Question

Please help me thanks.

A horizontal pipe is shown in the figure below. At the inlet (Point 1), the radius of the pipe is 4 cm and the velocity profile is given by: V = 16- y2 cm/s. At the outlet (Point 2), the radius is 2 cm and the velocity changes to a uniform profile, as shown in the figure. If the viscosity of the liquid inside the pipe is 0.07 [Pa s] and its density is 1030 kg/m3, calculate: у v= 16 - y2 2 8 cm V2 - 41 - (i) The shear stress in the inlet at -2cm [5 marks] Sheer stress = (ii) The average outlet velocity V2 , if the average inlet velocity is 7 cm/s. [5 marks] V2 = m/s (iii) The pressure at the outlet P2, if the inlet pressure is 100 Pa [5 marks] P2 = Pa

Answer 1

0 16.gh 21 8cm Given Radiu 8 ob pipe-4cm at inlet velocity profile=v=16-g2 cols at inlet Radius of pipe=2cm at outlet > viscosity of liquid inside=1=.07 Pa-s Density of liquid = 1030 kg/m3 مو حال بد به Shear 8tress at inlet - l=udv 15--2cm [=.07% (-2y) I= .07% (-9% (-9)) T.07X4 - •28 Pa t=.28 Pal Average velocity at inlet giren Vi=7cmis using continuity equation for same fluid AN,= Ag Vg

V 7 x 82 x 7 = 11 x 4² x vg -> 64x7 - 98 cmis 16 V=.28 mis Pressure at inlet given P, = 100 Pa using Bernoulli's equation - P +2= constant 33 2 + >loo Col 2 vi P2 Pi 33 vi + +2,- + + 2q as (2=22) 82 28 100 07 2 .28 11 Po so + 1030kg ag (v=7cmis 2x8 5.07) .07 .28 + lios P2 1030 . 1030 2 2 = 62.1475 Pa