Question

1. Design a suitable section for the overflow portion of a concrete gravity dam having the downstream face sloping at a slope of 0.7 H: 1 V. The design discharge for the spillway is 8,000 mº/s. The height of the spillway crest is kept at RL 204.0 m. The average river bed level at the site is 100.0 m. The spillway length consists of 6 spans having a clear width of 10 m each. Thickness of each pier may be taken to be 2.5 m. Take the coefficient of discharge may be assumed to be 2.2 2. Design a trapezoidal channel to pass a discharge of 10m3/s and the channel will be excavated through rock by blasting. The topography in the area is such that a bottom slope of 1 in 4000 will be suitable. Assume that n = 0.030 and select a value for the side slope, Z, as 1 horizontal to 4 verticals. Sketch the diagram for all calculated dimensions.

Answer 1

ANSWER :
Now i since the given spillway looks like a high welry the coefficient of discharge given 2.2 3 Q = Cole He where Le = L-2[ NK ptka He let us first work out the approximate value of He for a value of le=L- dear won ter way .6x 10 = 6om 1.8000 = 2.2x 60 H e 3 13/2 He 3/2 60-6 8000 2.2 x 60 He = (60-6143 -1505m. The height of spillway above the rives bed =h= 204-100 =104m Since h loy > 1033 Hd 15.5 It is a high spillway, the effect of velocity head can, therefore 1 be neglected Heth 15-5+104 since Hard >173 He 15.5 He The discharge coefficient B not affected boj boal water conditions and spill way remains or remains or highspilleay pels slope: The upstream face of the dam and spilway es proposed to

on the be kept vertical However, a better of 1310 will be provided from stability considerations in the lower part This better is small and will not have coefficient of discharge now be worked che) can Effective length of spillway out any effect as le=L-2 [Nike+ ka te Assuming 90° cut water nose piers and rounded abutments shall be provided we have kp=0001 ka=0.1 No. of pierse N=5 Also assume that the actual value of He is slightly more than the approximate value worked live, to 15.5m), say let it be 16.3m, we have le = 60-2 [ 5x0:01 +0.1]X 16.3 = 55-\m 3/2, 8000 = 2,2x55.1 x He 3/2 Hence a 2: 2x 55.18 He 8000 He 3/2 66 2,2 x 55.) He = (66) 2/3 - 16.4m 2 16.3 (assumed). Hence the assumed He for calculating te is all right

8000 72.5 X 12014 velocity head = Ha vah 10.91772 -0.043m. 2X 9,81 and dy The crest profile will be designated for Hy E16-4 Cheglecting relocity head) slote: The velocity head (Ha) can also be calculated velocity of approach = va = 8000 (60 +5X2.5)(104+16-4) = 0.917 mig. 29 This is very small and war, therefore neglected Downstream profile: The w.ersals profile for a vertical w/s face, is given by .y = Hence dy de 0.35 - у 1.85 -2.Hd 1.85 1.85 y = x ACHA )0185 2x (+6yjor85 1.85 & 21.6 Before we determine the various coordinates of the dls. profple, we shall first determine the tangent point. The dls slope of the damps to be oolH: IV given to ㅗ 6.7 ] 1.85-1 1852 21.6 Di

0085 2 21.6 1.85X0-74 :16. x = 22.4m Ye (22.4) 1.85 14.6m 21:6 The uls profile may be designated as per equation Y 0.724(x +0,27 Hd) 135 0.375 +0.126Hd -0.4315 Hd .X Hd 0.85 (2+0,2782 10:625 16.4m using Hd= eget 0.724 [e+ 0.27X16-4) 1.95 +0.126 x 16-4, (16.4)0085 X (2+0.27816.4)00625 -0.4315 (16.4)09375 y=0007 (x + 4.44) 685 +2-07-1.234 (2+4.48270-625 should go upto X = -0.27 Hd . or) X=-0.27 x 16.4 = -4°43m. This coure RiL 204.0 t (origin) C 2.01m x tangent pt (2214,14,8). -0.5 4.44800 011 0.02 0.063 0:27 0-65 Kaxis of -20 I spillway -3.0 - 4.0 1.34 -4.443 2007