Problems on trapezoidal channels Question A trapezoidal channel is to be designed to carry a discharge of 200 m3/sec. The channel slope is 0.0004 and manning coefficient is 0.012. the side slope is 1: 1, if the rate of linings is 100/m2, then the cost for most economical section per km length of channel is (in lakhs). Solution Given, Q=200m3/sec,S=0.0004,n=0.012 side slope = 1:1⇒m=1 For most economical section, R = y/2 and B+2my = 2y√1+m2 B+2y=2y√1+1 B = 0.828y Area of flow, A = (B+my)y =1.828y2 Using manning equation, Q=1nAR2/3S1/2 200=10.012×(1.828y2)(y2)2/3√0.0004 y = 5.71 m ∴ B = 0.828×5.71=4.73m Perimeter of the economical section = B+2y√1+m2 =4.73+2×5.71√2 = 20.88 m Cost per km =20.88×100×1000 = 2088031.9 = 2088032 = 20.88 lakhs...
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(Solved) A trapezoidal channel is to be designed to carry a
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