(a) Show that for Poiseuille flow in a tube of radius \(R\) the magnitude of the wall shearing stress, \(\tau_{r}\), can be obtained from the relationship $$\left|\left(\tau_{r_{z}}\right)_{\mathrm{will}}\right|=\frac{4 \mu Q}{\pi R^{3}}$$ for a Newtonian fluid of viscosity \(\mu .\) The volume rate of flow is \(Q\) (b) Determine the magnitude of the wall shearing stress for a fluid having a viscosity of \(0.004 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) flowing with an average velocity of \(130 \mathrm{mm} / \mathrm{s}\) in a \(2-\mathrm{mm}\) -diameter tube.

From the expression for velocity distribution in Pouseille's flow, \( V_{z} = - \frac{1}{4 \mu} r^2 + \frac{R^2}{4\mu} \), we can find the radial component of the shear stress tensor \(\tau _{r_{z}}\). Note that \( \tau _{r_{z}} = \mu \frac{\partial v}{\partial r} \). \n Computing the derivative we find that \(\tau _{r_{z}} = r \mu \). Notice how this stresses the fact that \(\tau _{r_{z}}\) is maximum when r=R, the radius of the pipe. Hence, \( \tau_{rz_{wall}} = \mu \frac{R}{2} \). \n Now, keep in mind that the flow rate, Q, is the integral of the velocity over the cross-sectional area, Q = \(\int_A v dA = \int_0^R \pi r dr (- \frac{1}{4 \mu} r^2 + \frac{R^2}{4\mu}) = \frac{1}{8 \mu} \pi R^4 \). \n Solving for \(\mu\) and substituting into the equation \(\tau_{rz_{wall}} = \mu \frac{R}{2} \) gives us \( \tau_{rz_{wall}} = \frac{4 \mu Q}{\pi R^3} \)

In part B, substituting the given values into the expression proved in part A would provide the magnitude of the wall shearing stress. The viscosity of the fluid (\(\mu\)) is 0.004 Ns/m^2, the flow rate (Q) can be determined from the average velocity (avg_v) and the tube radius (R). avg_v = 130mm/s = 0.130m/s and the diameter of the tube is 2mm, therefore the radius R = 2mm/2 = 0.001m. The volume flow rate Q = avg_v * cross sectional area = 0.130m/s * \(\pi (0.001m)^2\).\n Substituting these values into the equation for \(\tau_{rz_{wall}}\), the shearing stress at the wall of the tube is found.