# 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!

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:$$\frac{\gamma}{\gamma-1}d\left(\frac{p}{\rho}\right)+vdv=0$$