Question

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Based from Figure Q1 below, when water flows through trapezoidal channel section, the cross sectional area of the fluid (A) and the wetted perimeter of the fluid (P) are related as: 2yn Yn sin e Estimate the value of O for which P = 3 m using Secant Method, when A = 1 m2 and yn = 0.5 with 10° and 60° as the initial approximation for the e. Compute until the 8th iteration P= Yn cot@ + T Hoy H Figure Q1: Flow through trapezoidal channel

Answer 1

Pg. Noool Criven A that p= In cotot 24. sino yn we have to cakulate o for p=3m when A=1m² and In = o.s. flos = p-3 A - Incoto + 230 yo sino -0.5 cotot 2665) -3 0.5 sino 1-05 coto-1. sine Given Initial approximations are a= 10 and 6=60° f(a) = 0.5cot 10-1 sino 1.923 0.5 cotton -0.1339 son 60 first approximation, a. f(b)-bftas X = f(6)-f(a) to f-0.1339) – 60 (4-423) in W = -0.1339-1.923 1.4 = 56.7459 i du = +(56-745) = - 0.1320 (cod.

Pg. No.02 Hence the nearest roots are 56 745, 600 a= 56.745° b=600. M2 - a f(b)- bf(a) 56.745(-0.1339) 60 (0.1320) 2165-fa -0.1379-(0.1320) e le = - 169.3918 ° (not feasible). as 1 and b = 56.745 If then af (6)- b f(a) lo (0.1339)-- 56.745 (1923) f(6)-Has -0.1339 - 1.923 53.702. . f(x) = - 6.1265 i The new values a: 10°; b= 33.702°. are af (6) - bf (a) 10-01265) – 53702 (1.923) in z f(b)-f(a) - 0.1265- 1.923 x3 = 51.0046 f(x3) -0.118) new values a = 10; b=51.0046 I The one af(b) - bffat 10 (-0.118 - 510046(1.923) x 4 = (6) f(a) -0.1181- 1.923 48.632° in fcmus. 0.1078

Pg. No.03 New values are a=10°; 6 = 48.632 - a f(b).- bit(a) 10 (-0.1078) – 48.632(1923) xo -0.1878 - 1.923 46.5813 0.0964 b= 46.58139 New value are af (6)- bf(a) 10 F008964-46.5813(1.923! f[6)-f(a) -0.0964- 1.923 x = 44.8350° 0846 values are New a=10 b= 44-835° aflo-bflat 101-0.0846) - 44.835(1.923) : xy = -0.0846- 1.923 43.3671 if(x) = -0.0730 New values are a = 10° b= 43.36710 af(b)-bfcal lo100730). 43-3671 (1-9235 f(b)-f(a) 1.923 X8= 42.14680 & f(x) = -0.0622 Hence the value of o for P= 3m after 8th Iteration in sceand Method is Oq = 42.1468° [Ars). Ug = -0.0730