# Help with isentropic flow and energy equation

I'm trying to understand Wikipedia's derivation of the isentropic velocity-area relationship, here under "Flow analysis":

https://en.wikipedia.org/wiki/Isentropic_nozzle_flow

I don't quite understand what happens to the energy equation. I suppose $q$ and $w$ refer to energy added and the work done by the flow? Then later at "The energy equation is:" the sum of the work/energy and the $h$ (enthalpy?) becomes the ratio involving the specific heat ratio, pressure and density. How did this step occur? This is probably quite simple but it puzzles me and I couldn't really find another derivation that explains this step properly. If you could derive it in detail here or point to a derivation I would appreciate it, thank you!

The "energy equation" in this article refers to the steady state version of the open-system form of the first law of thermodynamics. Are you familiar with the derivation of the open system (control volume) version of the first law of thermodynamics(derived from the usual closed system version)? If not, you need to start by familiarizing yourself with this.

In the equation, q is the heat added to the system, and the w is the so-called "shaft work," which is equal to the total work minus the work required to push fluid into- and out of the control volume. If the control volume is adiabatic and no shaft work is done, then q = w = 0. So the equation reduces to: $$h+\frac{v^2}{2}=h_0+\frac{v_0^2}{2}$$ Taking the differential of this over a differential distance along the nozzle, this becomes: $$dh+vdv=0$$But, for an ideal gas, $$dh=C_pdT=\frac{C_p}{R}d\left(\frac{p}{\rho}\right)=\frac{\gamma}{\gamma-1}d\left(\frac{p}{\rho}\right)$$So the equation becomes:$$\frac{\gamma}{\gamma-1}d\left(\frac{p}{\rho}\right)+vdv=0$$