Question

An incompressible fluid is flowing through a vertical pipe with a constriction. The wide section is 2.00 cm in diameter and is at the top of the pipe. The pressure of the fluid in the wide section at the top is 200 kPa. The velocity of the fluid in the wide section is 4.00 m/s. The narrow section is located 4.00 m below the wide section. What is the diameter of the narrow section for the pressure of the fluid in the narrow section to equal the pressure in the wide section? (density of the fluid is 1,000 kg/m³)

Answer 1

Bernoulli's principle says,for a streamline,

1/2pv²+pgh+P= constant, where p=density of fluid,v= velocity of fluid,g is acceleration due to gravity,h is height of the fluid,P= pressure

Since, the fluid is incompressible, its density is constant. Let the density of fluid be d kg/m³.

For the wide section, v=4 m/s, height be h meters, given pressure=200 kPa=200000 Pa

For the narrow section, let velocity be v₂ m/s, height =(h-4) meters as narrow section is 4 meters below wide section, pressure=pressure at wide section=200 kPa=200000 Pa

So,using bernoulli's principle:1/2d*4²+dgh+200000=1/2dv₂²+dg(h-4)+200000=>1/2*4²+4g=1/2v₂²=>8+4g=1/2v₂²

=>v₂=(16+8g)^1/2 m/s=(16+8*9.8)^1/2=9.716 m/s

By continuity equation:A*v=constant,where A is cross sectional area, v is velocity

Let,area of wide section be A₁ m² and that of narrow section be A₂ m²

So,A₁(4)=A₂(9.716)=>A₁/A₂=9.716/4

The cross section is circular, so, Area=r² where r is radius of the cross section

=(d/2)² where d is diameter

=/4 d²

Diameter of wider cross section=2 cm, Let diameter of narrower cross section be D cm.

Do,/4 (2)²/[/4 D²]=9.716/4=>2/D=(9.716/4)^1/2=>D=2(4/9.716 )^1/2=1.283 cm.