A paradox when I was deriving Bernoulli's equation from energy equation

Question

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:

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.