Show that, for thermoelastic materials, the balance of energy
ρ
e
˙
−
σ
:
∇
v
+
∇
⋅
q
−
ρ
s
=
0
.
{\displaystyle \rho ~{\dot {e}}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\cdot \mathbf {q} -\rho ~s=0~.}
can be expressed as
ρ
T
η
˙
=
−
∇
⋅
q
+
ρ
s
.
{\displaystyle \rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}
Proof:
Since
e
=
e
(
F
,
T
)
{\displaystyle e=e({\boldsymbol {F}},T)}
, we have
e
˙
=
∂
e
∂
F
:
F
˙
+
∂
e
∂
η
η
˙
.
{\displaystyle {\dot {e}}={\frac {\partial e}{\partial {\boldsymbol {F}}}}:{\dot {\boldsymbol {F}}}+{\frac {\partial e}{\partial \eta }}~{\dot {\eta }}~.}
Plug into energy equation to get
ρ
∂
e
∂
F
:
F
˙
+
ρ
∂
e
∂
η
η
˙
−
σ
:
∇
v
+
∇
⋅
q
−
ρ
s
=
0
.
{\displaystyle \rho ~{\frac {\partial e}{\partial {\boldsymbol {F}}}}:{\dot {\boldsymbol {F}}}+\rho ~{\frac {\partial e}{\partial \eta }}~{\dot {\eta }}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\cdot \mathbf {q} -\rho ~s=0~.}
Recall,
∂
e
∂
η
=
T
and
ρ
∂
e
∂
F
=
σ
⋅
F
−
T
.
{\displaystyle {\frac {\partial e}{\partial \eta }}=T\qquad {\text{and}}\qquad \rho ~{\frac {\partial e}{\partial {\boldsymbol {F}}}}={\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}~.}
Hence,
(
σ
⋅
F
−
T
)
:
F
˙
+
ρ
T
η
˙
−
σ
:
∇
v
+
∇
⋅
q
−
ρ
s
=
0
.
{\displaystyle ({\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}):{\dot {\boldsymbol {F}}}+\rho ~T~{\dot {\eta }}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\cdot \mathbf {q} -\rho ~s=0~.}
Now,
∇
v
=
l
=
F
˙
⋅
F
−
1
{\displaystyle {\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {l}}={\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}}
. Therefore, using the
identity
A
:
(
B
⋅
C
)
=
(
A
⋅
C
T
)
:
B
{\displaystyle {\boldsymbol {A}}:({\boldsymbol {B}}\cdot {\boldsymbol {C}})=({\boldsymbol {A}}\cdot {\boldsymbol {C}}^{T}):{\boldsymbol {B}}}
, we have
σ
:
∇
v
=
σ
:
(
F
˙
⋅
F
−
1
)
=
(
σ
⋅
F
−
T
)
:
F
˙
.
{\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {\sigma }}:({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1})=({\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}):{\dot {\boldsymbol {F}}}~.}
Plugging into the energy equation, we have
σ
:
∇
v
+
ρ
T
η
˙
−
σ
:
∇
v
+
∇
⋅
q
−
ρ
s
=
0
{\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +\rho ~T~{\dot {\eta }}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\cdot \mathbf {q} -\rho ~s=0}
or,
ρ
T
η
˙
=
−
∇
⋅
q
+
ρ
s
.
{\displaystyle {\rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}}
Rate of internal energy/entropy for thermoelastic materials[ edit | edit source ]
Taking the material time derivative of the specific internal energy, we
get
e
˙
=
∂
e
¯
∂
E
:
E
˙
+
∂
e
¯
∂
η
η
˙
.
{\displaystyle {\dot {e}}={\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}:{\dot {\boldsymbol {E}}}+{\frac {\partial {\bar {e}}}{\partial \eta }}~{\dot {\eta }}~.}
Now, for thermoelastic materials,
T
=
∂
e
¯
∂
η
and
S
=
ρ
0
∂
e
¯
∂
E
.
{\displaystyle T={\frac {\partial {\bar {e}}}{\partial \eta }}\qquad {\text{and}}\qquad {\boldsymbol {S}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}~.}
Therefore,
e
˙
=
1
ρ
0
S
:
E
˙
+
T
η
˙
.
⟹
e
˙
−
T
η
˙
=
1
ρ
0
S
:
E
˙
.
{\displaystyle {\dot {e}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}+T~{\dot {\eta }}~.\qquad \implies \qquad {\dot {e}}-T~{\dot {\eta }}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}~.}
Now,
d
d
t
(
T
η
)
=
T
˙
η
+
T
η
˙
.
{\displaystyle {\cfrac {d}{dt}}(T~\eta )={\dot {T}}~\eta +T~{\dot {\eta }}~.}
Therefore,
e
˙
−
d
d
t
(
T
η
)
+
T
˙
η
=
1
ρ
0
S
:
E
˙
⟹
d
d
t
(
e
−
T
η
)
=
−
T
˙
η
+
1
ρ
0
S
:
E
˙
.
{\displaystyle {\dot {e}}-{\cfrac {d}{dt}}(T~\eta )+{\dot {T}}~\eta ={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}\qquad \implies \qquad {{\cfrac {d}{dt}}(e-T~\eta )=-{\dot {T}}~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}~.}}
Also,
d
d
t
(
1
ρ
0
S
:
E
)
=
1
ρ
0
S
:
E
˙
+
1
ρ
0
S
˙
:
E
.
{\displaystyle {\cfrac {d}{dt}}\left({\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}+{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}~.}
Hence,
e
˙
−
d
d
t
(
T
η
)
+
T
˙
η
=
d
d
t
(
1
ρ
0
S
:
E
)
−
1
ρ
0
S
˙
:
E
⟹
d
d
t
(
e
−
T
η
−
1
ρ
0
S
:
E
)
=
−
T
˙
η
−
1
ρ
0
S
˙
:
E
.
{\displaystyle {\dot {e}}-{\cfrac {d}{dt}}(T~\eta )+{\dot {T}}~\eta ={\cfrac {d}{dt}}\left({\cfrac {1}{\rho _{0}}}{\boldsymbol {S}}:{\boldsymbol {E}}\right)-{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}\qquad \implies \qquad {{\cfrac {d}{dt}}\left(e-T~\eta -{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}\right)=-{\dot {T}}~\eta -{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}~.}}
For thermoelastic materials, show that the balance of energy
equation
ρ
T
η
˙
=
−
∇
⋅
q
+
ρ
s
{\displaystyle \rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s}
can be expressed as either
ρ
C
v
T
˙
=
∇
⋅
(
κ
⋅
∇
T
)
+
ρ
s
+
ρ
ρ
0
T
∂
S
^
∂
T
:
E
˙
{\displaystyle \rho ~C_{v}~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s+{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}:{\dot {\boldsymbol {E}}}}
or
ρ
(
C
p
−
1
ρ
0
S
:
∂
E
~
∂
T
)
T
˙
=
∇
⋅
(
κ
⋅
∇
T
)
+
ρ
s
−
ρ
ρ
0
T
∂
E
~
∂
T
:
S
˙
{\displaystyle \rho ~\left(C_{p}-{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}\right)~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s-{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}}
where
C
v
=
∂
e
^
(
E
,
T
)
∂
T
and
C
p
=
∂
e
~
(
S
,
T
)
∂
T
.
{\displaystyle C_{v}={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}\qquad {\text{and}}\qquad C_{p}={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}~.}
For the special case where there are no sources and we can ignore heat conduction (for very fast processes), the energy equation simplifies to
ρ
(
C
p
−
1
ρ
0
S
:
α
)
T
˙
=
−
ρ
ρ
0
T
α
:
S
˙
{\displaystyle \rho ~\left(C_{p}-{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}\right)~{\dot {T}}=-{\cfrac {\rho }{\rho _{0}}}~T~{\boldsymbol {\alpha }}:{\dot {\boldsymbol {S}}}}
where
α
:=
∂
E
~
∂
T
{\displaystyle {\boldsymbol {\alpha }}:={\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}}
is the thermal expansion tensor which has the form
α
=
α
1
{\displaystyle {\boldsymbol {\alpha }}=\alpha {\boldsymbol {1}}}
for isotropic materials and
α
{\displaystyle \alpha \,}
is the coefficient of thermal expansion. The above equation can be used to calculate the change of temperature in thermoelasticity.
Proof:
If the independent variables are
E
{\displaystyle {\boldsymbol {E}}}
and
T
{\displaystyle T}
, then
η
=
η
^
(
E
,
T
)
⟹
η
˙
=
∂
η
^
∂
E
:
E
˙
+
∂
η
^
∂
T
T
˙
.
{\displaystyle \eta ={\hat {\eta }}({\boldsymbol {E}},T)\qquad \implies \qquad {\dot {\eta }}={\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\dot {\boldsymbol {E}}}+{\frac {\partial {\hat {\eta }}}{\partial T}}~{\dot {T}}~.}
On the other hand, if we consider
S
{\displaystyle {\boldsymbol {S}}}
and
T
{\displaystyle T}
to be the independent
variables
η
=
η
~
(
S
,
T
)
⟹
η
˙
=
∂
η
~
∂
S
:
S
˙
+
∂
η
~
∂
T
T
˙
.
{\displaystyle \eta ={\tilde {\eta }}({\boldsymbol {S}},T)\qquad \implies \qquad {\dot {\eta }}={\frac {\partial {\tilde {\eta }}}{\partial {\boldsymbol {S}}}}:{\dot {\boldsymbol {S}}}+{\frac {\partial {\tilde {\eta }}}{\partial T}}~{\dot {T}}~.}
Since
∂
η
^
∂
E
=
−
1
ρ
0
∂
S
^
∂
T
;
∂
η
^
∂
T
=
C
v
T
;
∂
η
~
∂
S
=
1
ρ
0
∂
E
~
∂
T
;
and
∂
η
~
∂
T
=
1
T
(
C
p
−
1
ρ
0
S
:
∂
E
~
∂
T
)
{\displaystyle {\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}=-{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}~;~~{\frac {\partial {\hat {\eta }}}{\partial T}}={\cfrac {C_{v}}{T}}~;~~{\frac {\partial {\tilde {\eta }}}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~;~~{\text{and}}~~{\frac {\partial {\tilde {\eta }}}{\partial T}}={\cfrac {1}{T}}\left(C_{p}-{\cfrac {1}{\rho _{0}}}{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}\right)}
we have, either
η
˙
=
−
1
ρ
0
∂
S
^
∂
T
:
E
˙
+
C
v
T
T
˙
{\displaystyle {\dot {\eta }}=-{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}:{\dot {\boldsymbol {E}}}+{\cfrac {C_{v}}{T}}~{\dot {T}}}
or
η
˙
=
1
ρ
0
∂
E
~
∂
T
:
S
˙
+
1
T
(
C
p
−
1
ρ
0
S
:
∂
E
~
∂
T
)
T
˙
.
{\displaystyle {\dot {\eta }}={\cfrac {1}{\rho _{0}}}~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}+{\cfrac {1}{T}}\left(C_{p}-{\cfrac {1}{\rho _{0}}}{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}\right)~{\dot {T}}~.}
The equation for balance of energy in terms of the specific entropy is
ρ
T
η
˙
=
−
∇
⋅
q
+
ρ
s
.
{\displaystyle \rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}
Using the two forms of
η
˙
{\displaystyle {\dot {\eta }}}
, we get two forms of the energy equation:
−
ρ
ρ
0
T
∂
S
^
∂
T
:
E
˙
+
ρ
C
v
T
˙
=
−
∇
⋅
q
+
ρ
s
{\displaystyle -{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}:{\dot {\boldsymbol {E}}}+\rho ~C_{v}~{\dot {T}}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s}
and
ρ
ρ
0
T
∂
E
~
∂
T
:
S
˙
+
ρ
C
p
T
˙
−
ρ
ρ
0
S
:
∂
E
~
∂
T
T
˙
=
−
∇
⋅
q
+
ρ
s
.
{\displaystyle {\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}+\rho ~C_{p}~{\dot {T}}-{\cfrac {\rho }{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~{\dot {T}}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}
From Fourier's law of heat conduction
q
=
−
κ
⋅
∇
T
.
{\displaystyle \mathbf {q} =-{\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla }}T~.}
Therefore,
−
ρ
ρ
0
T
∂
S
^
∂
T
:
E
˙
+
ρ
C
v
T
˙
=
∇
⋅
(
κ
⋅
∇
T
)
+
ρ
s
{\displaystyle -{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}:{\dot {\boldsymbol {E}}}+\rho ~C_{v}~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s}
and
ρ
ρ
0
T
∂
E
~
∂
T
:
S
˙
+
ρ
C
p
T
˙
−
ρ
ρ
0
S
:
∂
E
~
∂
T
T
˙
=
∇
⋅
(
κ
⋅
∇
T
)
+
ρ
s
.
{\displaystyle {\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}+\rho ~C_{p}~{\dot {T}}-{\cfrac {\rho }{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s~.}
Rearranging,
ρ
C
v
T
˙
=
∇
⋅
(
κ
⋅
∇
T
)
+
ρ
s
+
ρ
ρ
0
T
∂
S
^
∂
T
:
E
˙
{\displaystyle {\rho ~C_{v}~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s+{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}:{\dot {\boldsymbol {E}}}}}
or,
ρ
(
C
p
−
1
ρ
0
S
:
∂
E
~
∂
T
)
T
˙
=
∇
⋅
(
κ
⋅
∇
T
)
+
ρ
s
−
ρ
ρ
0
T
∂
E
~
∂
T
:
S
˙
.
{\displaystyle {\rho ~\left(C_{p}-{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}\right)~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s-{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}~.}}