# Bernoulli's equation and internal energy

I have come across an extension of Bernoulli's equation in the Feynman lectures.

It has another term in the energy equation describing the internal energy of the fluid:

$$\label{Eq:II:40:17} \frac{p_1}{\rho_1}+\frac{1}{2}\,v_1^2+\phi_1+U_1= \frac{p_2}{\rho_2}+\frac{1}{2}\,v_2^2+\phi_2+U_2$$

It then goes on to say that if the fluid is incompressible, internal energy is equal on both sides $U_1=U_2$, so it is removed.

But how does $U_1=U_2$ reconcile with the fact that a fluid gets colder as the pressure drops and the equipartition theorem which says that energy is shared equally amongst all degrees of freedom.

If static pressure drops then shouldn't internal energy also drop to maintain equipartition?

Why can we make this assumption?

• "the fact that a fluid gets colder as the pressure drops " Fact? Does water coming out of a tap cool down because it's depressurised? We're not talking about gases here. – Gert Oct 25 '16 at 2:48
• hi @Gert thanks for your response , No I was thinking about condensation in a vortex – Quentin Chester Oct 25 '16 at 2:51
• Did the reference say "incompressible" or did it say "inviscid?" – Chet Miller Oct 25 '16 at 3:10
• Hi @chestermiller it said Incompressible "which is the Bernoulli result with an additional term for the internal energy. If the fluid is incompressible, the internal energy term is the same on both sides, and we get again that Eq. (40.14) holds along any streamline. " – Quentin Chester Oct 25 '16 at 3:12
• Hi @chestermiller sorry I should add its from lecture 40 "flow of dry water" which deals purely in inviscid flow . Does that help ? – Quentin Chester Oct 25 '16 at 3:20