Compressible flow in a plate/plate rheometer: application to the experimental determination of reactive expansion’s models parameters for polyurethane foam

Abstract

The quality of polyurethane foam’s structure mainly depends on the manufacturing process, during what two concomitant principal exothermic chemical reactions take place: the first one creates CO₂ dissolved into the fluid matrix (initiating micro-bubbles, expansion and coarsening of the foam) and the second one leads to the polymerization. In order to follow the rheological evolution of the PU foam during the process, the evolutions of the rheological properties of the foam during the process have been measured in a plate/plate rheometer with imposed deformation (RMS 800) and imposed stress (Stresstech). An analytical solution of the mechanical balance equation for viscoelastic compressible fluid has been proposed to describe the flow in this plate/plate rheometer with radial expansion. The trace of the elastic modulus G′ and the loss modulus G′′ as a function of time highlights the different steps of the foaming and polymerization process. The temperature dependence of both reactions follows Arrhenius’ laws. The recording of G′ and G′′ allows to identify kinetic parameters of numerical models for describing the expansion of a polymer foam and taking into account both chemical reactions.

Appendix

Analytical resolution of steady state coupled shear/expansion flow in a plate/plate rheometer

The objective of this appendix is to determine stress and rate of strain fields during shear and radial expansion in a foaming sample within a parallel plate/plate rheometer submitted to sinusoidal oscillations. Due to the fact the degassing and conversion reactions, one has to consider that rheology and density are functions of time. As the temperature field can be considered as homogeneous in the plate-plate rheometer, the chemical reactions can be also considered as homogeneous, such that neither the density nor the rheological parameters depend on r or z.

We consider an upper convective Maxwell viscoelastic behaviour. The stress balance equation of the mixture is given by:

$$ \begin{array}{*{20}c} {\nabla \cdot \sigma + \rho (t)g = 0} \hfill & {where} \hfill & {\left\{ \begin{gathered} \sigma = \sigma \prime \, - p\prime I \hfill \\ \theta \frac{{\Im \sigma \prime }}{{\Im t}} + \sigma \prime = 2\eta (t)D \hfill \\ \frac{{\Im \sigma \prime }}{{\Im t}} = \frac{{\partial \sigma \prime }}{{\partial t}} + \nabla \sigma \prime \cdot u - \nabla u \cdot \sigma \prime - \sigma \prime \cdot \nabla^T u \hfill \\ \end{gathered} \right.} \hfill \\ \end{array} $$

(A1)

where σ,σ′ and p′ are respectively the Cauchy stress tensor, the viscoelastic extra-stress tensor and an arbitrary pressure. $\Im \sigma \prime /\Im t$ denotes the upper convective objective time derivative. θ is the characteristic time of the Maxwell model and η the viscosity.

One assumes the following velocity field:

$$ \left\{ {\begin{array}{*{20}c} {u_r \left( {t,r,z} \right) = V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)} \\ {u_{\theta } \left( {t,r,z} \right) = \Omega (t)rg(z)} \\ {u_z \left( {t,r,z} \right) = 0} \\ \end{array} } \right. $$

(A2)

The velocity gradient tensor is given by:

$$ \nabla u = \left[ {\begin{array}{*{20}c} {V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr}} & { - \Omega (t)g(z)} & { - V(t)f(r)\left( {\frac{2z}{{e^2 }}} \right)} \\ {\Omega (t)g(z)} & {V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{{f(r)}}{r}} & {\Omega (t)r\frac{dg}{dz}} \\ 0 & 0 & 0 \\ \end{array} } \right] $$

and the rate of strain tensor is equal to:

$$ D = sym\left( {\nabla u} \right) = \left[ {\begin{array}{*{20}c} {V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr}} & 0 & { - V(t)f(r)\left( {\frac{z}{{e^2 }}} \right)} \\ 0 & {V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{{f(r)}}{r}} & {\frac{1}{2}\Omega (t)r\frac{dg}{dz}} \\ { - V(t)f(r)\left( {\frac{z}{{e^2 }}} \right)} & {\frac{1}{2}\Omega (t)r\frac{dg}{dz}} & 0 \\ \end{array} } \right] $$

A-1 Solution for the continuity equation

We write the continuity equation at the macroscopic (on the volume V ₀ ) and microscopic levels.

• At the macroscopic level we writte a macroscopic balance equation on the volume V ₀ of the sample between the parallele plates V ₀ :

  $$ \frac{dM}{dt} = \int_{{V_0 }} {\frac{{\partial \rho }}{{\partial t}}} (t)dv + \int_{\Sigma } {\rho (t)u_r } \left( {t,r = R,z} \right)ds = 0 $$

  (A3)

The term $\int_{{V_0 }} {\frac{{\partial \rho }}{{\partial t}}dv}$ corresponds to the mass variation in the volume between the plates, and the term $\int_{\Sigma } {\rho (t)u_r } \left( {t,r = R,z} \right)ds$ denotes the mass of mixture which escapes from the rheometer through the periphery of the sample Σ (see Fig. 15). Then, taking into account that ρ is constant throughout the volume of the rheometer at each time we have: $\frac{{\partial \rho }}{{\partial t}}\int_{{V_0 }} {dv = - \rho (t)\int_{\Sigma } {u_r } \left( {t,r = R,z} \right)ds}$ such that:

$$ \frac{{\partial \rho }}{{\partial t}}2e\pi R^2 = - \rho (t)2\pi \int_{- e}^e {V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f\left( {r = R} \right)Rdz} $$

Fig. 15

Scheme of the radial expansion of the mixture out from the rheometer: V ₀ area between the two plates, Σ perimeter of this area. a frequency f ₀  = 10⁻² Hz, b frequency f ₀  = 5.10⁻² Hz, c frequency f ₀  = 10⁻¹ Hz, d frequency f ₀  = 1 Hz

By definition, f(r = R) = 1, such that:

$$ - \frac{1}{{\rho (t)}}\frac{{\partial \rho }}{{\partial t}} = \frac{{4V(t)}}{3R} $$

(A4)

• On the other and, at the microscale, the continuity equation is:

  $$ \frac{{\partial \rho }}{{\partial t}}(t) + \nabla \cdot \left( {\rho (t)u} \right)dv = 0 $$

  then, introducing the velocity field leads to:

  $$ \frac{{\partial \rho }}{{\partial t}}(t) + \rho (t)V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\left( {\frac{df}{dr} + \frac{{f(r)}}{r}} \right) = 0 $$

By integrating this equation in the volume V ₀ , one obtain:

$$ \frac{1}{{\rho (t)}}\frac{{\partial \rho }}{{\partial t}} + \frac{{4V(t)}}{{3R^2 }}\int_0^R {\left( {\frac{df}{dr} + \frac{{f(r)}}{r}} \right)rdr = 0} $$

Taking into account Eq. (A1) leads to:

$$ \int_0^R {\left( {\frac{df}{dr} + \frac{{f(r)}}{r}} \right)rdr = R} $$

and a trivial solution of this equation is:

$$ f(r) = \frac{r}{R} $$

(A5)

The radial velocity profile is then parabolic with respect to z and linear with respect to r.

A-2 Solution for the stress balance equation

Introducing the material derivative and the convective time derivative of the extra-stress tensor leads to the set of Maxwell equations:

$$ \left\{ \begin{gathered} \theta \left[ {\frac{{\partial \sigma \prime_{rr} }}{{\partial t}} + V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{rr} }}{{\partial r}} - 2V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr}\sigma \prime_{rr} + 2V(t)f(r)\left( {\frac{2z}{{e^2 }}} \right)\sigma \prime_{rz} } \right] + \sigma \prime_{rr} = 2\eta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr} \hfill \\ \theta \left[ {\frac{{\partial \sigma \prime_{{\theta \theta }} }}{{\partial t}} + V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{{\theta \theta }} }}{{\partial r}} - 2V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{{f(r)}}{r}\sigma \prime_{{\theta \theta }} - 2\Omega (t)r\frac{dg}{dz}\sigma \prime_{{\theta z}} } \right] + \sigma \prime_{{\theta \theta }} = 2\eta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{{f(r)}}{r} \hfill \\ \theta \left[ {\frac{{\partial \sigma \prime_{zz} }}{{\partial t}} + V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{zz} }}{{\partial r}}} \right] + \sigma \prime_{zz} = 0 \hfill \\ \theta \left[ {\frac{{\partial \sigma \prime_{{r\theta }} }}{{\partial t}} + V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{{r\theta }} }}{{\partial r}} - 2V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr}\sigma \prime_{{r\theta }} + V(t)f(r)\left( {\frac{2z}{{e^2 }}} \right)\sigma \prime_{{\theta z}} - \Omega (t)r\frac{dg}{dz}\sigma \prime_{rz} } \right] + \sigma \prime_{{r\theta }} = 0 \hfill \\ \theta \left[ {\frac{{\partial \sigma \prime_{rz} }}{{\partial t}} + V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{rz} }}{{\partial r}} - V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr}\sigma \prime_{rz} + V(t)f(r)\left( {\frac{2z}{{e^2 }}} \right)\sigma \prime_{zz} } \right] + \sigma \prime_{rz} = - 2\eta V(t)f(r)\left( {\frac{z}{{e^2 }}} \right) \hfill \\ \theta \left[ {\frac{{\partial \sigma \prime_{{\theta z}} }}{{\partial t}} + V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{{\theta z}} }}{{\partial r}} - V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{{f(r)}}{r}\sigma \prime_{{\theta z}} - \Omega (t)r\frac{dg}{dz}\sigma \prime_{zz} } \right] + \sigma \prime_{{\theta z}} = \eta \Omega (t)r\frac{dg}{dz} \hfill \\ \end{gathered} \right. $$

(A6)

Assuming sinusoïdal sollicitations for the upper plate of the rheometer, and introducing a complex notation for Eqs. (A1) and (A2), the partial time derivative reduces to $\partial /\partial t = i\omega = 2i\pi f_0$, where i denotes the pure imaginary number and f ₀ is the frequency of the sollicitation. Equation (A6) becomes:

$$ \left\{ \begin{gathered} \left[ {\left( {1 + i\omega \theta } \right)\sigma \prime_{rr} + \theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{rr} }}{{\partial r}} - 2\theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr}\sigma \prime_{rr} + 2\theta V(t)f(r)\left( {\frac{2z}{{e^2 }}} \right)\sigma \prime_{rz} } \right] = 2\eta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr} \hfill \\ \left[ {\left( {1 + i\omega \theta } \right)\sigma \prime_{{\theta \theta }} + \theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{{\theta \theta }} }}{{\partial r}} - 2\theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{{f(r)}}{r}\sigma \prime_{{\theta \theta }} - 2\theta \Omega (t)r\frac{dg}{dz}\sigma \prime_{{\theta z}} } \right] = 2\eta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{{f(r)}}{r} \hfill \\ \left[ {\left( {1 + i\omega \theta } \right)\sigma \prime_{zz} + \theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{zz} }}{{\partial r}}} \right] = 0 \hfill \\ \left[ {\left( {1 + i\omega \theta } \right)\sigma \prime_{{r\theta }} + \theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{{r\theta }} }}{{\partial r}} - 2\theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr}\sigma \prime_{{r\theta }} + \theta V(t)f(r)\left( {\frac{2z}{{e^2 }}} \right)\sigma \prime_{{\theta z}} - \theta \Omega (t)r\frac{dg}{dz}\sigma \prime_{rz} } \right] = 0 \hfill \\ \left[ {\left( {1 + i\omega \theta } \right)\sigma \prime_{rz} + \theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{rz} }}{{\partial r}} - \theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{df}{dr}\sigma \prime_{rz} + \theta V(t)f(r)\left( {\frac{2z}{{e^2 }}} \right)\sigma \prime_{zz} } \right] = - 2\eta V(t)f(r)\left( {\frac{z}{{e^2 }}} \right) \hfill \\ \left[ {\left( {1 + i\omega \theta } \right)\sigma \prime_{{\theta z}} + \theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)f(r)\frac{{\partial \sigma \prime_{{\theta z}} }}{{\partial r}} - \theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{{f(r)}}{r}\sigma \prime_{{\theta z}} - \theta \Omega (t)r\frac{dg}{dz}\sigma \prime_{zz} } \right] = \eta \Omega (t)r\frac{dg}{dz} \hfill \\ \end{gathered} \right. $$

(A7)

where Ω(t) = Ωexp(iωt) is imposed, and σ′(t) = σ′exp(i(ωt + φ)) and V(t) = Vexp(i(ωt + ψ)) are sinusoidal function with various time shift which are without interest in the following.

A-2.1 Particular case without expansion

When V(t) = 0, there is no radial expansion. Equation (A7) reduces to the well known equation of the plate/plate rheometer:

$$ \left\{ {\begin{array}{*{20}c} {\sigma \prime_{{\theta z}} = \eta \Omega (t)r\left( {\frac{{1 - i\omega \theta }}{{1 - \omega^2 \theta^2 }}} \right)\frac{dg}{dz}} \\ {\sigma \prime_{{\theta \theta }} = 2\eta \Omega^2 (t)r^2 \frac{{\theta \left( {1 - \omega^2 \theta^2 - 2i\omega \theta } \right)}}{{\left( {1 - \omega^2 \theta^2 } \right)^2 }}\left( {\frac{dg}{dz}} \right)^2 } \\ {\sigma \prime_{rr} = 0} \\ {\sigma \prime_{zz} = 0} \\ \end{array} } \right. $$

Then the mass balance equation implies that g = (z + e) / 2e [1]. In this case, the complex viscosity can be expressed by:

$$ \eta * = \frac{{\sigma_{{\theta z}} }}{{2\left( {\nabla_S u} \right)_{{\theta z}} }} = \frac{{\sigma \prime_{{\theta z}} }}{{2\left( {\nabla_S u} \right)_{{\theta z}} }} = \eta \prime - i\eta \prime \prime = \eta \left( {\frac{{1 - i\omega \theta }}{{1 + \omega^2 \theta^2 }}} \right) $$

(A8)

and the complex modulus by:

$$ G * = \omega \eta \prime \prime + i\omega \eta \prime = G\prime + iG\prime \prime $$

(A9)

such that:

$$ \begin{array}{*{20}c} {G\prime = \frac{\eta }{\theta }\left( {\frac{{\omega^2 \theta^2 }}{{1 + \omega^2 \theta^2 }}} \right)} \hfill & {\text{et}} \hfill & {G\prime \prime = \frac{\eta }{\theta }\left( {\frac{{\omega \theta }}{{1 + \omega^2 \theta^2 }}} \right)} \hfill \\ \end{array} $$

(A10)

A-2.2 General case

If V(t) is positive, the whole set of Eq. (A7) remains strongly coupled and we propose an analytical solution.

$$ \left\{ \begin{gathered} Z\left( {t,z} \right)\sigma \prime_{rr} + r\frac{{\partial \sigma \prime_{rr} }}{{\partial r}} - 2\sigma \prime_{rr} + 2\frac{{\left( {\frac{2z}{{e^2 }}} \right)}}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}r\sigma \prime_{rz} = \frac{{2\eta }}{\theta } \hfill \\ Z\left( {t,z} \right)\sigma \prime_{{\theta \theta }} + r\frac{{\partial \sigma \prime_{{\theta \theta }} }}{{\partial r}} - 2\sigma \prime_{{\theta \theta }} - 2\frac{{\Omega (t)}}{{V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}rR\frac{dg}{dz}\sigma \prime_{{\theta z}} = \frac{{2\eta }}{\theta } \hfill \\ Z\left( {t,z} \right)\sigma \prime_{zz} + r\frac{{\partial \sigma \prime_{zz} }}{{\partial r}} = 0 \hfill \\ Z\left( {t,z} \right)\sigma \prime_{{r\theta }} + r\frac{{\partial \sigma \prime_{{r\theta }} }}{{\partial r}} - 2\sigma \prime_{{r\theta }} + \frac{{\left( {\frac{2z}{{e^2 }}} \right)}}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}r\sigma \prime_{{\theta z}} - \frac{{\Omega (t)rR}}{{V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}\frac{dg}{dz}\sigma \prime_{rz} = 0 \hfill \\ Z(t,z)\sigma \prime_{rz} + r\frac{{\partial \sigma \prime_{rz} }}{{\partial r}} - \sigma \prime_{rz} + \frac{{\left( {\frac{2z}{{e^2 }}} \right)}}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}r\sigma \prime_{zz} = - \frac{\eta }{\theta }\frac{{\left( {\frac{2z}{{e^2 }}} \right)}}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}r \hfill \\ Z\left( {t,z} \right)\sigma \prime_{{\theta z}} + r\frac{{\partial \sigma \prime_{{\theta z}} }}{{\partial r}} - \sigma \prime_{{\theta z}} - \frac{{\Omega (t)}}{{V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}rR\frac{dg}{dz}\sigma \prime_{zz} = \frac{\eta }{\theta }\frac{{\Omega (t)}}{{V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}\frac{dg}{dz}rR \hfill \\ \end{gathered} \right. $$

where $Z\left( {t,z} \right) = \frac{{R\left( {1 + i\omega \theta } \right)}}{{\theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}} \ge 0$

• Eq. (A7) gives:

$$ \sigma \prime_{zz} = A\left( {\frac{r}{R}} \right)^{{ - Z\left( {t,z} \right)}} $$

such that A = 0 because the stress has to remain finite for r➔0 and then $\sigma \prime_{zz} = 0$.

• Eq. (A7) reduces in:

$$ \left( {Z\left( {t,z} \right) - 1} \right)\sigma \prime_{rz} + r\frac{{\partial \sigma \prime_{rz} }}{{\partial r}} = - \frac{\eta }{\theta }\frac{{\left( {\frac{2z}{{e^2 }}} \right)}}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}r $$

then

$$ \sigma \prime_{rz} = - \frac{\eta }{\theta }\frac{{\left( {\frac{2z}{{e^2 }}} \right)}}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}\frac{1}{{Z\left( {t,z} \right)}}r + B\left( {\frac{r}{R}} \right)^{{1 - Z\left( {t,z} \right)}} $$

such that B = 0 because this stress is zero on the symmetry axis.

• Eq. (A7) reduces in:

$$ \left( {Z\left( {t,z} \right) - 1} \right)\sigma \prime_{{\theta z}} + r\frac{{\partial \sigma \prime_{{\theta z}} }}{{\partial r}} = \frac{\eta }{\theta }\frac{{\Omega (t)}}{{V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}\frac{dg}{dz}rR $$

then:

$$ \sigma \prime_{{\theta z}} = \frac{\eta }{\theta }\frac{{\Omega (t)}}{{V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}\frac{dg}{dz}R\frac{1}{{Z\left( {t,z} \right)}}r + C\left( {\frac{r}{R}} \right)^{{1 - Z\left( {t,z} \right)}} $$

such that C = 0 because this stress is zero on the symmetry axis.

• Eq. (A7) becomes:

$$ \left( {Z\left( {t,z} \right) - 2} \right)\sigma \prime_{{r\theta }} + r\frac{{\partial \sigma \prime_{{r\theta }} }}{{\partial r}} = - \frac{{2\eta }}{\theta }\frac{{\Omega (t)}}{{V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}\frac{{\left( {\frac{2z}{{e^2 }}} \right)}}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}\frac{dg}{dz}R\frac{1}{{Z\left( {t,z} \right)}}r^2 $$

then

$$ \sigma \prime_{{r\theta }} = - \frac{{2\eta }}{\theta }\frac{{\Omega (t)}}{{V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}\frac{{\left( {\frac{2z}{{e^2 }}} \right)}}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}\frac{dg}{dz}R\frac{1}{{Z^2 \left( {t,z} \right)}}r^2 + D\left( {\frac{r}{R}} \right)^{{2 - Z\left( {t,z} \right)}} $$

such that D = 0 because this stress is zero on the symmetry axis.

• Eq. (A7) becomes:

$$ \left( {Z\left( {t,z} \right) - 2} \right)\sigma \prime_{rr} + r\frac{{\partial \sigma \prime_{rr} }}{{\partial r}} = \frac{{2\eta }}{\theta }\left[ {1 + \frac{1}{{Z\left( {t,z} \right)}}\frac{{\left( {\frac{2z}{{e^2 }}} \right)^2 }}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)^2 }}r^2 } \right] $$

then

$$ \sigma \prime_{rr} = \frac{{2\eta }}{\theta }\left[ {\frac{1}{{Z\left( {t,z} \right) - 2}} + \frac{1}{{Z^2 \left( {t,z} \right)}}\frac{{\left( {\frac{2z}{{e^2 }}} \right)^2 }}{{\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)^2 }}r^2 } \right] + E\left( {\frac{r}{R}} \right)^{{2 - Z\left( {t,z} \right)}} $$

such that E = 0 due to the symmetry axis in r = 0.

• Eq. (A7) becomes:

$$ \left( {Z\left( {t,z} \right) - 2} \right)\sigma \prime_{{\theta \theta }} + r\frac{{\partial \sigma \prime_{{\theta \theta }} }}{{\partial r}} = \frac{{2\eta }}{\theta }\left[ {1 + \frac{1}{{Z\left( {t,z} \right)}}\frac{{\Omega^2 (t)}}{{V^2 (t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)^2 }}\left( {\frac{dg}{dz}} \right)^2 R^2 r\,^2 } \right] $$

then

$$ \sigma \prime_{{\theta \theta }} = \frac{{2\eta }}{\theta }\left[ {\frac{1}{{Z\left( {t,z} \right) - 2}} + \frac{1}{{Z^2 \left( {t,z} \right)}}\frac{{\Omega^2 (t)}}{{V^2 (t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)^2 }}\left( {\frac{dg}{dz}} \right)^2 R^2 r^2 } \right] + F\left( {\frac{r}{R}} \right)^{{2 - Z\left( {t,z} \right)}} $$

such that F = 0 due to the symmetry axis in r = 0.

The stress tensor is then equal to:

$$ \sigma = \left[ {\begin{array}{*{20}c} {\frac{{2\eta }}{\theta }\left[ {\frac{{\theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}{{R\left( {1 + i\omega \theta } \right) - 2\theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}} + \frac{{\theta^2 }}{{\left( {1 + i\omega \theta } \right)^2 }}V(t)^2 \left( {\frac{2z}{{e^2 }}} \right)^2 \frac{{r^2 }}{{R^2 }}} \right] - p\prime } & { - 2\eta \theta \frac{{\Omega (t)}}{{\left( {1 + i\omega \theta } \right)^2 }}\left( {\frac{2z}{{e^2 }}} \right)\frac{dg}{dz}\frac{{V(t)}}{R}r^2 } & { - \frac{\eta }{{\left( {1 + i\omega \theta } \right)}}V(t)\left( {\frac{2z}{{e^2 }}} \right)\frac{r}{R}} \\ {sym} & {\frac{{2\eta }}{\theta }\left[ {\frac{{\theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}}{{R\left( {1 + i\omega \theta } \right) - 2\theta V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)}} + \theta^2 \frac{{\Omega^2 (t)}}{{\left( {1 + i\omega \theta } \right)^2 }}\left( {\frac{dg}{dz}} \right)^2 r^2 } \right] - p\prime } & {\frac{\eta }{{\left( {1 + i\omega \theta } \right)}}\Omega (t)\frac{dg}{dz}r} \\ {sym} & {sym} & { - p\prime } \\ \end{array} } \right] $$

the stress balance equation reduces to:

$$ \begin{array}{*{20}c} {\begin{array}{*{20}c} {\nabla \cdot \sigma + \rho (t)g = 0}{\text{then:}} \\ \end{array} } \\ {\left\{ {\begin{array}{*{20}c} { - \frac{{\partial p\prime }}{{\partial r}} + \frac{{2\eta }}{\theta }\frac{{\theta^2 }}{{\left( {1 + i\omega \theta } \right)^2 }}V(t)^2 \left( {\frac{2z}{{e^2 }}} \right)^2 \frac{2r}{{R^2 }} - \frac{\eta }{{\left( {1 + i\omega \theta } \right)}}V(t)\left( {\frac{2}{{e^2 }}} \right)\frac{r}{R} + \frac{{2\eta }}{\theta }\frac{{\theta^2 }}{{\left( {1 + i\omega \theta } \right)^2 }}\left[ {V(t)^2 \left( {\frac{2z}{{e^2 }}} \right)^2 - R^2 \Omega^2 (t)\left( {\frac{dg}{dz}} \right)^2 } \right]\frac{r}{{R^2 }} = 0} \hfill \\ {\frac{\eta }{{\left( {1 + i\omega \theta } \right)}}\Omega (t)\frac{{d^2 g}}{{dz^2 }}r - 8\eta \theta \frac{{\Omega (t)}}{{\left( {1 + i\omega \theta } \right)^2 }}\left( {\frac{2z}{{e^2 }}} \right)\frac{dg}{dz}\frac{{V(t)}}{R}r = 0} \hfill \\ { - \frac{{\partial p\prime }}{{\partial z}} - \frac{{2\eta }}{{\left( {1 + i\omega \theta } \right)}}V(t)\left( {\frac{2z}{{e^2 }}} \right)\frac{1}{R} - \rho (t)g = 0} \hfill \\ \end{array} } \right.} \\ \end{array} $$

where η, θ and ρ explicitely depend on time. From the second equation we obtain:

$$ \frac{{d^2 g}}{{dz^2 }} = \frac{{16\theta }}{{\left( {1 + i\omega \theta } \right)}}\frac{1}{{e^2 }}\frac{{V(t)}}{R}z\frac{dg}{dz} $$

which is equivalent to:

$$ \frac{dg}{dz} = Ke^{{\frac{{8\theta }}{{\left( {1 + i\omega \theta } \right)}}\frac{{V(t)}}{R}\frac{{z^2 }}{{e^2 }}}} $$

then:

$$ g(z) = K\int_{- e}^z {e^{{\frac{{8\theta }}{{\left( {1 + i\omega \theta } \right)}}\frac{{V(t)}}{R}\frac{{z^2 }}{{e^2 }}}} dz} $$

with g(z = −e) = 0 and g(z = e) = 1 we have:

$$ g(z) = \frac{{\int_{- e}^z {e^{{\frac{{8\theta }}{{\left( {1 + i\omega \theta } \right)}}\frac{{V(t)}}{R}\frac{{z^2 }}{{e^2 }}}} dz} }}{{\int_{- e}^e {e^{{\frac{{8\theta }}{{\left( {1 + i\omega \theta } \right)}}\frac{{V(t)}}{R}\frac{{z^2 }}{{e^2 }}}} dz} }} $$

(A11)

When V(t) = 0, one has g(z) = (z + e)/2e (see Section A-2.1).

From Eq. (A4), we have determined that the maximal value reached by the radial velocity is of about 2 mm.s⁻¹. We have then investigated the influence of V(t) in the range [0,10]mm.s⁻¹.

Figure (16) presents g(z) profile as a function of z/e for various V(t) values and for θ = 1 s, e = 1 mm and R = 50 mm (we have taken characteristic values for R and e in order to test the influence of the function g(z)).

Fig. 16

g(z) profile with respect to z/e as a function of V(t) for θ = 1 s, e = 1 mm and R = 50 mm: a f ₀ = 10⁻² Hz, b f ₀ = 5.10⁻² Hz, c f ₀ = 10⁻¹ Hz, d f ₀ = 1 Hz

It can be noticed that, in the range of experimental radial velocity (about 1 mm.s⁻¹), the profile g(z) is quasi linear such that the shear rate remains homogeneous with respect to z for θ = 1 s. This choice of θ is purely arbitrary but Eq. (A7) gives the following results:

• for f ₀ ≥ 1 Hz (corresponding to our experimental conditions), the profile g(z) remains linear for all value of θ,

• for 1 Hz ≥ f ₀ ≥ 10 ⁻⁵ Hz, the profile g(z) remains linear only if θ is lower than 10⁻¹ s.

As a consequence, for the experimental chosen frequency (2 Hz), the classical viscoelastic analysis can be used during the whole solidification process.

Equation (A7) proves that:

$$ p\prime \left( {r,z,t} \right) = P\left( {r,t} \right) - \frac{{2\eta }}{{\left( {1 + i\omega \theta } \right)}}V(t)\left( {\frac{{z^2 }}{{e^2 }}} \right)\frac{1}{R} - \rho (t)gz $$

where P is a function which need to be determined.

Equation (A7) then gives:

$$ \frac{{\partial p\prime }}{{\partial r}} = \frac{{\partial P}}{{\partial r}}\left( {r,t} \right) = - \frac{\eta }{{\left( {1 + i\omega \theta } \right)}}V(t)\left( {\frac{2}{{e^2 }}} \right)\frac{r}{R} + \frac{{2\eta }}{\theta }\frac{{\theta^2 }}{{\left( {1 + i\omega \theta } \right)^2 }}\left[ {3V(t)^2 \left( {\frac{2z}{{e^2 }}} \right)^2 - R^2 \Omega^2 (t)\left( {\frac{dg}{dz}} \right)^2 } \right]\frac{r}{{R^2 }} $$

then:

$$ P\left( {r,t} \right) = \frac{\eta }{{\left( {1 + i\omega \theta } \right)}}\left[ {\frac{\theta }{{\left( {1 + i\omega \theta } \right)}}\left( {3V(t)^2 \left( {\frac{2z}{{e^2 }}} \right)^2 - R^2 \Omega^2 (t)\left( {\frac{dg}{dz}} \right)^2 } \right)\frac{{r^2 }}{{R^2 }} - V(t)\frac{1}{{e^2 }}\frac{{r^2 }}{R}} \right] + Q(t) $$

then

$$ p\prime \left( {r,z,t} \right) = \frac{\eta }{{\left( {1 + i\omega \theta } \right)}}\left[ {\frac{\theta }{{\left( {1 + i\omega \theta } \right)}}\left( {3V(t)^2 \left( {\frac{2z}{{e^2 }}} \right)^2 - R^2 \Omega^2 (t)\left( {\frac{dg}{dz}} \right)^2 } \right)\frac{{r^2 }}{{R^2 }} - V(t)\frac{1}{{e^2 }}\frac{{r^2 }}{R}} \right] - \frac{{2\eta }}{{\left( {1 + i\omega \theta } \right)}}V(t)\left( {\frac{{z^2 }}{{e^2 }}} \right)\frac{1}{R} - \rho (t)gz + Q(t) $$

Identification of Q(t) : as a first approximation, we suppose that the mixture outside the rheometer can freely expand such that σ⋅n = 0 (n is the normal to the free surface of the mixture, which is varying between the midplane of the rheometer and the plates). We decide to impose the boundary condition at the plate periphery (see Fig. 15). So we have:

$$ \begin{array}{*{20}c} {\sigma_{zz} \left( {R,e} \right) = - \frac{\eta }{{\left( {1 + i\omega \theta } \right)}}\left[ {\frac{\theta }{{\left( {1 + i\omega \theta } \right)}}\left( {3V(t)^2 \left( {\frac{2}{e}} \right)^2 - R^2 \Omega^2 (t)\left( {\frac{dg}{dz}(e)} \right)^2 } \right) - V(t)\frac{1}{{e^2 }}R} \right] + \frac{{2\eta }}{{\left( {1 + i\omega \theta } \right)}}V(t)\frac{1}{R} + \rho (t)ge - Q(t) = 0} \hfill \\ {Q(t) = - \frac{\eta }{{\left( {1 + i\omega \theta } \right)}}\left[ {\frac{\theta }{{\left( {1 + i\omega \theta } \right)}}\left( {3V(t)^2 \left( {\frac{2}{e}} \right)^2 - R^2 \Omega^2 (t)\left( {\frac{dg}{dz}(e)} \right)^2 } \right) - V(t)\frac{1}{{e^2 }}R} \right] + \frac{{2\eta }}{{\left( {1 + i\omega \theta } \right)}}V(t)\frac{1}{R} + \rho (t)ge} \hfill \\ \end{array} $$

then:

$$ \begin{array}{*{20}c} {p\prime \left( {r,z,t} \right) = \frac{\eta }{{\left( {1 + i\omega \theta } \right)}}\left[ {\frac{\theta }{{\left( {1 + i\omega \theta } \right)}}\left( {3V(t)^2 \left( {\frac{2}{{e^2 }}} \right)^2 \left( {\frac{{z^2 r^2 }}{{R^2 }} - e^2 } \right) - \Omega^2 (t)\left( {\frac{dg}{dz}} \right)^2 r^2 + \Omega^2 (t)\left( {\frac{dg}{dz}(e)} \right)^2 R^2 } \right) - V(t)\frac{1}{{e^2 }}\frac{{R^2 - r^2 }}{R}} \right]} \\ { - \frac{{2\eta }}{{\left( {1 + i\omega \theta } \right)}}V(t)\left( {1 - \frac{{z^2 }}{{e^2 }}} \right)\frac{1}{R} - \rho (t)g\left( {z - e} \right)} \\ \end{array}$$

When θ = 0 (purely viscous case), p′ tends toward a limit value equal to:

$$ p\prime \left( {r,z,t} \right) = - \eta V(t)\frac{{R^2 - r^2 + 2\left( {e^2 - z^2 } \right)}}{{e^2 R}} - \rho (t)g\left( {z - e} \right) $$

in this case p′ depends only on V(t). If there is no radial expansion, this pressure is only equal to the gravity term (2 Pa) which may be neglectible.

A-2.3 Evaluation of flow and forces

Knowing the normal force and the torque at each time t, as well the flow rate Q _v escaping from the rheometer, it is possible to follow the evolution of viscosity η and relaxation time θ and then to have an estimation of the chemio-rheological couplings (Eqs. (8) and (9)):

• Flow

$$ Q_V (t) = \int\limits_{\Sigma } {u.nds} = \frac{{8\pi }}{3}eRV(t)f\left( {r = R} \right) = \frac{{8\pi }}{3}eRV(t) $$

(A12)

• Normal force

$$ F_N (t) = - \int\limits_{Base} {\sigma .nds} = - 2\pi \int_0^R {\sigma_{zz} \left( {z = + e} \right)rdr = 2\pi \int_0^R {p\prime \left( {r,z = + e,t} \right)rdr} } $$

Then:

$$ F_N (t) = \frac{\eta }{{\left( {1 + i\omega \theta } \right)}}\left[ {\frac{\theta }{{\left( {1 + i\omega \theta } \right)}}\left( {\frac{\pi }{2}R^4 \Omega^2 (t)\left( {\frac{1}{{\int_{- e}^e {e^{{\frac{{3\theta }}{{\left( {1 + i\omega \theta } \right)}}\frac{1}{{\pi eR^2 }}Q_V (t)\left( {\frac{{z^2 }}{{e^2 }} - 1} \right)}} dz} }}} \right)^2 - \frac{27}{{32\pi e^4 }}Q_V^2 (t)} \right) - \frac{3}{16e}Q_V (t)\frac{{R^2 }}{{e^2 }}} \right] $$

(A13)

Note also that in the case where Q _V (t) = 0, we find the well known expression of the normal force:

$$ F_N (t) = F_N^0 (t) = \frac{\pi }{2}\frac{{\eta \theta }}{{\left( {1 + i\omega \theta } \right)^2 }}\left( {\frac{{\Omega (t)}}{2e}} \right)^2 R^4 $$

or using real notations,

$$ F_N (t) = F_N^0 (t) = \frac{\pi }{2}\eta \theta \frac{{1 - \omega^2 \theta^2 }}{{\left( {1 + \omega^2 \theta^2 } \right)^2 }}\left( {\frac{{\Omega (t)}}{2e}} \right)^2 R^4 $$

The normal force is then non-zero if θ is not zero.

• Torque

$$ C(t) = \int\limits_{Base} {x \times \sigma .nds} = 2\pi \int_0^R {r\sigma_{{\theta z}} \left( {z = + e} \right)rdr} = 2\pi \int_0^R {\frac{\eta }{{\left( {1 + i\omega \theta } \right)}}\Omega (t)\frac{1}{{\int_{- e}^e {e^{{\frac{{3\theta }}{{\left( {1 + i\omega \theta } \right)}}\frac{{Q_V (t)}}{{\pi eR^2 }}\left( {\frac{{z^2 }}{{e^2 }} - 1} \right)}} dz} }}r^3 dr} $$

then:

$$ C(t) = \frac{\pi }{2}\frac{{\eta (t)}}{{\left( {1 + i\omega \theta } \right)}}\Omega (t)\frac{1}{{\int_{- e}^e {e^{{\frac{{3\theta }}{{\left( {1 + i\omega \theta } \right)}}\frac{{Q_V (t)}}{{\pi eR^2 }}\left( {\frac{{z^2 }}{{e^2 }} - 1} \right)}} dz} }}R^4 $$

(A14)

Note also that in the case where Q _V (t) = 0, we find the well known form of the torque:

$$ C(t) = C^0 (t) = \frac{\pi }{2}\frac{{\eta (t)}}{{\left( {1 + i\omega \theta } \right)}}\Omega (t)\frac{1}{2e}R^4 $$

or using real notations,

$$ C(t) = C^0 (t) = \frac{\pi }{2}\eta (t)\frac{{1 - \omega \theta }}{{\left( {1 + \omega^2 \theta^2 } \right)}}\Omega (t)\frac{1}{2e}R^4 $$

which is the classical equation used for incompressible materials.

Recording C and Fn has a function of time as well the flow rate escaping the rheometer, it would be then possible coupling Eqs. (A7), (A8) and (4) to derive viscosity and mean relaxation time as a function of time.