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Consider the shear flow in a pipe, Fig. (2), of diameter D. For laminar flow of a Newtonian fluid, the shear r= D/2 =0
Consider the shear flow in a pipe, Fig. (2), of diameter D. For laminar flow of a Newtonian fluid, the shear r= D/2 =0 r= - D/2 Figure 2: Shear flow in a pipe stress is simply proportional to the velocity gradient. du T = - T = dr 2Twr D where r is along the radial direction of the pipe, Tw is a constant and is the viscosity of the fluid. = 1. Derive the analytical expression for the velocity, u(r), with no-slip boundary condition (i.e. u(-D/2) = u(D/2) = 0) 2. Plot (by hand) the velocity profile and the shear stress profile in the radial direction. 3. Using N-point mesh (excluding boundaries) with second order central difference, write the discrete system Au = f 4. If you solve the above system numerically, do you expect the solution to have a numerical error? (In other words, will there be a difference between the exact and numerical solutions ?) If so, why? If not, why?
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