Question

A liquid of density 1.19 × 103 kg/m3 flows steadily through a pipe of varying diameter and height. At location 1 along the pipe the flow speed is 9.79 m/s and the pipe diameter is 10.7 cm. At location 2 the pipe diameter is 14.1 cm. At location 1 the pipe is 8.75 m higher than it is at location 2. Ignoring viscosity, calculate the difference between the fluid pressure at location 2 and the fluid pressure at location 1.

Answer 1

= density of liquid = 1190

V₁ = speed at location 1 = 9.79 m/s

A₁ = area at location 1 = (0.25) d₁² = (0.25) (3.14) (0.107)²

V₂ = speed at location 2 = ?

A₂ = area at location 2 = (0.25) d₂² = (0.25) (3.14) (0.141)²

using equation of continuity

A₁ V₁ = A₂ V₂

(0.25) (3.14) (0.107)² (9.79) = (0.25) (3.14) (0.141)² V₂

V₂ = 5.64 m/s

using bernoulli's theorem

P₁ + (0.5) V₁² + gh₁ = P₂ + (0.5) V₂² + gh₂

P₂ - P₁ = (0.5) V₁² + gh₁ - (0.5) V₂² - gh₂

P₂ - P₁ = (0.5) (1190) (9.79)² + (1190) (9.8) (8.75) - (0.5) (1190) (5.64)² - (1190) (9.8) (0)

P₂ - P₁ = 1.4 x 10⁵ Pa