Shock or Wave Drag

It is communally believed that regardless to the cause of the shock, the shock creates a drag (due to increase of entropy). In this section, the first touch of this phenomenon will be presented. The fact that it is assumed that the flow is frictionless does not change whether or not shock drag occur. This explanation is broken into two sections: one for stationary shock wave, two for moving shock shock wave. A better explanation should appear in the oblique shock chapter.

Consider a normal shock as shown in figure (5.5).

Figure: The diagram that reexplains the shock drag effect.

Gas flows in a supersonic velocity around a two-dimensional body and creates a shock. This shock is an oblique shock, however in this discussion, if the control volume is chosen close enough to the body is can be considered as almost a normal shock (in the oblique shock chapter a section on this issue will be presented that explains the fact that shock is oblique, to be irrelevant).

The control volume that is used here is along two stream lines. The other two boundaries are arbitrary but close enough to the body. Along the stream lines there is no mass exchange and therefore there is no momentum exchange. Moreover, it is assumed that the gas is frictionless, therefore no friction occurs along any stream line. The only change is two arbitrary surfaces since the pressure, velocity, and density are changing. The velocity is reduced U_x > U_y . However, the density is increasing, and in addition, the pressure is increasing. So what is the momentum net change in this situation? To answer this question, the momentum equation must be written and it will be similar to equation (). However, since F_y/F^* = F_x/F^* there is no net force acting on the body. For example, consider upstream of M_x=3. and for which

Normal Shock		Input: Mx		k = 1.4
Mx	My	Ty/Tx	ρy/ρx	Py/Px	P0y/P0x
3	0.475191	2.67901	3.85714	10.3333	0.328344

and the corespondent Isentropic information for the Mach numbers is

Isentropic Flow		Input: M			k = 1.4
M	T/T0	ρ/ρ0	A/A*	P/P0	PAR	F/F*
3	0.357143	0.0762263	4.23457	0.0272237	0.115281	0.653256
0.47519	0.95679	0.895451	1.3904	0.856759	1.19123	0.653257

Now, after it was established, it is not a surprising result. After all, the shock analysis started with the assumption that no momentum is change. As conclusion there is no shock drag at stationary shock. This is not true for moving shock as it will be discussed in section (5.3.1).