Question

Water (density = 1.00 ? 10 3 kg/m 3 ) flows at 10.0 m/s through a pipe with radius 0.030 m. The pipe goes up to the second floor of the building, 2.0 m higher, and thepressure remains unchanged. What is the radius of the pipe on the second floor?

Answer 1

By Bernoullis equation P/ρ + v²/2 + gh = constant for steady flow

therefore P₁/ρ + v₁²/2 + gh₁=P₂/ρ + v₂²/2 + gh₂

where 1 stands for bottom and 2 for top

given P₁= P₂

therefore v₁²-v₂²= 2gΔh

v₂²= 10²- 2x9.8x2

v₂= 7.8 m/s

Now by conservation of mass,ρAv = constant

A₁v₁=A₂v₂

πr₁²x 10 =πr₂²x 7.8

0.03²x 10 = r₂²x 7.8

r₂= 0.034 m

Answer 2

ussing the kinematic equation,

v^2 = u^2 + 2gh ( where v= speed of water at second floor),( u =speed of lighht in first floor_)

v^2 = 10^2 - 2(9.8)(2)

v^2 = 100 - 39.2

v= 7.797 m/s apprx

using the equation of conmtinity

A₁V₁= A₂V₂

π( 0.030)^2 ( 10 ) =πr²( 7.797)

r = 0.034 m apprx---------radius at second loor