5.7 Energy of the Explosion
The energy of the explosion, E 0 , comprises two parts; kinetic energy and heat
energy, the kinetic energy per unit volume is ρu
2 /2 and the heat energy per unit
volume is p/(γ À 1), so that the energy generated by the explosion is
E 0 ¼
Z R
0
1
2
ρu
2
þ
p
γ À 1
4πr
2 dr:
ð5:20Þ
We should in fact subtract from this the pre-shock internal energy of the air
engulfed by the shock, namely,
4πR
3
3
p 0
γ À 1
,
so that a more accurate expression for the energy of the explosion becomes
E 0 ¼
Z R
0
1
2
ρu
2
þ
p
γ À 1
4πr
2 dr À
4πR
3
3
p 0
γ À 1
:
ð5:21Þ
However, the similarity solution is based on the fact that the external atmospheric
pressure p 0 is negligible in comparison to the very high pressures generated by the
strong explosion, so that we will use the first expression for E 0 , namely, Eq. (5.20).
Substituting Eqs. (5.7a), (5.7b) and (5.7c) in Eq. (5.20) yields,
E 0 ¼ 4π
Z R
0
1
2
ρ 0 ψ
ð
ÞA
2 R
À3
ϕ
2
þ
ρ 0 A
2 R
À3 f
γ γ À 1
ð
Þ
r
2 dr
¼ 4πρ 0 A
2
Z R
0
1
2
ψR
À3
ϕ
2
þ
R
À3 f
γ γ À 1
ð
Þ
r
2 dr:
But r ¼ ηR, hence, dr ¼ Rdη and therefore, r
2 dr ¼ R
3
η
2 dη, consequently,
E 0 ¼ 4πρ 0 A
2
Z 1
0
1
2
ψR
À3
ϕ
2
þ
R
À3 f
γ γ À 1
ð
Þ
R
3
η
2 dη,
hence,
5.7 Energy of the Explosion
229
The energy of the explosion, E 0 , comprises two parts; kinetic energy and heat
energy, the kinetic energy per unit volume is ρu
2 /2 and the heat energy per unit
volume is p/(γ À 1), so that the energy generated by the explosion is
E 0 ¼
Z R
0
1
2
ρu
2
þ
p
γ À 1
4πr
2 dr:
ð5:20Þ
We should in fact subtract from this the pre-shock internal energy of the air
engulfed by the shock, namely,
4πR
3
3
p 0
γ À 1
,
so that a more accurate expression for the energy of the explosion becomes
E 0 ¼
Z R
0
1
2
ρu
2
þ
p
γ À 1
4πr
2 dr À
4πR
3
3
p 0
γ À 1
:
ð5:21Þ
However, the similarity solution is based on the fact that the external atmospheric
pressure p 0 is negligible in comparison to the very high pressures generated by the
strong explosion, so that we will use the first expression for E 0 , namely, Eq. (5.20).
Substituting Eqs. (5.7a), (5.7b) and (5.7c) in Eq. (5.20) yields,
E 0 ¼ 4π
Z R
0
1
2
ρ 0 ψ
ð
ÞA
2 R
À3
ϕ
2
þ
ρ 0 A
2 R
À3 f
γ γ À 1
ð
Þ
r
2 dr
¼ 4πρ 0 A
2
Z R
0
1
2
ψR
À3
ϕ
2
þ
R
À3 f
γ γ À 1
ð
Þ
r
2 dr:
But r ¼ ηR, hence, dr ¼ Rdη and therefore, r
2 dr ¼ R
3
η
2 dη, consequently,
E 0 ¼ 4πρ 0 A
2
Z 1
0
1
2
ψR
À3
ϕ
2
þ
R
À3 f
γ γ À 1
ð
Þ
R
3
η
2 dη,
hence,
5.7 Energy of the Explosion
229
