221 
Sec. 5 -10- Eq. 27-33 
with S constant) passing through the point has the same slope as the H curve. 
Then dS, = 0 there and, from Eq. (26) 
aP5\ /aP, Pee : 
( E : eS ata ot (27) 
Ss 
Pis the angle made by the line AZ and the negative V axis. Therefore, at 
this point, (J of Fig. 5-1), AZ is tangent to the H curve and the velocity 
of detonation is given by 
D= Vy (Py - P/V - Vy) = Vy yf (AP,/aVQ)q 
= (Vp )eo- 
From Eq. (7), Cal ‘4 = UAE = D/(D-U,) 
whence 
D= Uo + Co 
as Stated above, Furthermore, the entropy Sp is an extremal (since dSo = 0). 
Actually S5 is a minimm at this point, since by solving Eq. (26) for dS,/dVp 
and differentiating with respect to Vo at the point J one gets 
a2s | Vive fae 
a ee (29) 
aVo /H aT aVo~ /H 
as ‘as 
But 2 Slay we(a dP (30) 
/V 
so that 0 -(=) + (2) = 
aV/p aP/y \aV/g i (31) 
and ss) _(38) , fas _ 
qV/y- LAV Jp PWy \AV/y (32) 
Therefore, 
; fd \ 
ce ete = (eG N00 ae i= (3 33 
i ; ae (S)., =) TT AGYy ia KO ee) 
from which one finds that 
