A liquid of density 1370 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.67 m/s and the pipe diameter d 1 is 12.9 cm. At Location 2, the pipe diameter d 2 is 16.7 cm. At Location 1, the pipe is Δ y = 9.85 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.

P₂ - P₁=173.5kPa

Explanation:

The equation of continuity:

A₁v₁=A₂v₂

where A₁=πd₁²/4 and A₂=πd₂²/4

v₂ = (A₁/A₂) v₁

v₂={ (πd₁²/4) / (πd₂²/4) }v₁

v₂ = (d₁²/d₂²) v₁

Use Bernoulli's equation

P₂+pgz₂ + (pv₂²/2) = P₁+pgz₁ + (pv₁²/2)

The difference between the fluid pressure at location 2 and the fluid pressure at location 1

P₂ - P₁=pg (z₁-z₂) + {p (v₁²-v₂²) }/2=pg (z₁-z₂) + 1/2pv₁² (1 - (d₁/d₂) ⁴)

P₂ - P₁ = (1.370*10³*9.8*9.85) + (1/2) (1.370*10³ * (9.67) ²) { (1 - (0.129m/0.167m) ⁴}

P₂ - P₁=1.735*10⁵Pa

P₂ - P₁=173.5kPa