Question

Specific discharge of a wide rectangular channel is 10.0m2/s (Q 10.0m3/s/m). The channel consists of two long reaches having the following bed slope, So1 -0.0001 (for ABC), and So2 = 0.008 for (CDE), (see Figure 1) respectively. There are two gates to control the flow and the flow depth at the vena contractor after each gate is approximately 0.3m. Take the Manning's coefficient as 0.015 Locate every possible flow changing locations and, sketch and name the water surface 1. profiles along the channel. Indicate flow depth at each flow changing location using the notations (dni, dn2, de, dup, daown, d, d2 etc.) wwww Tabulate, flow depths at flow changing locations. If energy loss due to the gate is negligible, 2. plot specific energy variation along the channel (not to scale). Note: it is required either to tabulate or indicate energy values on the graph A B D E Figure 1 for Question 1

Answer 1

Given data Q=10.0 m /s / m So = 0.0001 S02 0.008 vena contractor = 0.3m meaning's coefficient n = 0.015 3 We know that critical depth (Y) of channel is g here g (acceleration due to gravity) 9.81 1023 YC =2.17m 9.81 Now we take first part ABC we calculate normal depth here we know manning equation Q=AV here A B*Y

2 1 and V 3 * n A Hydraulic (R) here A BY and P= B+2Y And we also know that its rectangular channel so 2Y will be negligible i.e 0 so it will be equal to Y BY A R = Y B 2Y Qy()(s) Y 5 10 0.015 О*п Y = 5.1m .......1 VO.0001 As just like equation 1 we now calculate Y, for the 2nd part CDE in a similar way as that 5 10 0.015 V0.008 О*п Y =1.36m .......2

We know that if Y,(ABC)Y it will be mild slop and if Y,(CDE) <Y^ it will be step slope now we calculate conjugate depth in mild slope it will be -Y,*Y,(Y+Y) g 2* 102 0.3*Y2 (0.3+Y,) 9.81 2 0.3*Y,Y 67.96 2 Y, 8.1m 2

After the c point water is rise due to steep channel Now in 2nd part of question (1)we calculate specific energy at A we know V Y Е 3D Y, + 2g 102 = 5.1 + '3D 'я |Е, 5.12 2 9.81 2 * 2g Y A 5.34m A Similarly we calculate all specific energy (2) we calculate specific energy at F 102 0.3 Еr 0.32 2 9.81 Ер 56.93m

(3) we calculate specific energy at H 102 E 8.1 6.122 9.81 8.20m (4) we calculate specific energy at C 3 Ec 2 3 *2.17 2 - C Ec 3.25m (5) we calculate specific energy at G 102 0.3 EG 0.32* 2 9.81 Ec 56.93m

(6) we calculate specific energy at E 102 Ef 1.36 1.362 2* 9.81 Ef 5.11m

M1 8.1m M2 NDL S2 M3 5.1n S3 ACDL 2,1m 0.3m CDL A /c NDL 0.3m D G Mild slope Ye=2.17m control point steep slope Yn1.36m NDL= normal depth level CDL critical depth level LL