Question

p=50 hPa 1 В 20 cm → B 1 2 20 W Angular Momentum Water is flowing through 2 10-cm-diameter pipe. into the atmosphere Determine the torque to anchor the attachment at A. A c Q = 150l/s P = 1000 kg /m²

Answer 1

In this problem we will apply impulse momentum equation on point B and the exit of pipe . Which will help us in finding the reaction exerted by fluid over the attachment A holding the bend. The torque is calculated about the centre of cuvature of bend'O'.

Soin PR = 50 200m o VB B 120cm 0,20 AR Given, Diameter of pipe = 10cm C 0,=90 = 150LIS Applying Continuity &a" a pe lovo kg/m2 Q = VA, 150x103 m3/s = vx = V x 7 x 10.1² 19.098 - 19.1 m/s TX 10.12 Now applying impulse momentum equ in x & y direction and finding Reaction forces in x &y direction | R & Ry) Vo= 150х4 x 10² - PB AB 6650 - Patmi AA Cos. 30°. x-dirn P, A, CASO, -P₂ Ae Cos O - Rx = salvos Oz - v, oso) In y-dion P, A, sino, & P2A, Sin O₂ thy = 89 [ Ve Sino y sing] For above PE AC CO - Pa An Cos goP - R = low x 15x16? [ Karasgo'r va o'] O XA (0.1)? -Rx - 150x 19.1 50XT Rx 2865.39 N.

In y direction: PBAB Sino: + PAAA singo" + Ry - 150 x 10 x 10²x [-va sin go-versing) → 50X10:12 Sino + 1x 10.02 x 1 + y = Ry A (0.112 - 150 x 19.1 - 150x 19.1 =-2865.00 78 N O Torque about 0 on boint A = Rx (0.2) + Ry (0.2) 2865.39x10-2) + 2865x02 = 1146.078 Nm 1.146 km Net torque 1.146 kNm A