Page:EB1911 - Volume 14.djvu/135

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HYDRODYNAMICS]
123
HYDROMECHANICS


With liquid of density ρ, this gives rise to a kinetic reaction to acceleration dU/dt, given by

πρb2 a2 + b2   d U = a2 + b2 M′ d U ,
a2 ~ b2 dt a2 ~ b2 dt
(14)

if M′ denotes the mass of liquid displaced by unit length of the cylinder r = b. In particular, when a = ∞, the extra inertia is M′.

When the cylinder r = a is moved with velocity U and r = b with velocity U1 along Ox,

φ = U a2 ( b2 + r ) cos θ − U1 b2 ( r + a2 ) cos θ,
b2a2 r b2a2 r
(15)
ψ = −U a2 ( b2 r ) sin θ − U1 b2 ( r a2 ) sin θ,
b2a2 r b2a2 r
(16)

and similarly, with velocity components V and V1 along Oy

φ = V a2 ( b2 + r ) cos θ − V1 b2 ( r + a2 ) cos θ,
b2a2 r b2a2 r
(17)
ψ = V a2 ( b2 r ) sin θ + V1 b2 ( r a2 ) sin θ,
b2a2 r b2a2 r
(18)

and then for the resultant motion

w = (U2 + V2) a2   z + a2b2   U + Vi
b2a2 U + Vi b2a2 z
−(U12 + V12) b2   z a2b2   U1 + V1i .
b2a2 U1 + V1i b2a2 z
(19)

The resultant impulse of the liquid on the cylinder is given by the component, over r = a (§ 36),

X = ρφ cos θ·adθ = πρa2 ( U b2 + a2 − U1 2b2 );
b2a2 b2a2
(20)

and over r = b

X1 = ρφ cos θ·bdθ = πρb2 ( U 2a2 − U1 b2 + a2 ),
b2a2 b2a2
(21)

and the difference X − X1 is the component momentum of the liquid in the interspace; with similar expressions for Y and Y1.

Then, if the outside cylinder is free to move

X1 = 0,  V1 = 2a2 ,   X = πρa2U b2a2 .
U b2 + a2 b2 + a2
(22)

But if the outside cylinder is moved with velocity U1, and the inside cylinder is solid or filled with liquid of density σ,

X = −πρa2U,   U1 = 2ρb2 ,
U ρ (b2 + a2) + σ (b2a2)
U − U1 = (ρσ) (b2a2) ,
U1 ρ (b2 + a2) + σ (b2a2)
(23)

and the inside cylinder starts forward or backward with respect to the outside cylinder, according as ρ > or < σ.

30. The expression for ω in (1) § 29 may be increased by the addition of the term

im log z = −mθ + im log r,
(1)

representing vortex motion circulating round the annulus of liquid.

Considered by itself, with the cylinders held fixed, the vortex sets up a circumferential velocity m/r on a radius r, so that the angular momentum of a circular filament of annular cross section dA is ρmdA, and of the whole vortex is ρmπ (b2a2).

Any circular filament can be started from rest by the application of a circumferential impulse πρmdr at each end of a diameter; so that a mechanism attached to the cylinders, which can set up a uniform distributed impulse πρm across the two parts of a diameter in the liquid, will generate the vortex motion, and react on the cylinder with an impulse couple −ρmπa2 and ρmπb2, having resultant ρmπ (b2a2), and this couple is infinite when b = ∞, as the angular momentum of the vortex is infinite. Round the cylinder r = a held fixed in the U current the liquid streams past with velocity

q′ = 2U sin θ + m/a;
(2)

and the loss of head due to this increase of velocity from U to q′ is

q2 − U2 = (2U sin θ + m/a)2 − U2 ,
2g 2g
(3)

so that cavitation will take place, unless the head at a great distance exceeds this loss.

The resultant hydrostatic thrust across any diametral plane of the cylinder will be modified, but the only term in the loss of head which exerts a resultant thrust on the whole cylinder is 2mU sin θ/ga, and its thrust is 2πρmU absolute units in the direction Cy, to be counteracted by a support at the centre C; the liquid is streaming past r = a with velocity U reversed, and the cylinder is surrounded by a vortex. Similarly, the streaming velocity V reversed will give rise to a thrust 2πρmV in the direction xC.

Now if the cylinder is released, and the components U and V are reversed so as to become the velocity of the cylinder with respect to space filled with liquid, and at rest at infinity, the cylinder will experience components of force per unit length

(i.) − 2πρmV, 2πρmU, due to the vortex motion;

(ii.) − πρa2 d U/dt, − πρa2 d V/dt, due to the kinetic reaction of the liquid;

(iii.) 0, −π(σρ) a2g, due to gravity,

taking Oy vertically upward, and denoting the density of the cylinder by σ; so that the equations of motion are

πσa2 dU = − πρa2 dU − 2πρmV,
dt dt
(4)
πσa2 dV = − πρa2 dV + 2πρmV − π (σρ) a2g,
dt dt
(5)

or, putting m = a2ω, so that the vortex velocity is due to an angular velocity ω at a radius a,

(σ + ρ) dU/dt + 2ρωV = 0,
(6)
(σ + ρ) dV/dt − 2ρωU + (σρ)g = 0.
(7)

Thus with g = 0, the cylinder will describe a circle with angular velocity 2ρω/(σ + ρ), so that the radius is (σ + ρ) v/2ρω, if the velocity is v. With σ = 0, the angular velocity of the cylinder is 2ω; in this way the velocity may be calculated of the propagation of ripples and waves on the surface of a vertical whirlpool in a sink.

Restoring σ will make the path of the cylinder a trochoid; and so the swerve can be explained of the ball in tennis, cricket, baseball, or golf.

Another explanation may be given of the sidelong force, arising from the velocity of liquid past a cylinder, which is encircled by a vortex. Taking two planes x = ± b, and considering the increase of momentum in the liquid between them, due to the entry and exit of liquid momentum, the increase across dy in the direction Oy, due to elements at P and P′ at opposite ends of the diameter PP′, is

ρdy (U − Ua2r−2 cos 2θ + mr−1 sin θ) (Ua2r−2 sin 2θ + mr−1 cos θ)

+ ρdy ( −U + Ua2r−2 cos 2θ + mr−1 sin θ) (Ua2r−2 sin 2θmr−1 cos θ)

= 2ρdymUr−1 (cos θa2r−2 cos 3θ),

(8)

and with y = b tan θ, r = b sec θ, this is

2ρm Udθ (1 − a2b−2 cos 3θ cos θ),
(9)

and integrating between the limits θ = ±1/2π, the resultant, as before, is 2πρmU.

31. Example 2.—Confocal Elliptic Cylinders.—Employ the elliptic coordinates η, ξ, and ζ = η + ξi, such that

z = c ch ζ, x = c ch η cos ξ, y = c sh η sin ζ;
(1)

then the curves for which η and ξ are constant are confocal ellipses and hyperbolas, and

J = d(x, y) = c2 (ch2 η − cos2 ξ)
d(η, ξ)
= 1/2c2 (ch2η − cos 2ξ) = r1r2 = OD2,
(2)

if OD is the semi-diameter conjugate to OP, and r1, r2 the focal distances,

r1, r2 = c (ch η ± cos ξ);
(3)
r2 = x2 + y2 = c2 (ch2 η − sin2 ξ)
= 1/2c2 (ch 2η + cos 2ξ).
(4)

Consider the streaming motion given by

w = m ch (ζγ), γ = α + βi,
(5)
φ = m ch (ηα) cos (ξβ), ψ = m sh (ηα) sin (ξβ).
(6)

Then ψ = 0 over the ellipse η = α, and the hyperbola ξ = β, so that these may be taken as fixed boundaries; and ψ is a constant on a C4.

Over any ellipse η, moving with components U and V of velocity,

ψ′ = ψ + Uy − Vx = [ m sh (ηα) cos β + Uc sh η ] sin ξ

- [ m sh (ηα) sin β + Vc ch η ] cos ξ;

(7)

so that ψ′ = 0, if

U = − m   sh (ηα) cos β, V = − m   sh (ηα) sin β,
c sh η c ch η
(8)

having a resultant in the direction PO, where P is the intersection of an ellipse η with the hyperbola β; and with this velocity the ellipse η can be swimming in the liquid, without distortion for an instant.

At infinity

U = − m ea cos β = − m cos β,
c ab
V = − m ea sin β = − m sin β,
c ab
(9)

a and b denoting the semi-axes of the ellipse α; so that the liquid is streaming at infinity with velocity Q = m/(a + b) in the direction of the asymptote of the hyperbola β.

An ellipse interior to η = α will move in a direction opposite to the exterior current; and when η = 0, U = ∞, but V = (m/c) sh α sin β.

Negative values of η must be interpreted by a streaming motion on a parallel plane at a level slightly different, as on a double Riemann sheet, the stream passing from one sheet to the other across a cut SS′ joining the foci S, S′. A diagram has been drawn by Col. R. L. Hippisley.