Question

An aqueous solution with a specific gravity of 1.12 flows through a channel with a variable cross section. data taken at two axial positions in the channel are shown below
                     Point 1              Point 2
Pgauge:    1.5x10^5 Pa 9.77x10^4 Pa
    u:               5 m/s                    ?
point 2 is 6 meters higher than point 1.
(a) neglecting friction, calculate the velocity at point 2
(b) If the pipe diameter at point 2 is 6cm, what is that diameter at point 1?

Answer 1

pressure at point 2 ( since the pressure is gauge pressure ) = 9.77*10⁴ pa +1.0123*10⁵ pa=198930 Pa

Pressure at point 1= 1.5*105+1.0123*10⁵= 2.5123*105 pa

since point is at 6m height, static pressure = pgh= 1.12*1000*9.8, Hence

Pressure difference between point 2 and point 1= 198930+1.12*9.8*6*100 (pgh) -(2.5123*10⁵)=13556

Flow takes place from point 2 to point 1 and applying Bernouli equation

P₂/p+ V22/2 + gZ2 = P1/p+ gZ1 + V12/2

{(P2+pgZ2)- {P1+pgZ1)}/p + V₂²/2= V1²/2

13556/1.12*1000 + V₂²/2 = 25/2

12.10+V₂²/2= 12.5

V₂²= 2*0.4

V2= 0.9 m/s

from mass flow rate= pVS

p is the density = specific gravity*density of water = 1.12*1000

p2V2S2= p1V1S1

S2,S1 are cross sections at point 2 and 1, V2 and V1 are velocities at 2 and 1

since density does not change

V1S1= V2S2

S1= 0.9*(PI/4)*6²/5

(PI/4)*d12= 0.9*(PI/4)* 36/5

d₁²= 0.9*36/5=

d1= 2.48 cm