Page:EB1911 - Volume 14.djvu/137

Changing to polar coordinates, x = r cos θ, y = r sin θ, the equation (2) becomes, with cos θ = μ,

r 2	d 2ψ	+ (1 − μ2)	d 2ψ	= 2 ζr 3 sin θ,
	dr 2		dμ

(8)

of which a solution, when ζ = 0, is

ψ = ( Ar n+1 +	B	) (1 − μ2)	dPn	= ( Ar n − 1 +	B	) y2	dPn	,
	r n		dμ		r n+2		dμ

(9)

φ = { (n + 1) Ar ⁿ − nBr ⁻ⁿ⁻¹ } P_n,

(10)

where P_n denotes the zonal harmonic of the nth order; also, in the exceptional case of

ψ = A₀ cos θ, φ = A₀/r ;
ψ = B₀r, φ = −B₀ log tan 1/2θ
= −1/2B₀ sh⁻¹ x/y.

(11)

Thus cos θ is the Stokes’ function of a point source at O, and PA − PB of a line source AB.

The stream function ψ of the liquid motion set up by the passage of a solid of revolution, moving with axial velocity U, is such that

1		dψ	= −U	dy	, ψ + 1/2Uy2 = constant,
y		ds		ds

(12)

over the surface of the solid; and ψ must be replaced by ψ′ = ψ + 1/2Uy² in the general equations of steady motion above to obtain the steady relative motion of the liquid past the solid.

For instance, with n = 1 in equation (9), the relative stream function is obtained for a sphere of radius a, by making it

ψ′ = ψ + 1/2Uy² = 1/2U (r ² − a³/r) sin² θ, ψ = −1/2Ua³ sin² θ/r;

(13)

and then

φ′ = Ux (1 + 1/2a³/r ²), φ = 1/2Ua³ cos θ/r ²,

(14)

−	dφ	= U	a3	cos θ,   −	dφ	= 1/2U	a3	sin θ,
	dr		r 3		rdθ		r 3

(15)

so that, if the direction of motion makes an angle ψ with Ox,

tan (ψ − θ) = 1/2 tan θ, tan ψ = 3 tan θ/(2 − tan² θ),

(16)

Along the path of a liquid particle ψ′ is constant, and putting it equal to 1/2Uc²,

(r ² − a³/r) sin² θ = c², sin² θ = c²r / (r ³ − a³),

(17)

the polar equation; or

y² = c²r ³ / (r ³ − a³), r ³ = a³y² / (y² − c²),

(18)

a curve of the 10th degree (C₁₀).

In the absolute path in space

cos ψ = (2 − 3 sin² θ) / √ (4 − sin² θ), and sin³ θ = (y³ − c²y) / a³,

(19)

which leads to no simple relation.

The velocity past the surface of the sphere is

1		dψ′	= 1/2U ( 2r +	a3	)	sin2 θ	= 3/2 U sin θ, when r = a;
r sin θ		dr		r 2		r sin θ

(20)

so that the loss of head is

(9/4 sin² θ − 1) U²/2g, having a maximum 5/4 U²/2g,

(21)

which must be less than the head at infinite distance to avoid cavitation at the surface of the sphere.

With n = 2, a state of motion is given by

ψ = −1/2 Uy²a⁴ μ/r ⁴,   ψ′ = 1/2 Uy² (1 − a⁴ μ/r ⁴),

(22)

φ′ = Ux + φ,   φ = −1/3 U (a⁴ / r ³) P₂,   P₂ = 3/2 μ² − 1/2,

(23)

representing a stream past the surface r ⁴ = a⁴μ.

35. A circular vortex, such as a smoke ring, will set up motion symmetrical about an axis, and provide an illustration; a half vortex ring can be generated in water by drawing a semicircular blade a short distance forward, the tip of a spoon for instance. The vortex advances with a certain velocity; and if an equal circular vortex is generated coaxially with the first, the mutual influence can be observed. The first vortex dilates and moves slower, while the second contracts and shoots through the first; after which the motion is reversed periodically, as if in a game of leap-frog. Projected perpendicularly against a plane boundary, the motion is determined by an equal opposite vortex ring, the optical image; the vortex ring spreads out and moves more slowly as it approaches the wall; at the same time the molecular rotation, inversely as the cross-section of the vortex, is seen to increase. The analytical treatment of such vortex rings is the same as for the electro-magnetic effect of a current circulating in each ring.

36. Irrotational Motion in General.—Liquid originally at rest in a singly-connected space cannot be set in motion by a field of force due to a single-valued potential function; any motion set up in the liquid must be due to a movement of the boundary, and the motion will be irrotational; for any small spherical element of the liquid may be considered a smooth solid sphere for a moment, and the normal pressure of the surrounding liquid cannot impart to it any rotation.

The kinetic energy of the liquid inside a surface S due to the velocity function φ is given by

T = 1/2ρ $\iiint$ [(

dφ

)

2

+ (

dφ

)

2

+ (

dφ

)

2

] dxdydz,

dx

dy

dz

= 1/2ρ $\iint$ φ	dφ	dS
	dν

(1)

by Green’s transformation, dν denoting an elementary step along the normal to the exterior of the surface; so that dφ/dν = 0 over the surface makes T = 0, and then

(

dφ

)

2

+ (

dφ

)

2

+ (

dφ

)

2

= 0,

dφ

= 0,

dφ

= 0,

dφ

= 0.

dx

dy

dz

dx

dy

dz

(2)

If the actual motion at any instant is supposed to be generated instantaneously from rest by the application of pressure impulse over the surface, or suddenly reduced to rest again, then, since no natural forces can act impulsively throughout the liquid, the pressure impulse ῶ satisfies the equations

1		dῶ	= −u,	1		dῶ	= −v,	1		dῶ	= ῶ,
ρ		dx		ρ		dy		ρ		dz

(3)

ῶ = ρφ + a constant,

(4)

and the constant may be ignored; and Green’s transformation of the energy T amounts to the theorem that the work done by an impulse is the product of the impulse and average velocity, or half the velocity from rest.

In a multiply connected space, like a ring, with a multiply valued velocity function φ, the liquid can circulate in the circuits independently of any motion of the surface; thus, for example,

φ = mθ = m tan⁻¹ y/x

(5)

will give motion to the liquid, circulating in any ring-shaped figure of revolution round Oz.

To find the kinetic energy of such motion in a multiply connected space, the channels must be supposed barred, and the space made acyclic by a membrane, moving with the velocity of the liquid; and then if k denotes the cyclic constant of φ in any circuit, or the value by which φ has increased in completing the circuit, the values of φ on the two sides of the membrane are taken as differing by k, so that the integral over the membrane

$\iint$ φ	dφ	dS = k $\iint$	dφ	dS,
	dν		dν

(6)

and this term is to be added to the terms in (1) to obtain the additional part in the kinetic energy; the continuity shows that the integral is independent of the shape of the barrier membrane, and its position. Thus, in (5), the cyclic constant k = 2πm .

In plane motion the kinetic energy per unit length parallel to Oz

T = 1/2ρ $\iint$ [

dφ

)

2

+ (

dφ

)

2

] dxdy = 1/2ρ $\iint$ [(

dψ

)

2

+ (

dψ

)

2

] dxdy

dx

dy

dx

dy

= 1/2ρ $\int$ φ	dφ	ds = 1/2ρ $\int$ ψ	dψ	ds.
	dν		dν

(7)

For example, in the equilateral triangle of (8) § 28, referred to coordinate axes made by the base and height,

ψ′ = −2Rαβγ/h = −1/2 Ry [ (h − y)² − 3x² ] /h

(8)

ψ = ψ′ − 1/2R [ ( 1/3 h − y)² + x² ]

= −1/2R [ 1/2h³ + 1/3 h²y + h) (x² − y²) − 3x²y + y³ ] /h

(9)

and over the base y = 0,

dx/dν = −dx/dy = + 1/2R ( 1/3 h² − 3x²) / h, ψ = −1/2R ( 1/9 h² + x²).

(10)

Integrating over the base, to obtain one-third of the kinetic energy T,

1/3T = 1/2ρ $\int _{-h/{\sqrt {3}}}^{h/{\sqrt {3}}}$ 1/4R² (3x⁴ − 1/27 h⁴) dx/h

= ρR² h⁴ / 135√3

(11)

so that the effective k² of the liquid filling the triangle is given by

k² = T / 1/2ρR²A = 2h² / 45
 = 2/5 (radius of the inscribed circle)²,

(12)

or two-fifths of the k² for the solid triangle.

Again, since

dφ/dν = dψ/ds,   dφ/ds = −dψ/dν,

(13)

T = 1/2ρ $\int$ φ dψ = −1/2ρ $\int$ ψ dφ.

(14)

With the Stokes’ function ψ for motion symmetrical about an axis.

T = 1/2ρ $\int$ φ	dψ	2πyds = πρ $\int$ φ dψ.
	yds

(15)

37. Flow, Circulation, and Vortex Motion.—The line integral of the tangential velocity along a curve from one point to another, defined by

$\int$ ( u	dx	+ v	dy	+ w	dz	) ds = ∫(udx + vdy + zdz),
	ds		ds		ds

(1)

is called the “flux” along the curve from the first to the second point; and if the curve closes in on itself the line integral round the curve is called the “circulation” in the curve.

With a velocity function φ, the flow

−∫dφ = φ₁ − φ₂,

(2)