∂
∂t
ρϖ
½ Šþ
∂
∂x
u ρϖ þ p
ð
Þ
½
Š¼0
and carrying out the differentiation we have
∂
∂t
ρϖ
ð Þþu
∂
∂x
ρϖ þ p
ð
Þþ ρϖ þ p
ð
Þ
∂u
∂x
¼ 0
and expanding we obtain,
∂
∂t
ρϖ
ð Þþu
∂
∂x
ρϖ
ð Þþu
∂p
∂x
þ ρϖ
∂u
∂x
þ p
∂u
∂x
¼ 0:
Hence,
ρ
∂ϖ
∂t
þ ϖ
∂ρ
∂t
þ uρ
∂ϖ
∂x
þ uϖ
∂ρ
∂x
þ u
∂p
∂x
þ ρϖ
∂u
∂x
þ p
∂u
∂x
¼ 0
and collecting terms we have
ρ
∂ϖ
∂t
þ u
∂ϖ
∂x
þ ϖ
∂ρ
∂t
þ u
∂ρ
∂x
þ u
∂p
∂x
þ ρϖ
∂u
∂x
þ p
∂u
∂x
¼ 0,
which can be written as
ρ
∂ϖ
∂t
þ u
∂ϖ
∂x
þ ϖ
∂ρ
∂t
þ u
∂ρ
∂x
þ ρ
∂u
∂x
!
þ u
∂p
∂x
þ p
∂u
∂x
¼ 0:
The quantity in square brackets in the latter equation goes to zero as we recognise
it as the terms appearing in the continuity equation, hence,
ρ
∂ϖ
∂t
þ u
∂ϖ
∂x
þ u
∂p
∂x
þ p
∂u
∂x
¼ 0
ð1:51Þ
Now using; ϖ ¼
u
2
2 þ e, we have
∂ϖ
∂t
¼ u
∂u
∂t
þ
∂e
∂t
and
∂ϖ
∂x
¼ u
∂u
∂x
þ
∂e
∂x
and substituting these relationships in Eq. (1.51) we have
ρ
∂e
∂t
þ u
∂e
∂x
þ u ρ
∂u
∂t
þ u
∂u
∂x
þ
∂p
∂x
!
þ p
∂u
∂x
¼ 0:
1.4 Conservation Equations in Plane Geometry
19