Question

9. Water at a pressure of 4.50 atm at street level flows into an office building at a speed of 0.80 m/s through a pipe 6.00 cm in diameter. The pipes taper down to 3.00 cm in diameter by the top floor, 27.0 m above. Calculate the water pressure in such a pipe on the top floor.

Answer 1

from equation of continuity

volume flow rate remains same in a fluid flow

A1*v1 = A2*v2

A1 = pI*r1^2

A2 = pi*r2^2

r1 = radius of larger pipe = 6/2 = 3 cm = 0.03 m

r2 = radius of smaller pipe = 3/2 = 1.5 cm = 0.015 m

v1 = 0.8 m/s

v2 = ?

pi*0.03^2*0.8 = pi*0.015^2*v2

v2 = 3.2 m/s

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from Bernoullis principle

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

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

h1 = 0

h2 = 27m

P1 = 4.5 atm = 4.5*10^5 Pa

P2 = 4.5*10^5 + (1/2)*1000*(0.8^2-3.2^2) + 1000*9.8*(0-27)

P2 = 1.806*10^5 Pa <<<<<----------ANSWER