Question

A liquid (ρ = 1.65 g/cm3) flows through a horizontal pipe of varying cross section as in the figure below. In the first section, the cross-sectional area is 10.0 cm2, the flow speed is 246 cm/s, and the pressure is 1.20 105 Pa. In the second section, the cross-sectional area is 4.50 cm2. (a) Calculate the smaller section's flow speed. m/s (b) Calculate the smaller section's pressure. Pa

Answer 1

(a)
Assume incompressible flow.
Conservation of mass requires that the volumetric flow rate in both sections is the same. Since volumetric flow is average velocity times cross sectional area:
v₁ · A₁ = v₂ · A₂
=>
v₁= v₂ · A₂/ A₁ = 2.46m/s · 10/4.5 =5.47m/s

(b)
Assume frictionless flow. Then the mechanical energy of the
fluid stay constant and you can apply Bernoulli's equation
p / ρ + g·h + v²/2 = constant
or
p₁ /ρ + g·h₁ + v₁²/2 = p₂ /ρ + g·h₂ + v₂²/2
Because the pipes are horizontal h₁= h₂
p₁ /ρ + v₁²/2 = p₂ /ρ + v₂²/2
=>
p₁= p₂+ ( v₂² - v₁²) · ρ/2
= 1.2·10⁵Pa + ((2.46m/s)² - (5.47m/s)²) · 1650kg/m³ / 2
= 10.03·10⁴Pa