Page:EB1911 - Volume 14.djvu/141

so that PT = c/1/2π, and the curve AP is the tractrix; and the coefficient of contraction, or

breadth of the jet	=	π	.
breadth of the orifice		π + 2

(4)

A change of Ω and θ into nΩ and nθ will give the solution for two walls converging symmetrically to the orifice AA₁ at an angle π/n. With n = 1/2, the reentrant walls are given of Borda’s mouthpiece, and the coefficient of contraction becomes 1/2. Generally, by making a′ = −∞, the line x′A′ may be taken as a straight stream line of infinite length, forming an axis of symmetry; and then by duplication the result can be obtained, with assigned n, a, and b, of the efflux from a symmetrical converging mouthpiece, or of the flow of water through the arches of a bridge, with wedge-shaped piers to divide the stream.

Fig. 5.	Fig. 6.

42. Other arrangements of the constants n, a, b, a′ will give the results of special problems considered by J. M. Michell, Phil. Trans. 1890.

Thus with a′ = 0, a stream is split symmetrically by a wedge of angle π/n as in Bobyleff’s problem; and, by making a = ∞, the wedge extends to infinity; then

ch nΩ = √	b	, sh nΩ = √	n	.
	b − u		b − u

(1)

Over the jet surface ψ = m, q = Q,

u = − eπφ/m = − be^πs/c,

ch Ω = cos nθ = √	1	, sh Ω = i sin nθ = i √	eπs/c	,
	eπs/c + 1		eπs/c + 1

(2)

e1/2πs/c = tan nθ,	1/2π		ds	=	2n	.
	c		dθ		sin 2nθ

(3)

For a jet impinging normally on an infinite plane, as in fig. 6, n = 1,

e1/2π²/c = tan θ, ch (1/2πs/c) sin 2θ = 1,

(4)

sh 1/2πx/c = cot θ, sh 1/2πy/c = tan θ,

sh 1/2πx/c sh 1/2πy/c = 1, e^{1/2π(x + y)/c} = e^1/2πx/c + e^{1/2 πy/c} + 1.

(5)

With n = 1/2, the jet is reversed in direction, and the profile is the catenary of equal strength.

In Bobyleff’s problem of the wedge of finite breadth,

ch nΩ = √	b	√	u − a	, sh nΩ = √	b − a	√	u	,
	a		u − b		a		u − b

(6)

cos nα = √	b	, sin nα = √	a − b	,
	a		a

(7)

and along the free surface APJ, q = Q, ψ = 0, u = e^−πφ/m = ae^πs/c,

cos nθ = cos nα √	eπ2/c − 1	,
	eπ2/c − cos2 nα

eπ2/c =	cos2 nα sin2 nθ	,
	sin2 nθ − sin2 nα

(8)

the intrinsic equation, the other free surface A′P′J′ being given by

eπ2/c =	cos2 nα sin2 nθ	,
	sin2 nα − sin2 nθ

(9)

Putting n = 1 gives the case of a stream of finite breadth disturbed by a transverse plane, a particular case of Fig. 7.

When a = b, α = 0, and the stream is very broad compared with the wedge or lamina; so, putting w = w′(a − b)/a in the penultimate case, and

u = ae−w ≈ a − (a − b)w′,

(10)

ch nΩ = ${\sqrt {}}$	w′ + 1	, sh nΩ = ${\sqrt {}}$	1	,
	w′		√ w′

(11)

in which we may write

w′ = φ + ψi.

(12)

Along the stream line xABPJ, ψ = 0; and along the jet surface APJ, −1 > φ > −∞; and putting φ = −πs/c − 1, the intrinsic equation is

πs/c = cot² nθ,

(13)

which for n = 1 is the evolute of a catenary.

Fig. 7.

43. When the barrier AA′ is held oblique to the current, the stream line xB is curved to the branch point B on AA′ (fig. 7), and so must be excluded from the boundary of u; the conformal representation is made now with

dΩ	= −	√ (b − a·b − a′)
du		(u − b) √ (u − a·u − a′)

(1)

dw	= −	m		1	−	m′		1	,
du		π		u − j		π		u − j

= −	m + m′	·	u − b	,
	π		u − j·u − j ′

b =	mj ′ + m′j	,
	m + m′

(2)

taking u = ∞ at the source where φ = ∞, u = b at the branch point B, u = j, j ′ at the end of the two diverging streams where φ = −∞; while ψ = 0 along the stream line which divides at B and passes through A, A′; and ψ = m, −m′ along the outside boundaries, so that m/Q, m′/Q is the final breadth of the jets, and (m + m′)/Q is the initial breadth, c₁ of the impinging stream. Then

ch 1/2Ω = ${\sqrt {}}$	b − a′	${\sqrt {}}$	u − b	, sh 1/2Ω = ${\sqrt {}}$	b − a	${\sqrt {}}$	u − a′	,
	a − a′		u − b		a − a′		u − b

(3)

ch Ω =	2b − a − a′	−	N	,
	a − a′		u − b

sh Ω = √ N	√ (2·a − u·u − a′)	,
	u − b

N = 2	a − b·b − a′	.
	a − a′

(4)

Along a jet surface, q = Q, and

ch Ω = cos θ = cos α − 1/2sin² α(a − a′) / (u − b),

(5)

if θ = α at the source x of the jet xB, where u = ∞; and supposing θ = β, β′ at the end of the streams where u = j, j ′,

u − b	=	1/2sin2 α	,	u − j	1/2sin2 α	cosθ − cosβ	,
a − a′		cos α − cos θ		a − a′		(cos α − cos β) (cos α − cos θ)

u − j ′	= 1/2sin2 α	cos θ − cos β′	;
a − a′		(cosα − cos β′) (cos α − cosθ)

(6)

and ψ being constant along a stream line

dφ	=	dw	, Q	ds	=	dφ	=	dw		du	,
du		du		dθ		dθ		du		dθ

πQ		ds	=	π		ds	=	(cos α − cos β) (cos α − cos β′) sin θ	,
m + m′		dθ		c		dθ		(cos α − cos θ) (cos θ − cos β) (cos θ − cos α′)

=	sin θ	+	cos α − cos β′	·	sin θ
	cos α − cos θ		cos β − cos β′		cos θ − cos β

cos α − cos β	·	sin θ	,
cos β − cos β′		cos θ − cos β′

(7)

giving the intrinsic, equation of the surface of a jet, with proper attention to the sign.

From A to B, a > u > b, θ = 0,

ch Ω = ch log	Q	= cos α − 1/2 sin2 α	a − a′
	q		a − b

sh Ω = sh log	Q	=	√ (a − u·u − a′)	sinα
	q		u − b

Q	=	(u − b) cos α − 1/2 (a − a′) sin2 α + √ (a − u·u − a′) sin α
q		u − b

(8)

Q	ds	= Q	ds		dφ	= −	Q		dw
	du		dφ		du		q		du

=	m + m′	·	(u − b) cos α − 1/2 (a − a′) sin2 α + √ (a − u·u − a′) sin α
	π		j − u·u − j ′

(9)

π	AB	= $\int _{b}^{a}$	(2b − a − a′) (u − b) − 2(a − b) (b − a′) + 2√ (a − b·b − a′·a − u·u − a′)	du,
	c		a − a′·j − u·u − j ′

(10)

with a similar expression for BA′.

The motion of a jet impinging on an infinite barrier is obtained by putting j = a, j ′ = a′; duplicated on the other side of the barrier, the motion reversed will represent the direct collision of two jets of unequal breadth and equal velocity. When the barrier is small compared with the jet, α = β = β′, and G. Kirchhoff’s solution is obtained of a barrier placed obliquely in an infinite stream.

Two corners B₁ and B₂ in the wall xA, with a′ = −∞, and n = 1, will give the solution, by duplication, of a jet issuing by a reentrant mouthpiece placed symmetrically in the end wall of the channel; or else of the channel blocked partially by a diaphragm across the middle, with edges turned back symmetrically, problems discussed by J. H. Michell, A. E. H. Love and M. Réthy.