Phan-Thien-Tanner fluid

The affine linear PTT fluid considered here uses a symmetric dimensionless tensor $A$ with $\stackrel{\nabla }{A}+(1+\alpha \mathrm{tr}A)A/\tau =2E$ and polymer stress $G_{0}A$. Here $\stackrel{\nabla }{A}$ is the upper-convected derivative, $G_0,\tau >0$ and the nonlinear relaxation parameter $\alpha$ is usually positive. Linearizing about rest gives the Maxwell fluid relaxation modulus $G(s)=G_0{e}^{-s/\tau }$. The PTT family also admits other relaxation functions and nonaffine derivatives; their nonlinear predictions need not equal those of this specified version.

For uniaxial extensional flow put $q=\tau \dot{\gamma }>0$, $f=1+\alpha \mathrm{tr}A$ and $T=f-2q$. The diagonal components are $a_{11}=2q/T$, $a_{22}=a_{33}=-q/(T+3q)$. The identity $f-1=\alpha \mathrm{tr}A$ gives the cubic equation in the title. For $\alpha >0$ its physical root has $T>\max (0,1-2q)$ and connects to $T=1$ at rest. Its extensional viscosity is ${\eta }_E=G_0\tau [2/T+1/(T+3q)]$, tending to $3G_0\tau$ at small $q$ and $2G_0\tau /\alpha$ at large $q$. At $\alpha =0$ the stable steady Maxwell extension instead ends at $q=1/2$; the finite high-rate limit must not be extended to that degenerate case.

In steady simple shear flow, the tensor equation gives $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ and the equation in the title. A circular pipe with axial pressure gradient $\Delta p$ has $G_0{a}_{rz}=\Delta pr/2$. Thus $w^{\prime }=\Delta pr/(2G_0\tau )+\alpha \Delta p^3{r}^3/(4G_0^3\tau )$, and no-slip boundary conditions givePositive $\alpha$ produces shear thinning; the sign of the pressure gradient determines the direction of the flow.