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!


1 Answer 1


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 $$\tau_{rz}=-\frac{1}{(\phi_o+\phi_1|\tau_{rz}|^{\alpha-1})}\frac{dv_z}{dr}=\frac{(P_0-P_L)}{2L}r$$

  • $\begingroup$ Thanks for your answer $\endgroup$ Commented Aug 28, 2020 at 20:42
  • $\begingroup$ I already said in the comments how to solve it. $\endgroup$ Commented Aug 29, 2020 at 0:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.