The balance of energy can be expressed as:
ρ
e
˙
−
σ
:
(
∇
v
)
+
∇
∙
q
−
ρ
s
=
0
{\displaystyle \rho ~{\dot {e}}-{\boldsymbol {\sigma }}:({\boldsymbol {\nabla }}\mathbf {v} )+{\boldsymbol {\nabla }}\bullet \mathbf {q} -\rho ~s=0}
where
ρ
(
x
,
t
)
{\displaystyle \rho (\mathbf {x} ,t)}
is the mass density,
e
(
x
,
t
)
{\displaystyle e(\mathbf {x} ,t)}
is the internal energy
per unit mass,
σ
(
x
,
t
)
{\displaystyle {\boldsymbol {\sigma }}(\mathbf {x} ,t)}
is the Cauchy stress,
v
(
x
,
t
)
{\displaystyle \mathbf {v} (\mathbf {x} ,t)}
is
the particle velocity,
q
{\displaystyle \mathbf {q} }
is the heat flux vector, and
s
{\displaystyle s}
is the
rate at which energy is generated by sources inside the volume
(per unit mass).
Recall the general balance equation
d
d
t
[
∫
Ω
f
(
x
,
t
)
dV
]
=
∫
∂
Ω
f
(
x
,
t
)
[
u
n
(
x
,
t
)
−
v
(
x
,
t
)
⋅
n
(
x
,
t
)
]
dA
+
∫
∂
Ω
g
(
x
,
t
)
dA
+
∫
Ω
h
(
x
,
t
)
dV
.
{\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }f(\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\partial {\Omega }}f(\mathbf {x} ,t)[u_{n}(\mathbf {x} ,t)-\mathbf {v} (\mathbf {x} ,t)\cdot \mathbf {n} (\mathbf {x} ,t)]~{\text{dA}}+\int _{\partial {\Omega }}g(\mathbf {x} ,t)~{\text{dA}}+\int _{\Omega }h(\mathbf {x} ,t)~{\text{dV}}~.}
In this case, the physical quantity to be conserved the total energy density
which is the sum of the internal energy density and the kinetic energy
density, i.e.,
f
=
ρ
e
+
1
/
2
ρ
|
v
⋅
v
|
{\displaystyle f=\rho ~e+1/2~\rho ~|\mathbf {v} \cdot \mathbf {v} |}
.
The energy source at the surface is a sum of the rate of work done by
the applied tractions and the rate of heat leaving the volume
(per unit area), i.e,
g
=
v
⋅
t
−
q
⋅
n
{\displaystyle g=\mathbf {v} \cdot \mathbf {t} -\mathbf {q} \cdot \mathbf {n} }
where
n
{\displaystyle \mathbf {n} }
is the
outward unit normal to the surface. The energy source inside the body
is the sum of the rate of work done by the body forces and the rate of
energy generated by internal sources, i.e.,
h
=
v
⋅
(
ρ
b
)
+
ρ
s
{\displaystyle h=\mathbf {v} \cdot (\rho \mathbf {b} )+\rho ~s}
.
Hence we have
d
d
t
[
∫
Ω
ρ
(
e
+
1
2
v
⋅
v
)
dV
]
=
∫
∂
Ω
ρ
(
e
+
1
2
v
⋅
v
)
(
u
n
−
v
⋅
n
)
dA
+
∫
∂
Ω
(
v
⋅
t
−
q
⋅
n
)
dA
+
∫
Ω
ρ
(
v
⋅
b
+
s
)
dV
.
{\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\text{dV}}\right]=\int _{\partial {\Omega }}\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)(u_{n}-\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}+\int _{\partial {\Omega }}(\mathbf {v} \cdot \mathbf {t} -\mathbf {q} \cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}
Let
Ω
{\displaystyle \Omega }
be a control volume that does not change with time. Then
we get
∫
Ω
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
dV
=
−
∫
∂
Ω
ρ
(
e
+
1
2
v
⋅
v
)
(
v
⋅
n
)
dA
+
∫
∂
Ω
(
v
⋅
t
−
q
⋅
n
)
dA
+
∫
Ω
ρ
(
v
⋅
b
+
s
)
dV
.
{\displaystyle \int _{\Omega }{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]~{\text{dV}}=-\int _{\partial {\Omega }}\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)(\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}+\int _{\partial {\Omega }}(\mathbf {v} \cdot \mathbf {t} -\mathbf {q} \cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}
Using the relation
t
=
σ
⋅
n
{\displaystyle \mathbf {t} ={\boldsymbol {\sigma }}\cdot \mathbf {n} }
, the identity
v
⋅
(
σ
⋅
n
)
=
(
σ
T
⋅
v
)
⋅
n
{\displaystyle \mathbf {v} \cdot ({\boldsymbol {\sigma }}\cdot \mathbf {n} )=({\boldsymbol {\sigma }}^{T}\cdot \mathbf {v} )\cdot \mathbf {n} }
, and invoking
the symmetry of the stress tensor, we get
∫
Ω
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
dV
=
−
∫
∂
Ω
ρ
(
e
+
1
2
v
⋅
v
)
(
v
⋅
n
)
dA
+
∫
∂
Ω
(
σ
⋅
v
−
q
)
⋅
n
dA
+
∫
Ω
ρ
(
v
⋅
b
+
s
)
dV
.
{\displaystyle \int _{\Omega }{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]~{\text{dV}}=-\int _{\partial {\Omega }}\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)(\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}+\int _{\partial {\Omega }}({\boldsymbol {\sigma }}\cdot \mathbf {v} -\mathbf {q} )\cdot \mathbf {n} ~{\text{dA}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}
We now apply the divergence theorem to the surface integrals to get
∫
Ω
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
dV
=
−
∫
Ω
∇
∙
[
ρ
(
e
+
1
2
v
⋅
v
)
v
]
dV
+
∫
Ω
∇
∙
(
σ
⋅
v
)
dV
−
∫
Ω
∇
∙
q
dV
+
∫
Ω
ρ
(
v
⋅
b
+
s
)
dV
.
{\displaystyle \int _{\Omega }{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]~{\text{dV}}=-\int _{\Omega }{\boldsymbol {\nabla }}\bullet \left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\mathbf {v} \right]~{\text{dV}}+\int _{\Omega }{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}\cdot \mathbf {v} )~{\text{dV}}-\int _{\Omega }{\boldsymbol {\nabla }}\bullet \mathbf {q} ~{\text{dV}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}
Since
Ω
{\displaystyle \Omega }
is arbitrary, we have
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
=
−
∇
∙
[
ρ
(
e
+
1
2
v
⋅
v
)
v
]
+
∇
∙
(
σ
⋅
v
)
−
∇
∙
q
+
ρ
(
v
⋅
b
+
s
)
.
{\displaystyle {\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]=-{\boldsymbol {\nabla }}\bullet \left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\mathbf {v} \right]+{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}\cdot \mathbf {v} )-{\boldsymbol {\nabla }}\bullet \mathbf {q} +\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~.}
Expanding out the left hand side, we have
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
=
∂
ρ
∂
t
(
e
+
1
2
v
⋅
v
)
+
ρ
(
∂
e
∂
t
+
1
2
∂
∂
t
(
v
⋅
v
)
)
=
∂
ρ
∂
t
(
e
+
1
2
v
⋅
v
)
+
ρ
∂
e
∂
t
+
ρ
∂
v
∂
t
⋅
v
.
{\displaystyle {\begin{aligned}{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]&={\frac {\partial \rho }{\partial t}}\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)+\rho ~\left({\frac {\partial e}{\partial t}}+{\frac {1}{2}}~{\frac {\partial }{\partial t}}(\mathbf {v} \cdot \mathbf {v} )\right)\\&={\frac {\partial \rho }{\partial t}}\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)+\rho ~{\frac {\partial e}{\partial t}}+\rho ~{\frac {\partial \mathbf {v} }{\partial t}}\cdot \mathbf {v} ~.\end{aligned}}}
For the first term on the right hand side, we use the identity
∇
∙
(
φ
v
)
=
φ
∇
∙
v
+
∇
φ
⋅
v
{\displaystyle {\boldsymbol {\nabla }}\bullet (\varphi ~\mathbf {v} )=\varphi ~{\boldsymbol {\nabla }}\bullet \mathbf {v} +{\boldsymbol {\nabla }}\varphi \cdot \mathbf {v} }
to get
∇
∙
[
ρ
(
e
+
1
2
v
⋅
v
)
v
]
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
∇
[
ρ
(
e
+
1
2
v
⋅
v
)
]
⋅
v
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
(
e
+
1
2
v
⋅
v
)
∇
ρ
⋅
v
+
ρ
∇
(
e
+
1
2
v
⋅
v
)
⋅
v
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
(
e
+
1
2
v
⋅
v
)
∇
ρ
⋅
v
+
ρ
∇
e
⋅
v
+
1
2
ρ
∇
(
v
⋅
v
)
⋅
v
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
(
e
+
1
2
v
⋅
v
)
∇
ρ
⋅
v
+
ρ
∇
e
⋅
v
+
ρ
(
∇
v
T
⋅
v
)
⋅
v
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
(
e
+
1
2
v
⋅
v
)
∇
ρ
⋅
v
+
ρ
∇
e
⋅
v
+
ρ
(
∇
v
⋅
v
)
⋅
v
.
{\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\bullet \left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\mathbf {v} \right]&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +{\boldsymbol {\nabla }}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }+{\frac {1}{2}}~\rho ~{\boldsymbol {\nabla }}(\mathbf {v} \cdot \mathbf {v} )\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }+\rho ~({\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {v} )\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }+\rho ~({\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )\cdot \mathbf {v} ~.\end{aligned}}}
For the second term on the right we use the identity
∇
∙
(
S
T
⋅
v
)
=
S
:
∇
v
+
(
∇
∙
S
)
⋅
v
{\displaystyle {\boldsymbol {\nabla }}\bullet ({\boldsymbol {S}}^{T}\cdot \mathbf {v} )={\boldsymbol {S}}:{\boldsymbol {\nabla }}\mathbf {v} +({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}})\cdot \mathbf {v} }
and the
symmetry of the Cauchy stress tensor to get
∇
∙
(
σ
⋅
v
)
=
σ
:
∇
v
+
(
∇
∙
σ
)
⋅
v
.
{\displaystyle {\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}\cdot \mathbf {v} )={\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})\cdot \mathbf {v} ~.}
After collecting terms and rearranging, we get
(
∂
ρ
∂
t
+
ρ
∇
∙
v
+
∇
ρ
⋅
v
)
(
e
+
1
2
v
⋅
v
)
+
(
ρ
∂
v
∂
t
+
ρ
∇
v
⋅
v
−
∇
∙
σ
−
ρ
b
)
⋅
v
+
ρ
(
∂
e
∂
t
+
∇
e
⋅
v
)
+
−
σ
:
∇
v
+
∇
∙
q
−
ρ
s
=
0
.
{\displaystyle \left({\frac {\partial \rho }{\partial t}}+\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} +{\boldsymbol {\nabla }}\rho \cdot \mathbf {v} \right)\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)+\left(\rho ~{\frac {\partial \mathbf {v} }{\partial t}}+\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} -{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right)\cdot \mathbf {v} +\rho ~\left({\frac {\partial e}{\partial t}}+{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }\right)+-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\bullet \mathbf {q} -\rho ~s=0~.}
Applying the balance of mass to the first term and the balance of linear
momentum to the second term, and using the material time derivative of
the internal energy
e
˙
=
∂
e
∂
t
+
∇
e
⋅
v
{\displaystyle {\dot {e}}={\frac {\partial e}{\partial t}}+{\boldsymbol {\nabla }}e\cdot \mathbf {v} }
we get the final form of the balance of energy:
ρ
e
˙
−
σ
:
∇
v
+
∇
∙
q
−
ρ
s
=
0
.
{\displaystyle {\rho ~{\dot {e}}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\bullet \mathbf {q} -\rho ~s=0~.}}