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Rouse rotational time

The second critical shear rate is much higher (i.e., 100 times Chauveteau, 1981) than the first one. The first critical shear rate is equal to the inverse of the longest rotational relaxation time k in the solution. Dilatancy starts as soon as the product of Rouse relaxation time and the maximum stretch rate, e, is greater than 4 (Chauveteau, 1981). The Rouse relaxation time demarcates the onset of entanglement effects (Roland et al., 2004). Chauveteau reported that the ratio of shear rate y to the maximum stretch rate e at the contraction was about 2.5 by laser anemometry for similar polymer solutions and flow geometries. Therefore, the second stretch rate (elongation rate) corresponds to the product of shear rate and Rouse relaxation time equal to 10. [Pg.216]

Doi and Edwards [ 1, Eq. 4.37 and parenthetical remark after Eq. 4.157] use the symbol T, for the Rouse reorientation time (or rotational relaxation time), which is twice the quantity defined by Eq. 6.3. Note that since N = M/Mq, % is proportional to for a given polymer. [Pg.196]

Rouse reorientation (rotational) time Rouse stress relaxation time... [Pg.491]

Characteristic Rouse times for chains with bending elasticity Terminal time for reptation Crossover time Rouse, local reptation Characteristic Rouse times for the all-rotational model Rouse time... [Pg.222]

As shown in Figure 27c, the rotational relaxation time varies as a power of length, zr N with /u = 2.6 0.4. The experimental result is in remarkable agreement with the scaling behavior of the rotational relaxation zr°c A/1 l2v with theory = 2.5, which follows from the Rouse model and the Flory exponent v = 3/4.189... [Pg.385]

In summary, we have used the Rouse chain model to obtain the diffusion constant of the center of mass and the time-correlation function of the end-to-end vector, which reflects the rotational motion of the whole polymer molecule. Since N is proportional to the molecular weight M, and K is independent of molecular weight, Eqs. (3.41) and (3.62) indicate that Dq and Tr depend on the molecular weight, respectively, as... [Pg.39]

Relaxation 67,70,96,99,111,155 Reptation model 1,24,42 Resolution 14 Resonance NSE 20 Rheology 35,55 Rotational isomeric state 118 Rotational transitions 117 Rouse diffusion coefficient 28,42, 175 Rouse model 24-26,30-35,38, 117, 119, 142, 193,200 —, generalized 47 Rouse time 27 RPA 162, 163, 199... [Pg.31]

Figure 7 summarizes the data for two key dynamic observables the mean-square monomer displacement gi,mid(0 and i,end(0 and the stress relaxation fimction. Hgures 7(a) and 7(b) show the data as a limction of time in units where f= l,feBT= 1 and unit length is b in Rouse and semiflexible models, Tq in freely jointed and freely rotating models, and a in the LJ+FENE-based models. Since these units are pretty arbitrary and the local details are very different (including presence of inertia), the results are scattered and it is not easy to see similarities and differences between the models. [Pg.147]

Figure 20 Mean-square displacement <(r(Q - i 0)f of segments (solid line.) and the rotational autocorrelation function C t) (dashed line). Regime transitions occur at the segmental (ts), entanglement (re). Rouse (tr). and disentanglement (i ) times according to the RouseAube model. The figure was adapted from Chavez, F. V. Saalwachter, K. Macromolecules 44,1549. ... Figure 20 Mean-square displacement <(r(Q - i 0)f of segments (solid line.) and the rotational autocorrelation function C t) (dashed line). Regime transitions occur at the segmental (ts), entanglement (re). Rouse (tr). and disentanglement (i ) times according to the RouseAube model. The figure was adapted from Chavez, F. V. Saalwachter, K. Macromolecules 44,1549. ...

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