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I have seen that term get defined as pressure energy (potential energy per volume).

But how does that make sense? because work is $F \cdot d$ so there must be a change of volume of the fluid that the pressure acts on and to store energy but in case of incompressible fluid that change is small almost negligible.

But if that term isn't energy how does the bernoulli's equation work if it is based on energy conservation how can I add a pressure term to other terms which are energy terms and apply the energy conservation rule?

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    $\begingroup$ Pressure, P=F/A, A= area $\endgroup$ Commented Dec 22, 2023 at 21:36
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    $\begingroup$ It's pressure, not pressure energy. The Bernoulli equation assumes incompressible fluid, change of pressure in such a fluid is not associated/does not cause change of internal energy. $\endgroup$ Commented Dec 22, 2023 at 22:33

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You are correct - the pressure term in Bernoulli's equation is not some kind of additional energy density contribution; it is just pressure. Calling $p$ "pressure energy" or "pressure energy density" is a misconception based on an attractive, widespread and wrong interpretation of the Bernoulli equation.

This wrong interpretation goes like this. The Bernoulli equation is

$$ \frac{1}{2}\rho v^2 +\rho g h + p = const. $$

This is a sum of terms with units of energy density, so the equation looks like it could be an equation of conservation of energy, if only the left-hand side was the total energy of the fluid. Intuitively we accept that increasing pressure in practice takes work, so without deep thought, we tend to call the term $p$ "pressure energy", and the misconception is born.

The problem with the intuition is, it takes zero work to increase pressure of a fluid element in this case, because we are dealing with incompressible flow. The element of fluid does not change its volume at all, and thus no work is consumed to increase pressure in it, even though pressure may increase during its motion.

The left-hand side is therefore not the total energy density of the fluid. Total energy in mechanics (including mechanics of incompressible fluids) is sum of kinetic and potential energy. For incompressible fluid in gravity field, density of total energy is sum of density of kinetic energy and density of potential energy:

$$ e = \frac{1}{2}\rho v^2 +\rho g h. $$

Where does does the Bernoulli equation come from then, if it is not a statement of conservation of energy?

It may be derived in different ways, e.g. from the Euler equation which is a form of Newton's equation $\mathbf F=m\mathbf a$ adapted to fluids. But the simplest derivation is from the variant of a work-energy theorem applied to the fluid element. This theorem (in contrast to the idea of conservation of energy of the element) says that kinetic energy of the fluid element changes as it undergoes a displacement, proportionally to net work of all forces acting on the element.

This can be adapted to the formulation: total energy of the fluid (kinetic energy + potential gravitational energy) changes in proportion to net work of the pressure forces. It is the work of the pressure forces which introduces the pressure term $p$ into our equation.

Let us consider a fluid element in shape of a cylinder, with its axis aligned with the flow. Let us consider a small displacement $\Delta x$ along its axis (much smaller than the cylinder length), while pressure force $p_1A$ acts on the back face (1 refers to the back face), and pressure force $-p_2A$ acts on the front face (2 refer to the front face). After the displacement, volume $A\Delta x$ of the element gets into a new region of the flow (near the front face of the cylinder) where total energy density is $e_2 = \frac{1}{2}\rho v_2^2 + \rho g h_2$, and the same volume is removed from the region where it had total energy density $e_1 = \frac{1}{2}\rho v_1^2 + \rho g h_1$ (near the back face of the cylinder). Thus increase of total energy of the cylinder is $$ A\Delta x \bigg(\frac{1}{2}\rho v_2^2 + \rho g h_2\bigg) - A\Delta x \bigg(\frac{1}{2}\rho v_1^2 + \rho g h_1\bigg). $$ Net work done by external pressure forces during this displacement is $$ A (p_1 - p_2) \Delta x. $$ Since change of energy equals this work, we have $$ \bigg(\frac{1}{2}\rho v_2^2 + \rho g h_2\bigg) - \bigg(\frac{1}{2}\rho v_1^2 + \rho g h_1\bigg) = p_1 - p_2. $$

Since we can imagine such cylinder anywhere on the streamline, the equation holds for any pair of points on the streamline. Denoting $$ \frac{1}{2}\rho v_1^2 + \rho g h_1 + p_1 =C $$ we can simply restate that for any point on the streamline, velocity $v$, height $h$ and pressure $p$ obey the equation $$ \frac{1}{2}\rho v^2 + \rho g h + p =C. $$

The pressure term comes from the expression of work on the fluid element; it does not come from expression of total energy of the fluid element.

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  • $\begingroup$ I was reading this... Just to clarify, would calling $p$ pressure work instead of pressure energy be descriptive? Basically if you write the equation in terms of differences it is evidently just the work-energy theorem so it seems appropriate $\endgroup$
    – HomoVafer
    Commented Dec 23, 2023 at 17:19
  • $\begingroup$ The pressure term $p$ comes from pressure work in the adapted work-energy theorem, but the term itself is not work - it's pressure. The work expression is $(p_1-p_2)A\Delta x$. $\endgroup$ Commented Dec 26, 2023 at 19:14
  • $\begingroup$ I get your point: the units of $p$ are $[N/m^2]$. Nonetheless I would rather think of this pressure term as $[J/m^3]$, which I think is closer to what the equation says. I've been a bit sloppy but what I meant was pressure work per unit volume... $\endgroup$
    – HomoVafer
    Commented Dec 27, 2023 at 16:19
  • $\begingroup$ That's the same unit. Pressure value is work per unit volume moved, but only the partial work done on one side of the motion. When all sides are taken into account, net work per unit volume moved is $p_1-p_2$. $\endgroup$ Commented Dec 27, 2023 at 17:27
  • $\begingroup$ Them having the same unit doesn't make them the same concept. I think we can all agree that the correct interpretation of $p$ is not force per unit surface but rather work per unit volume. I don't think anyone manages to get any deep understanding of this formula thinking about forces per unit surface. It's just terminology after all but better terminology makes better understanding. At least we agree that pressure energy doesn't make any sense (I still don't get why many engineers often use this expression, I guess they don't care about deep understanding...). $\endgroup$
    – HomoVafer
    Commented Dec 28, 2023 at 9:58
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What is the term 𝑃 in bernoulli's equation defined as?

Pressure energy.

because work is $F \cdot d$ so there must be a change of volume of the fluid that the pressure acts on and to store energy but in case of incompressible fluid that change is small almost negligible.

Keep in mind there are two types of work, boundary work (pressure volume work), $PdV$, and flow work, or $VdP$, the work required to push fluid into or out of a control volume. The former is used for a closed system. The latter is for an open system and is applicable to Bernoulli's equation.

$$W=F\cdot d$$

$$F=PA$$

$$W=PA \cdot d=PV$$

$$P=\frac{W}{V}$$

Ergo, Pressure = Energy per unit volume

In Bernoulli's equation:

$$P_{1} + \frac{1}{2}\rho v_{1}^{2}+\rho gh_{1}=P_{2} + \frac{1}{2}\rho v_{2}^{2}+\rho gh_{2}$$

where $P$ is pressure energy. See the figure below taken from the following website where you an find a more detailed discussion, http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html

Hope this helps.

enter image description here

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    $\begingroup$ There is no pressure energy in Bernoulli's equation , it's just pressure. Bernoulli's equation assumes incompressible fluid, where any change of pressure is for free (in terms of work). $\endgroup$ Commented Dec 22, 2023 at 22:36
  • $\begingroup$ @JánLalinský Pressure energy is in Bernoulli's equation. $\endgroup$
    – Bob D
    Commented Dec 22, 2023 at 22:45
  • $\begingroup$ How much work does it take to increase pressure of 1 liter of incompressible fluid from 1bar to 2bar? $\endgroup$ Commented Dec 22, 2023 at 22:46
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In an elementary volume flowing in $x$ direction, in a transition region of a pipe, where there is a change of area, we can use Newton's second law: $$-(P_{x+dx}-P_x)dydz = \rho dxdydz\frac{dv}{dt}$$

Multiplying both sides by $dx$ (the left side is then $dW = Fdx$) and dividing by the elementary volume:

$$-dP = \rho dx\frac{dv}{dt} = \rho vdv = d(\frac{1}{2}\rho v^2)$$

The minus sign comes from the fact that if $(P_{x+dx}-P_x)$ is negative, then there is a force to the right, increasing the velocity of the volume element.

Integrating: $$P + \frac{1}{2}\rho v^2 = cte$$

$P$ is related to the work necessary to increase the velocity of the fluid if the diameter of the pipe is reduced.

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