow of a point source at b > a on the x-axis, then f(z) = 2 Qπ ln(z − b) and
w(z) =
Q
2 π
(ln(z−b)+ln
( (^) a^2 z ¯
− b
Q
2 π
(ln(z−b)+ln(z−a^2 /b)−ln z)+C, (4.12)
where C is a constant. From this form it may be verified that the imaginary part of w is constant when z = aeiθ^. Note that the image system of the source, with singularities within the circle, consists of a source of strength Q at the image point a^2 /b, and a source of strength −Q at the origin.
Example 4.8: A point vortex at position zk of circulation Γk has the com- plex potential wk(z) = −i Γ 2 πk ln(z − zk ). A collection of N such vortices will have
the potential w(z) =
∑N
k=1 wk(z).^ Since vorticity is a material scalar in two- dimensional ideal flow, and the delta-function concentration may be regarded as the limit of a small circular patch of constant vorticity, we expect that each vortex must move with he harmonic flow created at the vortex by the other N − 1 vortices. Thus the positions zk(t) of the vortices under this law of motion is governed by the system of N equations,
dzj dt
−i 2 π
∑^ N
k=1,k 6 =j
Γk z − zk
Note the conjugation on the left coming from the identity w′^ = u − iv.
4.1.3 The theorem of Blasius
An important calculation in fluid dynamics is the force exerted by the fluid on a rigid body. In two dimensions and in a steady harmonic flow this calculation can be carried out elegantly using the complex potential.
Theorem 4 Let a steady uniform flow past a fixed two-dimensional body with bounding contour C be a harmonic flow with velocity potential w(z). Then, if no external body forces are present, the force (X, Y ) exerted by the fluid on the body is given by
X − iY =
iρ 2
C
( (^) dw dz
dz. (4.14)
4.2. FLOWS IN THREE DIMENSIONS 53
This introduction to the use of complex variables in the analysis of two- dimenisonal harmonic flows will provide the groundwork for a discussion of lift and airfoil design, to be taken up in chapter 5.
4.2 Flows in three dimensions
We live in three dimensions, not two, and the “flow past body” problem in two dimensions introduces a domain which is not simply connected, with important consequences. The relation between two and three-dimensional flows is partic- ularly significant in the generation of lift, as we shall see in chapter 5. In the present section we treat topics in three dimensions which are direct extensions of the two-dimensional results just given. They pertain to bodies, such as a sphere, which move in an irrotational, harmonic flow.
4.2.1 The simple source
The source of strength Q in three dimensions satisfies
div u = Qδ(x), u = ∇φ. (4.21)
Here δ(x) = δ(x)δ(y)δ(z) is the three-dimensional delta function. It has the following properties: (i) It vanishes if x 6 = 0. (ii) Any integral of δ(x) over an open region containing the origin yields unity. It is best to think of all relations involving delta functions and other distributions as limits of relations using smooth functions. In our case, integrating ∇^2 φ = Qδ(x) over a sphere of radius R 0 > 0 we get ∫
R=R 0
∂φ ∂n
dS = Q. (4.22)
Since ∇^2 φ = 0, x 6 = 0, and since the delta function must be regarded as an isotropic distribution, having no exceptional direction, we make the guess ( using now ∇^2 φ = R−^1 d^2 (Rφ)/dR^2 ) that φ = C/R, R^2 = x^2 + y^2 + z^2 for some constant C. Then (4.22) shows that C = − 4 Qπ. Thus the simple source in three dimensions, of strength Q, has the potential
φ = −
Q
4 π
R
Note that Q is equal to the volume of fluid per unit time crossing any deforma- tion of a spherical surface, assuming the deformed surface surounds the origin. 1
(^1) We indicate how to justify this calculation using a limit operation. Define the three- dimensional delta function by lim→ 0 δ(R) where δ = (^2) π^33 1+(R/^1 ) 3. Solving ∇^2 φ = δ =
R−^2 dRd
R^2 dφ dR
, under the condition that φ vanish at infinity, we obtain φ = − (^4) πR^1 + ∫ (^) ∞ R R
− 2 [tan− (^1) (R (^3) − (^3) ) − π/ 2 ]dR. For any positive R the integral tends to zero as → 0.
54 CHAPTER 4. POTENTIAL FLOW
(^31) 0. 5 0 0.5 1 1.5 2 2.5 3 3.
2
1
0
1
2
3
z/k
r/k
Figure 4.3: The Rankine fairing. All lengths are in units of k.
4.2.2 The Rankine fairing
We consider now a simple source of strength Q placed at the origin in a uniform flow W iz. The combined potential is then
φ = U z −
Q
4 π
R
The flow is clearly symmetric about the z-axis. In cylindrical polar coordinates (z, r, θ), r^2 = x^2 + y^2 we introduce again the Stokes stream function ψ:
uz = φz =
r
∂ψ ∂r
, ur = φr = −
r
∂ψ ∂z
Thus for (4.24) we have 1 r
∂ψ ∂r
= U +
Q
4 π
z R^3
Integrating,
ψ = U r^2 / 2 −
Q
4 π
( (^) z R
In (4.27) we have chosen the constant of integration to make ψ = 0 on the negative z-axis. We show the stream surface ψ = 0, as well as several stream surfaces ψ > 0, in figure 4.3. This gives a good example of a uniform flow over a semi-infinite body. An interesting question is whether or not such a body would experience a force. We will find below that D’Alembert’s paradox applies to finite bodies in three dimensions, that the drag force is zero, but it is not obvious that the result applies to bodies which are not finite. We will use this question to illustrate the use of conservation of momentum to calculate force on a distant contour. In figure 4.4 the large sphere S of radius R 0 is centered at the origin and intersects the fairing on the at a circle bounding
56 CHAPTER 4. POTENTIAL FLOW
100 150 200 250 300 350 400
100
150
200
250
300
Figure 4.5: Flow around an airship.
asymptotic radius of the airing as z → ∞. In this limit D′^ → D, the total drag of the fairing. Thus the momentum integral method gives
D + U Q − U 2 πr^2 ∞ = 0. (4.32)
But from (4.27) we see that the stream surface ψ = 0 is given by
z =
r^2 − 12 k^2 √ k^2 − r^2
, k^2 =
Q
πU
Thus r∞ = k, and (4.32) becomes
D + U Q − U Q = D = 0, (4.34)
so the drag of the fairing is zero.
Example 4.11: The flow considered now typifies the early attempts to model the pressure distribution of an airship. The model consists of a source of strength Q at position z = 0 on the z-axis, and a equalizing sink (source of strength −Q) at the pint z = 1 on the z-axis. Since the source strengths cancel, a finite body is so defined when the singularities are place in the uniform flow U iz. It can be shown (see problem 4.7 below), that stream surfaces for the flow are given by constant values of
U
R^2 sin^2 θ −
Q
4 π
cos θ + 1 − R cos θ √ R^2 − 2 R cos θ + 1
where R, θ are spherical polars at the origin, with axial symmetry. We show the stream surfaces in figure 4.4.
4.2. FLOWS IN THREE DIMENSIONS 57
4.2.3 The Butler sphere theorem.
The circle theorem for two-dimensional harmonic flows has a direct analog in three dimensions.
Theorem 5 Consider an axisymmetric harmonic flow in spherical polars (R, θ, ϕ), uϕ = 0, with Stokes stream function Ψ(R, θ) vanishing at the origin:
uR =
R^2 sin θ
∂θ
, uθ =
R sin θ
∂R
If a rigid sphere of radius a is introduced into the flow at the origin, and if the singularities of Ψ exceed a in distance from the origin, then the stream function of the resulting flow is
Ψs = Ψ(R, θ) −
R
a Ψ(a^2 /R, θ). (4.37)
It is clear that Ψs vanishes when R = a, so the surface of the sphere is a stream surface. Also the added term introduces no new singularities outside the sphere. Thus the theorrm is proved if we can verify that R a Ψ(a^2 /R, θ) represents a harmonic flow. In spherical polars with axial symmetry the only component of vorticity is
ωϕ =
R
[ (^) ∂(Ru θ) ∂R
∂uR ∂θ
]
Thus the condition on Ψ for an irrotational flow is
R^2
∂^2 Ψ
∂R^2
∂θ
sin θ
∂θ
≡ LRΨ = 0. (4.39)
If R a Ψ(a^2 /R, θ) is inserted into (4.39) we can show that the equation is satisfied provided it is satisfied by Ψ(R, θ), see problem 4.8. Finally, since Ψ(R, θ) van- ishes at the origin at least as R, RΨ(a^2 /R, θ is bounded at infinity and velocity component must decay as O(R−^2 ), so the uniform flow there is undisturbed. Example 4.12: A sphere in a uniform flow U iz has Stokes stream function
Ψ(R, θ) =
U
R^2 sin^2 θ
[
a^3 R^3
]
This translates into the following potential:
φ = U z
a^3 R^3
Example 4.13: Consider a source of strength Q place on the z axis at z = b and place a rigid sphere of radius a < b at the origin. The streamfunction for this source which vanishes at the origin is
Ψ(R, θ) = −
Q
4 π
(cos θ 1 + 1), (4.42)
4.3. APPARENT MASS AND THE DYNAMICS OF A SOLID BODY IN A FLUID 59
4.3 Apparent mass and the dynamics of a solid
body in a fluid
Although harmonic flow is an idealization never realized exactly in actual fluids (except in some cases of super fluid dynamics), it is a good approximation in many fluid problems, particularly when rapid changes occur. A good example is the abrupt movement of a solid body through a fluid, for example a swimming stroke of the hand. We know from experience that a abrupt movement of the hand through water gives rise to a force opposing the movement. It is easy to see why this must be, within the theory of harmonic flows. An abrupt movement of the hand through still water causes the fluid to move relative to a observer fixed with the still fluid at infinity. This observer would therefore compute at the instant the hand is moving a finite kinetic energy of the fluid, whereas before the movement began the kinetic energy was zero. To produce this kinetic energy work must have been done, and so a force with a finite component opposite to the direction of motion must have occurred. We are here dealing only with the fluid, but if the body has mass the clearly a force is also needed to accelerate that mass. Thus both the body mass and the fluid movement contribute to the force experienced. In a harmonic flow we shall show that, in the absence of external body forces, the force on a rigid body is proportional to its acceleration, and further the force contributed by the fluid can be expressed as an addition, apparent mass of the body. In other words the augmented force due to the presence of the surrounding fluid and the energy it acquires during motion of a body, can be explained as an inertial force associated with additional mass and the work done against that force. The term virtual mass is also used to denote this apparent mass. For a sphere, which has an isotropic geometry with no preferred direction, the apparent mass is just a scalar to be added to the physical mass. In general, however, the apparent mass associated with translation of a body in two or three dimensions will depend on the direction of the velocity vector. It thus must be a second order tensor, represented by the apparent mass matrix.
4.3.1 The kinetic energy of a moving body
Consider an ideal fluid at rest and introduce a moving rigid body, in two or three dimensions. An observer at rest relative to the fluid at infinity will se a disturbance of the flow which vanishes at infinity. It would be natural to compute compute the momentum of this motion by calculating the integral∫ ρudV of the region exterior to the body. The problem is that such harmonic flows have an expansion at infinity of the form
φ ∼ a ln r − A · rr−^2 + O(r−^2 ) (4.49)
in two dimensions and
φ ∼
a R
− A · RR−^3 + O(R−^3 ) (4.50)
60 CHAPTER 4. POTENTIAL FLOW
in three dimensions. Thus
ρ
∇φdV =
S
φndS, (4.51)
where S comprises both a surface in a neighborhood of infinity as well as the body surface, is not absolutely convergent as the distant surfaces recedes. We point out that a = 0 in two dimensions if the area of the body is fixed and there is no circulation about the body. In three dimensions a vanishes if the body has fixed volume, see problem 4.12. But even if a = 0 and φ = O(R−^1 ) the value of the integral is only condi- tionally convergent will depend on how one defines the distant surface. So the value attributed to the fluid momentum is ambiguous by this calculation. An unambiguous result is however possible, if we instead focus on the kinetic energy and from it determine the incremental momentum created by a change in velocity. Let us fix the orientation of the body and consider its movement through space, without rotation. This translation is completely determined by a velocity vector U(t). The, from the discussion of section 2.6 we know that a harmonic flow will satisfy the instantaneous boundary condition
∂φ ∂n = U(t) · n (4.52)
on the surface of the body. Now ∇^2 φ = 0 is a linear equation, and so we see that there must exist a vthe Φi as encoding the effect of the shape of the body from all possible harmonic flows associated with translation of the body. We may now compute the kinetic energy E of the fluid exterior to the body using u = Ui ∇Φi. (4.53)
Thus E(t) =
Mij UiUj , Mij = ρ
∇Φi · ∇Φj dV. (4.54)
the integral being over the fluid domain. Clearly the matrix Mij is symmetric, and thus dE = Mij Uj dUi. (4.55)
On the other hand the change of kinetic energy, dE, must equal, in the absence of external body forces, the work done by the force F which the body exerts on the fluid, dE = F · Udt. But according to Newton’s second law, the incremental momentum dP is given by dP = Fdt. Consequently dE = U · dP. From (4.54) we thus have dE = ρMij dUj Ui = dPiUi. (4.56)
Since this holds for arbitrary U we must have dPi = Mij dUj. Integrating and using the fact that Mij is independent of time and P = 0 when U = 0 we obtain
Pi = Mij Uj. (4.57)
62 CHAPTER 4. POTENTIAL FLOW
We thus can obtain the apparent mass of a body by a knowledge of the expansion of φ in a neighborhood of infinity. Given that we have computed a finite fluid momentum we are in a position to state
Theorem 6 (D’Alembert’s paradox) In a steady flow of a perfect fluid in three dimensions, and in steady flow in two dimensions for a body with zero circula- tion, the force experienced by the body is zero.
Clearly if the flow is steady dP/dt = F = 0, and we are done. Of course the proof hinges on the existence of a finite fluid momentum associated with a single-value potential function.
Example 4.14: To find the apparent mass matrix of a elliptic cylinder in two dimensions, we may use example 4.6. In the Z-plane the complex potential for uniform flow −Q(cos θ, sin θ) past the cylinder of radius a > b is W (Z) = −Qe−iθ^ Z − Qeiθ^ sa^ /Z. Since Z = 12 (z +
z^2 − 4 b^2 ) we may expand at infinity to get
w(z) ∼ −Qe−iθ^ z − Q
[ (^) a^2 eiθ^ − b^2 e−iθ z
]
so that A =
[
U (a^2 − b^2 ), V (a^2 + b^2 )
]
, (U, V ) = Q(cos θ, sin θ). (4.66)
Now the ellipse intersects the positive x-axis at its semi-major axis α = a
(^2) +b 2 a , and the positive y-axis at its semi-minor axis β = a
(^2) −b 2 a. From (4.66) we obtain the apparent mass matrix
M = 2πρ
a^2 − b^2 0 a^2 + b^2
− π
a^4 − b^4 a^2
= πρ
β^2 0 α^2
In particular for a circular cylinder the apparent mass is just the mass of of the fluid displaced by the body.
An alternative expression for the apparent mass matrix in terms of an inte- gral over the surface of the body rather than a distant surface is readily obtained in terms of the potential Φi. We have
E =
ρ 2
V
∇Φi · ∇Φj Ui Uj dV = ρ 2
V
∇ · Φj ∇Φi]dV UiUj. (4.68)
Applying the divergence theorem to the integral, surfaces So, sB , and observing that Φi∇Φj = O(|x|^1 −^2 N^ ), we see that the receding surface integral will give zero contribution. Recalling that ∂φ ∂n = U · n on the body surface, we see that ∂Φi ∂n =^ ni^ where the normal is directed out of the body surface. In applying the divergence theorem the normal at the body is into the body, with the result that (4.54) applies with
Mij = −ρ
Sb
Φj nidS, n directed out of the body. (4.69)
4.3. APPARENT MASS AND THE DYNAMICS OF A SOLID BODY IN A FLUID 63
It follows from (4.57) that the fluid momentum is given by
P = −ρ
Sb
φndS. (4.70)
We can verify the fact that (4.70) gives the fluid momentum by taking its time derivative, using the result of problem 1.6:
d dt
Sb
φndS =
Sb
∂φ ∂t ndS +
Sb
(u · n)∇φdS. (4.71)
Using the Bernoulli theorem for harmonic flow we have
d dt
Sb
φndS =
Sb
[
p ρ
|u|^2
]
n − (u · n)udS. (4.72)
Converting the terms on the right involving u to a volume integral, we observe that the latter converges absolutely at infinity, as so we have, for the integration over the domain exterior to Sb, ∫
V
[
u · ∇u −
∇|u|^2
]
dV = −
V
u × (∇ × u)dV = 0. (4.73)
Therefore
−
d dt
ρ
Sb
φndS =
Sb
pndS = F, (4.74)
where F is the force applied by the body to the fluid. Finally we note again that the inertial force required to accelerate a body in a perfect fluid will contain a contribution from the actual mass of the body, Mb. This mass appears as an additional term Mbδij in the expression (4.64) for the apparent mass matrix. The total momentum of the body including its apparent mass is thus Ii = Mij Uj + MbUi and Newton’s second law becomes
dIi dt
= Fi, (4.75)
where F is the force applied to the body, to accelerate it and the surrounding fluid.
4.3.2 Moment
We have so far restricted the motion of the body to translation, i.e. with no rotation relative to the fluid at infinity. In general a moment is experienced by a body in translational motion, so that in fact a free body will rotate and thereby give the apparent mass matrix a dependence upon time. The theory may be easily extended to include a time dependent apparent mass, due either to rotation and/or deformation of he body, see section 4.4. But even in steady translational motion of a body, a non-zero moment can result, see problem 4.14. (There is no D’Alembert paradox for moment.)
4.4. DEFORMABLE BODIES AND LOCOMOTION IN AN IDEAL FLUID 65
L 1
L 2
P (^) Q
(a) (^) (b)
Figure 4.7: Swimming in an ideal fluid.
the motion of the boundary of the body relative to the center of volume. If the potential of this harmonic flow of deformation is φD, then the deformation momentum is PD(t) = −ρ
Sb φDn^ ·^ idS. The total momentum if body and fluid is thus mUm(t) + M (t)Uv (t) + PD. If initially the fluid and body is at rest, then this momentum, which is conserved, must vanish, and it is for this reason that locomotion is possible. Consider first a body of uniform density. so the center of mass and of volume coincide. The Um = Uv = U and
U (t) =
PD(t) m + M (t)
There is no reason for the right-hand side o (4.78) to have non-zero time average, and when it does not, we call this locomotion by squirming. To see squirming in action it is best to treat an simple case, see example 4.15 below. Alternatively, we can imagine that the center of mass changes relative to the center of volume without and deformation. Then deformation occurs giving a new shape, then the center of mass again changes relative to the center of volume holding the boy fixed in the new shape. If the two shapes lead to different apparent masses, locomotion occurs by recoil swimming, see example 4.16.
Example 4.15: We show in figure 4.7(a) a squirming body of a simplified kind. The body consists of a then vertical strip of length L 1 (t), and a horizontal part of length L 2 (t). The length will change as a function of time, think of L 2 as being extruded from the material of L 1. We neglect the width w of the strips except when computing mass and volume. The later are constant, implying L 1 + L 2 = L is constant. The density of the material is taken as ρb, so the total mass is Mb = ρbwL and the total volume is wL. A cycle begins with L 1 = L, when L 2 begins to grow to the right. If X(t) denotes the position of the point P , then (ρbwL 1 + ρL^21 /4) dX dt is the momentum of the fluid and vertical segment, where we have used the formula for apparent mass of a flat plate in 2D. The velocity of the extruded strip varies linearly from dX dt at P to d(X dt+L 2 )at Q, so the momentum of the horizontal part is ρbwL 2 d(X + 12 L 2 )/dt, where we neglect the apparent mass of the extruded strip. The first half of the cycle stops when L 1 = 0. Assuming the start is from fluid and body at rest, the sum of these momenta remains zero throughout the
66 CHAPTER 4. POTENTIAL FLOW
0.17 0 20 40 60 80 100 120 140 160 180 200
1
Lambda
Delta X/
L
Figure 4.8: ∆X/L versus λ for the model squirmer of figure 4.7(a).
half-cycle:
(ρbwL + ρL^21 /4) dX dt
ρbwL 2 2
dL 2 dt
If we etL 2 = Lt/T, L 1 = L(1 − t/T ) where T is the half-period of the cycle, then we may obtain the change ∆X of X over the half-cycle by quadrature:
∆(X) =
λ
ln(1 + λ) −
λ
tan−^1 (
λ), λ =
ρL 4 ρbw
We show this relation in figure 4.8 So we see that at the end of the half-cycle the point P has moved a distance −∆X to the left. At this point, we imagine another half-cycle in which L 1 is created at the expense of L 2 , but at the point Q. Observe that at the start of the second half-cycle Q is located a distance L + ∆X from the initial position of P. It can be seen from considerations of symmetry that the point Q will move to the left a distance −∆X in time T over the second half-cycle. The the cycle is complete, L 2 = L, abnd the midpoint can be relabeled P. Thus the net advance to the right of the point P in time 2 T has been L − 2∆x, which from figure 4.8 always exceeds about. 68 L. Example 4.16: Recoil swimming can be illustrated by the 2D model of Figure (4.7)b. Let P denote the center of an elliptical surface of major,minor semi-axes α, β. Within this body is a mass M on a bar enabling it to be driven to the right or left. The weight of the shell and mechanism is m. Let the position of the center be X(t) and the position of the mass be x(t). At the beginning of the half-cycle the mass lies a distance β/2 to the right of P and the ellipse
