Question

Water flows through a pipe that gradually descends from a height of 6.78 m to the ground. Near the top, the cross-sectional area is 0.4 m2, and the pipe gradually widens so that its area near the ground is 0.8 m2. Water leaves the pipe at a speed of 16.8 m/s. What is the difference in the water pressure between the top and bottom of the pipe? In Pa

Answer 1

let h1 = 6.78 m
h2 = 0 m
A1 = 0.4 m^2
A2 = 0.8 m^2
let v1 and v2 are the speed of the water at h1 and h2 repsectively.
v2 = 16.8 m/s
use continuty equation,

A1*v1 = A2*v2

v1 = A2*v2/A1

= 0.8*16.8/0.4

= 33.6 m/s

now use Bernoulli's equation,

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

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

= 1000*9.8*(0 - 6.78) + (1/2)*1000*(16.8^2 - 33.6^2)

= -4.90*10^5 pa <<<<<<<-------------Answer

or

P2 - P1 = 4.90*10^5 pa <<<<<<<<----------Answer