PSET_lesson08_Simulator_1_base Magnitudes and Ameters – Tasks

Lesson 2 – Network Descrip2on Using Graph Theory

Semester 3 – Power Systems for Electrical Transporta2on

Lecturer: Pablo Arboleya Arboleya

Sustainable Transporta2on and Electrical Power

Systems

Universidad de Oviedo

Lesson 8 – Simulator 1 (base magnitudes and other parameters –tasks)

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Bases :Sb → Base Power V1b → Base Voltage in AC subsystemV2b → Base Voltage in link subsystemVdb → Base Voltage in DC subsystemfn → Base frequency

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Base Currents :

Ib1 =Sb3V1b

→ Base Current in AC subsystem

Ib2 =Sb3V2b

→ Base Current in link subsystem

Ib1 =SbVdb

→ Base Current in DC subsystem

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Base Impedances :

Zb1 =V1b

2

Sb → Base Impedance in AC subsystem

Zb2 =V2b

2

Sb → Base Impedance in link subsystem

Zdb =Vdb

2

Sb → Base Impedance in DC subsystem

3.2. Solución a la variación de dimensión del sistema 39

AC/DC AC/DCAC/DCAC/DC

G

G

1DC

2DC

3DC

4DC

5DC

6DC7DC

8link9link 10link

11AC

12AC

13AC 14AC15AC

16AC

Figura 3.1: Sistema AC/DC dimensiones constantes.

anterior el nt ahora es 3, luego los nodos de la topología de DC comenzarán en 4y así se seguirá con el criterio de numeración ya explicado.

Base Magnitudes

Other useful data to characterize the substa2ons

3.2. Solución a la variación de dimensión del sistema 39

AC/DC AC/DCAC/DCAC/DC

G

G

1DC

2DC

3DC

4DC

5DC

6DC7DC

8link9link 10link

11AC

12AC

13AC 14AC15AC

16AC

Figura 3.1: Sistema AC/DC dimensiones constantes.

anterior el nt ahora es 3, luego los nodos de la topología de DC comenzarán en 4y así se seguirá con el criterio de numeración ya explicado.

62 4. Formulación del modelo y resolución del flujo de cargas

vACNdj , vAC

Nqj

vDCNi

iLEdk, iLEqk

iLEk

edge k

nodo AC j

nodo DC i

Figura 4.9: Detalle de ecuaciones del rectificador.

Así pues la función que rige el frenado de un tren i es:

iDCti =

!"#

"$

Pi

vDCNi

vDCNi ! Vc

m(vDCNi " Vmax) Vc ! vDC

Ni ! Vmax

0 vDCNi # Vmax

(4.28)

Dondem = I0/(Vmax " V0).

4.2.4.2. Modelo matemático de la subestación de tracción

A lo largo de los apartados anteriores se plantearon cada una de las ecuacionesque rigen el funcionamiento por una lado del sistema AC que forma la red dealimentación y por otro del sistema DC formado por la red de tracción ferroviaria.

Como se comentó en capítulos anteriores, en esta tesis se tratará de resolver elsistema conjunto AC/DC de forma unificada, sin la necesidad de procesos iterati-vos entre los dos sistemas tratados de forma independiente. Queda por tanto clarala importancia de las ecuaciones que sirven como nexo de unión entre los dossistemas. Estas ecuaciones son las que afectan a las variables propias de la sub-estación de tracción y que tendrán por tanto una importancia vital en el correctofuncionamiento del sistema.

En este trabajo se modelan subestaciones de tracción formadas por rectifica-dos de 6 pulsos no controlados. Las ecuaciones (4.29) y (4.30) representan elbalance de potencias en un convertidor ideal sin perdidas, como el de la figura 4.9conectado entre el nodo de AC j y el nodo de DC i, representando el edge k.

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Sn → Rated Power of Power transformer (VA)V1 → Rated Voltage of Power transformer primary side (V)V2 → Rated Voltage of Power transformer secondary side (V)Pcu → Cooper Losses of Power transformer side (W)Vcc → Short circuit voltage of power transformer (%)Rdiode _ on → Conmutation reactance of the rectifier (Ω)Scc → Short circuit power at the AC point of coupling (VA)

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Transformer Parameters Estimation

Zcc _ t =V1

2Vcc

100SnI1 =

Sn3V1

I2 =Sn3V2

Rcc _ t =Pcu3I1

2 Xcc _ t = Zcc _ t2 − Rcc _ t

2

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Grid Parameters Estimation

Zeq = Xeq =V2

2

SccConmutation Reactance EstimationZeq _ total = Zeq + Zcc + Rdiode _ on

Input data

• All input data will be in a folder called input • Input folder will contain the next txt files: – bases.txt – lines_ac_IS.txt – lines_dc_IS.txt – links_acdc_IS.txt – nodes_ac_IS.txt

bases.txt 2e6 %Sb (Base Power) 24e3 %V1b (Base voltage, primary side) 1221 %V2b (Base voltage, secondary side) 1648.93 %Vdb=3*sqrt(2)*V2b/pi (Base Voltage, DC side) 500e6 %Scc (Short circuit power in the substa2ons) 50 %a (Rated frequency)

Depending of the kind of rec2fier

%number from to R (ohm) X (ohm) 1 1 2 7.21249 37.24256 2 1 4 7.69066 39.71168 3 1 6 6.47530 33.43600 4 2 5 1.99240 10.28800 5 3 4 4.00472 20.67888 6 3 5 1.89278 9.77360

lines_ac_IS.txt

lines_dc_IS.txt %number from to R(ohm) 1 1 2 0.59 2 2 3 1.18 3 2 4 0.354 4 3 4 0.59

%n dc_node ac_node Sn V1 V2 Pcu Vcc Rdiode_on % (from) (to) (VA) (V) (V) (W) (%) (ohm) 1 1 2 1315e3 24000 1221 13300 5.5 1e-‐3 2 3 3 1315e3 24000 1221 13300 5.5 1e-‐3 3 4 4 1315e3 24000 1221 13300 5.5 1e-‐3

links_acdc_IS.txt

nodes_ac_IS.txt %Type 1=PQ (P y Q (-‐) generated (+) consumed), 2=PV,3=slack. %Sub |Type | Vd | Vq | P | Q | |V| %num | | V | V | W | VAr | V 1 3 25000 0 0 0 0 2 1 0 0 0 0 0 3 1 0 0 0 0 0 4 1 0 0 0 0 0 5 1 0 0 400000 0 0 6 1 0 0 1000000 800000 0

Task 1 (45 min)

• Load all the data contained in the txt files (use the command “cd” to access to the input folder and “load” to read the data.

• Obtain all base magnitudes (currents, impedances) • Obtain the next variables – lines_dc same as lines_dc_IS but in p.u. – lines_ac same as lines_ac_IS but in p.u. – links_acdc same as links_acdc_IS but in p.u. in the first three columns, the next columns will be Rccpu Xccpu Zeqpu_total

– nodes_ac same as nodes_ac_IS but in p.u.

Task 2 (30 min)

• Calculate the number of nodes and edges: – ntrain=3; = number of trains (Now it will be a constant) – nlinks = number of substa2ons (links) – nnodes_dc = number of dc nodes without trains – nnodes_dc_t = number of dc nodes with trains – nnodes_ac = number of ac nodes without link nodes – nnodes_ac_t = number of ac nodes with link nodes – nnodes = total number of nodes – nedges_dc_n = edges connec2ng the trains – nedges_dc_ts = edges connec2ng the trains with the rest of DC nodes – nedges_dc_ss = edges connec2ng the DC nodes (the real ones) – nedges_dc = all DC edges – nedges_link = edges containing a rec2fier – nedges_ac = all AC edges including the power transformers – nedges = all AC and DC edges

Task 3 (45 min)

• Adapt the index of the nodes and the edges in the next variables to the described node and edge enumera2on criteria: – lines_ac – lines_dc – links_acdc – nodes_ac

• Create the next variables: – rec2fiers_acdc : each row represents a rec2fier:

• (edge number; from; to; Zpu)

– trafos_ac: each row represents a power transformer: • (edge_number; from; to; Rpu; Xpu)

Task 4 (20 min)

• Create vectors containing the indices of every kind of node:

• N_slack = slack nodes • N_PV = PV nodes • N_PQ = PQ nodes • N_AC = AC nodes (no trafo secondaries) • N_Train = train nodes • N_DC = DC nodes (no trains) • N_DCT = All DC nodes • N_link = Link nodes • N_DC_AC = DC nodes connected with the AC subsystem

Task 5 buid matrix M (1 hour)

• Build matrix M, and fill all posi2ons of RBDC with a parameter defined as infinite.

• Useful commands: – accumarray – ones – zeros – eye – transpose or ‘ – horzcat – vertcat – blkdiag – diag

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−RBDC ΓDC

−RBAC XB

AC ΓAC

−XBAC −RB

AC ΓAC

ΓDC( )T S

ΓAC( )T -I

ΓAC( )T -I

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

⋅

iEDC

iEdAC

iEqAC

itDC

iEL

iEdL

iNdAC

iEqL

iNqAC

vNDC

vNdAC

vNqAC

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

= 0€

nBDC

€

nBAC

€

nBAC

€

nNDC

€

nNAC

€

nNAC

€

nBDC

€

nBAC

€

nBAC

€

nNDC

€

nNAC

€

nNAC

€

nNDC

€

nNAC

€

nNAC

€

Num

ber of equ

a2on

s

Number of unknowns

train_pos defini2on:

train_pos= [1 4 5 0.5 443e3/Sb 2 4 5 0.8 380e3/Sb 3 4 6 0.7 -‐80e3/Sb];

• Each row represent a train, it must contain all the trains in the system, the columns will represent: – first -‐> the train index – Second -‐>from bus (always a real dc node) – third -‐> to bus (always a real dc node) – fourth -‐> distance in (%) from the train to the from node – fish -‐> train power in p.u. (+ = trac2on, -‐=braking)

• If al columns except the first one are zero, this means that the train is not ac2ve

Task 6 reallocate the trains in the system (2 hours)

• The trains must be reallocated if they are non ac2ve or if there are two or more in the same point.

• If they are non-‐ac2ve trains they will be reallocated connected to the lower index DC substa2on at a distance equal to a parameter called param=0.001. If there are more than one non-‐ac2ve trains, the second one will be at 2*param and so on.

• If more than on train are in the same electrical point they must be reallocated, the minimum distance between 2 trains must be equal to param.

• In the realloca2ng process, be careful of not obtaining distances greater or equal that 1 p.u. between the train and the from node.

• Useful commands: – find – Unique

• See the example in the next slide

Task 6 (trains realloca2on) train_pos= [1 0 0 0 443e3/Sb 2 10 11 0.9999 0 3 10 11 0.5 -‐80e3/Sb 4 11 12 0.5 -‐80e3/Sb 5 12 13 0 0 6 0 0 0 -‐80e3/Sb 7 10 11 0.9999 -‐80e3/Sb 8 12 13 0.5 0 9 12 13 0.3 -‐80e3/Sb];

10 12 13

11

1 6

3 2 7

4

5

9 8

• Trains 1 and 6 must be located near node 10.

• Trains 7 is in the same posi2on as2 a n d w h e n i t s r e a l l o c a t e d t h e distance will be equal or greater than one

• T r a i n 5 i s i n substa2on 12 so it should be reallocated near it

Suggested steps for task 6 • Define param=0.0001; it is the distance that separate the trains when they are in

the same point • Check if any train is out of the system and locate this trains in the first real DC line

next to the first DC substa2on at a distance of param, it doesn’t maner if there are more than one non-‐ac2ve trains in the same point at this moment. (The command “find” will be very useful)

• check if there is more than one train at the same posi2on, the command [a,b,c]=unique(train_pos(:,[2 3]),'rows'); use the parameters “a” and “c” to determine

• For each real dc line determine the index of the trains that are in that line. • Go through this indices, compare the posi2on of the train of each indices with the

rest of the trains and while exist another train in the same posi2on reallocate the actual one by adding param to the distance. (Two trains have the same posi2ons if the columns 2,3 and 4 of train_pos are the same)

• Then, check If there is some train at zero, adnif you find some one add param to all of them

• Check if there is some train with a distance greater than one, if you find someone divide all the distance by the maximum distance in per unit plus 2*param.

• Then use the command sortrows nested to obtain all the trains ordered in all lines.

Task 6 (train_pos with reallocated trains) train_pos =

1.0000 10.0000 11.0000 0.0002 0.2215 6.0000 10.0000 11.0000 0.0004 -‐0.0400 3.0000 10.0000 11.0000 0.5000 -‐0.0400 7.0000 10.0000 11.0000 0.9997 -‐0.0400 2.0000 10.0000 11.0000 0.9998 0 4.0000 11.0000 12.0000 0.5000 -‐0.0400 5.0000 12.0000 13.0000 0.0001 0 9.0000 12.0000 13.0000 0.3000 -‐0.0400 8.0000 12.0000 13.0000 0.5000 0

10 12 13

11

1 6

3 7 2

4

5

9 8

Task 7 (Ac2va2on of the DC edges) • In this task we will ac2vate the correct DC edges. • First we will construct this auxiliary matrix, very helpful to determine the correct

index of an edge. • Ady_dc_n_ts = 0 1 2 3 4 5 6 7 8 9 10 11 12 0 0 13 14 15 16 17 18 19 20 21 22 23 0 0 0 24 25 26 27 28 29 30 31 32 33 0 0 0 0 34 35 36 37 38 39 40 41 42 0 0 0 0 0 43 44 45 46 47 48 49 50 0 0 0 0 0 0 51 52 53 54 55 56 57 0 0 0 0 0 0 0 58 59 60 61 62 63 0 0 0 0 0 0 0 0 64 65 66 67 68 0 0 0 0 0 0 0 0 0 69 70 71 72 • This matrix has the same structure that the adjacency matrix between trains and

trains and substa2on, the number of rows is the number of trains (9 in this par2cular case of study) and the number of columns is the number of trains plus the number of substa2ons.

• If we want to know the index of the edge connec2ng the train “i” with the train or substa2on “j” we just have to extract the element Ady_dc_n_ts(i,j)

Task 7 (Ac2va2on of the DC edges) • For each real DC line, I extract the indices of the trains that are in that line

comparing that line with the train posi2on in train pos, for instance, the first DC line is (10,11)

• For each real DC line I create an auxiliary variable called aux with the indices of the trains in train pos located in that line, in this par2cular case aux=[1 2 3 4 5]

train_pos =

1 10 11 0.0002 0.2215 6 10 11 0.0004 -‐0.0400 3 10 11 0.5000 -‐0.0400 7 10 11 0.9997 -‐0.0400 2 10 11 0.9998 0 4 11 12 0.5000 -‐0.0400 5 12 13 0.0001 0 9 12 13 0.3000 -‐0.0400 8 12 13 0.5000 0

lines_dc =

73 10 11 0.4340 74 11 12 0.8680 75 11 13 0.2604 76 12 13 0.4340

For the 1st DC line (10,11): aux= [1 2 3 4 5] For the 2nd DC line (11,12): aux= [6] For the 3rd DC line (11,13): aux= empty matrix For the 4th DC line (12,13): aux= [7 8 9];

train_pos =

1 10 11 0.0002 0.2215 6 10 11 0.0004 -‐0.0400 3 10 11 0.5000 -‐0.0400 7 10 11 0.9997 -‐0.0400 2 10 11 0.9998 0 4 11 12 0.5000 -‐0.0400 5 12 13 0.0001 0 9 12 13 0.3000 -‐0.0400 8 12 13 0.5000 0

Task 7 (Ac2va2on of the DC edges) • Depending on the size of aux, we can have three different situa2ons:

– length(aux)=0; In this case we have to ac2vate the real DC line, this is the case of the third DC lline (11,13), so the edge 75 should be ac2vated.

– length(aux)=1; Just one train in the line, for instance the second line (11,12). Is the row number 6 in the train pos matrix, using the matrix Ady_dc_n_ts and the indices in the row number six I obtain the edges: Ady_dc_n_ts(train_pos(6,1), train_pos(6,2) ) is the first edge and the second one should be Ady_dc_n_ts(train_pos(6,1), train_pos(6,3) ) .

Ady_dc_n_ts(train_pos(6,1), train_pos(6,2) )= Ady_dc_n_ts(4,11)=40 Ady_dc_n_ts(train_pos(6,1), train_pos(6,3) )= Ady_dc_n_ts(4,11)=41

10 12 13

11

1 6

3 7 2

4

5

9 8

Edge 41

Edge 40

train_pos =

1 10 11 0.0002 0.2215 6 10 11 0.0004 -‐0.0400 3 10 11 0.5000 -‐0.0400 7 10 11 0.9997 -‐0.0400 2 10 11 0.9998 0 4 11 12 0.5000 -‐0.0400 5 12 13 0.0001 0 9 12 13 0.3000 -‐0.0400 8 12 13 0.5000 0

Task 7 (Ac2va2on of the DC edges) • Depending on the size of aux, we can have three different

situa2ons: – length(aux)>=2; that is the case of the line (10,11) for instance.

• We will ac2vate first the connec2on of the first substa2on with the first train (1,10) and the last substa2on with the last train (2,11).

• Then we will ac2vate the connec2ons between trains(1,6), (6,3), (3,7) and (7,2). Be careful because in Ady_dc_n_ts you must extract the index (3,6) because the index (6,3) is zero

10 12 13

11

1

6

3 7 2

4

5

9 8

9

5

26 27 17 21

train_pos = 1 10 11 0.0002 0.2215 6 10 11 0.0004 -‐0.0400 3 10 11 0.5000 -‐0.0400 7 10 11 0.9997 -‐0.0400 2 10 11 0.9998 0 4 11 12 0.5000 -‐0.0400 5 12 13 0.0001 0 9 12 13 0.3000 -‐0.0400 8 12 13 0.5000 0

Task 7 (Ac2va2on of the DC edges) • We have to obtain a variable with two rows called dc_ac2v, the first row will indicate the ac2ve edges,

and the second row the resistance of the edges. For calcula2ng the resistance we have to calculate the distances between the trains and the trains and substa2ons in p.u. and then mul2ply this value by the resistance of the DC lines in the matrix line_dc.

• For instance, the resistance of line 26 between trains 3 and 6 will be (0.5-‐0.0004)*0.4340 • For instance, the resistance of line 9 between train one and substa2on 10 will be (0.0002-‐0)*0.4340

lines_dc = 73 10 11 0.4340 74 11 12 0.8680 75 11 13 0.2604 76 12 13 0.4340

10 12 13

11

1

6

3 7 2

4

5

9 8

9

5

26 27 17 21

40

41 49

46 64 68

75

dc_ac2v = 9 21 5 26 27 17 40 41 75 49 68 46 64 0.0001 0.0001 0.0001 0.2168 0.2169 0.0001 0.4339 0.4340 0.2604 0.0001 0.2170 0.1302 0.0868

Finally we use these indices and values to correct M

TASK 8 SOLVER

€

min f (x) = XA2 + XB

2 + XC2

g1(x) = 2XA + 5XB ≤ 5g2(x) = XA + XC ≤ 2g3(x) = XA

2 + 2XB + XC ≤10h1(x) = 2XA − 2XB + XC = −2h2(x) =10XA + 8XB −14XC = 26h3(x) = −4XA + 5XB − 6XC

2 = 60 ≤ XA ≤101≤ XB ≤ 5−5 ≤ XC ≤ 8

XA 0 = XB 0 = XC 0 = −1

fmincon

€

A =2 1 01 0 1⎛

⎝ ⎜

⎞

⎠ ⎟ b =

52⎛

⎝ ⎜ ⎞

⎠ ⎟

Aeq =2 −2 1

10 8 −14⎛

⎝ ⎜

⎞

⎠ ⎟ beq =

-226⎛

⎝ ⎜

⎞

⎠ ⎟

lb = (0,1,−5)ub = (10,5,8)

[SOL,fval,exi{lag]=fmincon(@myfun,X0,A,b,Aeq,beq,lb,ub,@mycon,op2ons);

func2on example_fmincom param1=10; param2=6; XA=-‐1; XB=-‐1; XC=-‐1; X0=[XA XB XC]; A=[2 1 0; 1 0 1]; b=[5 2]; Aeq=[2 -‐2 1; 10 8 -‐14]; beq=[-‐2 26]; lb=[0 1 -‐5]; ub=[10 5 8];

op2ons=op2mset('Algorithm','interior-‐point','TolFun',1e-‐6,'TolCon',1e-‐6,'TolX',1e-‐10,'MaxFunEvals',5e6,'MaxIter',5e6); [SOL,fval,exi{lag,output,lambda]=fmincon(@myfun,X0,A,b,Aeq,beq,lb,ub,@mycon,op2ons);

func2on F = myfun(X) XA=X(1); XB=X(2); XC=X(3); F=[XA^2+XB^2+XC^2]; end func2on [c,ceq]=mycon(X) XA=X(1); XB=X(2); XC=X(3); c=[]; c=[c XA^2+2*XB+XC-‐param1]; ceq=[]; ceq=[ceq -‐4*XA+5*XB-‐6*XC^2-‐param2]; end end

TASK 8 SOLVER

TASK 8 SOLVER: Defining the parameters outside the solvers

%% Specified Values Needed to State the equa2ons %% AC nodes [a,b,c]=intersect(nodes_ac(:,1),N_slack'); V_spec_slack_d=nodes_ac(b,3); V_spec_slack_q=nodes_ac(b,4); [a,b,c]=intersect(nodes_ac(:,1),N_PQ'); P_spec_PQ=nodes_ac(b,5); Q_spec_PQ=nodes_ac(b,6); [a,b,c]=intersect(nodes_ac(:,1),N_PV'); P_spec_PV=nodes_ac(b,5); V_spec_PV=nodes_ac(b,7); %% Train aux=sortrows(train_pos,1); P_train=aux(:,5); %% Rec2fier R_eq=rec2fiers_acdc(:,4);

TASK 8 Ini2aliza2on of the unknowns %%%% Inizializa2on of the unknowns %%%%%%% X_I_E_DC=zeros(1,nedges_dc); X_I_Ed_AC=zeros(1,nedges_ac); X_I_Eq_AC=zeros(1,nedges_ac); X_I_t_DC=zeros(1,ntrain); X_I_E_L=zeros(1,nnodes_dc); X_I_Ed_L=zeros(1,nedges_link); X_I_Nd_AC=zeros(1,nnodes_ac); X_I_Eq_L=zeros(1,nedges_link); X_I_Nq_AC=zeros(1,nnodes_ac); X_V_N_DC=ones(1,nnodes_dc_t); X_V_Nd_AC=ones(1,nnodes_ac_t); X_V_Nq_AC=ones(1,nnodes_ac_t);

X0=[X_I_E_DC X_I_Ed_AC X_I_Eq_AC X_I_t_DC X_I_E_L X_I_Ed_L X_I_Nd_AC X_I_Eq_L X_I_Nq_AC X_V_N_DC X_V_Nd_AC X_V_Nq_AC];

%%% Solving procedure %%%%%%% A=[]; b=[]; Aeq=M; beq=zeros(size(M,1),1); ub=[]; lb=[]; op2ons=op2mset('Algorithm','interior-‐point','TolFun',1e-‐6,'TolCon',1e-‐6,'TolX',1e-‐10,'MaxFunEvals',5e6,'MaxIter',5e6) [SOL,fval,exi{lag,output,lambda]=fmincon(@myfun,X0,A,b,Aeq,beq,lb,ub,@mycon,op2ons)

TASK 8 Advice to use the variables inside myfun and mycon

%% Unknowns %%

%Currents I_E_DC=X(1:nedges_dc); %All edge currents in the DC subsystem I_E_d_AC=X(nedges_dc+1:nedges_dc+nedges_ac); %All the edge currents in the AC subsystem (d) I_E_q_AC=X(nedges_dc+nedges_ac+1:nedges_dc+2*nedges_ac); %All the edge currents in the AC subsystem (q) I_T_DC=X(nedges_dc+2*nedges_ac+1:nedges_dc+2*nedges_ac+ntrain); %Train Currents I_E_LR=X(nedges_dc+2*nedges_ac+N_DC_AC); %Links DC currents (Here I consider only the real currents, not the ghost currents) I_E_LG=X(nedges_dc+2*nedges_ac+N_GHOST); %Links ghost currents I_E_d_L=X(nedges_dc+2*nedges_ac+nnodes_dc_t+1:nedges_dc+2*nedges_ac+nnodes_dc_t+nlinks); %Links currents AC (d) I_N_d_AC=X(????); %Injected currents in AC nodes (d) I_E_d_L=X(????); %Links currents AC (q) I_N_q_AC=X(???????); %Injected currents in AC nodes (q) I_N_d_slack=X(??) %Injected current in slack node (AC) (d) I_N_d_PQ=X(??) %Injected current in PQ node (AC) (d) I_N_d_PV=X(??) %Injected current in PV node (AC) (d) I_N_q_slack=X(??) %Injected current in slack node (AC) (q) I_N_q_PQ=X(??) %Injected current in PQ node (AC) (q) I_N_q_PV=X(??) %Injected current in PV node (AC) (q)

%Voltages V_N_dc_t=X(nedges_dc+2*nedges_ac+nnodes_dc_t+2*nnodes_ac_t+N_Train %Voltage in train nodes (DC) V_N_dc_s=X(nedges_dc+2*nedges_ac+nnodes_dc_t+2*nnodes_ac_t+N_DC); %Voltage in substa2on nodes (DC) V_N_d_slack=X(nedges_dc+2*nedges_ac+nnodes_dc_t+2*nnodes_ac_t+N_slack); %Voltage in slack bus (AC) (d) V_N_d_PQ=X(???); %Voltage in PQ nodes (AC) (d) V_N_d_PV=X(???); %Voltage in PV nodes (AC) (d) V_N_d_link=X(???); %Voltage in link nodes (AC) (d) V_N_q_slack=X(???); %Voltage in slack bus (AC) (q) V_N_q_PQ=X(???); %Voltage in PQ nodes (AC) (q) V_N_q_PV=X(???); %Voltage in PV nodes (AC) (q) V_N_q_link=X(???); %Voltage in link nodes (AC) (q)

TASK 8 Advice to use the variables inside myfun and mycon

Task 9: Postprocesor

• Use a procedure similar to the one described in task 8 to be able to obtain directly from the solu2on all the varaibles described in the previous slides.

PSET_lesson08_Simulator_1_base Magnitudes and Ameters – Tasks

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