Page 279 - The Combined Finite-Discrete Element Method

262 FLUID COUPLING

By substituting temporarily ln(p c ) = y and ln(v c ) = x, equation (8.16) becomes

a (b−1)(ln(v o )−x) b−1
y =−x + ln 1 + e + ln(p o ) + ln(v o ) − ln(1 + a/v o ) (8.17)
b−1
v o
and by differentiation, the slope of the ln p − ln v curve is obtained:

a(b − 1) (b−1)(ln(v o )−x)
d(ln p) dy v o b−1 e
= =−1 − a (8.18)
d(ln v) dx (b−1)(ln(v o )−x)
1 + b−1 e
v o
In the limit when the specific volume becomes large
dy
lim =−1 (8.19)
x→∞ dx

and when the density becomes large

a(b − 1)
dy v o b−1
lim =−1 − (8.20)
x→− ln v o dx a
1 +
b−1
v o
3
For nitroglycerine expansion given in Figure 8.1, the initial density was 1600 kg/m , thus
3
initial specific volume is v = 1/1600 m /kg, which yields the initial slope of the log p
versus log v expansion curve:
10.99e − 10(4 − 1)
dy (1/1600) 4−1
lim =−1 − =−3.45 (8.21)
x→ln v o dx 10.99e − 10
1 +
(1/1600) 4−1
Thus, the slope of the curve changes from −1 at zero density of the detonation gas to
−3.45 at large densities of the detonation gas.

8.2.3 Isentropic adiabatic expansion of detonation gas

If the detonation gas is placed in a thermally insulating chamber, the walls of which
are able to move, and initially the chamber is completely filled with detonation gas, (i.e.
at initial density the volume of the gas equals the volume of the chamber), adiabatic
reversible expansion of the detonation gas takes place.
Because no heat exchange between the chamber walls and gas takes place, the change
in internal energy of the gas is due to the mechanical work only:

du =−pdv (8.22)