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73
HYDRAULICS

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§ III. Egg-Shaped Channels or Sewers.-In sewers for discharging storm water and house drainage the volume of flow is extremely variable; and there is a great liability for deposits to be left when the How is small, which are not removed during the short periods when the flow is large. The sewer in consequence becomes choked. could be found satisfying the foregoing conditions. To render the problem determinate, let it be remembered that, since for a given discharge Qoo Q/X, other things being the same, the amount of excavation will be least for that channelgwhich has the least wetted perimeter. Let d be the depth and b the bottom width of the channel, and let the sides slope n horizontal to I vertical 1

Di Cafting

l

(fig. 114), then

I 1' ~, . I, i /1 » 1 ~ 1 I -H, Q=(b+nd)d. u-s-14~9- -Q ~a~4<——33~|o—-~»-4<<s'-fo-»x-~—21~4—>, “ 1 1 28 1: 1 | f b x=b-l-2d/ (nz--1). 1 1 1 l l Wy, Both Sl and X are to be minima. 1 ' i i i l 'ww Differentiating, and equating to l' 1 1 I *W , ,, ... h a . Ze;-0 1 ~

ff Sf af 14: (db/dd +n>d +1 +114 = 0, 4455* Ein" 'gag ~s- /" “" " db/dd+2~/ (n2+1)=0; 1s ., —':=, ,1..;, ¢, .Wm

~1n'u1W*§§ })§ M eliminating db /dd, 1 ' ' 1 ' N pi-24 (n2+i>}d+b+nd=0rr b=2{, /<n2+1)-md.

FIG. III.*-SC3lé 20 ft. = 1 in. But S2/X=(b+hd)d/{b+2d/(n2+1)}.

I Inserting the value of b,

-, , . .1 . .. . .:

L ... .,20'.0!;-1: — -1 aa-o»~=m>g+~—— ~ 14-u-o—<~—f Slowking4Farmbrough's revenge»>?1f1+2a-oq<-»- — —120~o1——~~—»§ 1"=9/X=l2d/ ( 2+2I)'~"lll/ 1 2 1 5 = L, . .. 10p:.02— ... 4 5 = ' = Th wig/ (M +1>-§ 11dll=§ d, “ , ,, ,, ,, ,, , e tem ~, - -:.. .— — — —, - — -, -... ~ —1. ~¢ ¢, W-»-»-f-51553, ..-., .», ., . ' 1 ~ ' ~ - -§ (§ ®>°'lll|lllll“' Tf§ @)§§ "; § J=r- - "”l@%3§ W5'f"""*§ $77'?.'@:, § ¥s;» m U} - lMm~' vr at ls, Wlt given 51 e 5 QPeS» " ” ""' '”'“""" — - 4 ”?'>». §§ g~, , 1 ' W —~~ T § <~ % " at ' ' / I' ' the section is least for a given ~f" ~wf;° ' ;/V "

FIG. 112.—Scale 80 ft. =1 in.

To obtain uniform scouring action, the velocity of flow should be constant or nearly so; a complete uniformity of velocity cannot be obtained with any form of section suitable for sewers, but an approximation to uniform velocity is obtained by making the sewers of oval section. Various forms of oval have been suggested, the simplest being one in

which the radius of the

crown is double the radius

4- —~—- 4- —— ~o

discharge when the hydraulic mean depth is half the actual depth. A simple construction gives the form of the channel which fulfils this condition, for it can be shown that when m=§ <l the sides of the channel are tangential to a semicircle drawn on the water line.

I Since SZ/X = ed,

1 therefore S2 = Xd. (I)

Let ABCD be the channel (fig. 115); from E the centre of AD drop perpendiculars EF, EG, EH on the sides. Z E of the invert, and the i Let "“'§ '1" ' greatest width is two- AB=CD=a; BC=b; EF=EH=c; and EG=d. g » thirds the height. The 9=ar@a AEB~l-BEC-l-CED I ' } section of such a sewerl =aC+lbd 1 I ' is shown in fig. II3, the 1 x=2a+2 5 "2~~.> Y numbers marked on the § ' "'l" """°' " Z) /»' figure being proportional Pllttlflg fh€S€ V€1111€S 111 (I), ¥"' 6 I/" Problems on ac-I-%bd= (a+%b)d; and hence c=d. "~ ' ' ~~ hannels in which the T § "~ :;;><;f, Flow is Steady and at ...K .., . D 1 <»" '~, »' Uniform Velocity.-The 5 A 2 " /, /'x general equations given E '. "f7< 1 in §§ 96, 98 are H g, '4"' ' 5 r=<=<1+/S/m>. <1> § e “ ' ' ' tv-/2g =m1: (2) E

FIG- 113- Q=Qp (3) | 1 .-Problem I.-Given the transverse section of stream and dis- B G C charge, to find the slope. From the dimensions of the section FIG 115 Eng fl and m; from (1) find gf, from (3) find v, and lastly from (2) n 1.

Problem II.-Given the transverse section and slope, to find the discharge. Find v from (2), then Q from (5). Problem III.-Gi'en the discharge and slope, and either the breadth, depth, or general form of the section of the channel, to determine its remaining dimensions. This must generally be solved by approximations. A breadth or depth or both are chosen, and the discharge calculated. If this is greater than the given discharge, the dimensions are reduced and the discharge recalculated. Since m lies generally between the limits m=1l and m=§ d, where d is the depth of the st1'eam, and since, moreover, the velocity varies as xl (m) so that an error in the value of m leads only to a much less error in the value of the velocity calculated from it, we may proceed thus. Assume a value for m, and calculate '11 from it. Let v1 be this first approximation to U. Then Q/U1 is a first approximation to Q, say 91. With this value of S2 design the section of the channel; calculate a second value for Ta 'a r

fi 'av 1

1-1 4 =

T y —nd—sg. ...... (, , ., ..

FIG. 114.

trapezoidal in section (fig. 114), and m; calculate from it a second

value of 11, and from that a

second value for Sl. Repeat

the process till the successive values of m approximately

coincide.

§ 113. Problem IV. Most

Economical Form of Channel

for given Side Slopes.-Sup

pose the channel is to be

that the sides are to have a

given slope. Let the longitudinal slope of the stream be given, and also the mean velocity. An

infinite number of channels

That is, EF, EG, EH are all equal, hence a semicircle struck from E with radius equal to the depth of the stream will pass through F and H and be

tangential to the sides of

the channel.

To draw the channel,

describe a semicircle on

a' horizontal line with

radius=depth of channel.

""'" "H+ —» -6 —— >¢

FIG. 116.

The bottom will be a

horizontal tangent of that

| sizmicircle, and the sides tangents drawn at the required side s opes.

The above result may be obtained thus (fig. 116):- X=b-l-2d/sin B. (1)

Sl=d(b-Pd cot /3);

fl/d=b-l-d cot B; (2)

SZ/dz =b/d-I-cot B. (3)

From (I) and (2),

X =9/ol-d cot B-l~2d/sin B.

This will be a minimum for

dx, /dd = SZ/dz-|~cot, B-2/sin /3 =0, or SZ/d” =2 cosec. B-cot B. (4) or d =/ {S2 sin B/(2 -cos /3)}. From (3) and (4),

b/d=2(1 -cos B)/sin /S=2 tan é/3.