Ok, I am very rusty in thermodynamics. In fluid mechanics books, you can find the heat transport equation in terms of temperature $T$: $$ \rho c_p\left(\frac{\partial T}{\partial t}+\boldsymbol{u}\cdot\nabla T\right)=\nabla\cdot(\kappa\nabla T), $$ where $\rho$ is the density, $c_P$ the specific heat at constant pressure, $\boldsymbol{u}$ the flow field and $\kappa$ the thermal conductivity. Now I've seen that in terms of the specific entropy $S$, this equation can be rewritten as $$ \rho T\left(\frac{\partial S}{\partial t}+\boldsymbol{u}\cdot\nabla S\right)=\nabla\cdot(\kappa\nabla T), $$ From my understanding, we obtain this equation simply by setting $c_p\Delta T=T\Delta S$, and writing the derivatives as $$ \rho c_p\frac{\partial T}{\partial t}=\rho c_p\frac{\Delta T}{\Delta t}=\rho T\frac{\Delta S}{\Delta t}=\rho T\frac{\partial S}{\partial t}, $$ and similarly for the advective term.

Now, what I'd like to know, can't we do the same thing with the gradient term so that $$ \rho T\left(\frac{\partial S}{\partial t}+\boldsymbol{u}\cdot\nabla S\right)=\nabla\cdot\left(\frac{\kappa T}{c_p}\nabla S\right)? $$

  • $\begingroup$ Why do you think you can/cannot do so? $\endgroup$
    – Kyle Kanos
    Commented Feb 10 at 16:36
  • $\begingroup$ What you are trying to do is very interesting ;and inspired. Help in carrying out this specific analysis can be found in problem 11.D-1 Balance off Entropy Equation in Transport Phenomena by Bird, Stewart, and Lightfoot. $\endgroup$ Commented Feb 10 at 21:14

2 Answers 2


If you divide your 2nd equation by T, you get $$\rho\frac{DS}{Dt}=\frac{1}{T}\nabla \centerdot\left(k\nabla{T}\right)=\nabla \centerdot\left(\frac{k\nabla{T}}{T}\right)+\frac{k(\nabla{T})^2}{T^2}$$ The term in parenthesis in the divergenc\e terms on the right hand side can be written as $$\frac{k\nabla{T}}{T}=-\frac{\mathbf{q}}{T}=-\mathbf{\dot{S}}$$where $\mathbf{q}$ is the heat flux vector and $\mathbf{\dot{S}}$ is the entropy flux vector. The second term on the rhs can be identified as the positive definite rate of entropy generation per unit volume $g_s$. So the entropy balance equation becomes:$$\rho\frac{DS}{Dt}=-\nabla \centerdot{\mathbf{\dot{S}}}+g_s$$

  • 1
    $\begingroup$ In this case, it is interesting to note that, if I understood correctly, the starting point is to write $\frac{DS}{Dt}=- \frac{\vec{\nabla} \cdot \vec{j} }{T}=\frac{\delta Q }{T}$ although the transformation is not reversible. $\endgroup$ Commented Feb 11 at 16:00
  • $\begingroup$ No. Only the first equality is correct. Then, the product rule for differentiation is used: ud v= d(uv) - vdu $\endgroup$ Commented Feb 11 at 19:14
  • $\begingroup$ Can we express the entropy flux vector in terms of derivatives of the entropy? $\endgroup$ Commented Feb 11 at 20:40
  • $\begingroup$ No. Entropy transport is brought about by heat flow. $\endgroup$ Commented Feb 11 at 22:38
  • $\begingroup$ I forgot the $\rho$ but otherwise, the "first equality" is precisely what I wrote ? $\endgroup$ Commented Feb 12 at 7:37

Differentials. Once you know the relation $d a = f(\mathbf{z}) db$, where $f(\mathbf{z})$ is a known function, you can replace every differential operator acting $a$ with the corresponding differential operator acting on $b$, s.t.

$$\begin{aligned} \partial_t a & = f(\mathbf{z}) \partial_t b \\ \nabla a & = f(\mathbf{z}) \nabla b \ . \end{aligned}$$


  • the question is: is it true that $d T = \frac{T}{c_P} ds$?
  • and the answer is, in general, no.
  • a particular case for which the equations and manipulations above are true:
    • ideal gas: for an ideal gas, $$\begin{aligned} h & = c_P T \quad \text{with $c_P$=const. s.t.} \quad dh = c_P dT \\ ds & = c_P \frac{dT}{T} + R \frac{dP}{P} \end{aligned}$$
    • constant pressure, so that $dP = 0$, and thus $ ds = \frac{c_P}{T} dT$, and a term $\frac{DP}{Dt}$ appearing in the equations below is identically zero.

Differential equations in your question. I'd like to check the differential equations in your question, starting from the very physical principles, and highlighting the assumptions made to derive them.

I'd start with the total energy balance equation,

$$\rho \frac{D e^t}{Dt} = \rho \mathbf{g} +\nabla \cdot (\mathbb{T} \cdot \mathbf{u}) - \nabla \cdot \mathbf{q}$$

and the internal energy balance equation, derived from the difference of total energy and kinetic energy (derived from the dot product of the velocity field with the momentum equation)

$$\begin{aligned} & \rho \frac{D k}{Dt} = \rho \mathbf{g} \cdot \mathbf{u} + \nabla \cdot \mathbb{T} \cdot \mathbf{u} && \text{(kinetic energy)} \\ & \rho \frac{D e}{Dt} = \mathbb{T} : \nabla \mathbf{u} - \nabla \cdot \mathbf{q} && \text{(internal energy)} \end{aligned}$$

Now, for fluids, using the relation between the differentials of internal energy, entropy and density, $$ de = T ds + \frac{P}{\rho^2} d\rho \ ,$$ and the balance of mass $$ \frac{D \rho}{Dt} = - \rho \nabla \cdot \mathbf{u} $$, along with the constitutive equation for Newtonian fluids, $\mathbb{T} = -P\mathbb{I} + \mathbb{S}$, splitting the stress tensor as the sum of pressure and viscosity contribution, $\mathbb{S} = 2 \mu \mathbb{D} + \lambda (\nabla \cdot \mathbf{u}) \mathbb{I}$, with $\mathbb{D}$ the deformation velocity tensor, it's possible to derive the balance equation for entropy

$$\begin{aligned} \rho T \frac{Ds}{Dt} & = \rho \frac{De}{Dt} - \rho \frac{P}{\rho^2} \frac{D\rho}{Dt} = \\ & = -P \nabla \cdot \mathbf{u} + \mathbb{S} : \nabla \mathbf{u} - \nabla \cdot \mathbf{q} + \frac{P}{\rho} \rho \nabla \cdot \mathbf{u} = \\ & = \mathbb{S} : \nabla \mathbf{u} - \nabla \cdot \mathbf{q} \end{aligned}$$

Thus, using Fourier's law for conduction heat flux $\mathbf{q} = - k \nabla T$, the equations for the internal energy and for entropy can be written as $$\begin{aligned} \rho \frac{D e}{Dt} & = - P \nabla \cdot \mathbf{u} + \mathbb{S} : \nabla \mathbf{u} + \nabla \cdot ( k \nabla T ) && \text{(internal energy)} \\ \rho T \frac{D s}{Dt} & = \mathbb{S} : \nabla \mathbf{u} + \nabla \cdot ( k \nabla T ) && \text{(entropy)} \\ \end{aligned} \ ,$$

with viscosity contributions that are neglected in your question. The first equation can be expressed in terms of enthalpy, $h = e +\frac{P}{\rho}$, using the differential relations $dh = de + \frac{d P}{\rho} - \frac{P}{\rho^2} d \rho$, to make time derivative appear from the term $P \nabla \cdot \mathbf{u}$

$$P \nabla \cdot \mathbf{u} = -\frac{P}{\rho} \dfrac{D \rho}{Dt}$$

and manipulate the internal energy equation to get

$$ $$

For an ideal gas, or for any medium for which the diffenrential of entalpy is proportional to of the differential of $T$ through the constant-pressure heat capacity, $c_p$, $dh = c_p dT$, the balance equation of internal energy reads

$$\rho c_p \frac{D T}{D t} = \frac{D p}{Dt} + \mathbb{S} : \nabla \mathbf{u} + \nabla \cdot ( k \nabla T ) \ .$$

Now, this equation becomes the equation contained in your question if pressure is constant along the motion of the material particles $\frac{DP}{D t} = 0$, and viscous stresses are negligible.

  • $\begingroup$ fast anonymous downvoting. I'm waiting for a comment $\endgroup$
    – basics
    Commented Feb 10 at 16:28
  • $\begingroup$ I'd wager the "Editing... tomfoolery in addition to the post not really answering the question would do something for it. $\endgroup$
    – Kyle Kanos
    Commented Feb 10 at 16:38
  • $\begingroup$ Like if you aren't done writing your answer, you can delete the post, finish it & then undelete it. That's literally what most normal ppl do around here (though I imagine far more people write a complete & relevant answer before hitting the submit button). $\endgroup$
    – Kyle Kanos
    Commented Feb 10 at 16:39
  • $\begingroup$ The most relevant part, answering the question 100%, was already there. But you can keep downvoting $\endgroup$
    – basics
    Commented Feb 10 at 16:44

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