Page:EB1911 - Volume 14.djvu/140

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128
HYDROMECHANICS
[HYDRODYNAMICS


direction θ along a plane boundary, and to give a constant skin velocity over the surface of a jet, where the pressure is constant.

It is convenient to introduce the function

Ω = log ζ = log (Q/q) + θi
(4)
Fig. 4.

so that the polygon representing Ω conformally has a boundary given by straight lines parallel to the coordinate axes; and then to determine Ω and w as functions of a variable u (not to be confused with the velocity component of q), such that in the conformal representation the boundary of the Ω and w polygon is made to coincide with the real axis of u.

It will be sufficient to give a few illustrations.

Consider the motion where the liquid is coming from an infinite distance between two parallel walls at a distance xx′ (fig. 4), and issues in a jet between two edges A and A′; the wall xA being bent at a corner B, with the external angle β = ⁠1/2⁠π/n.

The theory of conformal representation shows that the motion is given by

ζ = [ √ (b − a′·u − a) + √(b − a·u − a′) ] 1/n , u = ae−πw/m;
√ (a − a′·u − b)  
(5)

where u = a, a′ at the edge A, A′; u = b at a corner B; u = 0 across xx′ where φ = ∞; and u = ∞, φ = ∞ across the end JJ′ of the jet, bounded by the curved lines APJ, A′P′J′, over which the skin velocity is Q. The stream lines xBAJ, xA′J′ are given by ψ = 0, m; so that if c denotes the ultimate breadth JJ′ of the jet, where the velocity may be supposed uniform and equal to the skin velocity Q,

m = Qc,   c = m/Q.

If there are more B corners than one, either on xA or x′A′, the expression for ζ is the product of corresponding factors, such as in (5).

Restricting the attention to a single corner B,

ζn = ( Q ) n (cos nθ + i sin nθ) = √ (b − a′·u − a) + √ (b − a·u − a′) ,
q   √ (a − a′·u − b)
(6)
ch nω = ch log ( Q ) n cos nθ + i sh log ( Q ) n sin nθ
q   q  
= ⁠1/2⁠(ζn + ζ−n) = √ b − a′ √ u − a ,
a − a′ u − b
(7)
sh nΩ = sh log ( Q ) cos nθ + i ch log ( Q ) n sin nθ
q q  
= ⁠1/2⁠(ζn + ζ−n) = √ b − a √ u − a′ ,
a − a′ u − b
(8)


∞ > a > b > 0 > a′ > −∞
(9)

and then

dΩ = − 1   √ (b − a′·b − a′) , dw = − m ,
du 2n (u − b) √ (a − a·u − a′) du πu
(10)

the formulas by which the conformal representation is obtained.

For the Ω polygon has a right angle at u = a, a′, and a zero angle at u = b, where θ changes from 0 to ⁠1/2⁠π/n and Ω increases by ⁠1/2⁠iπ/n; so that

dΩ = A , where A = √ (b − a·b − a′) .
du (u − b) √ (u − a·u − a′) 2n
(11)

And the w polygon has a zero angle at u = 0, ∞, where ψ changes from 0 to m and back again, so that w changes by im, and

dw = B , where B = − m .
du u π
(12)

Along the stream line xBAPJ,

ψ = 0,   u = ae−πφ/m;
(13)

and over the jet surface JPA, where the skin velocity is Q,

dφ = −q = −Q,   u = aeπsQ/m = aeπs/c,
ds
(14)

denoting the arc AP by s, starting at u = a;

ch nΩ = cos nθ = √ b − a′ √ u − a ,
a − a′ u − b
(15)
sh nΩ = i sin nθ = i √ a − b √ u − a′ ,
a − a′ u − b
(16)
∞ > u = aeπs/c > a,
(17)

and this gives the intrinsic equation of the jet, and then the radius of curvature

ρ = − ds = 1   dφ = i   dw = i   dw / dΩ
dθ Q dθ Q dΩ Q du du
= c · u − b   √ (u − a·u − a′) ,
π u √ (a − b·b − a′)
(18)

not requiring the integration of (11) and (12)

If θ = α across the end JJ′ of the jet, where u = ∞, q = Q,

ch nΩ = cos nα = √ b − a′ , sh nΩ = i sin nα= i √ a − b ,
a − a′ a − a′
(19)

Then

cos 2nα − cos 2nθ = 2 a − b·b − a′ = ⁠1/2⁠sin2 2nα a − a′
a − a′·u − b u − b
sin 2nθ = 2 √ (a − b.b − a′) √ (u − a·u − b′)
a − a′·u − b
(20)
= sin 2nα √ (a − a·b − a′) ;
u − b
2n   c ( 1 + b ) √ (a − b·b − a′)
φ ρ u − b √ (u − a·u − a′)
(21)
= a − a′ + (a + a′) cos 2nα − [ a + a′ + (a − a′) cos 2nα ] cos 2nθ × cos 2nα − cos 2nθ .
(a − a′) sin2 2nα sin 2nθ

Along the wall AB, cos nθ = 0, sin nθ = 1,

a > u > b,
(22)
ch nΩ = i sh log ( Q ) n = i √ b − a′ √ a − u ,
q   a − a′ u − b
(23)
sh nΩ = i ch log ( Q ) n = i √ a − b √ u − a′ ,
q   a − a′ u − b
(24)
ds = ds   dφ = m = c   Q
du dφ dt πqu π qu
(25)
π AB = ∫ab Q   du ∫ [ √ (a − b) √ (u − a′) + √ (b − a′) √ (a − u) ] 1/n   du .
c q u √ (a − a′) √ (u − b′)   u
(26)

Along the wall Bx, cos nθ = 1, sin nθ = 0,

b > u > 0
(27)
ch nΩ = ch log ( Q ) n = √ b − a′ √ a − u ,
q   a − a′ b − u
(28)
sh nΩ = sh log ( Q ) n = √ a − b √ u − a′ .
q   a − a′ b − u
(29)

At x where φ = ∞, u = 0, and q = q0,

( Q ) n = √ b − a′ √ a + √ a − b √ −a′ .
q0   a − a′ b a − a′ q
(30)

In crossing to the line of flow x′A′P′J′, ψ changes from 0 to m, so that with q = Q across JJ′, while across xx′ the velocity is q0, so that

m = q0·xx′ = Q·JJ′
(31)
JJ′ = q0 [ √ b − a′ √ a − √ a − b √ −a′ ] 1/n ,
xx′ Q a − a′ b a − a′ b  
(32)

giving the contraction of the jet compared with the initial breadth of the stream.

Along the line of flow x′A′P′J′, ψ = m, u = a′e−πφ/m, and from x′ to A′, cos nθ = 1, sin nθ = 0,

ch nΩ = ch log ( Q ) n = √ b − a′ √ a − u ,
q   a − a′ b − u
(33)
sh nΩ = sh log ( Q ) n = √ a − b √ u − a′ .
q   a − a′ b − u
(34)
0 > u > a′.
(35)

Along the jet surface A′J′, q = Q,

ch nΩ = cos nθ = √ b − a′ √ a − u ,
a − a′ b − u
(36)
sh nΩ = i sin nθ = i √ a − b √ u − a′ .
a − a′ b − u
(37)
a′ > u = a′eπ/sc > −∞,
(38)

giving the intrinsic equation.

41. The first problem of this kind, worked out by H. v. Helmholtz, of the efflux of a jet between two edges A and A1 in an infinite wall, is obtained by the symmetrical duplication of the above, with n = 1, b = 0, a′ = −∞, as in fig. 5,

ch Ω = √ u − a , sh Ω = √ − a ;
u u
(1)

and along the jet APJ, ∞ > u = aeπs/c > a,

sh Ω = i sin θ − i √ a = ie−1/2 πs/c,
u
(2)
PM = ∫∞s sinθ ds = ∫ e−⁠1/2⁠πs/c ds = c e−1/2 πs/c = c sin θ,
⁠1/2⁠π ⁠1/2⁠π
(3)
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