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

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