The Bernoulli equation states the following

$$ \frac{1}{2}u^2 + \frac{p}{\rho}+ gz = \text{constant along a streamline} $$


  • $u$ is the fluid flow speed at a point on a streamline,
  • $g$ is the acceleration due to gravity,
  • $z$ is the elevation of the point above a reference plane, with the positive z-direction pointing upward – so in the direction opposite to the gravitational acceleration,
  • $p$ is the pressure at the chosen point, and
  • $\rho$ is the density of the fluid at all points in the fluid.


  • friction by viscous must be negligible
  • flow must be steady, i.e. the flow parameters (velocity, density, etc...) at any point cannot change with
  • the flow must be incompressible: even though the pressure varies, the density must remain constant along a streamline

My question

Given that $\frac{1}{2}u^2$ is the dynamic component, $gz$ is the hydraulic head, how should one imagine $\frac{p}{\rho}$?

  • $\begingroup$ "Elastic potential energy" stored in squeezing the water to a given pressure. $\endgroup$
    – knzhou
    Commented Mar 17, 2019 at 15:43
  • $\begingroup$ @knzhou there is no potential energy stored in the water because we have incompressible flow. $\endgroup$ Commented Mar 17, 2019 at 18:34

2 Answers 2


Pressure work per unit mass of fluid.

  • $\begingroup$ How could $p/\rho$ be "pressure work"? Work done by pressure forces depends on how pressure varies in space, i.e. gradient of pressure, not its absolute magnitude. $\endgroup$ Commented Mar 17, 2019 at 18:38
  • $\begingroup$ @JánLalinský Well, for an inviscid fluid, the change in kinetic energy per unit mass plus the change in potential energy per unit mass must be equal to the work done on the fluid per unit mass. So, more precisely, $\frac{\Delta p}{\rho}$ would represent the work done by pressure forces per unit mass. I don't think there is an actual name for $p/\rho$ or a physical interpretation, but some people (not me) might refer to it as the "pressure energy." $\endgroup$ Commented Mar 17, 2019 at 21:28
  • $\begingroup$ Now I agree, that is a much better answer! $\endgroup$ Commented Mar 17, 2019 at 22:25

The Bernoulli equation may be derived from the conservation of energy. The conservation of energy is a wonderful property in physics where despite interactions where nearly everything we can measure changes: position, velocity, acceleration..., something stays the same over time. If something stays the same over time, then you can predict the future.

From your above equation, I'll multiply everything by $\rho$ so the Bernoulli equation looks more similar to conservation of energy. $$\frac{1}{2} \rho u^2 + p + \rho g z = Constant$$ I assume that density $\rho$ is constant and, on the right hand side of the equation, rather than writing $\rho * Constant$, I will call the term just $Constant$. Using $\rho$ instead of mass is like taking the standard energy conservation equation and dividing by volume.

The leftmost term, $\frac{1}{2}\rho u^2$, is as you said the dynamic component of conservation of energy. The potential energy term is $\rho g z$. These two terms have the same units but we detect them in different ways, so we can think of detecting energy in different ways. One way we notice that something has energy is that is have velocity, hence the dynamic component. From the potential term, we detect energy through a mass that can accelerate. Another way we detect energy is that a mass is exposed to a force. Force and acceleration go hand in hand, Newton's $F=ma$ can also be seen as a circular definition of force and acceleration. Force is acceleration and acceleration is force (scaled by mass). You cannot have force without acceleration and vice versa. So, in addition to velocity and acceleration, we can detect energy through force. Now, energy is a scalar, it has no direction, but Force is a vector (and it doesn't have the same units as energy). To know the energy that force contributes to the system we need to know what direction the being used, so we take the dot product and get work, $W$ in $\vec{F} \cdot \vec{d} = W$

Pressure is force divided by area, $p=\frac{F}{A}$

From the Bernoulli equation, let's multiply everything by volume (Volume is constant along the streamline), so density becomes mass $\rho V = m$. $$\frac{1}{2}mu^2 + p * V + m g z = Constant$$ For pressure, $$p * V=\frac{F}{A} V= \frac{F}{A} Ad = Fd$$ The volume is area times distance, the area cancels leaving force times distance, which is work. And work is energy. The pressure term in Bernoulli equation is our way of detecting energy through force.

  • $\begingroup$ +1 but pressure is not a vector, very nice answer $\endgroup$
    – veronika
    Commented Mar 18, 2019 at 18:15

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