Question

Problem #3 A rectangular channel is 25 ft wide and has a flow of 500 cfs at the velocity of 5 ft/sec a) Determine the specific energy of water in the channel b) What is the critical depth of flow in the channel c) What is the critical velocity

Answer 1

yft 11 solution : Given that rectangular channel is 25 ft wide (6) has a flow of 500 cifs (8) at the velocity of 5 ft/see (W) Assuming normal depth at velocity V = 5ft/see is wift So cross-sectional Area (A) = bxb A= bx25 A=254 ft 2 As we know s= VXA to b= 25ft 4 so that 500 ft /see = 5 ft/see x 250 ft 2 olt, de 500 on = 4 ft. @ Here we have to determine the specific enera of water in the channel. General Equation States: E= y + 7) where = nogmal depth of fluid (ft) 9= 32.2 ft /see? Ve velocity (felsee In our case; E = specific Energy (ift) 30 E= 4+ (592 2x 32.2 TE = 4.388 ft 5*25

of 6 Now we have to find the critical depth flow in the channel. . In general at critical depth (c) dE=0 ce when babe We have E= ut v E= bt 92 Q=NAT IT 2gA2 E=nt ar 29.6²22 E- + Derivating both side with respect to "y" we get 을 위해 한 은 은 2862 dy = 1- 282 2g6253 at th=0; 2=bc so that condition gives us O= 1 8² 962 ,3 ab?p3 I 962

V In our case, Critical depth will be b = / 500 2 % 32.2 x 252 I b = 2.316 ft. ③ we am Find the critical velocity in two ways het us discuss those ways one by one: Way 1: Using Fronde Number (For) At critical depth be; Fronde Number (Fr.) = 1 and we know Fr= vaiba at critical depth & be which implies - 1 = varde and v=vc V= √32:2x2.316 Tue 8.636 ft /see Way 2: Using 8-VXA relation I'm over Case & vxb xa but at critical depth road and Wave therefore = Vex bxe 09, 500- V2 x 25x 2.316 on Ves on = 8.636 ft/see hence 25x2. 316 (Answer)