% Clear variables

clear all;


% Set gas temperature and wall heat transfer coefficients at
% boundaries of the blade.  Note: Tcool(i) and hcool(i) are the
% values of Tcool and hcool for the ith boundary which are numbered
% as follows:  
%
%   1 = 1st internal cooling passage (from leading edge)
%   2 = 2nd internal cooling passage 
%   3 = 3rd internal cooling passage (including trailing edge slot)

Tcool = [600, 600, 600];    % ( K )
hcool = [1500, 1500, 1500]; % ( W/(K m^2) )

% Set blade material properties
kblade   =   30; % ( W/( K m) )

% Set blade chord
chord = 0.04; % (m)

% Load in the grid file

fname = input('Enter gridfile name: ','s');
load(fname);

fprintf('Number of nodes      = %i\n',Nv);
fprintf('Number of elements   = %i\n',Nt);

% NOTE:  after loading a gridfile using the load(fname) command,
%        three important grid variables and data arrays exist.  These are:
%
% Nt: Number of triangles (i.e. elements) in mesh
%
% Nv: Number of nodes (i.e. vertices) in mesh
%
% Nbc: Number of edges which lie on a boundary of the computational
%      domain.
% 
% tri2nod(3,Nt):  list of the 3 node numbers which form the current
%                 triangle.  Thus, tri2nod(1,i) is the 1st node of
%                 the i'th triangle, tri2nod(2,i) is the 2nd node
%                 of the i'th triangle, etc.
%
% xy(2,Nv): list of the x and y locations of each node.  Thus,
%           xy(1,i) is the x-location of the i'th node, xy(2,i)
%           is the y-location of the i'th node, etc.
%
% bedge(3,Nbc): For each boundary edge, bedge(1,i) and bedge(2,i) 
%               are the node numbers for the nodes at the end
%               points of the i'th boundary edge.  bedge(3,i) is an
%               integer which identifies which boundary the edge is
%               on. In this solver, the third value has the
%               following meaning:
%
%               bedge(3,i) = 0: edge is on the airfoil surface
%               bedge(3,i) = 1: edge is on the first cooling passage
%               bedge(3,i) = 2: edge is on the second cooling passage
%               bedge(3,i) = 3: edge is on the third cooling passage
% 

% Re-scale xy location because they currently have unit chord.
xy = chord*xy;

% Zero stiffness matrix and b vector
K = sparse(Nv, Nv);
b = zeros(Nv, 1);

% Loop over elements and calculate residual and stiffness matrix

for ii = 1:Nt,
  
  kn(1) = tri2nod(1,ii);
  kn(2) = tri2nod(2,ii);
  kn(3) = tri2nod(3,ii);
    
  xe(1) = xy(1,kn(1));
  xe(2) = xy(1,kn(2));
  xe(3) = xy(1,kn(3));

  ye(1) = xy(2,kn(1));
  ye(2) = xy(2,kn(2));
  ye(3) = xy(2,kn(3));

  % Calculate all of the necessary shape function derivatives, the
  % Jacobian of the element, etc.
  
  % Derivatives of node 1's interpolant 
  dNdxi(1,1) = -1.0; % with respect to xi1
  dNdxi(1,2) = -1.0; % with respect to xi2
  
  % Derivatives of node 2's interpolant
  dNdxi(2,1) =  1.0; % with respect to xi1
  dNdxi(2,2) =  0.0; % with respect to xi2

  % Derivatives of node 3's interpolant
  dNdxi(3,1) =  0.0; % with respect to xi1
  dNdxi(3,2) =  1.0; % with respect to xi2
  
  % Sum these to find dxdxi (note: these are constant within an element)
  dxdxi = zeros(2,2);
  for nn = 1:3,
    dxdxi(1,:) = dxdxi(1,:) + xe(nn)*dNdxi(nn,:);
    dxdxi(2,:) = dxdxi(2,:) + ye(nn)*dNdxi(nn,:);
  end
  
  % Calculate determinant for area weighting
  J = abs(dxdxi(1,1)*dxdxi(2,2) - dxdxi(1,2)*dxdxi(2,1));
  
  % Invert dxdxi to find dxidx using inversion rule for a 2x2 matrix
  dxidx = [ dxdxi(2,2)/J, -dxdxi(1,2)/J; ...
	   -dxdxi(2,1)/J,  dxdxi(1,1)/J];
  
  % Calculate dNdx 
  dNdx = dNdxi*dxidx;

  % Add contributions to stiffness matrix for node 1 weighted residual
  K(kn(1), kn(1)) = K(kn(1), kn(1)) - 0.5*kblade*(dNdx(1,1)*dNdx(1,1) + dNdx(1,2)*dNdx(1,2))*J;
  K(kn(1), kn(2)) = K(kn(1), kn(2)) - 0.5*kblade*(dNdx(1,1)*dNdx(2,1) + dNdx(1,2)*dNdx(2,2))*J;
  K(kn(1), kn(3)) = K(kn(1), kn(3)) - 0.5*kblade*(dNdx(1,1)*dNdx(3,1) + dNdx(1,2)*dNdx(3,2))*J;
  
  % Add contributions to stiffness matrix for node 2 weighted residual
  K(kn(2), kn(1)) = K(kn(2), kn(1)) - 0.5*kblade*(dNdx(2,1)*dNdx(1,1) + dNdx(2,2)*dNdx(1,2))*J;
  K(kn(2), kn(2)) = K(kn(2), kn(2)) - 0.5*kblade*(dNdx(2,1)*dNdx(2,1) + dNdx(2,2)*dNdx(2,2))*J;
  K(kn(2), kn(3)) = K(kn(2), kn(3)) - 0.5*kblade*(dNdx(2,1)*dNdx(3,1) + dNdx(2,2)*dNdx(3,2))*J;
  
  % Add contributions to stiffness matrix for node 3 weighted residual
  K(kn(3), kn(1)) = K(kn(3), kn(1)) - 0.5*kblade*(dNdx(3,1)*dNdx(1,1) + dNdx(3,2)*dNdx(1,2))*J;
  K(kn(3), kn(2)) = K(kn(3), kn(2)) - 0.5*kblade*(dNdx(3,1)*dNdx(2,1) + dNdx(3,2)*dNdx(2,2))*J;
  K(kn(3), kn(3)) = K(kn(3), kn(3)) - 0.5*kblade*(dNdx(3,1)*dNdx(3,1) + dNdx(3,2)*dNdx(3,2))*J;

end

% Loop over boundary edges and account for bc's
% Note: the bc's are all convective heat transfer coefficient bc's
% so they are of 'Robin' form.  This requires modification of the
% stiffness matrix as well as b.

for ii = 1:Nbc,
  
  % Get node numbers on edge
  kn(1) = bedge(1,ii);
  kn(2) = bedge(2,ii);
  
  % Get node coordinates
  xe(1) = xy(1,kn(1));
  xe(2) = xy(1,kn(2));
  
  ye(1) = xy(2,kn(1));
  ye(2) = xy(2,kn(2));
  
  % Calculate edge length
  ds = sqrt((xe(1)-xe(2))^2 + (ye(1)-ye(2))^2);
  
  % Determine the boundary number
  bnum = bedge(3,ii);
  if (bnum == 0), % Blade surface
    
    [Text, hext] = Thgas( 0.5*(xe(1)+xe(2)), 0.5*(ye(1)+ye(2)), chord );
    
  else, % cooling passage
    Text = Tcool(bnum);
    hext = hcool(bnum);
  end
  
  % Based on boundary number, set heat transfer bc
  K(kn(1), kn(1)) = K(kn(1), kn(1)) - hext*ds*(1/3);
  K(kn(1), kn(2)) = K(kn(1), kn(2)) - hext*ds*(1/6);
  b(kn(1))        = b(kn(1))        + hext*ds*0.5*Text;
  
  K(kn(2), kn(1)) = K(kn(2), kn(1)) - hext*ds*(1/6);
  K(kn(2), kn(2)) = K(kn(2), kn(2)) - hext*ds*(1/3);
  b(kn(2))        = b(kn(2))        + hext*ds*0.5*Text;
  
end

% Solve for temperature
T = -K\b;

% Plot solution
bladeplot;

% Report outputs
Tmax = max(T);
Tmin = min(T);

fprintf('Tmin = %7.2f, Tmax = %7.2f\n',Tmin,Tmax);