Energy balance

The free energy of the viscoelastic fluid is the sum of kinetic and elastic contributions:

$$\mathcal{F} = \int d^3 r \left\{ {1\over 2} \rho u^2 + {\et... ...sigma$}} - \log \det {\mbox{\boldmath$\sigma$}} ] \right\} \;.$$ (3.30)

where the last term represents the entropy of polymer molecules. The rate of change of the different components of the free energy can be obtained from Eqs. (3.24,3.20)

$${\partial \over \partial t} {1\over 2} u^2 = {\mbox{\boldma... ...{\boldmath$u$}_i + \sigma_{ij} \nabla_i \mbox{\boldmath$u$}_j]$$ (3.31)

$${\partial \over \partial t} \textrm{tr}{\mbox{\boldmath$\sig... ...extrm{tr} [{\mbox{\boldmath$\sigma$}} - {\mbox{\boldmath$1$}}]$$ (3.32)

$${\partial \over \partial t} \log \det {\mbox{\boldmath$\sigm... ...r}[{\mbox{\boldmath$\sigma$}}^{-1} - {\mbox{\boldmath$1$}}]\;.$$ (3.33)

The forcing provides the input of kinetic energy which is then partially dissipated due to viscosity and relaxation of polymers. The term $\sigma_{ij} \nabla_j \mbox{\boldmath$u$}_i + \sigma_{ij} \nabla_i \mbox{\boldmath$u$}_j$ has not a definite sign, and represents the exchange between kinetic and elastic energy which can goes in both direction. Summing together the different contributions one obtain the rate of change of the free energy:

$${\partial \mathcal{F} \over \partial t} = \rho \int d^3r \... ...\boldmath$1$}} + {\mbox{\boldmath$\sigma$}}^{-1}] \right\} \;.$$ (3.34)

Since the conformation tensor ${\mbox{\boldmath$\sigma$}}$ is positive definite and symmetric, it can always be decomposed as the product of two symmetric matrices ${\mbox{\boldmath$S$}}$:

$$\sigma_{ij} = S_{ik} S_{kj}$$ (3.35)

so that the last term, which represent the energy dissipation rate due to polymers, can be rewritten as:

$$\textrm{tr} [{\mbox{\boldmath$\sigma$}} - 2 {\mbox{\boldmath... ...r} [ ({\mbox{\boldmath$S$}} - {\mbox{\boldmath$S$}}^{-1})^2 ]$$ (3.36)

showing that it has a definite sign.

In the statistically steady state the average values of the free energy $\mathcal{F}$ is constant and the energy balance reads:

$$F = \nu \langle (\nabla_i u_j)^2 \rangle + { 2 \eta \nu \ove... ...+ \langle \textrm{tr}{\mbox{\boldmath$\sigma$}}^{-1}) \rangle$$ (3.37)

where $F$ is the average energy input per unit mass. Assuming that the average kinetic energy and polymer elongation have statistically constant values, it follows that the average entropy production rate vanishes

$$\langle \textrm{tr}{\mbox{\boldmath$\sigma$}}^{-1} \rangle - \textrm{tr}{\mbox{\boldmath$1$}} = 0$$ (3.38)

and the entropy of polymer molecule is conserved.