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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 twodimensional 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).
Next: The Moving Shocks
Up: Operating Equations and Analysis
Previous: Shock Thickness
Index
Created by:Genick BarMeir, Ph.D.
On:
20071121