Question

2. Water flow is measured by a 4 ft wide rectangular weir. The weir crest is 2.5 ft above the channel basement. The minimum and maximum water depths in the weir are measured as 0.4 ft and 2 ft, respectively. Calculate the minimum and maximum flowrates for a contracted rectangular weir with the length of 2 ft. (Note: Consider the relation between h and b/3, and choose appropriate equation to calculate the flowrate). B-4 ft b=2 ft Th=0.4 and 2 f+ P=2.5 ft

Please help me to solve these two questions. Thank you.

Answer 1

Please refer the following images for the solution.

2) Given: b=2 ft hmin = 0.4ft bmon=2ft p=2.5ft B = 4 ft solution : Minimum flowrate of water Qmin = 2 Cd b ( eg hmin 13/2 3/2 Qmain - 70.5715037 2812x32.174 x 0.4 where, c d = coefficient of discharge of rectangular weir CJ = 0.571503 (from data table) Smin = 1:546374 67% Maximum flowpate of water, 3/2 Qman: 3 cd b / 2g Iman Qman - 3 2 x - 2 x 0.571503x2x2x32.174 x 2 Qman = 17.288995 ft /

3) Given. dp = 304.8 mm = 0.3048 m do = 238.1 mm = 0.2381m Py = 2.41 bara AP=0.032 157 bar = 3215.7 N/m2 Solution. Discharge through the pipe, Q = Cd Ao 2AP V SCI-dor काम where, Cd = 0.6 for orifice plate Ix0.23812 2 Ao = I do 0.044525 m² 4 O. 6 x 0.044525x 2x3215.7 11 1-344250x(1- V 0.23819 0.30484 2. 9 = 2.332474 m² : Mass flow rate of steam, m = sx = 1.344250 x 2.332474 ..m= 3.135429 kg/s