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4.1 Design a sarda type fall

Given problem:

Design a srada type fall for the data given below:

Full supply discharge u.s./d.s. = 10 cumecs

Drop = 1 m

Full supply level u.s./d.s. = 101.5/100.5 m

Full supply depth u.s./d.s. = 1.50/1.50 m

ed level u.s./d.s. = 100/!! m

ed width u.s./d.s. = "/" m

ligh#s coefficient = $

Design all protection wor% and estimate the pro&ect.

'afe e(it gradient may be ta%en e)ual as *.

Solution:

4.1.1 Calculation of H and d

'ince the discharge is less than 1+ cumecs, so rectangular crest will be adopted.

'o, discharge is given by

-= 1."5/2/31/4 213

here,

  = length of crest = d.s. bed width = " m

- = 10 cumecs

  = top width of crest = 0."m 2assumed3

6ut the values of -,, in e)213:

  10= 1."57"7/72/0."31/4

  'o, = 20.4543/5 = 0.$" m

8d = D8 drop in level

8d = 1.508 1 = .5 m

d = .5090.$" = 1.$ m

eight of crest above bed 2h3= D9 = 1.590.$" = 0.$ m

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4.1.2 Design of crest

rectangular crest will be adopted.

;op width of crest = 0." mase width of crest = 28d3/< = 20.$"81.$3/ = 1.5 m

'o, velocity of approach 2a3 = discharge/area

  = 10/2"81.5371.5 = 0.$0 m/sec

elocity head = a/g = 0.$0/27!."13 = 0.05 m

'o, u/s ;> = u/s F'8 velocity head = 101.58 0.05 = 101.55 m

?.. of crest = u/s F'9 = 101.59 0.$" = 100.$ m

@se crest level of 100.$ m.

4.1.3 Dsign of cistern

alue of > = u/s ;>9 ?.. of crest = 101.559 100.$ = 0."05 m

'o, depression of cistern = ( = 1/+2>3/ = *20."05713/ = 0.14 m

ength of cistern = lc = 52>3 = 520."057131/ = +.5m

?.. of bed of cistern = ?.. of d/s bed A ( = !!9 0.14 = !".$" m

4.1.4 Design of impervious floor

Ba(. seepage head = crest level A d/s bed level = 100.$9 !! = 1.$ m

'o, total floor length re)uired = C7 = $71.$ = 1.0+ m

Binimum d/s floor length re)uired = 2water depth 8 1.3 8 E

  = 21.581.3 8 1E = 4.+ m

et us provide u/s cutoff d1 = 0."m and d/s cutoff d = 1 m.

'o, total vertical length of creep = 20."813 = .4m

ength of horiontal impervious floor = 1.0+9 .4 = ".++ m

'o, 4.+ m floor length is provide at d/s and balance ".++9 4.+ = .0+m is to be provided under

u/s of the crest.

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4.1. Calculation for uplift pressure and t!ic"ness

;otal creep length = 1+82181.43 = 1!. m

;he uplift pressure under the d/s floor will be counter balanced by the wt. of water and henceno thic%ness is re)uired. owever, provide a minimum thic%ness 0f 0.+ m.

For these points the ma(. vertical ordinate between the ligh#s .G. line and floor level gives

the uplift pressure.

Ba(. unbalance head under the toe of the crest = 1.4 m

;his can also be calculated as under :

'tatic head = 1.$19 2.0+8713/1.0+E8(

  = 1.$190.4E80. = 1.4 m'o, thic%ness re)uired = 1.4/2

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  & $!osla%s e'it gradient curve(

4.1.) Calculation of pressure

*+S cutoff:

d1 = 0."m , / b = 1.0+m

1/L = d1/b = 0."/1.0+ = 0.044+

From curve given below, M> = +N, MD = 1$N

'o, MC1 = 1009+ = $4N

MD1 = 10091$ = "N

ssume thic%ness of floor = 0.+ at the u/s.

'o, corrected MC1