Since the internal energy per unit mass e is given by e ¼ p/(γ À 1)ρ, we can write
this latter energy equation as
ÀU s
d
dω
1
2
ρu
2
þ
p
γ À 1
!
þ
d
dω
u
1
2
ρu
2
þ
p
γ À 1
þ u p À σ x
ð
Þ
!
¼ 0
and on integrating, we obtain,
ÀU s
1
2
ρu
2
þ
p
γ À 1
!
þ u
1
2
ρu
2
þ
p
γ À 1
þ u p À σ x
ð
Þ
!
¼ constant:
Applying the boundary conditions at x ¼ + 1, namely, u ¼ 0, p ¼ p 0 , ρ ¼ ρ 0 and
σ x ¼ 0, accordingly, the constant of integration is ÀU s p 0 /(γ À 1). Hence, this energy
equation becomes
À
1
2
ρU s u
2
À
U s p
γ À 1
þ
U s p 0
γ À 1
þ
1
2
ρu
3
þ
pu
γ À 1
þ u p À σ x
ð
Þ¼0:
Substituting the relationship for p À σ x according to Eq. (3.65) in this latter
equation, we find that
À
1
2
ρU s u
2
À
p
γ À 1
U s À u
ð
Þþ
U s p 0
γ À 1
þ
1
2
ρu
3
þ up 0 þ U s ρ 0 u
2
¼ 0:
Let us now eliminate the pressure p in the above equation by using Eq. (3.65),
hence, after a few algebraic steps the following equation is obtained;
À
1
2
ρU s u
2
À
U s À u
γ À 1
4
3
μ
du
dω
þ
γup 0
γ À 1
À
U
2
s ρ 0 u
γ À 1
þ
γρ 0 U s u
2
γ À 1
þ
1
2
ρu
3
¼ 0, ð3:66Þ
where the relationship, σ x ¼ (4μ/3)(∂u/∂x), has been substituted. By dividing
Eq. (3.66) by ρ 0 and using Eq. (3.64), we find that
À
1
2
U
2
s u
2
U s À u
ð
Þ
À
4 U s À u
ð
Þμ
3 γ À 1
ð
Þρ 0
du
dω
þ
uc
2
0
γ À 1
À
U
2
s u
γ À 1
þ
γU s u
2
γ À 1
þ
1
2
U s
U s À u
u
3
¼ 0,
where we have used the equation for the speed of sound c 0 , namely, c
2
0 ¼ γp 0 =ρ 0 . By
multiplying the previous equation by (U s À u)(γ À 1) and after grouping various
terms, the following equation results,
U s À u
ð
Þ
2
ρ 0
4
3
μ
du
dω
¼
1
2
γ þ 1
ð
ÞU s u
2 U s À u
ð
Þþuc
2
0 U s À u
ð
ÞÀU
2
s u U s À u
ð
Þ,
hence, we can finally write it in the following manner,
126
3 Conditions Across the Shock: The Rankine-Hugoniot Equations