Question

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.

Answer 1

pressure in the second point is P₂ = 165833.75 Pa

Step-by-step 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

`P_2 = P_1 + (1)/(2)\rho (v_1^2-v_2^2) + \rho g h`

`P_2 = 51500 + 0.5 * 1000*(1.8^2-0.45^2)+ 1000* 9.81* 11.5`

P₂ = 165833.75 Pa

pressure in the second point is P₂ = 165833.75 Pa