Expressing the internal energy e as a function of s and ρ, we can write
e ¼ e s, ρ
ð Þ,
so that
de ¼
∂e
∂s
ρ
ds þ
∂e
∂ρ
s
dρ:
ð1:56Þ
Hence, by comparing Eqs. (1.55) and (1.56) we have,
T ¼
∂e
∂s
ρ
and
p
ρ 2 ¼
∂e
∂ρ
s
ð1:57Þ
From Eq. (1.56) we can write the following two equations;
∂e
∂t
¼
∂e
∂s
ρ
∂s
∂t
þ
∂e
∂ρ
s
∂ρ
∂t
and
∂e
∂x
¼
∂e
∂s
ρ
∂s
∂x
þ
∂e
∂ρ
s
∂ρ
∂x
,
hence
∂e
∂t
¼ T
∂s
∂t
þ
p
ρ 2
∂ρ
∂t
and
∂e
∂x
¼ T
∂s
∂x
þ
p
ρ 2
∂ρ
∂x
,
after using Eq. (1.57). Substituting these latter two equations in Eq. (1.52) we obtain,
ρ T
∂s
∂t
þ
p
ρ 2
∂ρ
∂t
þ uT
∂s
∂x
þ
up
ρ 2
∂ρ
∂x
À
p
ρ
∂ρ
∂t
þ u
∂ρ
∂x
¼ 0,
so that
22
1 Brief Outline of the Equations of Fluid Flow