6
$\begingroup$

I consider a spherical nanoparticle (NP) that is shrinking, for instance due to a melting process. I know that the pressure $p$ is proportional to $1/R(t)$, where $R(t)$ is the radius of the NP at time $t$. Conservation of energy states \begin{equation} \rho\frac{\partial u}{\partial t}=\nabla\cdot\vec q, \end{equation} where $\rho$ is the density, $u$ is the internal energy and $\vec q$ is the heat flux given by Fourier's law, \begin{equation} \vec q=-k\nabla T. \end{equation} The classical heat equation is obtained by assuming that the volume is constant along the whole process, which leads to $du=c_vdT$. Under constant pressure, a similar equation holds, derived from \begin{equation} dh=du+d(pv), \end{equation} where $h$ is the specific enthalpy. In this case, where neither the volume nor the pressure is constant, in which way could the internal energy be related to the temperature to obtain an equation involving only $T$? The only quantity which is kept constant in this case is the density, but I don't know how to go on... Any help would be appreciated!

$\endgroup$
2
  • $\begingroup$ Have you considered the answer? $\endgroup$ – Solar Mike Apr 29 '19 at 8:52
  • $\begingroup$ If the density is constant, then the specific volume must, by definition, also be constant, and it's specific volume (not total volume) that appears in that formula for the specific enthalpy. $\endgroup$ – Daniel Hatton Dec 19 '20 at 15:30
0
$\begingroup$

Have you considered an elastic control volume around the solid? The boundary shrinks as the particle melts, and calculations can be made with extrinsic (total internal energy) rather than intrinsic variables (specific internal energy). If you draw your control volume only around the solid phase, you'll have to estimate the mass leaving the control volume as liquid: $$ \left. \frac{dm}{dt} \right|_{NP} = -\dot{m}_{liq} $$ and estimate the internal energy leaving the control volume as liquid as well as the internal energy used to melt the particle: $$\left. \frac{dU}{dt} \right|_{NP} = -\dot{m}_{liq}(u_{ls} +u_{l}) + \int_{A} \nabla \cdot \vec{q} $$ Where $u_{l}$ is the specific internal energy at which the NP melts and $u_{ls}$ is the latent heat of melting I integrated the heat flux in this case, over the surface area of the particle. The equations can be given in terms of temperature and mass if you know the specific heats ,$C_{s}$ and $C_{l}$ for solid and liquid, and . $$\left. c_{s}\left( m\frac{dT}{dt} + T\frac{dm}{dt} \right) \right|_{NP} = \int_{A}\nabla \cdot \vec{q} -\dot{m}_{liq}\left(T_{sat}c_{l} + u_{ls} \right) $$
I'm sorry to say, there's still a lot of loose ends. How fast does liquid flow off the particle surface? How will you model the change in surface area of the particle? I assumed that liquid leaves the control volume at exactly the melting temperature, and that there was no material in the control volume which was "mushy"(saturated solid-liquid). best of luck!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.