Question

Water at a pressure of 3.30 atm at street level flows into an office building at a speed of 0.80 m/s through a pipe 6.40 cm in diameter. The pipes taper down to 2.80 cm in diameter by the top floor, 30.0 m above. Calculate the water pressure in such a pipe on the top floor.

Answer 1

from equation of continuity

volume flow rate is constant

A1*v1 = A2*v2

v2 = (A1/A2)*v1

A1 = pi*r1^2 pi*d1^2/4

A2 = pi*r2^2 = pi*d2^2/4

v2 = speed of water at the top floor

v1 = speed of water at the street level

v2 = (pi*d1^2/(pi*d2^^2))*v1

v2 = (6.4/2.8)^2*0.8 = 4.18 m/s

from Bernoulli's principle

P1 + (1/2)*rho*v1^2 + rho*g*h1 = P2 + (1/2)*rho*v2^2 + rho*g*h2

P1 = 3.3 atm = 3.3*10^5 Pa

P2 = ?

h1 = 0 m

h2 = 30.0 m

rho = density of water = 1000 kg / m^3

P2 = P1 + (1/2)*rho*( (v1 ^2 - v2^22 ) + rho*g*(h1-h2)

P2 = (3.3*1.013*10^5) + (1/2)*1000*( 0.8^2 - 4.18^2) + 1000*(0-30)

P2 = 2.96*10^5 pa

P2 = (2.96*10^5)/(1.013*10^5)

P2 = 2.92 atm