Viscous Flow Assignment 7 — Computational Viscous Flow  
Consider the following two-dimensional channel with smooth walls:
duct.png  ./download/file.php?id=3147&sid=a205d37c23dad1ad9a838a2b9f5104f0  ./download/file.php?id=3147&t=1&sid=a205d37c23dad1ad9a838a2b9f5104f0
Part #1
Starting from the Navier-Stokes equations, demonstrate that the $x$-momentum equation for laminar fully-developed flow in a channel collapses to: $$ \frac{\partial}{\partial y}\left(\mu \frac{\partial u}{\partial y} \right) = \frac{\partial P}{\partial x} $$
Part #2
Starting from the $x$-momentum equation derived in part #1, demonstrate that the exact solution to this problem for constant $\mu$ and constant $\rho$ is of: $$ u=-\frac{3\mu {\rm Re}_{D_H}}{\rho D_H H^2}\left(\frac{y^2}{2} - Hy \right) $$ with $$ {\rm Re}_{D_H}=-\frac{\rho D_H H^2}{3\mu^2} \frac{\partial P}{\partial x} $$
Part #3
Using the finite-volume approach, derive a set of discretization equations for the physical model outlined in Part 1 and for a 1D grid with uniform grid spacing. Also write down the discretization equations in coefficient form for both inner nodes and boundary nodes.
Part #4
Using the C code template attached and with $H=0.1$ m, $\rho=1000$ kg/m$^3$, $\mu=10^{-3}$ kg/ms, and $dP/dx=-2$ Pa/m, do the following:
(a)  Modify the function “find_coefficients_and_rhs” so that it uses the coefficients derived in Part #3:
/* finds the discretization coefficients a,b,c in kg/m3s and the RHS rhs in Pa/m for all nodes given
   N -> number of nodes
   H -> half height of the channel in m
   mu -> viscosity of the fluid in kg/ms
   dPdx -> pressure gradient along x in Pa/m
void find_coefficients_and_rhs(long N, double H, double mu, double dPdx, double *a, double *b, double *c, double *rhs){

(b)  Compare graphically the solution obtained to the exact solution in Part #2 for the number of nodes $N$ set to 5, 10, and 50 nodes. Are you surprised at the results?
(c)  Print out the C code you used to obtain results.
Due on Wednesday November 28th at 9:00. Do all parts.
PDF 1✕1 2✕1 2✕2