Engineer's Notebook §1.2

Bernoulli's equation

Along a streamline in steady, frictionless flow, pressure, kinetic energy and height trade off against each other, but their sum stays fixed:

p + ½ρv² + ρgh = C
p
pressure
Pa
ρ
density
kg/m³
v
flow speed
m/s
g
gravitational acceleration
m/s²
h
height above a datum
m
C
constant along the streamline
Pa
D
pipe diameter
m
Δh
rise from inlet to outlet
m
H
total head
m
A pipe that narrows and changes height between an inlet on the left and an outlet on the right. A standpipe over each section shows the gauge pressure as a column height. p₁ = 200 kPa p₂ = 167 kPa Δh = 2 m (drawn 4 units/m)
Fig. 1.2-1Standpipe columns show gauge pressure to scale. The pipe's rise is drawn to a fixed scale.
v2 = v1(D1/D2)²5.56 m/s
½ρ(v1² − v2²)−13.4 kPa
−ρg Δh−19.6 kPa
p2167 kPa

Worked example

Uses the values in Fig. 1.2-1. Change them and the working updates.

Given
  • p1inlet gauge pressure200 kPa
  • v1inlet speed2 m/s
  • D1inlet diameter100 mm
  • D2outlet diameter60 mm
  • Δhoutlet height above inlet2 m
  • ρwater998 kg/m³
Find

The gauge pressure p2 at the outlet.

Solution
v2 = v1(D1/D2)² = 2 m/s × (100/60)² = 5.56 m/s p2 = p1 + ½ρ(v1² − v2²) − ρg Δh = 200 kPa + ½ × 998 kg/m³ × ((2 m/s)² − (5.56 m/s)²) − 998 kg/m³ × 9.81 m/s² × 2 m = 200 − 13.4 − 19.6 = 167 kPa
Answer
p2 = 167 kPa

That is gauge pressure, about 268 kPa absolute.

Related forms

  • Draining tank (Torricelli)

    v = √(2gh)

  • Pitot tube

    v = √(2Δpρ)

  • Head form, in metres

    pρg + v²2g + z = H

Assumptions
  • Flow is steady and incompressible.
  • There is no friction. Real pipes lose head to friction, so the real outlet pressure is lower.
  • The equation applies along one streamline, or between sections where the flow is uniform.
  • No pump or turbine sits between the two points.
  • Pressures here are gauge. Absolute pressure is gauge plus about 101.3 kPa, and water starts to boil (cavitate) near 2.3 kPa absolute at 20 °C.

Derivation

Follow a small volume V of fluid from section 1 to section 2.

  1. The pressure difference does work on it: (p1 − p2)V.
  2. Its kinetic energy rises by ½ρV(v2² − v1²), and its potential energy by ρVg(h2 − h1).
  3. With no friction, work in equals energy gained. Divide by V and collect each section's terms:
  4. p1 + ½ρv1² + ρgh1 = p2 + ½ρv2² + ρgh2.