MATLAB牛頓拉夫遜法算潮流分析

1、matlab牛頓拉夫遜法算潮流分析the function of this program is to use newton, the rafson methodthe % bl matrix: 1, the first branch of the branch; no. 2, terminal; branch resistsnce; 4, the circuit (or transformer admittance);% 5, the change of the branch; the branch head is at the k side of 1, 1 side 0;the % 7,

2、line/transformer id (0/1) transformer parameter is calculated to the end of the branch when the end of the branch is on the k side, 0 to the first end% b2 matrix: 1, power of the node generator; the load power of the node:% 3, pq node voltage initial value, or a given value of the balanee node and p

3、v node voltagethe connection of the capacitor (inductance) of the shunt capacitance (inductance) is received by the node% 5, node classification label: 1 is the balanee node (should be number 1): 2 for the pq node; 3 for pv nodes;the clear;isb = 1; % input (' please enter the balanee bus node nu

4、mber: isb 二')；pr = 1 e - 5; % in put (' please in put the error accuracy: pr 二')；n 二 10; % input (' please enter the number of nodes: n 二')；nl = 10; % input ')；('pleaseenter a number of directions: nlbl1,0,1, 2, 4, 4,1, 01, 4, 1,1, 0, 0, 1, 1, 1, 1, 4, 1, 1, 1, 0,1,4,5. 1

5、 + 19.2i 2. le-4110 0;2,3,4.25 + 16i1. 75e-4 i 1, 0, 0;4,5,4.25 + 16i7e minus 4i 1 0;1,3,5. 1 + 19.2i 2.le-4 1100;6, 4, 5. 95 + 22.4 i 9. 8e4 i 1, 0, 0;2, 7, 1. 78 + 53.89 i 0, 38.5/231, 0, 1;3, 8, 1.49 + 48.02 i 0, 11/231, 0, 1;4, 9, 9, 9, 9, 4, 9, 1, 9, 1, 0, 11, 0, 1.5 10 2. 46 + 70. 17 i 0 38. 5

6、/231 0 1% input (' please enter the matrix formed by the branch parameter: bl 二')；b2 isequal to 0,0, 25.5, 0, 1;2200;2200;2200;2200;120,2313;0,61. 1137. 8735,0,2;0,47. 5329.4610,0,2;0,54. 3233.6610,0,2;2)0 40. 74 + 25. 25 i 35 0% input (' please enter the matrix of the parameter paramete

7、rs of each node: b2 二')；% n 二 4; % input (' please enter the number of nodes: n 二')；% nl = 4; % input (' please enter a number of directions: nl二)；% bl is equal to 1, 2, 4 + 16i 0, 0, 0;% 1, 3, 4 + 16i 0, 0, 0;% 2, 3, 2 + 8i 0, 0, 0;% 2, 41. 49 + 4& 02 i 0, 11/110, 1 % input (

8、9; please enter the matrix formed by the branch parameter: bl 二')；% b2 is equal to 0, 0, 115, 0, 1;% 0, 0, 110, 0, 2;% 0, 20 + 4i 110, 0, 2;% 0, 10 + 6i 100 2 % input (' please input the matrix of the parameters of each node: b2 二')；y 二 zeros (n); e 二 zeros (1, n); f = zeros (1, n): v 二

9、zeros (1, n) ; sida 二 zeros (1, n) ; si 二 zeros (nl);the admittance matrix 一for 1=1: nl % from 1 to nlif bl (i, 7)二 1 %is on one sideif bl (i, 6)二二 0 % left (the first end)p = bl (i, 1); q = bl (i, 2);the else %left node(the first end)is on the k sidep = bl (i, 2); q 二 bl (i, 1);the endy (p, q)二 y (

10、p, q) - 1 / (bl (i, 3) * bl (i, 5). % the diagonal elementy of q, p is equal to y of p, q. % the diagonal elementy (q, q)二 y (q, q) + 1)/(b1 (i, 3) * bl (i, 5)八 2) ; % diagonal element k sidey (p)二 y (p)二 y (p) + 1. / bl (i, 3) + bl (i, 4) ; % diagonal element 1 + excitation admittance"else&quo

11、t;p = bl (i, 1); q = bl (i, 2);y (p,q)二 y(p, q) -1./bl (i,3) ; % thediagonal elementy of q,pisequal toyofp, q.%thediagonal elementy (q)二y(q)二 y (q)+1./ bl(i,3)+ bl(i, 4). / 2.0;%diagonally equal to half of the liney (p)二y(p)二 y (p)+1./ bl(i,3)+ bl(i, 4). / 2.0;%diagonally, half of the power of the l

12、inethe endthe enddisp (' the admittance matrix y 二')；disp (y);%given the initial node voltageand given each node power injection g 二 real (y); b 二 imag (y); the % decomposes the real and imaginary parts of the admittance matrixfor 1 = 1: n % given the real and imaginary parts of the initial

13、voltage of each nodee (i)二 real (b2 (i, 3);f (i)二 imag (b2 (i, 3);v (i)二 abs (b2 (i, 3) ; % pv, balance node, and pq node voltage modulusthe endfor i = 1: n % given each node injection powers (i)二 b2 (i, 1) -b2 (i, 2) ; the % i node is injected with the power sg - slb (i, i) = b (i, i) + b2 (i, 4) ;

14、 the % i node has no amount of work compensationthe end% 二二二二二二二二二二二二二二二二二二 with newton iteration 一 ralph monson method is used to solve the nonlinear algebraic equation (power equation)二二二二二二二二二二二二二二二二二二p 二 real (s) ; q 二 imag (s) ; the % decomposition of the active and reactive power of each nodei

15、ct1 二 0; it2 二 1; no 二 2 * n; n1 二 no + 1; a 二 0; % iterations, ict1, a; the number of nodes that are not satisfied with the convergence is it2while it2 =0%n0=2 *n jacobian; n1 + 1 二 no extensionjz 二' jacobi matrix (', num2str (a), ') cancellation operations，; jz1 二'jacobi matrix c ,

16、 num2str (a), ') the second iteration ' ; jzo 二'power equation number c , num2str (a), ') the time difference:%to calculatethe power of each node and power deviation and voltage deviation of pv nodes for 1 = 1: n % n nodes 2n rows (two equations p and q or u for each node)p 二 2 * i -

17、 1; m 二 p + 1; c 二 0; d 二 0;for jl = 1: n % i row n columns (n nodes interadmittance and node voltage multiplied by the current)c (i)二 c+ g* e(jl)_ b(i,jl)* f(jl)；% s(ej- bijgij* *fj)d (i)二 d+ g(i,jl)* f(jl)+ b(i,jl)* e(jl);% s(fj +bijgij* *ej)the endthe calcullate(d vallueof the work and the :react

18、;ivepowerp qpl 二 c (i) * e (i) + f (i) * d (i) ; power p % node calculation ei s (ej - bij gij * * fj) + fi 藝(fj + bij gij * * ej)q1 二 c (i) * f (i) -e (i) * d (i) ; power q % node calculation fi 2 (ej - bij gij * * fj) ei s (fj + bij gij * * ej)v2 二 e (i)八 2 + f (i)2; % voltage die squaredthe diffe

19、rence between the power difference and the pv node is equal to 二二if i 二 isb % non-equilibrium node (pq or pv node)if b2 (i, 5) 二 3 % non-pv nodes (only pq nodes)j (m, nl)二 p (i) -pl; the % pq node has the power difference j (m, nl) to extend column delta pj (p, nl)二 q (i) -ql; the % pq node has no w

20、ork power difference j (p, nl) expands column delta qelse % pv nodes 二二二二二二j (m, nl)二 p (i) -pl; the % pv node has the power difference j (m, nl) to extend column delta pj (p, nl)二 v (i)2 - v2. the % pv node voltage module (p,nl) extends column delta uthe endend % (if i = isb) non-equilibrium nodes

21、(pq or pv nodes)end % (fori 二 1:n)nnodes 2n rows (two equations p dnd q oru for eachnode)for m = 1:nojjn1 (m)二j (m,nl);the enddisp (jzo); disp (jjn1);%deviation value judgmentpower and voltage deviation value whether meet the requirements of pv nodes for k = 3: no % removes all nodes other than the

22、balanced node1 and 2det 二 abs (k, nl);if det > = pr % pq node，s power deviation and the voltagedeviation of the pv node are metit2 二 it2 + 1; the number of nodes that does not meet the requirement is plus 1the endthe endict2 (a)二 it2; the number of nodes that do not meet therequirement; a is the

23、number of iterationsict1 二 ict1 + 1; % the number of iterationsif ict2 (a)二二 0; the number of nodes currently not satisfied is zerobreak % exits the iteration operationthe end%above to calculate the various nodes of power and power deviation and voltage deviation of pv nodes the jacobi matrix formed

24、 the correct equation for the jacobi matrixfor i 二 2: n % n nodes 2n rows (two equations p and q or u for each node)if i 二 isb % non-equilibrium node (pq or pv node)if b2 (i, 5) 二 3%, the element of the jacobi matrix for the pq node is equal to 二二二c (i)二 0; d (i)二 0;for jl = 1: n % i row n columns (

25、n nodes interadmittance and node voltage multiplied by the current)c 二 c (i) + g (i) * e (jl) - b (i, jl) * f (jl) ; % s (ej -bij gij * * fj)d (i)二 d (i) + g (i, jl) * f (jl) + b (i, jl) * e (jl); % s (fj + bij gij * * ej)the endfor jl = 2: n % i row n columns (2n jacobi matrix elements dp/de and dp

26、/df or dq/de and dq/df)if jl 二 isb&j 1 二 i % unbalanced node &non-diagonal elementxi 二-g (i, jl) * e - b (i, jl) * f (i) ; % xi 二 dp/de 二-dq/df 二-x4x2 = b (i, jl) * e (i) - g (i, jl) * f (i) ； x2 二 dp/df 二 dq/de 二 x3the x3 二 x2; % x2 二 dp/df x3 二 dq/dex4 二-xi; % xi 二 dp/de x4 二 dq/dfp=2*i- 1

27、; q=2 * jl-1;j (p, q) = x3; m 二 p + 1; so it's going to be dq/de j (p, n) 二 dq.j (m, q)二 xi; q 二 q + 1; the dp/de j (m, n)二 dp.j (p, q)二 x4; j (m, q)二 x2; % x4 二 dq/df x2 二 dp/dfelseif jl 二二二 i&j 1 二 isb % non-equilibrium node & diagonal elementxi二-c(i)-g(i) * e (i)- b (i) * f (i) ;% dp/

28、dex2二-d(i)+b(i) * e (i)-g (i) * f (i) ;% dp/dfx3 二 d (i) + b (i) * e (i) -g (i) * f (i) ; % dq/dex4二-c(i)+g(i) * e (i)+ b (i) * f (i); % dq/dfp 二 2*1 - 1;q二 2* jl-1;j (p, q)二 x3; %expansion columndelta q j of p,n)二 dq;m 二 p + 1;j (m, q)二 xi; q 二 q + 1; j (p, q)二 x4; % expansion column delta p j (m,

29、n)二 dp;j (m, q)二 x2;the endthe endelse % if b2 (i, 5) 二 3 otherwise (that is the pv node)the element of the jacobi matrix is equal to 二二 二二二for jl = 1: nif jl 二 isb&j 1 二 i % unbalanced node &non-diagonal elementxi 二-g (i, jl) * e (i) - b (i, jl) * f (i); % dp/dex2 二 b (i, jl) * e (i) - g (i

30、, jl) * f (i); % dp/dfx5 =0; x6 二 0:p 二 2 * i -error.1; q 二 2 * jl-1; j (p, q)二 x 5; % pv node voltagem 二 p + 1;j (m, q)二 xi; q 二 q + 1; j (p, q)二 x6; the % pv node has the work error j (m, n)二 dp;j (m, q)二 x2;elseif jl 二二二 i&j 1 二 isb % non-equilibrium node & diagonale1ementxi 二-c (i) -g (i

31、) * e (i) - b (i) * f (i); % dp/dex2 二-d (i) + b (i) * e (i) -g (i) * f (i) ; % dp/dfx6 is equal to minus 2 times f of i.p = 2*i-1; q = 2* jl-1; j (p, q)二 x 5; % pv node voltage errorm 二 p + 1;j (m, q)二 xi; q 二 q + 1; j (p, q)二 x6; the % pv node has the work error j (m, n)二 dp;j (m, q)二 x2;the endth

32、e endend % (if b2 (i, 5) = 3 else)end % (if i 二 isb)end % (for 1 = 1: n) n nodes 2n rows (two equations p and q or u for each node)jzo 二'formed the first (', num2str (a), ') the subjacobi matrix:% disp (jzo); disp (j);% 二二二二二二二二二二二二二二二二二二二二二二二二二二二二二二 二 above to form a complete jacobi mat

33、rix =the elimination of the modified equation formed by the jacobi matrix is solved by the gaussian elimination method (by the column elimination and the return of the matrix)for k = 3: no % no = 2 * n (from the third row, the first, second line is the equilibrium node)for kl 二 k + 1: m % from the j

34、acobi element of k + 1 to the extension delta p, delta q or delta uj (k, kl)二 j (k, kl). / j (k, k) ; the non-diagonal elements of k row k columns are normalized by the k line k column diagonal elementthe endj (k, k) = 1; the % diagonally normalized k lines k column diagonal elements assignment 1% =

35、 in columns elimination operation 二二二二二二二二二二二二二二二二二二二二二二二二二二for k2 二 k + 1: no % from k + 1 to 2 * n last rowfor k3 二 k + 1: n1 % from k2 + 1 to the extension column cancels out the triangle elements after the k + 1 rowj (k2, k3)二 j (k2, k3) -j (k2, k) * j (k, k3); % elimination operationend % uses

36、the current row k3 column element minus the current row k column element times the k row k3 column elementj (k2, k) = 0; the k column element of the current row is 0the endthe end% jz = ' jacobi matrix (', num2str (a), ') cancellation operations，; jz1 二'jacobi matrix (', num2str

37、(a), ') the second iteration '% disp (jz); disp (j);% = = = = = = = = = = = = = = = = = = = = on column generation algorithm = = = = = = = = = = = = = = = = = = = = = = = = = =for k = no: - 1:3for kl = k - 1: -1:3j (kl, nl)二 j (kl, nl) - j (kl, k) * j (k, nl);j (kl, k)二 0;the endfor m 二 1: n

38、ojjnl (m)二 j (m, nl);the enddisp (jz1); disp (jjnl); % disp (j);%modify the node voltagefor k 二 3:2: no-1l is equal to k plus 1.e (l) = e (l) -j (k, nl) ; % modification node voltage real partki 二 k + 1;f (l) = f (l) -j (kl, nl) ; % modifies the voltage virtual partof the nodeu (l)二 sqrt (e (l) &quo

39、t; 2 + f (l)八 2):the enddisp (' each node voltage module '); disp (u):%二二二二二二二二二 end of an iterationthe end%* * * * * the following for the end of the iterative calculation of the output process *disp (' iteration:');disp (ict1-1);disp (' not the number of precision requirements:

40、');disp (ict2);for k = 1: nv (k)二 sqrt (e (k)八 2 + f (k)八 2) ; percent calculates the magnitude of each node's voltagesida (k)二 atan (f (k). / e (k) the percent calculates the angle of each node，s voltagee (k)二 e (k) + f (k) * j; % will represent each node voltage in the pluralcalculated out

41、putdisp (“ the voltage complex value of the nodes is (the node number is from small to large): “);disp (e) ; the actual voltage value of each node is shown in the pluraldisp (')disp (' the value of the voltage modulus of each node is the size of the node number of nodes from small to large:&

42、#39;);disp (v); the % shows the magnitude of v for each nodedisp (')disp (' the voltage phase of each node is sida (the node number is from small to large):disp (sida) ; the % display the voltage phase angle of each nodec (p)二 0;c (p)二 c (p) + conj (p, q). % calculate the conjugate value of

43、the injection current of each nodethe ends (p)二 e (p) * c (p) ; % calculates the conjugate value of the power s 二 voltage x injection current of each nodethe enddisp (“ power s of each node (node number from small to large): )；disp (s); the % shows the injection power of each nodedisp (')disp (&

44、#39; the first power si on each branch of the branch is in the same order as you enter bl:');for i 二 1: nlp = bl (i, 1); q = bl (i, 2);if bl (i, 7)二二 0si (p, q)二 e (p) * conj (e (p) * bl (i, 4). / 2 + (e (p) -e (q) / bl (i, 3).siz (i)二 si (p, q)；the elseif bl (i, 6)二二 0si (p, q)二 e (p) * (e (p)

45、* bl (i, 4)(p) * bl (i, 5) -e (q) * (1. / (bl (i, 3) * bl (i, 5) siz (i)二 si (p, q);the elsesi (p, q)二 e (p) * conj (e (p) - e (q) * bl (i, 5) * (1)/(b1 (i, 3) * bl (i, 5)八 2);siz (i)二 si (p, q);the endthe end's (' num2str， (p),? num2str (q),')二'，num2str (si (p, q)disp (zf);disp (

46、9;the enddisp (' the end power sj of each branch is the same as when you enter bl):');for i = 1: nlp = bl (i, 1); q = bl (i, 2);if bl (i, 7)二二 0(q, p)二 e (q) * conj (e (q) * bl (i, 4). / 2 + (e (q) -e (p). / bl (i, 3).sjy (i)二 sj (q, p);the elseif bl (i, 6)二二 0sj (q) (q, p)e * conj (e (q) -

47、e (p) * bl (i, 5) * (1)/(b1 (i, 3) * bl (i, 5)八 2);sjy (i) = sj (q, p);the else (q, p)二 e (q) * (e (q) * bl (i, 4).+ (e (q) * bl (i, 5) -e (p) * (1. / (bl (i, 3) * bl (i, 5)sjy (i)二 sj (q, p);the endthe end's (' s' (' num2str '), '? num2str (p),，)二disp (zf);disp (')；the e

48、nddisp (the power loss ds for each branch of the branch is the same as when you enter bl in the order);for i = 1: nlp 二 bl (i, 1); q 二 bl (i, 2);ds (i) = si (p) + sj (q, p);"the zf 二'ds (', num2str (p), ，, num2str (q),')二 num2str (ds (i)disp (zf);disp ('')；the endzws 二 0; jd

49、dy 二;jdp 二;jdq 二;jddyjd 二;for 1 = 1: n % total net loss for all nodes injected with power algebra andzws 二 zws + s (i);jddyjd = strcat (jddyjd, num2str (i),(', num2str (' i)'),')jddy 二 strcat (jddy, num2str (i), ' (', num2str (v (i),');jdp 二 strcat (jdp, num2str (i), ' c , num2str),j);(jdq, num2str (i),z，);the endjddyjd 二 strcat c node voltage angle: jddy