03_PowerFlows

Electrical microgrids designSession 3

Power flows

Luis Ismael Minchala Avila

Universidad de CuencaDepartamento de Elctrica, Electrnica y Telecomunicaciones

[email protected]

Ismael Minchala A. (UCuenca) Power flows April 15th, 2015 1 / 52

Agenda

1 Introduction

2 Gauss elimination

3 Jacobi and Gauss-Seidel

4 Newton-Raphson

5 The power-flow problem

6 Control of power-flow problem

7 Summary and questions


Introduction

Introduction

Successful power system operation under normal balanced three-phasesteady-state conditions requires the following:

Generation supplies the demand (load) plus losses.Bus voltage magnitudes remain close to rated values.Generators operate within specified real and reactive power limits.Transmission lines and transformers are not overloaded

The power-flow (PF) computer program (sometimes called load flow) is thebasic tool for investigating these requirements.


Introduction

Introduction

The PF program computes the voltage magnitude and angle at eachbus in a power system under balanced three-phase steady-stateconditions.

It also computes real and reactive power flows for all equipmentinterconnecting the buses, as well as equipment losses.

The PF problem is formulated as a set of nonlinear algebraic equationssuitable for computer solution.


Introduction

Characteristics of WTG for WPP

Wind turbine generators (WTG) have developed from small machineswith output power ratings on the order of kilowatts to severalmegawatts.

The principle of wind turbine (WT) operation is based on twowell-known processes.

1 The conversion of kinetic energy of moving air into mechanical energy.This is accomplished by using aerodynamic rotor blades and a varietyof methodologies for mechanical power control.

2 The electromechanical energy conversion through a generator that istransmitted to the electrical grid.


Introduction

Characteristics of WTG for WPP

WTs can be classified by their mechanical power control, and furtherdivided by their speed control.

At the top level turbines can be classified as either stall regulated orpitch regulated.

1 Stall regulation is achieved by shaping the turbine blades such that theairfoil generates less aerodynamic force at high wind speed..

2 Pitch regulation is achieved through the use of pitching devices in theturbine hub, which twist the blades around their own axes.

WTs are further divided into fixed speed (Type 1), limited variablespeed (Type 2), or variable speed with either partial (Type 3) or full(Type 4) power electronic conversion. In a Type 5 WT the mechanicaltorque converter between the rotors shaft and the generators shaftcontrols the generator speed to the electrical synchronous speed.


Introduction

Type 1 WTG: Limited Variable Speed

Squirrel-cage Induction Generator (SCIG) connected directly to thestep up transformer.The turbine speed is fixed (or nearly fixed) to the electrical gridfrequency.It generates real power (P) when the turbine shaft rotates faster thanthe electrical grid frequency creating a negative slip (positive slip andpower is motoring convention).For a given wind speed, the operating speed of the turbine understeady conditions is a nearly linear function of torque.For sudden changes in wind speed, the mechanical inertia of the drivetrain will limit the rate of change in electrical output


Introduction



Introduction


Wound rotor induction generators are connected directly to the WTGstep-up transformer in a fashion similar to Type 1 with regards to themachines stator circuit, but also include a variable resistor in the rotorcircuit.This can be accomplished with a set of resistors and power electronicsexternal to the rotor with currents flowing between the resistors androtor via slip rings.The variable resistors are connected into the rotor circuit softly andcan control the rotor currents quite rapidly so as to keep constantpower even during gusting conditions, and can influence the machinesdynamic response during grid disturbances.


Introduction



Introduction

Type 3: Variable speed partial power electronics conversion

Known commonly as the doubly fed induction generator (DFIG) ordoubly fed asynchronous generator (DFAG), takes the Type 2 designto the next level, by adding variable frequency ac excitation (instead ofsimply resistance) to the rotor circuit.The additional rotor excitation is supplied via slip rings by a currentregulated, voltage-source converter, which can adjust the rotorcurrents magnitude and phase nearly instantaneously.This rotor-side converter is connected back-to-back with a grid sideconverter, which exchanges power directly with the grid.


Introduction

Type 4: Variable speed with full power electronics conversion

Offers a great deal of flexibility in design and operation as the outputof the rotating machine is sent to the grid through a full-scaleback-to-back frequency converter.The turbine is allowed to rotate at its optimal aerodynamic speed,resulting in a wild ac output from the machine.In addition, the gearbox may be eliminated, such that the machinespins at the slow turbine speed and generates an electrical frequencywell below that of the grid.


Introduction

Type 5 WTG

Consist of a typical WTG variable-speed drive train connected to atorque/speed converter coupled with a synchronous generator (SG).The closely coupled SG, operating at a fixed speed can then bedirectly connected to the grid through a synchronizing circuit breaker.The SG can be designed appropriately for any desired speed (typically6 pole or 4 pole) and voltage.This approach requires speed and torque control of the torque/speedconverter along with the AVR, synchronizing system, and generatorprotection system inherent with a grid-connected SG.


Introduction

Voltage control capabilities

The voltage control capabilities of a WTG depend on the wind turbinetype.

Type 1 and Type 2 WTGs can typically not control voltage. Instead,these WTGs typically use power factor correction capacitors (PFCCs)to maintain the power factor or reactive power output on the lowvoltage terminals of the machine to a setpoint.

Types 3 through 5 WTGs can control voltage. These WTGs arecapable of varying the reactive power at a given active power andterminal voltage, which enables voltage control


Gauss elimination

Solutions to LAE: Gauss elimination

Lets consider:

A11 A12 . . . A1NA21 A22 . . . A2N...

.... . .

...AN1 AN2 . . . ANN

x1x2...xN

=

y1y2...yN

(1)Ax = y

If A is an upper triangular matrix with nonzero diagonal elements, we have:

A11 A12 . . . A1,N1 A1N0 A22 . . . A2,N1 A2N...

.... . .

......

0 0 . . . 0 ANN

x1x2...xN

=

y1y2...yN

(2)Ismael Minchala A. (UCuenca) Power flows April 15th, 2015 15 / 52

Gauss elimination

Back substitution

xN =yNANN

xN1 =yN1 AN1,NxN

AN1,N1

xk =yk

Nn=k+1 AknxnAkk

(3)


Gauss elimination

Gauss elimination

A11 A12 . . . A1N0

(A22 A21A11A12

). . .

(A2N A21A11A1N

)...

.... . .

...0

(AN2 A21A11A12

). . .

(ANN A21A11A1N

)

x1x2...xN

=

y1y2 A21A11 y1

...yN AN1A11 y1

(4)

A(2)11 A(2)12 . . . A

(2)1N

0 (2)A22 . . . A(2)2N

......

. . ....

0 0 . . . A(2)NN

x1x2...xN

=

y1y (2)2...

y (2)N

(5)


Gauss elimination

Example: Gauss elimination and back substitution

Solve using Gauss elimination and back substitution:[10 52 9

] [x1x2

] [63

]Solution:

[10 50 9 2105

] [x1x2

] [6

3 2106]

=

[6

3 2106]

[10 50 8

] [x1x2

] [68

]=

[61.8

]x2 =

y (1)2A(1)22

=1.88

= 0.225

x1 =y (1)1 A(1)12 x2

A(1)11=

6 (5)(0.225)10

= 0.4875


Gauss elimination

Exercise: Triangularizing a matrix

Use Gauss elimination to triagularize: 2 3 14 6 810 12 14

x1x2x3

579


Jacobi and Gauss-Seidel

Iterative solutions to LAE

A general iterative solution to (1) proceeds as follows:

1 Select an initial guess x(0)2 x(i + 1) = g[x(i)] i = 0, 1, 2, . . .

3 Continue until xk(i+1)xk(i)xk(i) <

x(i) is the ith guess, g is an N vector of functions that specify the iterationmethod, and is a specified tolerance.



The Jacobi method

yk = Ak1x1 + Ak2x2 + + Akkxk + + AkNxN (6)xk =

1Akk

[yk (Ak1x1 + + Ak,k1xk1 + Ak,k+1xk+1 + + AkNxN)]

xk =1Akk

[yk

k1n=1

Aknxn N

n=k+1

Aknxn

](7)

The Jacobi method uses the old values of x(i) at iteration i to generate the newvalue of xk(i + 1). That is:

xk(i + 1) =1Akk

[yk

k1n=1

Aknxn(i)N

n=k+1

Aknxn(i)

]k = 1, 2, . . . ,N (8)



The Jacobi method

x(i + 1) = Mx(i) + D1y (9)M = D1 (D A) (10)

D =

A11 0 0 . . . 00 A22 0 . . . 0

0...

.... . .

......

......

. . . 00 0 0 . . . ANN

(11)



Example: Jacobi method

Solve using the the Jacobi method.[10 52 9

] [x1x2

] [63

]Following the methodology just presented, we have:

D1 =[

10 00 9

]1=

[ 110 00 19

]M =

[ 110 00 19

] [0 510 29 0

]=

[0 510 29 0

][x1(i + 1)x2(i + 1)

]=

[0 510 29 0

] [x1(i)x2(i)

]+

[ 110 00 19

] [63

]



Example: Jacobi method

Starting with x1(0) = 0 and x2(0) = 0, the iterative solution is given in thefollowing table:

i 0 1 2 3 4 5 . . . 10x1(i) 0 0.6000 0.4333 0.5000 0.4814 0.4888 . . . 0.48749x2(i) 0 0.3333 0.2000 0.2370 0.2222 0.2263 . . . 0.22500

The convergence criterion is satisfied at the 10th iteration, since:

x1(10) x1(9)x1(9) = 0.48749 0.487520.48749

= 6.2 105 < x2(10) x2(9)x2(9) = 0.22500 0.225020.22502

= 8.9 105 < Ismael Minchala A. (UCuenca) Power flows April 15th, 2015 24 / 52

Newton-Raphson

Iterative solutions to NLAE: Newton-Raphson

Considering a set of nonlinear algebraic equations in matrix form:

f(x) =

f1(x)f2(x)...

fN(x)

= y (12)0 = y f(x)

Adding Dx to both sides of (12), where D is a square N N invertiblematrix,

Dx = Dx + y f(x)x = x + D1 [y f(x)]

x(i + 1) = x(i) + D1{y f[x(i)]} (13)Ismael Minchala A. (UCuenca) Power flows April 15th, 2015 25 / 52

Newton-Raphson

Iterative solutions to NLAE: Newton-Raphson

One method for speciying D, called Newton-Raphson, is based on thefollowing Taylor series expansion of f(x) about an operating point x0:

y = f(x0) +fx

x=x0

. . .

x = x0 +

[fx

x=x0

]1(x x0)

x(i + 1) = x(i) + J1(i) (y f[x(i)]) (14)

J(i) =fx

x=x(i)

=

f1x1

f1x2 . . .

f1xN

f2x1

f2x2 . . .

f2xN

......

. . ....

fNx1

fNx2 . . .

fNxN

(15)


Newton-Raphson

Example: Solution to polynomial equations

Solve the scalar equation f (x) = y , where y = 9 and f (x) = x2. Startingwith x(0) = 1, use the Newton-Raphson approach.

J(i) =ddx(x2)x=x(i) = 2x |x=x(i) = 2x(i)

x(i + 1) = x(i) +1

2x(i)[9 x2(i)]

Starting with x(0) = 1, successive calculations of the NR equations gives:

i 0 1 2 3 4 5x(i) 1 5.00000 3.40000 3.02353 3.00009 3.00000


Newton-Raphson

Exercise: Newton-Raphson method: solution to NLAE

Solve: [x1 + x2x1x2

]=

[1550

]x0 =

[49

]Defining f1 = (x1 + x2) and f2 = x1x2, we have:

J(i)1 =

[f1x1

f1x2

f2x1

f2x2

]x=x(i)

=

[1 1

x2(i) x1(i)

]1=

1x1(i) x2(i)

[x1(i) 1x2(i) 1

]

[x1(i + 1)x2(i + 1)

]=

[x1(i)x2(i)

]+

1x1(i) x2(i)

[x1(i) 1x2(i) 1

] [15 x1(i) x2(i)50 x1(i)x2(i)

]

i 0 1 2 3 4x1(i) 4 5.20000 4.99130 4.99998 5.00000x2(i) 9 9.80000 10.00870 10.00002 10.00000


Newton-Raphson

Newton-Raphson method

Equation (14) can be rewritten as follows:

J(i)x(i) = y(i) (16)x(i) = x(i + 1) x(i) (17)y(i) = y f[x(i)] (18)

The following 4 steps complete the NR algorithm:

1 Compute y(i) from (18)2 Compute J(i) from (15)3 Using Gauss elimination and back substitution, solve (16)4 Compute x(i + 1) from (17)


Newton-Raphson

Exercise: Newton-Raphson in four steps

Complete the aforementioned four steps for solving the previos exercise.[x1 + x2x1x2

]=

[1550

]x0 =

[49

]


The power-flow problem

The power flow problem

The computation of voltage magnitude and phase angle at each bus ina power system under balanced three-phase steady-state conditions.

The starting point for a power-flow problem is a single-line diagram ofthe power system, from which the input data for computer solutionscan be obtained.The following variables are associated with each bus k :

1 Voltage magnitude Vk ;2 Phase angle k ;3 Net real power Pk ; and4 Reactive power Qk supplied to the bus.

At each bus, two of these variables are specified as input data, and theother two are unknowns to be computed by the power-flow program




The power delivered to bus k is separated into generator and load terms:

Pk = PGk PLk (19)Qk = QGk QLk (20)

Each bus k is categorized into one of the following three bus types:1 Swing bus (or slack bus). The swing bus is a reference bus for which

V1, tipically 1.00o2 Load (PQ) bus. Pk and Qk are input data. The power-flow program

computes Vk and k .3 Voltage controlled (PV) bus. Pk and Vk are input data. The

power-flow program computes Qk and k . Maximum and mininum varlimits of the equipment in these kind of nodes are also input data.



Node equations

The node equations can be summarized into three steps:

1 Consider a circuit with N + 1 nodes, called also buses. Select onenode as the reference bus and define all the remaining node voltageswith respect to this node.

2 Transform every voltage source and its respective series impedanceinto an equivalent current source in parallel with the correspondingadmittance.

3 Write node equations as follows:Y11 Y12 Y1NY21 Y22 Y2N...

.... . .

...YN1 YN2 . . . YNN

V10V20...

VN0

=

I10I20...IN0

(21)where YNN is the bus admittance matrix, and VN1 and IN1 arethe voltage and current vectors.



Node equations

The elements Yij of the admittance matrix are formed as follows:

Ykk =Nk=1

Admittances connected to bus k (22)

Ykn = N

k=1k 6=n

Admittances connected between buses k and n (23)

The off-diagonal elements Ykn of Y are called mutual admittances and thediagonal elements Ykk are called self-admitances. The node equations areactually current equations written for each node in compliance withKirchhoffs current law.



Example: Power-flow input data and Ybus

Bus Type V (deg) PG QG PL QL QGmax QGmin1 Swing 1.0 0 - - 0.0 0.0 - -2 Load - - 0.0 0.0 8.0 2.8 - -3 Const. voltage 1.05 - 5.2 - 0.8 0.4 4.0 -2.84 Load - - 0.0 0.0 0.0 0.0 - -5 Load - - 0.0 0.0 0.0 0.0 - -

Sbase = 100 MVA, Vbase = 15 kV at buses 1, 3, and 345 kV at buses, 2, 4, 5



Example: Power-flow input data and YbusTable : Line input data

Bus-to-bus R X G B max MVA2-4 0.0090 0.100 0.0 1.72 12.02-5 0.0045 0.050 0.0 0.88 12.04-5 0.00225 0.025 0.0 0.44 12.0

Table : Transformer input dataBus-to-bus R X Gc Bm max MVA max TAP sett1-5 0.00150 0.02 0.0 0.0 6.0 -3-4 0.00075 0.01 0.0 0.0 10.0 -

Table : Input data and unknowsBus Input data Unknows1 V1 = 1.0, 1 = 0.0 P1, Q12 P2 = PG2 PL2 = 8 V2, 2

Q2 = QG2 QL2 = 2.83 V3 = 1.05 Q3, 3

P3 = PG3 PL3 = 4.44 P4 = 0, Q4 = 0 V4, 45 P5 = 0, Q5 = 0 V5, 5



Example: Power-flow input data and Ybus

The elements of Ybus are computed as follows:

Y21 = Y23 = 0

Y24 =1

R 24 + jX24

=1

0.009 + j0.1= 0.89276 + j9.91964

Y25 =1

R 25 + jX25

=1

0.0045 + j0.05= 1.78552 + j19.8393

Y22 =1

R 24 + jX24

+1

R 25 + jX25

+ jB 242

+ jB 252

= (0.892 j9.91964) + (1.78552 j19.83932) + j 1.722

+ j0.882

= 2.67828 j28.4590



Power-flow equations

For power system and load flow analysis, a set of power balance equationsis used:

Pk = VkNn=1

YknVn cos (k n kn) (24)

Qk = VkNn=1

YknVn sin (k n kn) (25)

where Vk and k are the magnitude and angle of node k voltage; Ykn andkn are the magnitude and angle of the element of the k-th row and n-thcolumn of the admittance matrix Y; and Pk and Qk are the active andreactive power at node k .



Power-flow solution by Newton-Raphson

x =[V

]=

2...NV2...VN

; y =

[PQ

]=

P2...PNQ2...

QN

f(x) =[P(x)Q(x)

]=

P2(x)...

PN(x)Q2(x)

...QN(x)

where all V, P and Q terms are in per-unit and terms are in radians.Ismael Minchala A. (UCuenca) Power flows April 15th, 2015 40 / 52


Power-flow solution by Newton-Raphson

yk = Pk = Pk(x) = VkNn=1

YknVn cos (k n kn) (26)

yk+N = Qk = Qk(x) = VkNn=1

YknVn sin (k n kn) (27)

k = 2, 3, . . . ,N

J =

P22

. . . P2NP2V2 . . .

P2VN

.... . .

......

. . ....

PN2

. . . PNNPNV2 . . .

PNVN

Q22

. . . Q2NQ2V2 . . .

Q2VN

.... . .

......

. . ....

QN2

. . . QNNQNV2 . . .

QNVN

(28)



Power-flow solution by Newton-Raphsonn 6= k

J1kn =Pkn

= VkYknVn sin (k n kn)

J2kn =Pkn

= VkYknVn cos (k n kn)

J3kn =Qkn

= VkYknVn cos (k n kn)

J4kn =Qkn

= VkYknVn sin (k n kn)n = k

J1kk =Pkk

= VkN

n=1;n 6=kYknVn sin (k n kn)

J2kk =Pkk

= VkYkk cos kk +Nn=1

YknVn cos (k n kn)

J3kk =Qkk

= VkN

n=1;n 6=kYknVn cos (k n kn)

J4kk =Qkk

= VkYkk sin kk +Nn=1

YknVn sin (k n kn)Ismael Minchala A. (UCuenca) Power flows April 15th, 2015 42 / 52


Power-flow solution by Newton-Raphson (algorithm)

1 Use (13) and (16) to compute:

y(i) =[

P(i)Q(i)

]=

[P P[x(i)]QQ[x(i)]

](29)

2 Calculate the Jacobian3 Use Gauss elimination and back substitution to solve[

J1(i) J2(i)J3(i) J4(i)

] [(i)V(i)

]=

[P(i)Q(i)

](30)

4 Compute

x(i + 1) =[(i + 1)V(i + 1)

]=

[(i)V(i)

]+

[(i)V(i)

](31)

Starting with initial value x(0), the procedure continues until convergenceis obtained or until the number of iterations exceeds a specified maximum.Ismael Minchala A. (UCuenca) Power flows April 15th, 2015 43 / 52


Power-flow solution by Newton-Raphson (algorithm)

For each voltage-controlled bus, the magnitude Vk is already known,and the function Qk(x) is not needed.We could omit Vk from the x vector and Qk from the y vector. Wecould also omit from the Jacobian matrix the column corresponding topartial derivatives with respect to Vk and the row corresponding topartial derivatives of Qk(x).Rows and corresponding columns for voltage-controlled buses can beretained in the Jacobian matrix. Then during each iteration, thevoltage magnitude Vk(i + 1) of each voltage-controlled bus is reset toVk , which is input data for that bus.


Control of power-flow problem

Control of power flow

The following means are used to control system power flows:

Prime mover and excitation control of generators;

Switching of shunt capacitor banks, shunt reactors, and static varsystems;

Control of tap-changing and regulating transformers.



Control of power flow (generator control)

Vt is the generator terminal voltage, Eg is the excitation voltage, is thepower angle, and Xg is the positive-sequence synchronous reactance.

I =Ege j

Vt

S = P + jQ = Vt I = Vt(Ege j VtjXg

)P = Re S =

VtEgXg

sin (32)

Q = Im S =VtXg

(Eg cos Vt) (33)Ismael Minchala A. (UCuenca) Power flows April 15th, 2015 46 / 52



Eq. (32) shows that the real power P increases when the power angle increases.

When the prime mover increases the power input to the generatorwhile the excitation voltage is held constant, the rotor speed increases.

As the rotor speed increases, the power angle also increases, causingan increase in generator real power output P. There is also a decreasein reactive power output Q, given by (33).

From the power-flow standpoint, an increase in prime-move powercorresponds to an increase in P at the constant-voltage bus to whichthe generator is connected. The power-flow program computes theincrease in along with the small change in Q.




Equation (33) shows that Q increases when the excitation voltage Egincreases. From the operational standpoint, when the generatorexciter output increases while holding the prime-mover powerconstant, the rotor current increases.

As the rotor current increases, the excitation voltage Eg also increases,causing an increase in generator reactive power output Q. There isalso a small decrease in d required to hold P constant in (32).

From the power-flow standpoint, an increase in generator excitationcorresponds to an increase in voltage magnitude at theconstant-voltage bus to which the generator is connected. Thepower-flow program computes the increase in reactive power Qsupplied by the generator along with the small change in .



Control of power flow (shunt capacitor)

Before the capacitor bank is connected, the switch SW is open and the busvoltage equals ETh.

After the bank is connected, SW is closed, and the capacitor current ICleads the bus voltage Vt by 90o. The phasor diagram shows that Vt is largerthan ETh when SW is closed.

From the power-flow standpoint, the addition of a shunt capacitor bank to aload bus corresponds to the addition of a negative reactive load, since acapacitor absorbs negative reactive power.

The power-flow program computes the increase in bus voltage magnitudealong with the small change in .



Power-flow solution in MATLAB


Summary and questions

Summary

In this session we have studied:

Different wind turbines configurations;Numerical methods for solving linear an nonlinear algebraic equations;The power-flow problem;The Newton-Raphson algorithm for solving the power-flow problem.


Appendix Questions

Preguntas


IntroductionGauss eliminationJacobi and Gauss-SeidelNewton-RaphsonThe power-flow problemControl of power-flow problemSummary and questionsAppendix

Documents

03_PowerFlows