Question

DESCRIPTION T-shaped open channel has two sub-channels more rough surfaces than the main channel. If the longitudinal slope of the channel is 0.0004 and y-2.5m, y=1m, b;=5m, b -8m, n, -0.018m, 12-0.036m. ASKED • find the total discharge flowing in the channel DESIGN • If this channel falls into a rectangular cross-section at the downstream side, and it is planned to have normal flow depth, y=1.5m, calculate the required base width, b. • If there is a 40cm smooth step up at the last part of the channel, what will be de flow depth above the step. SKETCH b b=? yu2=1.5m (b;= b

Answer 1

Solo Given im h1=0.036? - 16,20.035 8m fm 192-00381 S=0.0004 sm calculation of Discharge Area- area = 8x18m? Perimeter = 8+1=9m R = 8/9 2+ (%)"&C.000) "3 28 191=4.1058 m3 /sec) 0.036 area of Izarea al 3 and Call data are same] Q3 24.1088 m3/sec

Area 2 area = 5*(3.57217.5 m² Perimeter = 2.5+5+2.5=10m Q=1 / 17.51% 0.018 Clo J x (0.ooo4) X 17°5 Q2:20.234 molle Discharge = Q,+QztQ3 24.1088+28.237 +4.688 Discharge (q) = 36.45 m3/sec i) Rectangulas Channel - assuming no energy loss and upstream energy-downstream energy and mannings Constant an,20.018 30=0.004,4=1.5m Area=BX 1.5 Perimeter = B+3 Q = + R229A SR7² 36.45 = RO Kts 30us 5 o die Christ J Ro y Wax (150) 120.854 (0+3)*43 = (158)*%

1765184.603 (B+3)2 = (153) 5 35303.69237 +317733.229 +211822.15 B=7.685 -B+4645.22B? + 27871.333 +41807=0 B: 18.uusm) hence I width of rectangular section 18.448m iii) fo r a 12 at et port y at last part 10.4m