Question

A liquid of density 1230 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.31 m/s and the pipe diameter d1 is 12.9 cm . At Location 2, the pipe diameter d2 is 17.7 cm . At Location 1, the pipe is Δ y = 8.09 m higher than it is at Location 2. Ignoring viscosity, calculate the difference Δ P between the fluid pressure at Location 2 and the fluid pressure at Location 1.

Answer 1

given
rho = 1230 kg/m^3
v1 = 9.31 m/s
d1 = 12.9 cm
d2 = 17.7 cm
y1 - y2 = 8.09 m

delta_P = ?

let v2 is the speed of fluid at location 2.

use continuity equation, A2*v2 = A1*v1

(pi*d2^2/4)*v2 = (pi*d1^2/4)*v1

v2 = (d1/d2)^2*v1

= (12.9/17.7)^2*9.31

= 4.94 m/s

now use, Bernoulli's equation,

rho*g*y1 + P1 + (1/2)*rho*v1^2 = rho*g*y2 + P2 + (1/2)*rho*v2^2

P2 - P1 = rho*g*(y1 - y2) + (1/2)*rho*(v1^2 - v2^2)

= 1230*9.8*8.09 + (1/2)*1230*(9.31^2 + 4.94^2)

= 1.66*10^5 Pa <<<<<<<<<-------------------------Answer