A Semi-Implicit, Three-Dimensional Model for Estuarine ... - USGS

5. Numerical Experiments 117
Each test simulation for the tidal problem is run for a total of six days (144 hours). This allows sufficient time for any solution
instabilities to appear. The results of each simulation are presented at the location x = 180,000 ft (55 km) from the seaward bound-
ary of the tidal channel. A solution for water surface elevation and velocity is shown in figure 5.15. After simulating ten cycles of
the M 2 tide, results from the 11th cycle are examined in detail and used in the computation of error measures.
5.2.2.1 Two-Level Scheme
Figure 5.16 shows three solutions of the tidal problem using the two-level scheme; one is the base solution included in both
graphs. The solid lines present the base solution obtained by using the two-level scheme with a very small grid size of Δt = 0.5
minutes, Δx = 400 ft (122m) and with five iterations (niter = 5). The base solution is assumed to be free of numerical error for the
purposes of computing error measures for the solutions with coarser grid sizes. The two solutions indicated by the dashed curves
in figure 5.16 are computed using large time steps of 60 minutes (fig. 5.16A) and 30 minutes (fig. 5.16B.); both solutions use a
space step (Δx) of 6,000 ft (1,828 m) and two iterations (niter = 2). Figure 5.17 shows a plot of the volumetric transport (U = uh)
from the solution using Δt = 60 minutes compared with the volumetric transport of the base solution. The root mean square error
(RMS e) of the solution is computed relative to the base solution using the volumetric transport variable defined in table 5.5. The
RMS e is 8.7 percent for the 60-minute solution and 2.3 percent for the 30-minute solution. The other error measures defined in are
computed as A e = 5.2 percent, P e = 0.35 percent, and M e = 4.7 percent for the solution using Δt = 60 minutes, and
A e = – 0.8 percent, P e = 0.16 percent, and M e = 1.0 percent for the solution using Δt = 30 minutes.
WATER SURFACE ELEVATION, IN FEET
5
4
3
2
1
0
−1
−2
−3
−4
−5
Water surface elevation
Velocity
0 1 2
3
4 5 6
TIME, IN DAYS
11th cycle
Figure 5.15. Solution of the tidal problem for water surface elevation and velocity at the location x = 180,000 feet.
The results of simulations for the 11th cycle of the M 2 tide are examined in detail in figure 5.16.
VELOCITY, IN FEET PER SECOND