At one point in a pipeline, the water speed is 1.8 m/s, and the gauge pressure is 51500 pa. find the gauge pressure (in pa) at a second point in the line 11.5 m lower than the first if the pipe diameter at the second point is twice that at the first. use a density of water 1000 kg/m3.

pressure in the second point is P₂ = 165833.75 Pa

Explanation:

given,

water speed = 1.8 m/s

gauge pressure(P₁) = 51500 Pa

gauge pressure at another point (P₂) = ?

distance of the point at 11.5 m

density of the water = 1000 kg/m³

diameter of the second pipe  is double than first.

A₁ V₁ = A₂ V₂

π r² ×  1.8 = π × (2r)² ×  V₂

V₂ = 0.45 m/s

now, gauge pressure at P₂

P₂ = 165833.75 Pa

pressure in the second point is P₂ = 165833.75 Pa

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

inverse v=i*r

nice nice