Why is wave drag non-zero for a flow without shockwaves? Wave drag is often defined as the component of drag that arises due to the presence of shockwaves. Shocks result in an entropy rise and a corresponding increase in drag. However, when using Supersonic Potential Flow Theory (Ackeret Theory), the flow is assumed to be fully isentropic (i.e. no shockwaves) and it can be shown (see Bertin, Houghton, Anderson for reference) that the wave drag is non-zero for such flows. So, what is the mechanism behind why the wave drag arises in such a flow without shocks?

More Detail (from Aerodynamics for Engineers by Bertin)
The drag can be computed from Navier-Stokes momentum equations as:

$$D=-\oint_{S_\infty}p\hat{i}\cdot d\vec{S} - \oint_{S_\infty}\rho (\hat{i}\cdot\vec{u})(\vec{u}\cdot d\vec{S})$$

Usually, all boundaries other than the rear boundary (Trefftz plane) would have the same flow properties as the free-stream, and they fall out of the integral. We are left with:

$$D=\int_{S_T}{\rho u(V_\infty-u)dS}$$

As you correctly pointed out, the momentum defect of a potential flow should be exactly zero, which means that drag should also be zero.

However, for oblique shock in small-disturbance potential flow, the oblique shock (or rather, each constant potential wave characteristic) extends to infinity, carrying energy with it. If you want to calculate the far-field drag, you would need to do the integral around the entire far-field. This is different from energy dissipation in viscous boundary layer or subsonic normal shock. 



Picture ref. https://nikander.github.io/compflow/Anderson/Chapter4/print/