# How does pressure energy convert to thermal in incompressible flow?

I am thinking of a straight thermally isolated pipe, and incompressible flow through it. Due to friction, pressure energy drop of fluid, (increase of internal energy), can be calculated via the Darcy–Weisbach equation. If I look at the internal energy of fluid as the average velocity of molecular movement, and pressure energy as impact of molecules on the pipe: how come that one of these energies can raise and another drop, in the incompressible flow? Because in an other situation where we start to apply external heat to a pipe, we expect both pressure and internal energy of the fluid to rise?

Edit 1:

As pressure energy I am referring to $\frac{p}{\rho}$, and as thermal (internal) energy to $c_v T$. All for a mass of 1 kg. By using the Kinetic Molecular Theory principles, I thought that I can visualise the process of flow as I mentioned: temperature increase - molecule speed increase, which leads to a higher pressure exerted on walls of pipe. But in reality, as the pressure drops, the temperature increases which messes my approach. Can I take in consideration that the density of any fluid (for example water), isn't constant (with drop of pressure and temperature rise, density lowers at some decimal), and therefore the pressure drops, even with temperature rise, which fits my visualising process, or is this all too much of trying to simplify the situation? Any answer appreciated.

• The part of your question starting at "How come that one of these energy can raise and another drop" is not very clear. Please provide a more specific example (with equations) to clarify what you're asking. – Han-Kwang Nienhuys May 29 '16 at 19:45
• It still unclear what you're asking. – Han-Kwang Nienhuys May 31 '16 at 6:54