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Инженерия / Гидравлика

Bernoulli equation (basic)

The basic Bernoulli equation links pressure head, velocity head, and elevation head along a streamline for frictionless, steady, incompressible flow.

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Гидравлика

Формула

$$\frac{p_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{v_2^2}{2g} + z_2$$
diagram Two-point Bernoulli balance

Energy heads at two stations on the same streamline.

Pressure, velocity, and elevation heads sum to a constant without losses.

Обозначения

$p_1, p_2$
Pressure at stations 1 and 2, Pa
$\rho$
Fluid density, kg/m^3
$v_1, v_2$
Velocities at stations 1 and 2, m/s
$z_1, z_2$
Elevations of stations 1 and 2, m
$g$
Gravitational acceleration, m/s^2

Условия применения

  • Steady incompressible flow along one streamline.
  • Negligible shaft work and heat transfer between stations.
  • No significant head loss due to friction.

Ограничения

  • Not exact in real pipes with friction, local losses, or pumps/turbines unless additional terms are added.
  • Viscosity and turbulence losses are not represented in this basic form.
  • Assumes single-phase flow without cavitation.

Подробное объяснение

Bernoulli combines different head components derived from mechanical energy conservation for fluid particles.

Как пользоваться формулой

  1. Set up station 1 and station 2 values.
  2. Convert all terms to head (meters) when possible.
  3. Solve for the unknown variable and verify reasonableness with geometry.

Историческая справка

The Bernoulli relation evolved from early conservation principles and became foundational in classical hydraulics.

Пример

If p1/(ρg)=40 m, v1²/2g=2 m, z1=10 m and v2²/2g + z2 = 7 m, then p2/(ρg)=45 m and p2 = 441,000 Pa for water.

Частая ошибка

Adding major losses and ignoring them as negligible in long rough pipes leads to overprediction of pressure.

Практика

Задачи с решением

Find downstream pressure head

Условие. p1/(ρg)=50 m, v1=3 m/s, z1=5 m, v2=2 m/s, z2=8 m, ρ=1000 kg/m^3, g=9.81.

Решение. LHS = 50 + 3^2/(2·9.81) + 5 ≈ 55.46. RHS kinetic+elevation at 2 = 2^2/(19.62)+8 = 8.20. So p2/(ρg) = 47.26.

Ответ. p2/(ρg) ≈ 47.3 m (p2 ≈ 463 kPa).

Find upstream velocity

Условие. p1/(ρg)=30 m, z1=2 m, p2/(ρg)=20 m, z2=5 m, v2=1 m/s, g=9.81.

Решение. Using Bernoulli: v1^2/(2g) = 20 + 1^2/(19.62) + 5 - (30+2) = -7.95 m. Negative value means assumptions violated or input inconsistency.

Ответ. Given data are inconsistent with frictionless Bernoulli.

Дополнительные источники

  • Çengel, Y. A., & Cimbala, J. M. (2015). Fluid Mechanics: Fundamentals and Applications.
  • Fox, R. W., Pritchard, P. J., & McDonald, A. T. (2011). Introduction to Fluid Mechanics, 8th ed.

Связанные формулы

Инженерия

Pressure head

$h_p = \frac{p}{\rho g}$

Pressure head converts absolute pressure to an equivalent water column height, useful for balancing energies in flow systems.

Инженерия

Darcy–Weisbach head loss

$h_f = f\frac{L}{D_h}\frac{v^2}{2g}$

The Darcy–Weisbach equation estimates major head loss in fully developed pipe flow using friction factor and geometric ratio.

Инженерия

Pump power

$P = \frac{\rho g Q H}{\eta}$

Pump input power equals hydraulic power raised by the pump divided by pump efficiency.