ABSTRACT
This is the simplest case to treat. To develop the equations, consider the conditions on either side of a shock wave. In order to view these conditions, it is convenient to use co-ordinates which move with the shock wave. From this viewpoint, velocities are as shown in Figure 6.2(b). Because of the compression, it is imperative that the equation of mass continuity (i.e. mass conservation) be used:
mass entering control volume = mass leaving control volume
pAc = (p + op) (c - uo)A which may be rewritten as
pUo = op(c - uo) Similarly, applying the momentum equation
force = mass flow x change in velocity
(p + op)A - pA = pAc [c - (c - uo)]
Hence
op = pcuo
(6.3)
(6.4)
By definition,
therefore
mass m p=--=~
volume V
dp m P dV V2 V
so
therefore
From (6.3),
From (6.4),
From (6.5),
op oV -=--
K=~ Op/p
op uo p (c - uo)
Op uo=-pc
Substituting for Uo and Op/p in (6.6), Op (Op/pc) K [c - (Op/pc)]
Rearranging,
If Op ~ K, then pc2/K = 1
therefore
(6.5)
(6.6)
(6.7) Equations (6.4) and (6.7) may then be solved simultaneously to obtain a value for Op. Note that no allowance has been made for straining of the pipe material, since the pipe is assumed to be rigid (i.e. E ~ 00). It should be further noted that the passage of events is extremely rapid, as will become apparent in the following example.
