Similarly, the quantity in square brackets in this latter equation goes to zero
according to the momentum equation; hence,
ρ
∂e
∂t
þ u
∂e
∂x
þ p
∂u
∂x
¼ 0:
By substituting for ∂u/∂x using the continuity equation this latter equation can be
written in the following form;
ρ
∂e
∂t
þ u
∂e
∂x
À
p
ρ
∂ρ
∂t
þ u
∂ρ
∂x
¼ 0
ð1:52Þ
or writing it in Lagrangian form as
ρ
De
Dt
À
p
ρ
Dρ
Dt
¼ 0:
ð1:53Þ
As ρ ¼ 1/υ, where υ is the specific volume (that is, the volume per unit mass of
material), then
Dρ
Dt
¼ Àρ
2 Dυ
Dt
,
so that the energy balance equation in Lagrangian form becomes,
De
Dt
¼ Àp
Dυ
Dt
:
ð1:54Þ
If the caloric equation for an ideal gas in the form,
e ¼
pυ
γ À 1
is substituted in Eq. (1.54) it is easy to verify that
D pυ
γ
ð Þ
Dt
¼ 0,
so that
∂
∂t
þ u
∂
∂x
pυ
γ
ð Þ ¼ 0:
It is important to appreciate that this result has been obtained by assuming that the
fluid element is non-conducting, devoid of viscosity and obeys the ideal gas equation
20
1 Brief Outline of the Equations of Fluid Flow