Question

: A 23.62 inch pipe and 1968.50 ft long has a head loss of 47.24 inch. Find the discharge flowing in the pipe using f = 0.012, the discharge flowing in the pipe using n = 0.014, and the discharge flowing in the pipe using C = 120

Answer 1

= 0.012 Note : 1ft = 12 inch Manning's constant 1 Given data : For a pipe : diameter = 23.62 inch - d-] = 1.968 ft Length = 1968.50 ft - (1968.50 x 12) inch - L head loss - 47.24 inch - H = 3.936 Ft friction factor=f Hazen's Constant = C = 120 gravity - 386.08 ine/sec Solution : @ Discharge using d, L, H, & f H, f.Liv? 0.012 * (1968:50)x12 XV => 47.24 = 29. 2X386.08x 23.62 8615 82.92 => r? 283.46 V = 55.13 inch/sec x 23.62? - A A= 438.17 inch? = 3.04 ft? Using continuity equation : 9 = A.V Q - 438.17 x 55.13 = 24156 70 inch/sec = 13.98 ft /sec 2 Now, area 0 f pipe = I xd 2 TT ** 9 •> Hydraulic radius = A P A

(В a 1.49 А A Tel n 3.04 TX 1.968 t 3 /sec p-> Welted Perimeter of pipe= id ® Discharge using a lie, manning's formula): V-K (9.)%3.52 [P+ wetted perimeter] / [:K= 1.49 in British units] 1.49 313 3.936 V2 3.04 0.014 1968.5 [ put all the values in feet] 9 - 106.42 0.623 x 0.044X3.04 9 = 9 ft/sec 9 X123 inch 9 = 15573 inch sec © Discharge using c (ie, Hazen- Williams eg') V=K.CR063 90.54 a A K.A.C πd 13.04 TT X1.9 68 g = 4.52 x 0.64 x 0.035 X120 - 12.149 ft/sec 9 = 12.15 x 12 inchº/sec 20994.78 inch/sec =k.c(4)°*? (, 30-54 S = 0.63 3.936 O 54 X120 1968.5 3 с 9 he/sec