forked from eacdamal/simpar
102 lines
2.8 KiB
C++
102 lines
2.8 KiB
C++
#include <iostream>
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#include <vector>
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#include <math.h>
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#include <iomanip>
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using namespace std;
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int main()
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{
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int l = 1925; // Edge length
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int dx = 5; // Spatial step
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int kmax = l / dx; // Number of spatial steps
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/**
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* Coefficients alpha and gamma are different for different
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* edges. It depends on the edge's lenght, cross-sectional area etc.
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*/
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double alpha = 0.744;
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double beta = 0.004;
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double gamma = 5829;
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/**
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* dt - the number of approximation in Euler method
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* h - discretization step in Euler method
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*
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* Those two values work well with all existing steps and give us
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* result that satisfies the initial condition pretty close.
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*
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* Changing of any of them might result in NaN error.
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*/
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int dt = 100000;
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double h = 0.007;
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// Stores the approximated values for each step
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double Q[kmax][dt];
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double P[kmax][dt];
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// Prepare initial condition
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for (int i = 0; i < dt; i++)
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{
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// Loop over spatial steps
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for (int k = 0; k < kmax; k++)
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{
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// Boundary condition for the last spatial step
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if (k == kmax - 1)
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{
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Q[k][i] = 0;
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P[k][i] = -202.0; // P = -202 when Q = 10
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} else
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{
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Q[k][i] = 0;
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P[k][i] = 0;
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}
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}
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}
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/**
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* In fact, we have two 'steps'. The first step represents the approximation
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* criteria in Euler's method (dt) and the second one represents the actual
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* spatial step along the edge (dx).
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*/
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for (int i = 2; i < dt; i++)
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{
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// Loop over spatial steps
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for (int k = 1; k < kmax; k++)
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{
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Q[k][i + 1] = Q[k][i] + h * (alpha * (P[k - 1][i] - P[k][i]) - beta * Q[k][i] * abs(Q[k][i]));
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// DO NOT calculate pressure for the last element such as it is given by default
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if (k != kmax - 1)
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P[k][i + 1] = P[k][i - 2] + h * (gamma * (Q[k][i] - Q[k + 1][i]));
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}
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}
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// Print out the first 10 values of Q for each spatial step
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std::cout << "FIRST 10 VALUES\n";
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for(int i = 0; i < kmax; i++)
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{
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std::cout.width(5);
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std::cout.flags(std::ios::left);
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std::cout << i << ": ";
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for(int k = 0; k < 10; k++)
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std::cout << std::setw(10) << Q[i][k];
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std::cout << std::endl;
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}
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// Print out the last 10 values of Q for each spatial step
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std::cout << "LAST 10 VALUES:\n";
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for(int i = 1; i < kmax; i++)
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{
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std::cout.width(5);
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std::cout.flags(std::ios::left);
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std::cout << i << ": ";
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for(int k = 0; k < 10; k++)
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std::cout << std::setw(10) << Q[i][dt - 10 + k];
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std::cout << std::endl;
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}
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return 0;
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}
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