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A viscous fluid has velocity that depends on position in a pipe: Viscous fluid The speed is maximum at the center of the tube. It decreases away from the center. The speed is zero on the walls of the tube..
= The function v(r), for fluid speed in a pipe as a function of distance from the center of the pipe, r, should be maximal at r 0 and zero at the walls. In other words, v(0) = Umax and v(R)= 0, where R is the radius of the pipe. The goal of this problem is to solve for the function v(r). (a) Show that explain why, for steady, laminar flow in a horizontal pipe (so that we can ignore gravity), the Navier-Stokes equation reduces to 0= -Vp+n² (b) We'd like to solve the above equation for 7 subject to the boundary conditions discussed at the beginning of the problem. In cylindrical coordinates (where the liquid is traveling in the z direction), and with symmetry considerations allowing us to write 7 = v(r) 2, the above equation simplifies to 8² v 18v 1 Ap = Ər² rər η Δε (you don't have to show this). Take Ap/Az to be constant. Solve this equation for v(r) in terms of r, Umax, and R. What is Umax in terms of n, Ap/Az, and R? Hint: two derivatives of something equals a constant...