WLF model for the pressure dependence of zero shear viscosity of polycarbonate

Abstract

Zero shear viscosity data of an amorphous polycarbonate (PC) were obtained by pressure-sweep tests at seven pressures from 1 to 700 bar, at 160, 180, 200, 220, and 240 °C above the glass transition temperature T _g . Independent shifts of the logarithmic zero shear viscosity and of the pressure were performed in order to build an empirical master curve at the reference temperature T ₀ = 240 °C. On the basis of a Vogel-type model, analytical expressions which fit the empirical logarithmic zero shear viscosity and the pressure shift factors were obtained. Additionally, a general equation of the zero shear viscosity as a function of pressure at any reference temperature was deduced. It is shown that the temperature dependence of the zero shear viscosity follows the Williams-Landel-Ferry (WLF) equation, with a pressure dependence parameter B(P) = B(0)$\left(1-4.4\times {10}^{-4}\frac{1}{\mathrm{bar}}P\right)$, B(0) = 02.87 from which it is obtained that the pressure dependence of fractional free volume of PC at the glass transition temperature is f(T _g , P) = 0.022 B(P).

Appendices

Appendix 1 Derivation of the fractional free volume at T_g (Eq. 21)

From the value $\frac{f\left({T}_0,0\right)}{B(0)}=0.17$ we can obtain the fractional free volume $\frac{f\left({T}_g,0\right)}{B(0)}$ at the glass transition temperature T _g . This can be done by taking into account that the temperature dependence of the free volume is

$$ \frac{f\left(T,0\right)}{B(0)}=\frac{f\left({T}_0,0\right)}{B(0)}+\frac{\alpha_f(0)}{B(0)}\left(T-{T}_0\right) $$

(A1)

If T=T _g , then

$$ \frac{f\left({T}_g,0\right)}{B(0)}=\frac{f\left({T}_0,0\right)}{B(0)}+\frac{\alpha_f(0)}{B(0)}\left({T}_g-{T}_0\right)=0.17+\left[\frac{1}{698.4}\frac{1}{{}^{\circ}\mathrm{C}}\right]\left(136.5-240\right)=0,022 $$

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Appendix 2: Derivation of the master curve (Eq. 26)

Equation 3 is written for the reference temperature T ₀:${a}_{\eta}\left(T,P;{T}_0\right)=\frac{\eta \left(T,P\right)}{\eta \left({T}_0,P\right)}$. For a new reference temperature T _ref  ≠ T ₀ we get

$$ {a}_{\eta}\left(T,P;{T}_{ref}\right)=\frac{\eta \left(T,P\right)}{\eta \left({T}_{ref},P\right)} $$

(A2)

By dividing these expressions, we obtain

$$ \frac{a_{\eta}\left(T,P;{T}_{ref}\right)}{a_{\eta}\left(T,P;{T}_0\right)}=\frac{\frac{\eta \left(T,P\right)}{\eta \left({T}_{ref},P\right)}}{\frac{\eta \left(T,P\right)}{\eta \left({T}_0,P\right)}}\kern0.36em \to \frac{a_{\eta}\left(T,P;{T}_{ref}\right)}{a_{\eta}\left(T,P;{T}_0\right)}=\frac{1}{\frac{\eta \left({T}_{ref},P\right)}{\eta \left({T}_0,P\right)}} $$

(A3)

The right-hand side of the last relation is $\frac{\eta \left({T}_{ref},P\right)}{\eta \left({T}_0,P\right)}={a}_{\eta}\left({T}_{ref},P;{T}_0\right)$. Therefore, we obtain

$$ \frac{a_{\eta}\left(T,P;{T}_{ref}\right)}{a_{\eta}\left(T,P;{T}_0\right)}=\frac{1}{a_{\eta}\left({T}_{ref},P;{T}_0\right)} $$

(A4)

From this equation, we have ${a}_{\eta}\left(T,P;{T}_{ref}\right)=\frac{a_{\eta}\left(T,P;{T}_0\right)}{a_{\eta}\left({T}_{ref},P;{T}_0\right)}$ and therefore

$$ \log {a}_{\eta}\left(T,P;{T}_{ref}\right)= \log {a}_{\eta}\left(T,P;{T}_0\right)- \log {a}_{\eta}\left({T}_{ref},P;{T}_0\right). $$

(A5)

In particular, for zero pressure we have

$$ \log {a}_{\eta}\left(T,0;{T}_{ref}\right)= \log {a}_{\eta}\left(T,0;{T}_0\right)- \log {a}_{\eta}\left({T}_{ref},0;{T}_0\right). $$

(A6)

Taking into account that $\log {a}_{\eta}\left(T,0;{T}_0\right)=-2.56\left[\frac{T-{T}_0}{\left(T-{T}_0\right)+118.46{}^{\circ}\mathrm{C}}\right]$ and $\log {a}_{\eta}\left({T}_{ref},0;{T}_0\right)=-2.56\left[\frac{T_{ref}-{T}_0}{\left({T}_{ref}-{T}_0\right)+118.46{}^{\circ}\mathrm{C}}\right]$

we obtain

$$ \log {a}_{\eta}\left(T,0;{T}_{ref}\right)=-2.56\left[\frac{T-{T}_0}{\left(T-{T}_0\right)+118.46{}^{\circ}\mathrm{C}}\right]+2.56\left[\frac{T_{ref}-{T}_0}{\left({T}_{ref}-{T}_0\right)+118.46{}^{\circ}\mathrm{C}}\right] $$

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