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fixed Euler's step and number of iteration for dx = 5

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mejjete 2024-03-12 21:34:40 +01:00
parent 5807ba4b64
commit 6a4ded9ca7
1 changed files with 24 additions and 44 deletions
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62
sequential.cpp
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@ -1,39 +1,29 @@
#include <iostream>
#include <vector>
#include <iomanip>
#include <math.h>
#include <iomanip>
using namespace std;
int main()
{
// Length
int l = 1925;
// Spatial step
int dx = 40;
// Temporal step
int dt = 2500;
// Number of spatial steps
int kmax = l / dx;
// Initializing parameters
vector<double> list(dt);
// Arrays to store data
double Q[kmax][dt];
double P[kmax][dt];
int i;
int l = 1925; // Edge length
int dx = 40; // Spatial step
int dt = 100000; // Euler's step
int kmax = l / dx; // Number of spatial steps
// Parameters
double alpha = 0.093;
double beta = 0.004;
double gamma = 728.6;
double h = 0.05;
double sum_of_h = 0;
double h = 0.007;
// Arrays to store data
double Q[kmax][dt];
double P[kmax][dt];
// Loop over time steps
for (i = 0; i < dt; i++)
for (int i = 0; i < dt; i++)
{
// Loop over spatial steps
for (int k = 0; k < kmax; k++)
@ -52,31 +42,20 @@ int main()
}
// Main calculation loop
for (i = 2; i < dt - 2; i++)
for (int i = 2; i < dt; i++)
{
// Add current time step to the list
sum_of_h = sum_of_h + h;
list.push_back(sum_of_h);
std::cout << "Q at step " << i;
// Loop over spatial steps
for (int k = 1; k < kmax; k++)
{
// Boundary condition for the last spatial step
if (k == kmax - 1)
{
Q[k][i + 1] = Q[k][i] + h * (alpha * (P[k - 1][i] - P[k][i]) - beta * Q[k][i] * abs(Q[k][i]));
} else
{
// Calculate Q and P for non-boundary spatial steps
Q[k][i + 1] = Q[k][i] + h * (alpha * (P[k - 1][i] - P[k][i]) - beta * Q[k][i] * abs(Q[k][i]));
// DO NOT calculate pressure for the last element such as it is given by default
if (k != kmax - 1)
P[k][i + 1] = P[k][i - 2] + h * (gamma * (Q[k][i] - Q[k + 1][i]));
}
}
}
// Print out the first 10 values of the timestemp
// Print out the first 10 of Q for each spatial step
std::cout << "FIRST 10 VALUES\n";
for(int i = 0; i < kmax; i++)
{
@ -89,18 +68,19 @@ int main()
std::cout << std::endl;
}
// Print out the last 10 values of the timestemp
std::cout << "LAST 10 VALUES\n";
for(int i = 0; i < kmax; i++)
// Print out the first 10 of Q for each spatial step
std::cout << "LAST 10 VALUES:\n";
for(int i = 1; i < kmax; i++)
{
std::cout.width(5);
std::cout.flags(std::ios::left);
std::cout << i << ": ";
for(int k = 0; k < 10; k++)
std::cout << std::setw(10) << Q[i][2489 + k];
std::cout << std::setw(10) << Q[i][dt - 10 + k];
std::cout << std::endl;
}
return 0;
}