16) A non-viscous fluid flows through a horizontal pipe of varying cross-sectional area. Find the pressure variation between the pipe´s extremes in function of the entry velocity, v, and the density , ρ, if:
a) The exit area of the pipe is triple the entry’s area
b) The exit diameter is triple the entry´s diameter

Here another classic problem of energy conservation in a fluid (Bernoulli). Let´s go with the first part in which the exit area is triple the entry´s.


Fluídos - Ricardo Cabrera


Written as an equation, the condition of this problem says:

SS = 3 SE

That has its consequence in the velocity, and in the velocity squared, given that:


SS . vS = SE . vE

3 SE . vS = SE . vE

3 vS = vE

And the squares:

9 vS² = vE²

vS² = vE² / 9

Now we can write Bernoulli´s equation (without the terms of potential energy, as everything happens at an equal height) to calculate the pressure variation, ΔP.

ΔP = ½ ρ (vE² vS²)

ΔP = ½ ρ (vE² vE²/ 9) = ½ ρ (8/9) vE²


          ΔP = (4/9) ρ vE²                              a)


The new condition of this problem relates the pipes´ diameters, not their sections:

dS = 3 dE

dS² = 9 dE²

Yet, we can relate all the sections, remember that a circular section is  equal to S = (π/4) d²

(π/4) dS² = 9 (π/4) dE²

SS = 9 SE

This has a consequence in the velocity and in the velocity squared, as:


SS . vS = SE . vE

9 SE . vS = SE . vE

9 vS = vE

81 vS² = vE²

vS² = vE² / 81

Now we write  Bernoulli’s equation:

ΔP = ½ ρ (vE² vS²)

ΔP = ½ ρ (vE² vE²/ 81 ) = ½ ρ (80/81) vE²


Fluídos - Ricardo Cabrera

          ΔP = (40/81) ρ vE²                              b)


In both cases we face a pressure rise, owning to the fact that at the exit the velocity is less than at the entrance, and at an equal height, higher pressure regions have lower fluid speed.


CHALLENGE: What would be the pressure variation if the density of the fluid was 1.200 kg/m3 and the entry velocity 0,082 m/s?


Fluídos - Ricardo Cabrera


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