In fluid dynamics , the general equation of heat transfer is a nonlinear partial differential equation describing specific entropy production in a Newtonian fluid subject to thermal conduction and viscous forces :[ 1] [ 2]
ρ
T
D
s
D
t
⏟
Heat Gain
=
∇
⋅
(
κ
∇
T
)
⏟
Thermal Conduction
+
μ
2
(
∂
v
i
∂
x
j
+
∂
v
j
∂
x
i
−
2
3
δ
i
j
∇
⋅
v
)
2
+
ζ
(
∇
⋅
v
)
2
⏟
Viscous Dissipation
{\displaystyle \underbrace {\rho T{Ds \over {Dt}}} _{\text{Heat Gain}}=\underbrace {\nabla \cdot (\kappa \nabla T)} _{\text{Thermal Conduction}}+\underbrace {{\mu \over {2}}\left({\partial v_{i} \over {\partial x_{j}}}+{\partial v_{j} \over {\partial x_{i}}}-{2 \over {3}}\delta _{ij}\nabla \cdot {\bf {v}}\right)^{2}+\zeta (\nabla \cdot {\bf {v}})^{2}} _{\text{Viscous Dissipation}}}
where
s
{\displaystyle s}
is the specific entropy,
ρ
{\displaystyle \rho }
is the fluid's density ,
T
{\displaystyle T}
is the fluid's temperature ,
D
/
D
t
{\displaystyle D/Dt}
is the material derivative ,
κ
{\displaystyle \kappa }
is the thermal conductivity ,
μ
{\displaystyle \mu }
is the dynamic viscosity ,
ζ
{\displaystyle \zeta }
is the second Lamé parameter ,
v
{\displaystyle {\bf {v}}}
is the flow velocity ,
∇
{\displaystyle \nabla }
is the del operator used to characterize the gradient and divergence , and
δ
i
j
{\displaystyle \delta _{ij}}
is the Kronecker delta .
If the flow velocity is negligible, the general equation of heat transfer reduces to the standard heat equation . It may also be extended to rotating , stratified flows , such as those encountered in geophysical fluid dynamics .[ 3]
Extension of the ideal fluid energy equation
edit
For a viscous, Newtonian fluid, the governing equations for mass conservation and momentum conservation are the continuity equation and the Navier-Stokes equations :
∂
ρ
∂
t
=
−
∇
⋅
(
ρ
v
)
ρ
D
v
D
t
=
−
∇
p
+
∇
⋅
σ
{\displaystyle {\begin{aligned}{\partial \rho \over {\partial t}}&=-\nabla \cdot (\rho {\bf {v}})\\\rho {D{\bf {v}} \over {Dt}}&=-\nabla p+\nabla \cdot \sigma \end{aligned}}}
where
p
{\displaystyle p}
is the pressure and
σ
{\displaystyle \sigma }
is the viscous stress tensor , with the components of the viscous stress tensor given by:
σ
i
j
=
μ
(
∂
v
i
∂
x
j
+
∂
v
j
∂
x
i
−
2
3
δ
i
j
∇
⋅
v
)
+
ζ
δ
i
j
∇
⋅
v
{\displaystyle \sigma _{ij}=\mu \left({\partial v_{i} \over {\partial x_{j}}}+{\partial v_{j} \over {\partial x_{i}}}-{2 \over {3}}\delta _{ij}\nabla \cdot {\bf {v}}\right)+\zeta \delta _{ij}\nabla \cdot {\bf {v}}}
The energy of a unit volume of the fluid is the sum of the kinetic energy
ρ
v
2
/
2
≡
ρ
k
{\displaystyle \rho v^{2}/2\equiv \rho k}
and the internal energy
ρ
ε
{\displaystyle \rho \varepsilon }
, where
ε
{\displaystyle \varepsilon }
is the specific internal energy. In an ideal fluid, as described by the Euler equations , the conservation of energy is defined by the equation:
∂
∂
t
[
ρ
(
k
+
ε
)
]
+
∇
⋅
[
ρ
v
(
k
+
h
)
]
=
0
{\displaystyle {\partial \over {\partial t}}\left[\rho (k+\varepsilon )\right]+\nabla \cdot \left[\rho {\bf {v}}(k+h)\right]=0}
where
h
{\displaystyle h}
is the specific enthalpy . However, for conservation of energy to hold in a viscous fluid subject to thermal conduction, the energy flux due to advection
ρ
v
(
k
+
h
)
{\displaystyle \rho {\bf {v}}(k+h)}
must be supplemented by a heat flux given by Fourier's law
q
=
−
κ
∇
T
{\displaystyle {\bf {q}}=-\kappa \nabla T}
and a flux due to internal friction
−
σ
⋅
v
{\displaystyle -\sigma \cdot {\bf {v}}}
. Then the general equation for conservation of energy is:
∂
∂
t
[
ρ
(
k
+
ε
)
]
+
∇
⋅
[
ρ
v
(
k
+
h
)
−
κ
∇
T
−
σ
⋅
v
]
=
0
{\displaystyle {\partial \over {\partial t}}\left[\rho (k+\varepsilon )\right]+\nabla \cdot \left[\rho {\bf {v}}(k+h)-\kappa \nabla T-\sigma \cdot {\bf {v}}\right]=0}
Equation for entropy production
edit
Note that the thermodynamic relations for the internal energy and enthalpy are given by:
ρ
d
ε
=
ρ
T
d
s
+
p
ρ
d
ρ
ρ
d
h
=
ρ
T
d
s
+
d
p
{\displaystyle {\begin{aligned}\rho d\varepsilon &=\rho Tds+{p \over {\rho }}d\rho \\\rho dh&=\rho Tds+dp\end{aligned}}}
We may also obtain an equation for the kinetic energy by taking the dot product of the Navier-Stokes equation with the flow velocity
v
{\displaystyle {\bf {v}}}
to yield:
ρ
D
k
D
t
=
−
v
⋅
∇
p
+
v
i
∂
σ
i
j
∂
x
j
{\displaystyle \rho {Dk \over {Dt}}=-{\bf {v}}\cdot \nabla p+v_{i}{\partial \sigma _{ij} \over {\partial x_{j}}}}
The second term on the righthand side may be expanded to read:
v
i
∂
σ
i
j
∂
x
j
=
∂
∂
x
j
(
σ
i
j
v
i
)
−
σ
i
j
∂
v
i
∂
x
j
≡
∇
⋅
(
σ
⋅
v
)
−
σ
i
j
∂
v
i
∂
x
j
{\displaystyle {\begin{aligned}v_{i}{\partial \sigma _{ij} \over {\partial x_{j}}}&={\partial \over {\partial x_{j}}}\left(\sigma _{ij}v_{i}\right)-\sigma _{ij}{\partial v_{i} \over {\partial x_{j}}}\\&\equiv \nabla \cdot (\sigma \cdot {\bf {v}})-\sigma _{ij}{\partial v_{i} \over {\partial x_{j}}}\end{aligned}}}
With the aid of the thermodynamic relation for enthalpy and the last result, we may then put the kinetic energy equation into the form:
ρ
D
k
D
t
=
−
ρ
v
⋅
∇
h
+
ρ
T
v
⋅
∇
s
+
∇
⋅
(
σ
⋅
v
)
−
σ
i
j
∂
v
i
∂
x
j
{\displaystyle \rho {Dk \over {Dt}}=-\rho {\bf {v}}\cdot \nabla h+\rho T{\bf {v}}\cdot \nabla s+\nabla \cdot (\sigma \cdot {\bf {v}})-\sigma _{ij}{\partial v_{i} \over {\partial x_{j}}}}
Now expanding the time derivative of the total energy, we have:
∂
∂
t
[
ρ
(
k
+
ε
)
]
=
ρ
∂
k
∂
t
+
ρ
∂
ε
∂
t
+
(
k
+
ε
)
∂
ρ
∂
t
{\displaystyle {\partial \over {\partial t}}\left[\rho (k+\varepsilon )\right]=\rho {\partial k \over {\partial t}}+\rho {\partial \varepsilon \over {\partial t}}+(k+\varepsilon ){\partial \rho \over {\partial t}}}
Then by expanding each of these terms, we find that:
ρ
∂
k
∂
t
=
−
ρ
v
⋅
∇
k
−
ρ
v
⋅
∇
h
+
ρ
T
v
⋅
∇
s
+
∇
⋅
(
σ
⋅
v
)
−
σ
i
j
∂
v
i
∂
x
j
ρ
∂
ε
∂
t
=
ρ
T
∂
s
∂
t
−
p
ρ
∇
⋅
(
ρ
v
)
(
k
+
ε
)
∂
ρ
∂
t
=
−
(
k
+
ε
)
∇
⋅
(
ρ
v
)
{\displaystyle {\begin{aligned}\rho {\partial k \over {\partial t}}&=-\rho {\bf {v}}\cdot \nabla k-\rho {\bf {v}}\cdot \nabla h+\rho T{\bf {v}}\cdot \nabla s+\nabla \cdot (\sigma \cdot {\bf {v}})-\sigma _{ij}{\partial v_{i} \over {\partial x_{j}}}\\\rho {\partial \varepsilon \over {\partial t}}&=\rho T{\partial s \over {\partial t}}-{p \over {\rho }}\nabla \cdot (\rho {\bf {v}})\\(k+\varepsilon ){\partial \rho \over {\partial t}}&=-(k+\varepsilon )\nabla \cdot (\rho {\bf {v}})\end{aligned}}}
And collecting terms, we are left with:
∂
∂
t
[
ρ
(
k
+
ε
)
]
+
∇
⋅
[
ρ
v
(
k
+
h
)
−
σ
⋅
v
]
=
ρ
T
D
s
D
t
−
σ
i
j
∂
v
i
∂
x
j
{\displaystyle {\partial \over {\partial t}}\left[\rho (k+\varepsilon )\right]+\nabla \cdot \left[\rho {\bf {v}}(k+h)-\sigma \cdot {\bf {v}}\right]=\rho T{Ds \over {Dt}}-\sigma _{ij}{\partial v_{i} \over {\partial x_{j}}}}
Now adding the divergence of the heat flux due to thermal conduction to each side, we have that:
∂
∂
t
[
ρ
(
k
+
ε
)
]
+
∇
⋅
[
ρ
v
(
k
+
h
)
−
κ
∇
T
−
σ
⋅
v
]
=
ρ
T
D
s
D
t
−
∇
⋅
(
κ
∇
T
)
−
σ
i
j
∂
v
i
∂
x
j
{\displaystyle {\partial \over {\partial t}}\left[\rho (k+\varepsilon )\right]+\nabla \cdot \left[\rho {\bf {v}}(k+h)-\kappa \nabla T-\sigma \cdot {\bf {v}}\right]=\rho T{Ds \over {Dt}}-\nabla \cdot (\kappa \nabla T)-\sigma _{ij}{\partial v_{i} \over {\partial x_{j}}}}
However, we know that by the conservation of energy on the lefthand side is equal to zero, leaving us with:
ρ
T
D
s
D
t
=
∇
⋅
(
κ
∇
T
)
+
σ
i
j
∂
v
i
∂
x
j
{\displaystyle \rho T{Ds \over {Dt}}=\nabla \cdot (\kappa \nabla T)+\sigma _{ij}{\partial v_{i} \over {\partial x_{j}}}}
The product of the viscous stress tensor and the velocity gradient can be expanded as:
σ
i
j
∂
v
i
∂
x
j
=
μ
(
∂
v
i
∂
x
j
+
∂
v
j
∂
x
i
−
2
3
δ
i
j
∇
⋅
v
)
∂
v
i
∂
x
j
+
ζ
δ
i
j
∂
v
i
∂
x
j
∇
⋅
v
=
μ
2
(
∂
v
i
∂
x
j
+
∂
v
j
∂
x
i
−
2
3
δ
i
j
∇
⋅
v
)
2
+
ζ
(
∇
⋅
v
)
2
{\displaystyle {\begin{aligned}\sigma _{ij}{\partial v_{i} \over {\partial x_{j}}}&=\mu \left({\partial v_{i} \over {\partial x_{j}}}+{\partial v_{j} \over {\partial x_{i}}}-{2 \over {3}}\delta _{ij}\nabla \cdot {\bf {v}}\right){\partial v_{i} \over {\partial x_{j}}}+\zeta \delta _{ij}{\partial v_{i} \over {\partial x_{j}}}\nabla \cdot {\bf {v}}\\&={\mu \over {2}}\left({\partial v_{i} \over {\partial x_{j}}}+{\partial v_{j} \over {\partial x_{i}}}-{2 \over {3}}\delta _{ij}\nabla \cdot {\bf {v}}\right)^{2}+\zeta (\nabla \cdot {\bf {v}})^{2}\end{aligned}}}
Thus leading to the final form of the equation for specific entropy production:
ρ
T
D
s
D
t
=
∇
⋅
(
κ
∇
T
)
+
μ
2
(
∂
v
i
∂
x
j
+
∂
v
j
∂
x
i
−
2
3
δ
i
j
∇
⋅
v
)
2
+
ζ
(
∇
⋅
v
)
2
{\displaystyle \rho T{Ds \over {Dt}}=\nabla \cdot (\kappa \nabla T)+{\mu \over {2}}\left({\partial v_{i} \over {\partial x_{j}}}+{\partial v_{j} \over {\partial x_{i}}}-{2 \over {3}}\delta _{ij}\nabla \cdot {\bf {v}}\right)^{2}+\zeta (\nabla \cdot {\bf {v}})^{2}}
In the case where thermal conduction and viscous forces are absent, the equation for entropy production collapses to
D
s
/
D
t
=
0
{\displaystyle Ds/Dt=0}
- showing that ideal fluid flow is isentropic .
^ a b Landau, L.D. ; Lifshitz, E.M. (1987). Fluid Mechanics (PDF) . Course of Theoretical Physics . Vol. 6 (2nd ed.). Butterworth-Heinemann. pp. 192–194. ISBN 978-0-7506-2767-2 . OCLC 936858705 .
^ Kundu, P.K.; Cohen, I.M.; Dowling, D.R. (2012). Fluid Mechanics (5th ed.). Academic Press. pp. 123–125. ISBN 978-0-12-382100-3 .
^ Pedlosky, J. (2003). Waves in the Ocean and Atmosphere: Introduction to Wave Dynamics . Springer. p. 19. ISBN 978-3540003403 .
^ Laguerre, Onrawee (2010-05-21), Farid, Mohammed M. (ed.), "Heat Transfer and Air Flow in a Domestic Refrigerator" , Mathematical Modeling of Food Processing (1 ed.), CRC Press, pp. 453–482, doi :10.1201/9781420053548-20 , ISBN 978-0-429-14217-8 , retrieved 2023-05-07
^ Swift, G. W.; Wardt, W. C. (October–December 1996). "Simple Harmonic Analysis of Regenerators" . Journal of Thermophysics and Heat Transfer . 10 (4): 652–662. doi :10.2514/3.842 .
^ Cuffey, K. M. (2010). The physics of glaciers . W. S. B. Paterson (4th ed.). Burlington, MA. ISBN 978-0-12-369461-4 . OCLC 488732494 . {{cite book }}
: CS1 maint: location missing publisher (link )