Assignment 1 for 923

1. Example 1:

In this example, the task is to calculate the inductance per meter per phase as well as the

inductance of the double circuit

Figure 1. Double-circuit conductor configuration

Figure 2. Distance between phases in double-circuit

In this task, Matlab will be used to provide the results

A.Calculation by hand

a)Determine the inductance per meter per phase in henries:

D12=D1’2’=9 m; D11’=D22’=D33’=18 m;

D12’=D1’2=D13=D1’3’=9*2*sin60º=9√3 m;

D13’=D1’3=9 m;

DABeq=4√ D12∗D1 '2 '∗D12 '∗D1 ' 2=

4√9∗9∗9√3∗9√3=11.8447 m;

In similar way, DBCeq=DABeq=11.8447 m;

DACeq=4√ D13∗D1 '3 '∗D 13'∗D 1' 3=

4√9∗9∗9√3∗9√3=11.8447m;

GMD=3√ DABeq∗DBCeq∗DACeq=11.8447m

GMR=r’=1.8 cm=0.018 m;

GMRA=GMRB=GMRC=√r '∗D11 '=√0.018∗18=0.5692 m

Inductance per meter phase in henries:

L=¿2∗10−7∗lnGMDGMR

=2∗10−7∗ln11.84470.5692

=6.0708∗10−7 H /m

b)Determine the inductance of the double circuit:

Inductance of one circuit is calculated below:

Deq=3√ DAB∗DBC∗DAC=

3√9∗9∗9√3=10.8084 m; r’=0.018 m;

L=2∗10−7∗lnGMDGMR

=2∗10−7∗ln10.8084

0.018=1.28∗10−6 H /m

Inductance of double circuit: Ltotal=L2=1.28∗10−6

2=0.64∗10−6 H /m

c)Relative error:

Relative error=0.607−0.64

0.607∗100 %=−5.43 %

B.Calculation by Matlab

function [L Ltotal Error]=inductance(D12,D15,D14,GMR) % this is M-file to calculate

inductance

D45=D12; D42=D15; D23=D12; D13=D15; % Define the value for the distance between

conductors

DABeq=sqrt(D12*D45*D15*D42);

DABeq=sqrt(DABeq); DBCeq=DABeq; DACeq=DABeq;

GMD=(DABeq*DBCeq*DACeq)^(1/3);

GMRA=sqrt(GMR*D14);

GMRB=GMRA; GMRC=GMRA;

L=2*10^(-7)*log(GMD/GMRA); % calculate inductance in question a

Deq=(D12*D23*D13)^(1/3); % Calculate GMD for one circuit

L1=2*10^(-7)*log(Deq/GMR); % Calculate inductance for one circuit

Ltotal=L1/2; % Calculate the inductance of the double circuit in question b

Error=(L-Ltotal)*100/L; % Calculate relative error

2. Example 2:

a)Develop a bus admittance matrix for example

Y11=(3.815629-j*19.078144)+(5.169561-j*25.847809)=8.98519-j*44.925953

Y12=Y21=-3.815629+j*19.078144

Y13=Y31=-5.169561+j*25.847809

Y14=Y41=0

Y22=(3.815629-j*19.078144)+( 5.169561-j*25.847809)=8.98519-j*44.925953

Y23=Y32=0

Y24=Y42=-5.169561+j*25.847809

Y33=(5.169561-j*25.847809)+(3.023705-j*15.118528)=8.193266-j*40.966337

Y34=Y43=-3.023705+j*15.118528

Y44=( 5.169561-j*25.847809)+ (3.023705-j*15.118528) =8.193266-j*40.966337

As a result of that, the bus admittance matrix is:

Y= Y11 Y12 Y13 Y14

Y21 Y22 Y23 Y24

Y31 Y32 Y33 Y34

Y41 Y42 Y43 Y44

With Yij (i,j=1÷4) are received above values

b)Using Matpower to run power flow

*Run power flow using Newton’s method through list of Matlab commands below

>>mpopt=mpoption(‘PF_ALG’,1); % use Newtons’ method

>>result=runpf(‘case4gs’,mpopt);

*Run power flow using Gauss Seidel method through list of Matlab commands below:

>>mpopt=mpoption(‘PF_ALG’,4); % use Gauss Seidel method


Comment: Basing on the results obtained from the calculation, it is easy to see that two

method give us the same value for this circuit.

The results can be listed in the table below:

Method Newton Gauss Seidel

Voltage magnitude

and angle

(at load bus)

Bus 1: 1.00 pu (0 deg)

Bus 2: 0.982 pu (-0.976 deg)

Bus 3: 0.969 pu (-1.872 deg)

Bus 4: 1.020 pu (1.523 deg)

Generator dispatch Generator 1: P=186.81 MW; Q=114.50 MVAr

Generator 2: P=318.00 MW; Q=181.43 MVAr

Table. 1 Voltage (at load bus) and generator dispatch in two methods

Branch From To From bus injection To bus injection Loss

bus bus

P(MW) Q(MVAr) P(MW) Q(MVAr) P(MW) Q(MVAr)

1 1 2 38.69 22.3 -38.46 -31.24 0.227 1.13

2 1 3 98.12 61.21 -97.09 -63.57 1.031 5.16

3 2 4 -131.4 -74.11 133.25 74.92 1.715 8.58

4 3 4 -102.91 -60.37 104.75 56.93 1.835 9.18

Table. 2 Power flow in the line and power loss in two methods

Furthermore, the number of iteration are required for obtaining converged solution using

+Gauss Seidel method: 28 (converged in 0.05 seconds)

+Newton’s method: 3 (converged in 0.01 seconds)

Comparing with Gauss Seidel method, Newton’s method is faster converged.

c)Investigate the effect of changing reactive power at bus 3 and 4 independently

To do that, we need to change the value of reactive power which is used for each bus by

using this command:

*If the reactive power at bus 3 was changed:

>>define_constants;

>>mpc=loadcase(‘case4gs’);

>>mpc.bus(3,QD)=X % X is the value we want to use for reactive power at bus 3

>>runpf(mpc);

The results obtained from three cases (Q3=123.94 MVAr, 200 MVAr and 0 MVAr

respectively) are followed:

Q3=123.94 MVAr Q3=200 MVAr Q3=0 MVAr

Voltage at load bus

Bus 1 1.00 pu (0 deg) 1.00 pu (0 deg) 1.00 pu (0 deg)

Bus 2 0.982 pu (-0.976 deg) 0.982 pu (-0.976 deg) 0.982 pu (-0.984 deg)

Bus 3 0.969 pu (-1.872 deg) 0.949 pu (-1.707 deg) 0.999 pu (-2.137deg)

Bus 4 1.020 pu (1.523 deg) 1.020 pu (1.524 deg) 1.020 pu (1.509 deg)

Generation dispatch

Generator1 P= 186.21 MW

Q= 114.5 MVAr

P= 188.13 MW

Q= 166.53 MVAr

P= 186.04 MW

Q= 34 MVAr

Generator 2 P= 318 MW

Q= 181.43 MVAr

P= 318 MW

Q= 212.47 MVAr

P= 318 MW

Q= 133.52 MVAr

Table. 3 Voltage at load bus and generation dispatch when reactive power at bus 3 was changed

Q3 Branch From

bus

To

bus

From bus injection To bus injection Loss


123.94 1 1 2 38.69 22.3 -38.46 -31.24 0.227 1.13

MVAr 2 1 3 98.12 61.21 -97.09 -63.57 1.031 5.16

3 2 4 -131.4 -74.11 133.25 74.92 1.715 8.58

4 3 4 -102.91 -60.37 104.75 56.93 1.835 9.18

200

MVAr

1 1 1 38.68 22.3 -38.45 -31.24 0.227 1.13

2 1 3 99.45 113.24 -97.7 -111.82 1.756 8.78

3 2 4 -131.55 -74.11 133.26 74.92 1.716 8.58

4 3 4 -102.30 -88.18 104.74 87.98 2.436 12.18

0

MVAr

1 1 2 38.96 22.25 -38.73 -31.18 0.229 1.14

2 1 3 97.08 -19.24 -96.36 15.09 0.719 3.59

3 2 4 -131.27 -74.17 132.98 74.96 1.711 8.55

4 3 4 -103.64 -15.09 105.02 8.98 1.378 6.89

Table. 4 Power flow in the line and power loss when reactive power at bus 3 was changed

Comment:

+The voltage magnitude at bus 1, 2 and 4 was showed the same value when Q3 was changed

(increased or decreased) while the voltage magnitude at bus 3 was increased (it means that is

was improved) if the reactive power requirement from this bus decreased and vice versa.

+For two generators: the same value for real power was provided by them in three cases

whereas the reactive power generators supplied decreased if Q3 reduced and increased if Q3

increased.

+Power loss in the branch 1 from bus 1 to bus 2 remained constant. However the power loss

in other branches reduced when Q3 reduced and rise when Q3 increased. In detail, power loss

in branches (from 1 to 3 and from 3 to 4) were affected directly by the change in the value of

reactive power at bus 3.

*If the reactive power at bus 4 was changed:

>>define_constants;


>>mpc.bus(4,QD)=Y; % Y is the value we want to change for reactive power at bus 4

>>runpf(mpc);

The results obtained from the calculation in three cases (Q4=49.58 MVAr, Q4=100 MVAr and

Q4=0 MVAr respectively) are:

Q4=49.58 MVAr Q4=100 MVAr Q4=0 MVAr

Voltage at load bus

Bus 1 1.00 pu (0 deg) 1.00 pu (0 deg) 1.00 pu (0 deg)



Bus 4 1.020 pu (1.523 deg) 1.020 pu (1.523 deg) 1.020 pu (1.523 deg)

Generation dispatch

Generator1 P= 186.81 MW P= 186.81 MW P= 186.81 MW

Q= 114.5 MVAr Q= 114.5 MVAr Q= 114.5 MVAr

Generator 2 P= 318 MW

Q= 181.43 MVAr

P= 318 MW

Q= 231.85 MVAr

P= 318 MW

Q= 131.85 MVAr

Table. 5 Voltage at load bus and generation dispatch when reactive power at bus 4 was changed

Q4 Branch From

bus

To

bus

From bus injection To bus injection Loss


49.58

MVAr

1 1 2 38.69 22.3 -38.46 -31.24 0.227 1.13

2 1 3 98.12 61.21 -97.09 -63.57 1.031 5.16

3 2 4 -131.4 -74.11 133.25 74.92 1.715 8.58

4 3 4 -102.91 -60.37 104.75 56.93 1.835 9.18

100

MVAr

1 1 2 38.69 22.3 -38.46 -31.24 0.227 1.13

2 1 3 98.12 61.21 -97.07 -63.57 1.031 5.16

3 2 3 -131.54 -74.11 133.25 74.92 1.715 8.58

4 3 4 -102.91 -60.37 104.75 56.93 1.835 9.18

0

MVAr

1 1 2 38.69 22.3 -38.46 -31.24 0.227 1.13

2 1 3 98.12 61.21 -97.07 -63.57 1.031 5.16

3 2 4 -131.54 -74.11 133.25 74.92 1.715 8.58

4 3 4 -102.91 -60.37 104.75 56.93 1.835 9.18

Table. 6 Power flow in the line and power loss when reactive power at bus 4 was changed

Comment:

+ The voltage magnitude at bus 1, 2 and 3 was presented the same value when Q4 was

changed (increased or decreased) while the voltage magnitude at bus 4 also remained

constant because bus 4 is voltage controlled bus.

+For two generators: the same value for real and reactive power was provided by generator 1

in three cases. However, the reactive power generators 2 supplied decreased if Q4 reduced

and increased if Q4 increased whereas the real power remained constant.

+Power loss in the all branches keep constant. Contrary to the case where Q3 was adjusted,

the value of Q4 does not affect to power loss in every branch.

3. Example 3:

In this example, the number of 60 Hz three-phase, uncompensated transmission line required

to transmit this power with one out of service in three different cases will be determined in

case it is desired to transmit 2200 MW from a power plant to a load center located 300 km

from the plant.

Assuming: Vs=1 pu, Vr=0.95 pu, and δmax=35º

a. For 345 kV lines and Zc=300 Ω

SIL =3452

300=397 MW

Neglecting losses with l=300 km and δ=35º

P=1.0∗0.95∗397∗sin (35 °)

sin (2∗π∗3005000

)=397∗1.48=588 MW / line

=1.48*SIL

The practical line loadability curve in figure 5.12 (Glover, Sarma and Overbye, third edition,

pp. 231) can be used to obtain P=1.48*SIL for typical 300-km overhead 60 Hz

uncompensated lines.

In order to transmit 2200 MW with one line out of service:

Number of 345-kV line=2200 MW

588 MW / line+1=4.74 ≈ 6

b. For 500-kV lines and Zc=275 Ω

SIL =5002

275=909 MW

P=1.0∗0.95∗909∗sin (35 °)

sin (2∗π∗3005000

)=909∗1.48=1345 MW / line

=1.48*SIL

Number of 345-kV line=22001345

+1=2.64 ≈ 4

c. For 765-kV lines and Zc=260 Ω

SIL =7652

260=2251 MW

P=1.0∗0.95∗2251∗sin (35 °)

sin ( 2∗π∗3005000

)=2251∗1.48=3331 MW / line

=1.48*SIL

Number of 345-kV line=22003331

+1=1.66 ≈ 2

4. Example 4

4a. Wind turbine generation

*There are four major wind turbine generators:

Type 1: induction generator

Type 2: induction generator with variable rotor resistance

Type 3: doubly-fed asynchronous generator

Type 4: asynchronous or synchronous generator with full converter interface

The principle of wind turbine operation is based on two well-known processes. The first one

involves the conversion of kinetic energy of moving air into mechanical energy. This is

accomplished by using aerodynamic rotor blades and a variety of methodologies for

mechanical power control. The second process is the electro-mechanical energy conversion

through a generator that is transmitted to the electrical grid.

The difference between Type 1, Type 2, Type 3 and Type 4 wind turbine generator is that

mechanical power regulation: fixed speed (type 1), limited variable speed (type 2), variable

speed with either partial (type 3) or full (type 4) power electronic conversion. The different

speed control types are implemented via different rotating AC machines and the power

electronics.

*The characteristic of each type of wind turbine generator (WTG)

+Type 1:

The type 1 WTG is implemented with a squirrel-cage induction generator (SCIG) and is

connected to the step-up transformer directly. See Figure 3. The turbine speed is fixed (or

nearly fixed) to the electrical grid’s frequency, and generates real power (P) when the turbine

shaft rotates faster than electrical grid frequency creating a negative slip (positive slip and

power is motoring convention). Figure 4 shows the power flow at the SCIG terminals. While

there is a bit of variability in output with the slip of the machine, Type 1 turbine typically

operate at or very close to a rated speed. A major drawback of the induction machine is the

reactive power that it consumes for its excitation field and the large currents the machine can

draw when started “across-the-line”. To ameliorate these effects the turbine typically employs

a soft starter and discrete steps of capacitor banks within the turbine [2].

Figure 3. Typical configuration of a Type 1 WTG [2]

Figure 4. Variation of real and reactive power for SCIG [2]

+Type 2:

In this type, wound rotor induction generators are connected directly to the WTG step-up

transformer in a fashion similar to Type 1 with regards to the machines stator circuit, but also

include a variable resistor in the rotor circuit. See Figure 5. This can be accomplished with a

set of resistors and power electronics external to the rotor with current flowing between the

resistors and rotor via slip rings. Alternatively, the resistors and electronics can be mounted

on the rotor, eliminating the slip rings – this is the Weier design. The variable resistors are

connected into the rotor currents quite rapidly so as to keep constant power even during

gusting conditions, and can influence the machine’s dynamic response during grid

disturbances [2].


By adding resistance to the rotor circuit, the real power curve, which was shown in Figure 4,

can be “stretched” to the higher slip and higher speed ranges. See Figure 6. That is to say that

the turbine would have to spin faster to create the same output power, for an added rotor

resistance. This allows some ability to control the speed, with the blades’ pitching

mechanisms and move the turbines operation to a tip speed ratio (ration of tip speed to the

ambient wind speed) to achieve the best energy capture. It is typical that speed variations of

up to 10% are possible, allowing for some degree of freedom in energy capture and self

protection torque control [2].

Figure 6. Variation of real and reactive power with external rotor resistor in a Type 2 WTG [2]

*Type 3:

The Type 3 turbine takes the Type 2 design to the next level, by adding variable frequency

AC excitation (instead of simply resistance) to the rotor circuit. The additional rotor

excitation is supplied via slip rings by a current regulated, voltage-source converter, which

can adjust the rotor current’s magnitude and phase nearly instantaneously. This rotor-side

converter is connected back-to-back with a grid side converter, which exchanges power

directly with the grid [2].


A small amount of power injected into the rotor circuit can affect a large control of power in

the stator circuit. This is a major advantage of the DFIG – a great deal of control of the output

is available with the presence of a set of converters that typically are only 30% of the rating

of the machine. In addition to the real power that is delivered to the grid fro, the generator’s

stator circuit, power is delivered to the grid through the grid-connected inverter when the

generator is moving faster than synchronous speed. When the generator is moving slower

than synchronous speed, real power flows from the grid, through both converters, and from

rotor to stator. These two models, made possible by the four-quadrant nature of the

converters, allows a much wider speed range, both above and below synchronous speed by up

to 50%, although narrower ranges are more common [2].

*Type 4:

The Type 4 (Figure 8) offers a great deal of flexibility in design and operation as the output

of the rotating machine is sent to the grid through a full-scale back-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 gear-box may be eliminated, such that

the machine spins at the slow turbine speed and generates an electrical frequency well below

that of the grid. This is no problem for this type, as the inverter, and offer the possibility of

reactive power supply to the grid, much like a STATCOM. The rotating machines of this type

have been constructed as wound rotor synchronous machines, similar to conventional

generators found in hydroelectric plants with control of the field current and high pole

numbers, as permanent magnet synchronous machines, or as squirrel cage induction

machines. However, based upon the ability of the machine side inverter to control real and

reactive power flow, any type of machine could be used. Advances in power electronic

devices and controls in the last decade have made the converters both responsive and efficient

[2].


4b. (Glover, Sarma and Overbye, problem 6.1)

Install power world simulator and run the problem 6_61_PQ and 6_61_PV in case the line

between buses 2 and 4 is opened.

Figure 9. Simulation results in case Problem 6_61_PQ when the line between buses 2 and 4 is closed

Figure 10. Simulation results in case Problem 6_61_PQ when the line between buses 2 and 4 is

opened

Comment: as can be seen from the figures 9 and 10, when the line between buses 2 and 4 is

opened, the voltage at bus 4 decreased while the power flow in the line (from buses 1 to 3 and

from buses 2 to 5) increased. In addition, the reactive power which was supplied by shunt

capacitor was reduced whereas the change in the real and reactive power generated by

generators 1, 2, 6, and 7 was experienced.

Figure 11. Simulation results in case Problem 6_61_PV when the line between buses 2 and 4 is closed

Figure 12. Simulation results in case Problem 6_61_PV when the line between buses 2 and 4 is

opened

Comment: In this case, the voltage at bus 4 remained constant (1 pu) while only line between

buses 2 and 4 showed the increase in power flow. The real and reactive power provided by

generators 1, 6 and 7 remained nearly old value.

In summary:

+In term of voltage at bus 4, the case where the generation at bus 4 is modelled as a Type 3 or

Type 4 showed the better result than the case generation at bus 4 is modelled as a Type 1 or

Type 2 because in the first case (6_61_PQ) the voltage at bus 4 decreased when the line

between 2 and 4 is opened whereas it keep constant (1 pu) in the second case (6_61_PV).

+The advantage of a Type 3 or Type 4 wind turbine is the ability to keep the voltage at the

bus where it is located at the desired value (in this problem, that value is equal to 1 pu). In

addition, the reactive power which Type 3 or Type 4 wind turbine supplies can be adjusted to

regulate the voltage bus while for Type 1 or Type 2 it is impossible (keep the same value of -

48 MVAr in both cases when the line between 2 and 4 is closed and opened). Therefore, the

appearance of a shunt capacitor is not necessary if we have a Type 3 or Type 4 wind turbine.

+As can be seen from the figures 9 and 10, the shunt capacitor affect directly to the voltage

control. If the reactive requirement of the load does not change, the voltage at bus relates

closely to the reactive power a shunt capacitor can supplies. For example, if the reactive

power which a shunt capacitor supplies increases, the reactive power flow in the line will

decrease then the voltage drop decreases and the voltage at the bus where the shunt capacitor

connected to rises because the voltage at the source remains constant. As we can see from the

figures 9 and 10, the reactive provided by shunt capacitor varies from 62.44 to 65.94 MVAr.

REFERENCE

1. “Description and Technical Specification for Generic WTG Models – A Status

Report”, a Working Group Joint Report between the WECC Working Group on

Dynamic Performance of Wind Power Generation of the IEEE PES Power Stability

Controls Subcommittee of the IEEE PES Power System Dynamic Performance

Committee, in the conference proceedings of the 2011 IEEE PES Power System

Conference and Exhibition, Phoenix, Arizona.

2. “Characteristics of wind turbine generators for wind power plants”, an IEEE Pes

Wind Plant Collector System Design Working Group, in Power and Energy Society

General Meeting, 2009, pp.1-5

3. “Power System Analysis and Design” by J. D. Glover, M. S. Sarma and T. J.

Overbye, fourth Edition, Thomson publisher 2008.

4. Matpower 4.1 User Manual, Ray D.Zimmerman and Carlos E. Murillo-Sánchez, Dec.

14, 2011.

APPENDIX

A. The simulation results when using Newton’s method and Gauss Seidel method

*Run power flow using Newton’s method

>>mpopt=mpoption(‘PF_ALG’,1); % use Newtons’ method


*Run power flow using Gauss Seidel method:

>>mpopt=mpoption(‘PF_ALG’,4); % use Gauss Seidel method


B. Simulation results when changing the reactive power at bus 3 and bus 4 independently

*Changes at bus 3:

+Q3=123.94

>>define_constants;


>>mpc.bus(3,QD)=123.94;

>>runpf(mpc);

+Q3=200 MVAr

+Q3=0 MVAr

*Changes at bus 4:

+Q4=49.58 MVAr

>>define_constants;


>>mpc.bus(4,QD)=49.58;

>>runpf(mpc);

+Q4=100 MVAr

+Q4= 0 MVAr;

Assignment 1 for 923

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