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Show that the velocity profile of a fluid that obeys Ellis's viscosity law and that flows through a tube of radius R and length L is given by expression:

$$ v(i)-v(i-1)+\Delta r\left(\phi_{0}+\phi_{1}\left|\tau_{r z}\right|^{\alpha-1}\right) \tau(i)=0 $$

and the shear stress is given by the expression:

$$ \tau(i)-\tau(i-1)-\frac{1}{i \Delta r} \cdot \frac{v(i)-v(i-1)}{\phi_{0}+\phi_{1}\left|\tau_{r z}\right|^{\alpha-1}}-\Delta r\left(\frac{P_{0}-P_{L}}{L}\right)=0 $$

I know that the Ellis's viscosity law is given as a differential equiation given by the expression:

$$ -\frac{d v_{x}}{d y}=\left(\phi_{0}+\phi_{1}\left|\tau_{y x}\right|^{\alpha-1}\right) \tau_{y x} $$

but I don't know how to prove it Thanks!

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The viscosity of the Ellis fluid is related to the shear stress by $$\eta=\frac{1}{(\phi_o+\phi_1|\tau_{rz}|^{\alpha-1})}$$The shear stress is related to the radial derivative of the axial velocity by:$$\tau_{rz}=-\eta\frac{dv_z}{dr}$$From a force balance on the fluid plug between r = 0 and arbitrary r, and between z = 0 and z = L, $$-2\pi r L\tau_{rz}+\pi r^2(P_0-P_L)=0$$From this force balance, it follows that $$\tau_{rz}=\frac{(P_0-P_L)}{2L}r$$ So combining previous equations, we have $$\tau_{rz}=-\frac{1}{(\phi_o+\phi_1|\tau_{rz}|^{\alpha-1})}\frac{dv_z}{dr}=\frac{(P_0-P_L)}{2L}r$$

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  • $\begingroup$ Thanks for your answer $\endgroup$ Aug 28 '20 at 20:42
  • $\begingroup$ I already said in the comments how to solve it. $\endgroup$ Aug 29 '20 at 0:25

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