Question

You have seen an example of solving Navier-Stokes equations for flow in a circular tube of radius R (Poiseuille flow). In those instances, the tube is horizontal, and flow is along the horizontal axis (z-axis in cylindrical coordinates). Now imagine if the tube is turned 90° so that the flow is downward and is now parallel to gravity g (A) Start with the Navier-Stokes equations, set up the problem, and clearly indicate what assumptions are used in your simplification of the equations Solve for an expression for as a function or radius and other constants Does your answer make sense? Compare your answer above to the Poiseuille flow solution for a flat, horizontal tube flow What happens in both cases above if there is no pressure applied? What is the driving force for fluid flow now? (B) (C) (D)

Answer 1

Calculating the two-dimensional flow around a cylinder (radius a, located at x = y = 0) in a uniform stream U involves solving Partialdifferential u^-/dt + (u^- middot nabla) u^- = -1/rho nabla P + v nabla^2 u^-, nabla middot u^- = 0, with the Boundary conditions u^- = 0 on x^2 + y^2 = a^2, u^- (U, 0) as x^2 + y^2 rightarrow infinity Rewrite this problem in non-dimensional form using the dimensionless variables x'^- = x^-/a, u'^- = u^-/U, P' = P/rho V^2, t' = tU/a Note that the scaling x'^- = x^-/a implies nabla' = a nabla and t' = tV/a gives Partialdifferential/Partialdifferential t' = a/V Partialdifferential/Partialdifferential t change to dimensionless variables V^2/a Partialdifferential u'^-/Partialdifferential t' + v^2/a (u'^- middot nabla')u'^- = -rho v^2/rho a nabla' p' + gamma U/a^2 nabla'^2 u'^- and divide by v^2/a Partialdifferential u'^-/Partialdifferential t' + (u'^- middot nabla') u'^- = -rho v^2/rho a nabla' p'