Question

1. Water flows in a rectangular channel in which the width changes smoothly from b, to b2 = k*b, with no change in bed level. Prove that the force on the side wall transition in the direction of flow is equal to 1+kr where r=yz/y1.

Answer 1

ANSWER:

Warer flows for a rectangular channel in which the width Chonger swithly from be to be = kabe, with no change in bed level. The force on the side wall transition in the direction of How is equal to 1/2 l be ye (1-f ² - 360(1-7)) I+KT where r-Yolgo on Flow = 9 derrying ibalence) Energy beulamde ekt O and @ Yint of vine Yz+Az + vino Area of Section 6 = Bxy, Area of section = BX Y Z Where Be width Pur x, and va value in equation o You (eing = 7a+ 4z + ) Yet = Yat D2+ er et Yint egy - 12 42-= - Vi -Y2 – 14)»ge @ 2

I low Applyin low Applying momportum equatform (Pa+M1) – (P + Me) - F - Po= (presiune force at © = Y YX BY Per pressure foole at @= r Y x BY2 Mes Momentum force at 0 = P 932 Ma: Momentum foru at @ P03 Putting Hi & me values in experten TB (y?- Ys?) - pohled $) -F Now putting value of or from eq@ PB (v?-Y2) + P 298 (71-92-4a) y Yz-Y = F (+10) * " YY MP-43 ** . Y, 2.42 (:-eg) YA [ 143-9 + 90 manat). 38-]F 14+ 739844) + (4 %+412 -4,658) yta) -F 18+Y2] 1 9 1 - x^+xy - v2144982-44,5-439280] =+ (Y1+12) 78 (y2 - Y+ 3%, %%2 – 34, 8z? - 4 YiY2A2)F 1 YI+Y2 as we know , (9-b} = q3_63+ 3a²b-3ab² TB [W-Ya) 3- 4 V1 V242 7-F 2 1 9 +42 -

Taking y, common ros (83-4 mm, hk) er o - Yi+Y2 ,94 and Y.Y242 . Ya x Aty farting these values in eq 0 By (10892 473").r slow force per unit width Flo -378*(1-67,387(19)