Page 159 - Introduction to Computational Fluid Dynamics

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Table 5.1: Porous medium – Problem 2. 2D CONVECTION – CARTESIAN GRIDS
l x/L 0.0 0.25 0.50 0.75 1.0
0 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
1 0.266E+01 0.266E+01 0.106E−02 0.284E−06 0.284E−06
2 0.737E+00 0.737E+00 0.705E+00 0.146E+01 0.146E+01
3 0.114E+01 0.114E+01 0.913E+00 0.919E+00 0.919E+00
4 0.982E+00 0.982E+00 0.974E+00 0.103E+01 0.103E+01
5 0.101E+01 0.101E+01 0.992E+00 0.992E+00 0.992E+00
6 U 0.999E+00 0.999E+00 0.998E+00 0.100E+01 0.100E+01
7 0.100E+01 0.100E+01 0.999E+00 0.999E+00 0.999E+00
8 0.100E+01 0.100E+01 0.100E+01 0.100E+01 0.100E+01
9 0.100E+01 0.100E+01 0.100E+01 0.100E+01 0.100E+01
10 0.100E+01 0.100E+01 0.100E+01 0.100E+01 0.100E+01
11 0.100E+01 0.100E+01 0.100E+01 0.100E+01 0.100E+01
12 0.100E+01 0.100E+01 0.100E+01 0.100E+01 0.100E+01
0 0.400E+06 0.000E+00 0.000E+00 0.000E+00 0.000E+00
1 0.400E+06 0.295E+06 0.283E+06 0.155E+06 0.000E+00
2 0.400E+06 0.266E+06 0.192E+06 0.837E+05 0.000E+00
3 0.400E+06 0.294E+06 0.211E+06 0.993E+05 0.000E+00
4 0.400E+06 0.294E+06 0.202E+06 0.958E+05 0.000E+00
5 0.400E+06 0.298E+06 0.202E+06 0.982E+05 0.000E+00
6 P 0.400E+06 0.299E+06 0.201E+06 0.987E+05 0.000E+00
7 0.400E+06 0.299E+06 0.201E+06 0.993E+05 0.000E+00
8 0.400E+06 0.300E+06 0.200E+06 0.996E+05 0.000E+00
9 0.400E+06 0.300E+06 0.200E+06 0.998E+05 0.000E+00
10 0.400E+06 0.300E+06 0.200E+06 0.999E+05 0.000E+00
11 0.400E+06 0.300E+06 0.200E+06 0.999E+05 0.000E+00
12 0.400E+06 0.300E+06 0.200E+06 0.100E+06 0.000E+00
12 p /p 0.000E+00 0.114E−03 −0.150E−03 0.341E−03 0.000E+00

m

12 p /p 0.000E+00 −0.963E−04 0.188E−03 −0.288E−03 0.000E+00
sm

solution given here, but velocity u = 0 at all nodes. In Problem 2, p(1) = 4 × 10 5
and p(IN ) = 0, but p = 0 at all interior nodes of the domain. Again u = 0at
all nodes. Thus, in both problems, the guessed velocity is zero and the boundary

pressures are held fixed so that p (1) = p (IN) = 0. Relaxation parameters are

taken as α = β = 1.

For Problem 1, by solving for u and p , the exact solutions (not shown here)
for p and u are obtained in just one iteration although the initial guess for u was
zero. This is because the initial guess for pressure was itself the exact solution and,
therefore, required no correction.
Table 5.1 shows evolutions with iteration number l for Problem 2. Notice that
because of the poor initial guess for pressure, the exact velocity solution is obtained
in eight iterations whereas the correct pressure prediction requires twelve iterations.