# Table 1 The calculation of the apparent viscosityη S and the shear rate$\stackrel{˙}{\mathbf{\gamma }}$,

Measuring system Equation Constant
Double gab measuring system $\tau =\frac{1+{\delta }^{2}}{\left({\delta }^{2}·{R}_{3}^{2}+{R}_{2}^{2}\right)}·\frac{M}{4000·\mathit{\pi L}{C}_{L}}$ C L = 1.10
δ = 1.0245
RMKi = 20.25 mm
$\stackrel{˙}{\mathrm{\gamma }}=\omega ·\frac{1+{\delta }^{2}}{{\delta }^{2}-1}$
RMKa = 21.00 mm
L = 78.7 mm
Cylinder measuring system $\tau =\frac{1+{\delta }^{2}}{2000·{\delta }^{2}}·\frac{M}{2\mathit{\pi }\mathit{L}·{R}_{i}^{2}·{C}_{L}}$ C L = 1.10
δ = 1.0848
$\stackrel{˙}{\mathrm{\gamma }}=\omega ·\frac{1+{\delta }^{2}}{{\delta }^{2}-1}$ RMKi = 13.33
L = 40.003 mm
Ball measuring cell τ = C ss ·M ${C}_{\mathrm{SS}}=15.0\frac{\mathrm{Pa}}{\text{mNm}}$
${C}_{\mathrm{SR}}=0.427\frac{\text{min}}{\mathrm{s}}$
$\stackrel{˙}{\mathrm{\gamma }}={C}_{\mathit{SR}}·n$
d K = 12.0 mm
Paddle mixer  $\mathit{Ne}=\frac{P}{p·{n}^{3}·{d}^{5}}=\frac{{C}_{\text{lam}}}{\mathit{Re}}{C}_{\text{turb}}$ CMO = 10.94
Clam = 189.6
Cturb = 9.4
$\stackrel{˙}{\mathrm{\gamma }}={C}_{\mathrm{MO}}·n$
d R = 30 mm
d B = 143 mm
Pipe viscosimeter  ${\tau }_{w}=\frac{d·△p}{4·L}$ Lpipe = 2,500 mm
dpipe = 43.2
${\left(\frac{\mathit{dw}}{\mathit{dr}}\right)}_{w}=\frac{3n\text{'}+1}{4n\text{'}}·\frac{8·{w}_{m}}{d}$
$n\text{'}=\frac{\mathit{dln}\left(\frac{d·△p}{4·L}\right)}{\mathit{dln}\phantom{\rule{0.5em}{0ex}}\frac{\left(8·{w}_{m}\right)}{d}}$ 