How is Bernoulli's equation a statement of conservation of energy?

Question

In the derivation of Bernoulli's equation

$$W_{nc} = (P_1 - P_2)V $$ $$W_{nc} = \Delta KE + \Delta PE $$ $$(P_1 - P_2)V = 1/2mv_2^2 - 1/2mv_1^2 + mgh_2 - mgh_1 $$ $$P_1V + 1/2mv_1^2 + mgh_1 = P_2V + 1/2mv_2^2 + mgh_2$$ $$P_1 + 1/2\rho v_1^2 + \rho gh_1 = P_2 + 1/2\rho v_2^2 + \rho gh_2$$

I have read that this is just conservation of energy, but how is energy conserved even though a non conservative force ((P1-P2)V) is applied on the system? Won't just the existence of the non conservative work not make the energy of the system conserved? Bernoulli's equation is definitely mathematically true(and very useful), but I'm finding it hard to accept that 'energy is conserved'.

Answer 1

It depends on the energies you are considering.

You're right in the "introductory mechanics" sense, energy is conserved when $\Delta E=\Delta K+\Delta U=0$ for a system.

However, in this case the work is being done by the force(s) associated with the pressure. So one can include this in a change in total "energy" of the system. Then we have a conserved quantity: $$\Delta E=\Delta K+\Delta U-(P_1-P_2)V=0$$

This quantity is conserved because the work done by the fluid pressure goes into changing its kinetic and potential energy.

Of course this means that the claim that Bernoulli's principle is equivalent to energy conservation is not entirely true, but one can still fudge the wording around a bit and people will usually still know what you mean by it.

Answer 2

For a compressible fluid in steady flow Bernoulli says that $$ \frac 12 |{\bf v}|^2 +h $$ is constant along streamlines. Here $h$ is the specific enthalpy i.e the enthalpy $H=U+PV$ per unit mass.

Bernoulli is therefore a generalization of the thermodynamic statement that the enthalpy of a fluid (gas or liquid) is conserved when it flows though an orifice to include the kinetic energy of the fluid. For a unit mass $V=\rho^{-1}$ and, for the adiabtic ($dS=0$) flow of an incompressible fluid, the internal energy $U$ is an ignorable constant --- so the specific enthalpy reduces to $h=P/\rho$. I think it is useful, however, to realize that the usually-presented incompressible Bernoulli equation generalises the energy conservation in throttling processes that is explained in intro thermodymamics.

Answer 3

I am going to explain the factors only intuitively withot any kind of mathematics.

According to Bernoulli's theorem the sum of total of these three kinds of energies kinetic energy, potential energy and pressure energy remains constant along a streamline in a steady flow of an ideal fluid. But before you start reading Bernoulli's equation you should consider the following things$-$

•The flow is assumed to be steady, i.e,there is no change in pressure, velocity and density of the fluid at any point wrt time.However, in the problems of unsteady flow(not turbulent) with gradually changing velocity, Bernoulli's equation can be applied without considerable error.

• The fluid is assumed to be incompressible. Since liquids are incompressible, Bernoulli's equation can be applied to all liquids. However,it can be applied to the gas flow where there is little variation of pressure, velocity and temperature so that density of gas can be assumed to be constant.

• The fluid is assumed to be irrotational. The irrotationality means the net angular momentum at any point in the fluid flow is zero.

• The fluid is assumed to be ideal, i.e,the energy loss due to friction is assumed to be absent. For more details ideal fluid is inviscid and incompressible. In other words, there is no dissipation of energy due to internal friction between adjacent layers of the fluid and density of the fluid remains constant.

To quote Wikipedia:

In a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline. This requires that the sum of kinetic energy, potential energy and internal energy remains constant.Thus an increase in the speed of the fluid – implying an increase in its kinetic energy (dynamic pressure) – occurs with a simultaneous decrease in (the sum of) its potential energy (including the static pressure) and internal energy. If the fluid is flowing out of a reservoir, the sum of all forms of energy is the same on all streamlines because in a reservoir the energy per unit volume (the sum of pressure and gravitational potential (ρ g h) is the same everywhere.

Answer 4

I have read that this is just conservation of energy, but how is energy conserved even though a non conservative force ((P1-P2)V) is applied on the system?

Bernoulli equation is not a statement of conservation of energy, nor is it a result of just this statement. This is because energy of any liquid element we can apply this equation to does change, as a result of work of pressure forces from the side where the pressure is higher. The work of pressure needs to be taken into account.

Bernoulli equation is a useful equation that follows from the so-called work-energy theorem (not exactly energy conservation). The theorem says: work of external forces acting on a liquid element equals increase of kinetic energy of that element.

In the usual setup for Bernoulli equation, there are two forces that can do work on the element: gravity force, and pressure force. Together these change kinetic energy of the liquid.

Let the element have pressure $P_1$ upstream and $P_2$ downstream, and let $P_1 > P_2$ so the pressure forces upstream accelerate the liquid element by pushing on one its side of area $S$. Total work by those forces when the element moves along by one its length $\Delta L$ and by $\Delta h$ upwards is

$$ \text{work of gravity} + \text{work of pressure forces} = \text{increase in kinetic energy} $$ $$ - \Delta mg\Delta h + P_1 S \Delta L - P_2 S \Delta L = \frac{1}{2}\Delta m\Delta (v^2) $$ Dividing both sides by volume of the element $S\Delta L$, and introducing density $\rho = \Delta m/\Delta V$ we obtain

$$ -\rho g \Delta h + (P_1 - P_2)= \frac{1}{2}\rho \Delta(v^2) $$

This still has the form "work equals increase in kinetic energy", but it is more customary to put everything on one side of the equation:

$$ \rho g \Delta h + (P_2 - P_1) + \frac{1}{2}\rho \Delta(v^2) = 0. $$

Expression $P_2-P_1$ is increase in pressure $\Delta P$ (although in our case pressure decreases, so $\Delta P < 0$. We can write the result $$ \rho g \Delta h + \Delta P + \frac{1}{2}\rho \Delta(v^2) = 0 $$ or $$ \rho g h + P + \frac{1}{2}\rho v^2 = \text{const.} $$

The first term is potential energy of unit volume in gravity field, the last term is kinetic energy of unit volume, and the middle term... is not energy at all. Pressure has no associated potential energy to it. The reason for this is the pressure term depends only on local pressure of the liquid, but the liquid is incompressible, which means the liquid does not have any additional energy due to being compressed. It takes zero energy to increase pressure in incompressible liquid. Thus the Bernoulli equation (the "const version") is not an energy conservation equation. It is a "work-energy theorem" equation, which is a bit obscured in the "const version" but completely obvious from the "Delta version". Pressure term is there because of work of external pressure forces. These forces do not have potential energy associated with them.