Question

A liquid is flowing through a horizontal pipe whose radius is 0.00752 m. The pipe bends straight upward through a height of 4.40 m and joins another horizontal pipe whose radius is 0.0487 m. What volume flow rate will keep the pressures in the two horizontal pipes the same?

Answer 1

R1 = 0.00752 m
R2 = 0.0487 m

A1 = pi r^2 = 3.14 x 0.00752^2 = 0.0001775 m^2
A2 = pi r^2 = 3.14 x 0.0487^2 = 0.007447 m^2

volume flow rate = A1 V1 = A2 V2
V1 = A2 V2 / A1 = 0.007447 V2 / 0.0001775 = 41.954 V2

Bernoulli equation:
P1 + 1/2 rho V1^2 + rho g h1 = P2 + 1/2 rho V2^2 + rho g h2
P1 = P2
1/2 rho V1^2 + rho g h1 = 1/2 rho V2^2 + rho g h2
1/2 (41.954 V2)^2 + 10 .0 = 1/2 V2^2 + 10. 4.40
25.4599 V2^2 = 0.5 V2^2 + 99
890.069 V2^2 = 44
V2^2 = 0.0494
V2 = 0.2222 m/sec

volume flow rate = A2 V2 = 0.007447 x 0.2222 = 0.03351m^3/sec