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
>>result=runpf(‘case4gs’,mpopt);
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
P(MW) Q(MVAr) P(MW) Q(MVAr) P(MW) Q(MVAr)
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=loadcase(‘case4gs’);
>>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 2 0.982 pu (-0.976 deg) 0.982 pu (-0.976 deg) 0.982 pu (-0.976 deg)
Bus 3 0.969 pu (-1.872 deg) 0.969 pu (-1.872 deg) 0.969 pu (-1.872 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
P(MW) Q(MVAr) P(MW) Q(MVAr) P(MW) Q(MVAr)
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].
Figure 5. Typical configuration of a Type 2 WTG [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].
Figure 7. Typical configuration of a Type 3 WTG [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].
Figure 8. Typical configuration of a Type 4 WTG [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
>>result=runpf(‘case4gs’,mpopt);
*Run power flow using Gauss Seidel method:
>>mpopt=mpoption(‘PF_ALG’,4); % use Gauss Seidel method
>>result=runpf(‘case4gs’,mpopt);
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=loadcase(‘case4gs’);
>>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=loadcase(‘case4gs’);
>>mpc.bus(4,QD)=49.58;
>>runpf(mpc);
+Q4=100 MVAr
+Q4= 0 MVAr;