Problem statement: Thermal conductivity of a liquid increase with the density according to the expression: $$ \lambda = \dfrac{\lambda_0}{1-b(\rho-\rho_0)}$$ where $\lambda_0, b, \rho_0 >0$. Suppose also that this material can be considered as incompressible, in such a way that the equation of state is $\text{d}\rho = -\alpha \rho \text{d}T$ (pressure is decoupled from the density, $\kappa_T \sim 0$). A not too deep layer of that liquid is heated on its free surface ($z=0$), in such a way that: $$ \lambda \left(\dfrac{\text{d}T}{\text{d}z}\right)_{z=0} = \phi_0 >0.$$ With these hypothesis, propose and solve the differential equations that give us the temperature and density as a function of depth $\rho(z), T(z)$. Check that at the bottom ($z \rightarrow -\infty$), both density and temperature reach constant values. Determine the difference between the surface temperature and the bottom temperature. For simplicity, take $\rho(z=0)=\rho_0$.

My solution: Suppose we have a liquid with a thermal conductivity that is $$ \lambda =\dfrac{\lambda_0}{1-b(\rho-\rho_0)}, \qquad \lambda_0, b, \rho_0 >0. \tag{a}$$ (I don't know what liquids verify this experimental equation, if anyone knows...)

I do some approximations:

  • It's incompressible, so the state equation is $$\text{d}\rho = -\alpha \rho \text{d}T \tag{b}$$ and of course, $\kappa_T \sim 0$. A not too deep layer of this liquid is heated on its free surface ($z=0$) so $$ \lambda \left( \dfrac{\text{d}T}{\text{d}z}\right)_{z=0} = \phi_0 >0. \tag{c}$$

Let be $$ A \equiv -\dfrac{\alpha b \phi_0}{\lambda_0} <0, \quad B\equiv \dfrac{1+b\rho_0}{b} >0, \quad C\equiv \dfrac{\rho_0 +B}{\rho_0} >0, \quad D \equiv AB <0.$$

I know that (from entropy balance): $$ \dfrac{\text{d}}{\text{d}z} \left(\lambda \dfrac{\text{d}T}{\text{d}z}\right)=0 \tag{1}$$ so $$ \lambda \dfrac{\text{d}T}{\text{d}z} = C, \quad \forall z \in (-\infty,0]. \tag{2}$$ From (c) and (2) we have $$ \lambda \dfrac{\text{d}T}{\text{d}z} = \phi_0 >0, \qquad \forall z \in (-\infty,0].\tag{3}$$ Using the chain rule in (3) and (b), we write $$ \dfrac{\text{d}T}{\text{d}z} =\dfrac{\phi_0}{\lambda} =\dfrac{\text{d}T}{\text{d}\rho} \dfrac{\text{d}\rho}{\text{d}z} = -\dfrac{1}{\alpha \rho}\dfrac{\text{d}\rho}{\text{d}z}$$ so $$ \boxed{\rho'(z) = -\dfrac{\alpha \phi_0}{\lambda_0} \rho (z) \bigl( 1- b (\rho(z)-\rho_0)\bigr)}.\tag{4}$$ The solution of Eqn. (4) is $$ \boxed{\rho(z) = \dfrac{B}{C \text{e}^{-Dz}-1}}. \tag{5}$$ Substituting the equation (5) in (3) $$ \dfrac{\text{d}T(z)}{\text{d}z} =\dfrac{\phi_0}{\lambda \bigl( \rho(z)\bigr)}$$ so $$ \boxed{-\dfrac{\alpha}{D}\text{d}T = \left( \dfrac{C \text{e}^{-Dz}-2}{C\text{e}^{-Dz}-1}\right) \text{d}z}, \tag{6}$$ and integrating this, we have $$\boxed{T(z) = E-\dfrac{1}{\alpha} \left[ Dz +\log |\text{e}^{Dz}-C|\right]}. \tag{7} $$

  • Density at the free surface: $$ \rho (0) = \rho_0$$ but at the bottom $$ \underset{z \longrightarrow -\infty}{\lim} \rho(z) = 0. \tag{8}$$

The solution of (8) at the bottom is zero and it's not physically acceptable, where's my mistake? $$ \underset{z\longrightarrow -\infty}{\lim} T(z) = \infty$$ and diverges. What's wrong?

From Non-Equilibrium Thermodynamics Theory (for example Jan Sengers, 2006):

  • Mass balance: $$ \partial_t \rho=-\nabla (\rho \textbf{v})$$ and since the liquid is motionless $\textbf{v}\equiv 0$, so the deviatoric stress tensor is $\Pi \equiv 0$ and $$ \partial_t \rho = 0 \quad \Longrightarrow \quad \rho=\rho(z).$$
  • Momentum balance: $$\partial_t (\rho \textbf{v})= -\nabla (\rho \textbf{v} \textbf{v}-\Pi)= +\textbf{f}_v-\nabla P.$$ There is hydrostatic equilibrium, then $\textbf{f}_v -\nabla P =0$. Equation (1) is from here.
  • Entropy balance: $$ \partial_t (\rho s)= -\nabla \bigl(\rho s \textbf{v} +\dfrac{1}{T} (\textbf{Q}-\sum \mu_k \textbf{J}_k)\bigr) + \dot{S} $$ where the production of entropy is $$\dot{S} =\textbf{Q} \cdot \nabla (1/T) +\dfrac{1}{T} \Phi : (\nabla \textbf{v})^{(s)} -\sum \textbf{J}_k \cdot \nabla (\mu_k /T) -\dfrac{1}{T}\sum \mu_k \xi_k$$. The equation (2) comes from here.
  • Fenomenological relation: Fourier's law $$\textbf{Q} = -\lambda \nabla T, $$ where $\lambda>0$ is the thermal conductivity.
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    $\begingroup$ Where did you get Eqn. 1 from? Why do you think it is necessary for this equation to apply to this situation? $\endgroup$ Commented May 10, 2017 at 0:04
  • $\begingroup$ @ChesterMiller, I add more information about. $\endgroup$ Commented May 10, 2017 at 6:01
  • $\begingroup$ For hydrostatic equilibrium of an incompressible fluid (which is allowed to expand thermally only), $\frac{d\rho}{dz}=-\rho \alpha T(z)$, so $\rho=\rho_0\exp{(-\alpha\int_0^z{T(z')dz'})}$ $\endgroup$ Commented May 10, 2017 at 11:55
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    $\begingroup$ Please check Eqn. 5. $\endgroup$ Commented May 11, 2017 at 14:25
  • 1
    $\begingroup$ Thank you @ChesterMiller, this are the mistake, the Egn. 5. $\endgroup$ Commented May 11, 2017 at 20:32

1 Answer 1


Chester Miller (@ChesterMiller) found the mistake: the Egn. (5).

The equation (5) is $$ \boxed{\rho(z)= \dfrac{B}{(C-2)\text{e}^{-Dz}+1} = \rho_0 \dfrac{1+b\rho_0}{\text{e}^{\frac{\alpha \phi_0}{\lambda_0} (1+b\rho_0)z}+b\rho_0}}, \tag{*}$$ so, now the limits are right $$ \rho(0)= \rho_0 >0$$ and $$\underset{z\rightarrow -\infty}{\lim} \rho(z)=\rho_0 +\dfrac{1}{b}\equiv \rho_F >0.$$ Substituting the equation (*) in (b) we obtain $$ \int_{\rho_0}^\rho \text{d}(\log \rho) =-\alpha \int_{T_0}^T \text{d}T$$ and we have $$ \boxed{T(z) = T_0 +\dfrac{1}{\alpha} \log \left( \dfrac{\text{e}^{-Dz}+ b\rho_0}{1+ b\rho_0}\right)= T_0 +\dfrac{1}{\alpha} \log \left[ \dfrac{\text{e}^{\frac{\alpha \phi_0}{\lambda_0} (1+b\rho_0)z}+ b\rho_0}{1+b\rho_0 }\right]}.$$ These limits also fit very well $$ T(0)= T_0 >0$$ and $$ \underset{z \rightarrow -\infty}{\lim}T(z) = T_0 +\dfrac{1}{\alpha} \log \left(\dfrac{b\rho_0}{1+b\rho_0}\right) = T_0 +\Delta T\equiv T_F, \quad 0 < T_F < T_0,$$ so $\Delta T <0.$

  • 1
    $\begingroup$ It seems to me your equation says $T_F<T_0$, which is correct. $\endgroup$ Commented May 11, 2017 at 23:13
  • $\begingroup$ Yes, you're right. My bad. $\endgroup$ Commented May 11, 2017 at 23:28

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