# Why system approach does not consider flow energy (flow work)?

The total energy of a simple compressible system consists of 3 parts namely the kinetic energy, potential energy and the internal energy. The total energy of a control volume consists of an additional form of energy called the flow energy.

Why in system approach to fluid flow don't we consider the flow energy (Pv) even though the fluid element in the system has a pressure which can change (without the volume changing)?

Here I am trying to understand the energy equation with reference to the system approach and control volume approach to fluid flow (Lagrangian and Eulerian description).

• Isn't $pV$ just work done by a pressure? It is only included when added to the system. Not if it happens within the system. Just as heat. I believe this is already a part of internal energy. In what connection have you seen it defined seperately for a control volume and called "flow energy" Dec 19, 2017 at 11:11
• @Steeven PV is the flow work the is going into (or out of depending on the sign) control volume and not a system. PdV is the work done on the system. The internal energy of a system is a function of temperature and not the pressure. Dec 19, 2017 at 11:22
• @Steeven Flow energy is always defined in connection with control volume I think. Dec 19, 2017 at 11:28
• I was chasing formulas yesterday and found $V\sigma \mathrm{d}\epsilon$, is that what you are looking for?
– Emil
Jun 25, 2018 at 6:53
• Before you can understand the open system (control volume) version of the first law of thermodynamics, you must first understand the closed system (no mass flow in or out) version. Are you sure you understand that (and why it includes only the internal energy)? Once you understand the closed system version, you need to then review the derivation of the open system version to get comfortable with the idea that the two versions are equivalent. Jun 25, 2018 at 12:55

Consider a fluid flowing though a pipe of Area $A$, lets assume thin section of fluid $dX$ which has to flow through a distance $L$ through that pipe. Now for the liquid to move that distance of $L$, the liquid behind that thin section will be pushing it ($dX$) to move through that length $L$. So this is work done by fluid behind on the thin section ($dX$) to move a length $L$. Which can be written as : $Force \times Distance = (Pressure \times Area) \times (L) = (P \times A ) \times l = PV$
Now in the above example the thin section($dX$) is analogous to Piston in a control mass system and fluid behind it is pushing just like the fluid in a cylinder piston device. Now suppose the liquid is not flowing through the pipe, then there is no work being done on the thin section $dX$ by the liquid behind to move, so work done is zero which is flow work, so stationary liquids don't have flow work.