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Introduction to MIKE 11 by Bunchingiv Bazartseren Cottbus May 22, 2001
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May 22, 2001Introduction to MIKE 11 Outline General Hydrodynamics within MIKE 11 flow types numerical solution Modelling with MIKE 11 Example demonstration input preparation simulation visualization
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May 22, 2001Introduction to MIKE 11 General 1 1D flow (wave) simulation Application into water system for what purpose? design management operation where? river estuaries irrigation systems
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May 22, 2001Introduction to MIKE 11 General 2 Main modules Rainfall-runoff NAM, UHM Hydrodynamics governing equations for different flow types Advection-dispersion and cohesive sediment 1D mass balance equation Water quality AD coupled for BOD, DO, nitrification etc Non cohesive sediment transport transport material and morphology
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May 22, 2001Introduction to MIKE 11 Saint Venant equation 1 Unsteady, nearly horizontal flow 0 q 2 2 ARC QgQ x h gA x A Q t Q t A x Q where, Q - discharge, m 3 s -1 A - flow area, m 2 q - lateral flow, m 2 s -1 h - depth above datum, m C - Chezy resistance coefficient, m 1/2 s -1 R - hydraulic radius, m - momentum distribution coefficient
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May 22, 2001Introduction to MIKE 11 Saint Venant equation 2 Variables two independent (x, t) two dependent (Q, h) Conditions for solution 2 point initial (Q, h) 1 point up/downstream h Q Q=f(h)
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May 22, 2001Introduction to MIKE 11 Flow types Fully dynamic 0 2 RAC QQ ggAi x h gA Diffusive wave - no inertia Kinematic wave - pure convective 0 i x h
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May 22, 2001Introduction to MIKE 11 Finite difference method Discretization into time and space t xx t x nn 1 Difference between explicit and implicit scheme
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May 22, 2001Introduction to MIKE 11 Solution scheme 1 Structured, cartesian grid Implicit scheme (Abbott-Ionescu) Continuity equation - h centered Momentum equation - Q centered j n j n j n j n j x QQQQ x Q 2 22 1 1 11 1 1 Example discretization:
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May 22, 2001Introduction to MIKE 11 Transformation into linear equations Solution scheme 2 j n jj n jj n jj j n jj n jj n jj DhCQB1B1hA DQChBQA 111 1111 1 1 11 1 1 1 11 1 j n jj n jj n jj DCBA1111 1 1 11 1 Tri-diagonal matrix form of equation A 0 B 0 C 0 A 1 B 1 C 1 A 2 B 2 C 2... A jj B jj C jj 0 1 2. jj D 0 D 1 D 2. D jj n+1n =. all zeros (mass) (momentum)
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May 22, 2001Introduction to MIKE 11 Less equation than unknowns Use of suitable boundary conditions Introducing additional variables Solution scheme 3 Substitution of into the linear equations Derivation of recurrence relations jjj jjj j jjj j j BEA CAD F BEA C E 1 1 j n jj n j FE 11 1
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May 22, 2001Introduction to MIKE 11 Double sweep algorithm calculate the coefficients A-D obtain E jj, F jj from right hand boundary sweep forward to calculate E j, F j sweep back to calculate j n+1 for all grid Solution scheme 4
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May 22, 2001Introduction to MIKE 11 Network of open channels 1 Use of graph theory Set of vertices and edges edges - channels nodes - river confluence
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May 22, 2001Introduction to MIKE 11 Incidence matrix from the network Confluence nodes - h boundary Each channel - diagonal matrix Consideration of lateral flow Network of open channels 2 1 1 1 1 1 1 1 1 1 edges nodes 113131113131
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