Question

# A liquid (ρ = 1.65 g/cm3) flows through a horizontal pipe of varying cross section as...

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

(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

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