Question

Water is being ejected at the rate 0.2 mil min from a live hydrant with a diameter of 0.15m at the inlet (flange). The diameter redures to 0.02 m at the outlet. The inlet pressure is 200 kPa (absolute) and the exit pressure is atmospheric. Neglect the weight of the pipe and fluid and assume a steady-state flow. Determine (a) inlet and outlet velocities and (b) the torque acting on the flange. Assume g=9.8 mish, Pwater = 980 kg/m², Patm = 101.32 kPa 10m ₂ L treat-

Answer 1

Given that

the rate 0.2 m^3/min

diameter of 0.15 m

the diameter reduces to 0.02 m

inlet pressure is 200 KPa

assume g=9.8 m/s^2

pwater=980 kg/m^3

patm=101.32 KPa

Answen Tom iz < a - - 0,= 0.157 12 -0.02m Pi + 200 spa 1 0-0.2mimin (9) Aiv (022) = n/y (0.155v, TV, 0.18863915 a A2 V2 e 2) - (944)10. 073 v2 V2 = 10.6103 mis. (b) Apply momentum conservation ecuation a long in I x-anis pini - txt P2 A2. - SO (N2-V1) +161 32% IDXT/4 (0.62) - 200 *108114 015) . (015) - =980x (0-2) (10.6103-015563]

f :3502.02N Parault=fR = f (1/2) I shox 20-20SE Wins9.292 •