Question

A liquid is flowing through a horizontal pipe whose radius is 0.0346 m. The pipe bends straight upward through a height of 11.0 m and joins another horizontal pipe whose radius is 0.0625 m. What volume flow rate will keep the pressures in the two horizontal pipes the same?

2 2 2 2

Answer 1

v^2 /2 + gh + p /ρ is a constant
v1^2 /2 + gh1+ p1 /ρ = v2^2 /2 + gh2+ p2 /ρ
Since p1 = p2
v1^2 /2 + gh1 = v2^2 /2 + gh2
v1^2 ─ v2^2 = 2 g (h2 – h1)
v1^2 ─ v2^2 = 2 x 11 x9.8
v1^2 ─ v2^2 = 215.6
v1^2 {1- (A1 /A2)^2 } since A1v1 (= rate of fluid flow = A2v2 as the fluid is incompressible):
(A1/ A2) ^2 = (r1 / r2) ^4 = (0.0346 /0.0625) ^4 = 0.0939
0.9061*v1^2 = 215.6
v1 = 15.43 m/ s
A1 v1 = Л r^2 v1 = Л x 0.0346^2 x 15.43 = 0.058 m^3 /s