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

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

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