Page 280 - The Combined Finite-Discrete Element Method

EXPANSION OF THE DETONATION GAS 263

From the assumption that at constant temperature T the internal energy u does not depend
upon the specific volume v, it follows that any changes in internal energy are due to the
temperature change:
du = c v dT =−pdv (8.23)

From equation of state, it follows that the pressure of detonation gas is given by

b
p = (ρ + aρ )RT (8.24)
which, when substituted into (8.23), yields

b
c v dT =−(ρ + aρ )RT dv (8.25)

Substituting
1 1
v = and thus dv =− dρ (8.26)
ρ ρ 2
into (8.25) gives
c v dT
1 b−2
= + aρ dρ (8.27)
R T ρ

and after integration
1 R 1 b−2
T v v
dT = + aρ dρ (8.28)
T ρ
T c v c c v
with assumption that R and c v are constants and that R/c v = k − 1, one obtains

a b−1 b−1
k−1
ln(T /T c ) = ln(ρ/ρ c ) + (k − 1) (ρ − ρ c ) (8.29)
b − 1
b−1
= ln(ρ/ρ c ) k−1 − (k − 1) aρ c [1 − (ρ/ρ c ) b−1 ]
b − 1
 b−1 
aρ c
−(k−1) [1−(ρ/ρ c ) b−1 ]
= ln (ρ/ρ c ) k−1 e b − 1 


or
b−1
aρ c
−(k−1) [1−(ρ/ρ c ) b−1 ]
T/T c = (ρ/ρ c ) k−1 e b − 1 (8.30)

From equation of state, it follows that

b−1 b−1
ρ 1 + aρ c (ρ/ρ c ) T
p/p c = (8.31)
b−1
ρ c 1 + aρ c T c