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

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96 A Semi-Implicit, Three-Dimensional Model for Estuarine Circulation 5.2.1.1 Two-Level Scheme Figure 5.2 shows three computed hydrographs using the two-level, semi-implicit scheme described in section 4.2.1. Two of the solutions are based on computations using a coarse numerical grid size of Δx = 5,000 ft (1,524 m) and Δt = 5 minutes; the third solution is considered a “base solution” and is computed using an extremely small grid size that keeps the numerical error in the solution negligible. The base solution agrees to a high degree of accuracy with the solutions to this problem obtained using the other flow models tested by the USGS. All three solutions were computed with the number of iterations (niter) of the numerical scheme fixed throughout the simulation; the coarse grid solutions used values of niter = 1 and niter = 2, the base solution used niter = 5. Figure 5.2 indicates that a significant improvement in accuracy is gained by solving the coarse grid case using two iterations rather than one. The root mean square error (RMS e in table 5.1) of the solution measured against the base solution improved from 7.6 percent using niter = 1 to 2.7 percent using niter = 2. The improvement is due to a centering in time of the nonlinear coefficients of the finite-difference momentum equation by the iteration process, as discussed in section 4.2.1. Because the frictional attenuation of the routed hydrograph is significant, the backward-in-time evaluation of the frictional coefficient γ in equation 4.7 is mostly responsible for the large underestimation of the wave attenuation in the solution using niter = 1; also affecting the accuracy, but to a lesser degree, is the backward-in-time evaluation of the coefficient H in the pressure (water surface slope) and friction terms in equation 4.7. The centering in time of both γ and H by the iteration process raises the accuracy of the time integration scheme from first to second order. DISCHARGE, IN CUBIC FEET PER SECOND 800 700 600 500 400 300 200 Upstream Boundary 50,000 feet 100 0 100 200 300 400 500 TIME, IN MINUTES Δt = 5 seconds, Δ x = 50 feet, niter = 5 Δt = 5 minutes, Δ x = 5,000 feet, niter = 1 Δt = 5 minutes, Δ x = 5,000 feet, niter = 2 RMS error niter = 1 7.6 percent niter = 2 2.7 percent Figure 5.2. Hydrographs computed with the two-level, semi-implicit scheme using a large spatial step size of Δx = 5,000 feet (1,524 meters) and a time step size of Δt = 5 minutes. Solutions using one iteration (niter = 1) and two iterations (niter = 2) are compared against a base solution using a very small time and space step and five iterations (niter = 5). The hydrograph at the upstream boundary is also plotted. RMS, root mean square.
Table 5.1. Definition of error measures (expressed as percentages) for the hydrograph test problem. Description Definition Root mean square error: Root mean square error of the computed hydrograph, normalized on the peak discharge of the base solution hydrograph Amplitude error: Error in the peak discharge of the computed hydrograph, normalized on the peak discharge of the base solution hydrograph RMS e Phase error: Error in the time associated with the center of gravity of the computed hydrograph, normalized on the time associated with the cen- Pe = ter of gravity of the base solution hydrograph where A e 100 Q n nts n ( ( x) – Qb ( x) ) 2 ∑ n = 1 ( nts) 1 = ------------------------------------------------------------------------ ⁄ max 2 Qb ( x) 100 Q max max ( ( x) – Qb ( x) ) = ------------------------------------------------------------max Qb ( x) 100( T( x) – Tb( x) ) -------------------------------------------- Tb( x) x = 50,000 feet Tx ( ) = tQxt ( ( , ) – Q0)dt -------------------------------------------------- TL ∫0 ( Qxt ( , ) – Q0)dt Mass preservation error: Error in mass preservation for the computed solution, normalized on the total mass from the base solution Me = ⎛ L ⎞ ⎜ ∫0 ( Hxt ( , ) – H0)dx⎟ 100⎜1– --------------------------------------------------- ⎟ L ⎜ ⎟ ⎝ ∫0 ( Hb( x, t) – H0)dx⎠ t = 500 minutes nts = number of time steps in 500 minute simulation n Qb ( x) = discharge for the base solution at location x and time t = nΔt Qmax b ( x) = peak discharge for the base solution at location x (= 510.25 cubic feet per second at x = 50,000 feet) Tb(x) = time associated with center of gravity of base solution at location x (= 362.85 minutes for x = 50,000 feet) Hb(x, t) = depth of flow for the base solution at location x and time t Q0 = initial steady discharge (= 250 cubic feet per second) H0 = initial water depth determined by Manning’s formula for a wide rectangular channel (= 1.688464 feet) TL = time of simulation (= 500 minutes) L = length of channel (= 150,000 feet) Integrations were computed numerically using Simpson’s Rule. TL ∫0 5. Numerical Experiments 97 1⁄ 2 x = 50,000 feet x = 50,000 feet
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In cooperation with the Interagency
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vi 4.2 Semi-Implicit One-Dimensiona
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A Semi-Implicit, Three-Dimensional Model for Estuarine ... - USGS

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