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The Bernoulli's equation is usually thought to be applied to an incompressible fluid (without potential energy change) as

$$\frac12 v_1^2 + \rho P_1 = \frac12 v_2^2 + \rho P_2 $$

Where, v is velocity, $\rho$ is the density (which is constant), P is the pressure.

Q1: Here is there any change in the temperature (internal energy)?

The Bernoulli's equation applied to an compressible fluid (without potential energy change) as

$$h_1+\frac12v_1^2 = h_2+\frac12v_2^2$$ $$U_1+P_1V_1+\frac12v_1^2 = U_2+P_2V_2+\frac12v_2^2$$

Where, h is the enthalpy, U is the internal energy, V is the volume.

If we consider the flow to incompressible, then $V_1=V_2$. But here for the incompressible flow of a compressible fluid, the temperature (internal energy and also enthalpy) changes, as can be seen by using the ideal gas equation ($PV=RT$, if we consider $V$ to be constant then as pressure $P$ changes the temperature $T$ changes).

Q2: Is it correct to state that when Bernoulli's equation is applied to an incompressible fluid, there is NO internal energy change, but when applying Bernoulli's equation to a compressible fluid (say ideal gas) undergoing an incompressible flow, there IS internal energy change?

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  • $\begingroup$ See Transport Phenomena by Bird et al, Section 3.3 and especially Example 3.5-1, and Chapter 11. $\endgroup$ Jul 16, 2023 at 10:44

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Q1: Here is there any change in the temperature (internal energy)?

The Bernoulli equation does not involve temperature at all. So the temperature may change or may not change, the Bernoulli equation does not imply either way.

Usually in applications of the Bernoulli equation, we assume the temperature does not change, because changing temperature of water requires lots of heat or dissipated work (pressure work can't change its temperature by much, due to its low compressibility) and source of those are usually not present. However, if there is a heat source heating up incompressible liquid, then the Bernoulli equation will hold, while temperature of the liquid will change with position.

Q2: Is it correct to state that when Bernoulli's equation is applied to an incompressible fluid, there is NO internal energy change, but when applying Bernoulli's equation to a compressible fluid (say ideal gas) undergoing an incompressible flow, there IS internal energy change?

No. For incompressible fluid, the original Bernoulli equation does not imply anything on its internal energy, except it can't change via volume work.

In the case of the generalized Bernoulli equation, volume work is allowed.

Both variants of the Bernoulli equation do not imply anything on changes of internal energy that are due to different type of work or heat exchange than volume work.

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