Function: Series addition implementation

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Hello
I am trying to solve an equation of which i need to be using the new values of V and delta as an update in equations (1).The closed equation (1) when expanded will be in the form of equation (2) for i = 2 and j is between 1 -5.
% P(i) = sum(j=1->n) |Vi||Vj|(Gij * cos(delta_i - delta_j) + Bij * sin(delta_i - delta_j) EQUATION (1) % The formula for calculating the P
Implementation of P(2) for example:
P(2) = P(2) + V(2)*V(1)*(G(2,1)*cos(delta(2)-delta(1)) + B(2,1)*sin(delta(2)-delta(1))) + V(2)*V(2)*(G(2,2)*cos(delta(2)-delta(2)) + B(2,2)*sin(delta(2)-delta(2))) + V(2)*V(3)*(G(2,3)*cos(delta(2)-delta(3)) + B(2,3)*sin(delta(2)-delta(3))) + V(2)*V(4)*(G(2,4)*cos(delta(2)-delta(4)) + B(2,4)*sin(delta(2)-delta(4))) + V(2)*V(5)*(G(2,5)*cos(delta(2)-delta(5)) + B(2,5)*sin(delta(2)-delta(5))) EQUATION (2)
Please is the below program good enough to represent the expansion of P(2) in equation (2) as an example ?
% Program
P = zeros(nbus,1);
Q = zeros(nbus,1);
V = zeros(nbus,1);
% B and G are constants
% Computing P
% nbus = 5
for i = 1:nbus
for j = 1:nbus
if type(i) == 2 % Computing Pi & Qi for Droop bus
P(i) = V(i)*V(j)*(G(i,j)*cos(delta(i)-delta(j)) + ...
B(i,j)*sin(delta(i)-delta(j)))+ PO(i)-PL(i);
Q(i) = V(i)*V(j)*(G(i,j)*sin(delta(i)-delta(j)) - ...
B(i,j)*cos(delta(i)-delta(j)))+ QO(i)-QL(i);
end
end
end
  5 个评论
Dyuman Joshi
Dyuman Joshi 2023-1-31
编辑:Dyuman Joshi 2023-1-31
Okay. But still have not answered my other queries.
The equation in the link do not contain PO, PL, QO and QL, where as your code does.
Walter Roberson
Walter Roberson 2023-2-2
for i = 1:nbus
for j = 1:nbus
if type(i) == 2 % Computing Pi & Qi for Droop bus
That is inefficient. type(i) does not change according to changes in j, so you should be testing type(i) outside the for j loop, only executing the for j loop if type(i) == 2
P(i) = V(i)*V(j)*(G(i,j)*cos(delta(i)-delta(j)) + ...
B(i,j)*sin(delta(i)-delta(j)))+ PO(i)-PL(i);
Q(i) = V(i)*V(j)*(G(i,j)*sin(delta(i)-delta(j)) - ...
B(i,j)*cos(delta(i)-delta(j)))+ QO(i)-QL(i);
P(i) does not depend upon P(i) or Q(i) so at each different j you completely overwrite P(i) and Q(i) and the final result will be the same as if you had only done the final j=nbus .

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