1D heat transfer

[nesca87]

Hello. I am newbie in c++.

I got an assignment that asked me to make a one dimensional heat transfer problem by using finite difference explicit method with particular boundary condition.

The discretization should be u[i,j] = ((1 - 2 * 0.1)* u(i-1,j-1) + ((0.1)*(u(i+1,j-1)) + ((0.1)* (u(i - 1,j-1))

which i is counter of the percentage length and j is counter of time. I know it would be easier if I made the program into 2D vector i.e %length and time.

But, I was asked to make it just 1D, and put the counter of time as looping process. I supposed to save the iteration every 20 loops

I tried the program below. It seemed working, but the .dat file didn't occur.
So, I couldn't see the result.

could somebody help me finding the mistakes?

thank you in advance

  /*
* One Dimensional Heat Transfer
*/

#include <iostream>
#include <fstream>
#include <vector>
using namespace std;

void saveArray(vector<double>&u, int nx, int nt, double dt, double dx, double e);

int main()
{
	double t = 0;
	int nx = 11;
	double dt = 0.001;
	double dx = 0.1;
	double e = 0;
	vector<double> u(nx);
	vector<double> un(nx);
		

	// Initial condition:
	for (int i = 1; i <= nx - 1; ++i)
		{
			if (i*dx >= 0.0 && i*dx <= 0.5)	 //u(i,0)=2*dx, 0<=dx<=0.5
			{
				u[i] = 2 * i*dx;
			}
			else if (i*dx >= 0.5&&i*dx <= 1.0) //u(i,0)=2*(1-dx), 0.5<=dx<=1.0
			{
				u[i] = 2 * (1 - (i*dx));
			}
			cout << u[i] << endl;
		}

		// Finite-difference loop:

		u[0] = un[0];		// Boundary Condition

		for (int k = 0; k <= nx - 1; ++k)
		{
			un[k] = u[k];
		}

		for (int i = 1; i <= nx - 2; i++)
		{
			u[i] = ((1 - 2 * 0.1)* un[i]) + ((0.1)*(un[i + 1])) + ((0.1)* (un[i - 1]));
			e = u[i] - un[i];
			cout << u[i] << endl;
		}
	
	
		if (e >= 0.0001)
		{
			int nt = 0; nt++;
		}
		return 0;
}


// Save array to file:
void saveArray(vector<double>&u, int nx, int nt, double dt, double dx, double e)
{
	ofstream myfile;
	myfile.open("1d_convection02.dat");
	myfile << "%\t" << "nx = " << nx << "\tnt = " << nt <<
		"\tdt = " << dt << endl;

	for (int i = 0; i <= nx - 1; i++)
	{
		myfile << i*dx << "\t\t";
		for (int j = 0; j%nt==20; j++)
		{
			myfile << u[i] << "\t\t";
		}
		myfile << endl;
	}
}

[mutexe]

You could try actually calling the method you've written :)))

[nesca87]

I'm sorry, I can't catch up what you mean

[Ispil]

Well, you created the function saveArray, but never actually called it. In other words, the function isn't being used.

[nesca87]

ah ya, I forgot about it. Thank you.

I've placed it into main loop function as

saveArray(u, nx, nt, dx, dt, e)

before "return", but apparently still doesn't work :(

it said that

Error 1 error LNK2019: unresolved external symbol "void __cdecl saveArray(class std::vector<double,class std::allocator<double> >
&,int,double,double)" (?saveArray@@YAXAAV?$vector@NV?$allocator@N@std@@@std@@HNN@Z) referenced in function _main C:\Users\Hicha Aquino\Documents\Hicha\EFS Research Things\Programming\HeatTransfer1D\HeatTransfer1D\main.obj HeatTransfer1D

[Zhuge]

Could you show the updated code? The error indicates you seem to be trying to call saveArray() with only 4 parameters, but your definition has more than that.

[nesca87]

Sure. Here they are.

#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
using namespace std;

void saveArray(vector<double>  &u, int nx, double dx, double dt);

void main()
{
	//Displaying welcome message
	std::cout << "1D Heat Transfer!\n";

	//Declaring variable
	int nx = 11;
	double dt = 0.001;
	double dx = 0.1;
	double r = dt / dx / dx;			//Constant of discretization
	double error = 0;


	vector<double> u(nx);
	vector<double> u_old(nx);

	//Initializing u
	for (int i = 1; i <= nx - 1; ++i)
	{
		if (i*dx >= 0.0 && i*dx <= 0.5)		//u(i,0)=2*dx, 0<=dx<=0.5
		{
			u[i] = 2 * i*dx;
		}
		else if (i*dx >= 0.5&&i*dx <= 1.0)	//u(i,0)=2*(1-dx), 0.5<=dx<=1.0
		{
			u[i] = 2 * (1 - (i*dx));
		}

	}

	error = 1000;
	//Executing iteration
	while (error >= 0.0001)
	{
		//Copying u to temporary array
		u_old = u;

		//Calculating new u
		for (int i = 1; i <= nx - 2; i++)
		{
			u[i] = (r*u_old[i - 1]) + ((1 - (2 * r))*u_old[i]) + (r*u_old[i + 1]);
		}

		//Calculating error

		for (int i = 0; i <= nx - 1; i++)
		{
			double de = abs(u[i] - u_old[i]);	//Absolute Error
			error = max(error, de);			//Maximum error

		}
		
	}

	//Saving result to output file
	saveArray(u, nx, dx, dt);

	//Displaying exit message
	std::cout << "Program ended correctly!\n";
	system("PAUSE");
}

void saveArray(vector<double>  &u, int nx, double dx)
{
	
	ofstream myfile;
	myfile.open("1d_convection02.dat");
	myfile << "%\t" << "nx = " << nx << endl;
	int j = 0; j++;
	for (int nt = 0; nt = nt + 1;)
	while (j <= nt - 1)
	{
		if (j%nt == 20)		//save data every 20 time step
			{
				for (int i = 0; i <= nx - 1; i++)
			{
				myfile << i*dx << "\t\t";
				myfile << u[i] << "\t\t";
				myfile << endl;
			}
			}
				else return;
			}
			myfile.close();
}

[Zhuge]

Your definition of saveArray() on line 71 takes only 3 parameters, but your call on line 64 and your prototype on line 7 give it 4 parameters.

[Kikiman]

Look at your function definition :

void saveArray(vector<double> &u, int nx, double dx, double dt);

And your function declaration :

void saveArray(vector<double> &u, int nx, double dx)
{
   //...
}

They are completely different. You forgot double dt in your function declaration.

[TwilightSpectre]

The problem is that your saveArray declaration is different to your definition of it, so the linker treats them as 2 different functions. You may want to change the definition to take a 4th variable as input.

[nesca87]

thank you so much for the answer. I've synchronized the definition and the declaration function. It seems work. But still there is no .dat file getting out.

I am sorry for asking to much.

[mutexe]

while (error >= 0.0001)
it's stuck in this loop, because
error = max(error, de);

is always evaluating to 1000.

de is always 0. do you have your logic correct when you subract one element from the other?

edit: in fact it doesnt matter what value de takes, that loop will always be infinite.