I am having an exercise: Deriving the Bernoulli's equation ($\space p_1+\frac{1}{2}\rho V_1^2 = p_2+\frac{1}{2}\rho V_2^2 $ ) from the energy equation: $$\rho \frac{D(e+V^2/2)}{Dt} = \nabla(pV)$$To make it clear: $\rho$ is the density, e is the internal energy of one infinitesimal element, p and V are the pressure and the velocity, respectively. with the conditions: steady, incompressible, invisid flow and no body forces.

Here is what it was going: Because of the invisid flow, then I thought $$\frac{De}{Dt}=0 \space(*)$$(maybe I was wrong at this point)

Then I had the equation: $$\rho \frac{D(V^2/2)}{Dt} = \nabla(pV)$$ It was straightforward for me to derive the Bernoulli's equation from above equation and I had done the job, but then a thing arise: Bernoulli's equation hold along a streamline, consider a streamline of the flow below: enter image description here

Let $V_1 \neq V_2$, then from Bernoulli's equation, we have $p_1 \neq p_2$. Assume the flow is perfect gas, then from the perfect gas's equation: $p = \rho RT$, we will have $T_1 \neq T_2$ (because $\rho$, R are constant). We also know that e = $c_vT$ , $c_v$ is the specific heat at constant volume. We point out: $e_1 \neq e_2$, that means the element at 1 has different internal energy from the element at 2 at an instant time. But after amount of time, the element at 1 (has internal engery $e_1$) will go to 2 and achive the internal energy $e_2$ so we can say $De/Dt \neq 0$. This result is contrast with the above (*) equation. Can someone point out my mistake?

  • $\begingroup$ Indeed if you are going to consider temperature changes then it is incorrect to assume $De/Dt=0$ in the beginning. $\endgroup$
    – Deep
    Commented Aug 21, 2017 at 3:51
  • $\begingroup$ We usually use Bernoulli's equation to specify different values of pressure, then we must have different values of temperature because of the equation: $p = \rho RT$. If we assume temperature not change, what is the meaning of the Bernoulli equation?, T not change, then p not change, then V not change, assuming T not change leads to a similar paradox. $\endgroup$
    – Dat
    Commented Aug 21, 2017 at 4:48
  • $\begingroup$ Your assumption that internal energy does not change is valid only for incompressible flow, as soon as a fluid particle gets compressed, its energy, in general, changes. $\endgroup$ Commented Jan 2, 2019 at 20:35

3 Answers 3


You had to assume the fluid is incompressible to write Bernoulli. The equation of state of such a fluid is definitively not the perfect's gas law. Or conversely, a perfect gas is not incompressible in general. To apply Bernoulli's principle to it, you need at least to assume the pressure at the beginning and the end of the streamline are equal. And then that no heat transfer took place along the streamline.

  • $\begingroup$ Why assuming pressure at the beginning and the end of the steamline are equal while pressure is supposed to change when apply Bernoulli's principle? $\endgroup$
    – Dat
    Commented Jan 7, 2019 at 10:11

The Bernoulli equation for the steady flow of compressible fluids is $$ h+ \frac 12 |{\bf v}|^2=const. $$ along streamlines. Here $h$ is the enthalpy $U+PV$ per unit mass. It only reduces to the OP's expression when $\rho=1/V$ is a constant.

  • $\begingroup$ With $\rho$ is constant, does $\gamma$ have to go to infinity too ? $\endgroup$
    – Dat
    Commented Jan 1, 2019 at 17:35
  • 1
    $\begingroup$ @Dat. In incompressible flow the pressure is not a very useful quantity because it takes whatever value is needed to keep the volume of a bit of fluid constant. In other words Euler's equation should be regarded as an equation determining $P$ rather than one in which $P$ determines ${\bf v}$. As an expample, this is why water hammers are bad for plumbing systems -- a sudden stop in a flow cause by shutting a valve can prodce very large pressures that can rupture the piping. Thus $\gamma$ is pretty useless too. $\endgroup$
    – mike stone
    Commented Jan 1, 2019 at 17:41
  • $\begingroup$ I don't think so. Pressure is quite important if you have an incompressible flow over an airfoil. Howerver, could you edit your answer to manipulate how your equation reduce to my expression when $\rho$ is a constant ? It would really help me $\endgroup$
    – Dat
    Commented Jan 2, 2019 at 2:12
  • $\begingroup$ @Dat Ideal gases are not ideal fluids (i.e. not incompressible), see cfd-online.com/Forums/main/… and facstaff.cbu.edu/rprice/lectures/idealgas.html. $\endgroup$ Commented Jan 2, 2019 at 12:44
  • $\begingroup$ Also, what mike stone states is correct. Velocities are evaluated with boundary layer approximations for incompressible flows across airfoils, whose pressure and shear stress distributions are then solved for to obtain lift and drag coefficients. $\endgroup$ Commented Jan 2, 2019 at 12:56

After much deliberation, I think I've figured out the issue. The assumption $De/Dt = 0$:

$$ \implies \frac{\partial e}{\partial t} + \vec v \cdot \nabla e = 0$$ is correct for an incompressible flow with the trivial solution $\partial e/\partial t = \nabla e = 0$, and only applies for steady, inviscid, adiabatic flow. The conservation equation you're looking for is:

$$\frac{D}{Dt}\left(e + \frac{P}{\rho} + \frac{v^2}{2}\right) = 0$$

Integrating this gives you a constant, which is considered to be the total enthalpy:

$$ \implies e + \frac{P}{\rho} + \frac{v^2}{2} = h_0 $$ Physically, the internal energy of the gas undergoes exchanges with its kinetic energy and pressure, leaving its total enthalpy $h_0$ (a condition in which a fluid element is adiabatically brought to rest, with enthalpy defined as $h = e + PV$, where $V = 1/\rho$ is the specific volume) is constant along a streamline.

To explain this in the case of an incompressible flow, consider the compressible Bernoulli equation in your described case where the specific volume (and hence the density) isn't changing:

$$ h_2 - h_1 = c_P(T_2 - T_1) = c_V(T_2 - T_1) + \frac{P_2 - P_1}{\rho} = \frac{v_1^2 - v_2^2}{2}, \quad c_P = \frac{\gamma R}{\gamma - 1}, \,c_V = \frac{R}{\gamma - 1}$$

To approximate an incompressible gas in this situation, let $\gamma \to \infty$, and you will obtain Bernoulli's equation for incompressible flows along a streamline, which is equivalent to Anderson's solution with $c_P$ and the equation of state for an ideal gas.

Since the temperatures are different at different points in space, the internal energies are different (which in fact results in changes in internal energy over time along the streamline by the conservation law $De/Dt = 0$ [directional derivative in the direction $\vec v$]). Hence your changes in enthalpy result in changes in temperature at constant pressure (almost by definition), or changes in pressure at constant volume), when the gas is considered to be incompressible. Note that the flow itself is not incompressible in the former case.

  • $\begingroup$ I have so much questions, first look at page 210, avionicsengineering.files.wordpress.com/2016/11/… Author said "Indeed, Bernoulli’s equation can be derived from the general energy equation, such as Equation (2.114)" I can not derive Bernoulli equation from equation (2.114). $\endgroup$
    – Dat
    Commented Jan 4, 2019 at 7:55
  • $\begingroup$ Second, why do $\gamma$ have to be infinite ? You force $\gamma$ to be infinite to drive Bernoulli equation from energy equation, that is you add extra assumption. That means energy equation only reduce to Bernoulli equation when you force $\gamma$ to be infinite and $\rho$ to be constant. But consider when a flow over an airfoil with M<<1, although $\rho$ is considered to be constant in this case, $\gamma$ is still 1.4. That means energy equation can not reduce to Bernoulli in this case. $\endgroup$
    – Dat
    Commented Jan 4, 2019 at 8:05
  • $\begingroup$ 1. Some assumptions reduce the energy equation to: $\rho D(e + \vec v^2/2)/Dt = -\nabla \cdot (p \vec v)$. For incompressible flow, $\nabla \cdot \vec v = 0$, so expand the terms in the previous expression and start cancelling some. Anderson also covers this derivation more generally for compressible flow in Eq. 7.49. 2. For incompressible flow $M \to \infty$ and $M = \sqrt{\gamma RT}$. Since the gas constant and temperature are fixed, $\gamma \to \infty$. Thus the energy equation reduces to Bernoulli's equation in incompressible flow, as your question requested. $\endgroup$ Commented Jan 4, 2019 at 8:38
  • $\begingroup$ Based on your above comment, we only have incompressible flow when $\gamma \to \infty$. But consider a flow over an airfoil with M something like 0.1 at sea level. With that condition, I believe $\gamma$ is still 1.4, but why do people still consider the flow to be incompressible? $\endgroup$
    – Dat
    Commented Jan 4, 2019 at 10:29
  • $\begingroup$ Incompressibility when $M < 0.3$ is an approximation made in real-world experiments/simulations so the math is less cumbersome. I think the difference in results vary by like 2% within that bound. To derive Bernoulli's equation, certain conditions of the flow must be fulfilled, which are assumed when deriving it in via the momentum equation as well. $\endgroup$ Commented Jan 4, 2019 at 11:31

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