Report_AKbar_PDF

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Report
Power System Engineering (POSE)
Power Flow Studies
Akbar Pamungkas Sukasdi
329335
27-05-2014
Bachelor of Electronic Electrical Engineering
Report
Saxion University
M. H. Troomplan 28
Enschede, 27th
May 2014

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Akbar Pamungkas Sukasdi
Bachelor of Electronic Electrical Engineering
329335@student.saxion.nl
Saxion University
M. H. Troomplan 28
Enschede, 27th
May 2014

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POSE Assignment and Answers
Instruction 1 : Power flow with loads of the constant Z type
Part one:
Loads in general can be modelled in 3 different ways: a. Constant Power Load (kVA)
b. Constant Current (Amps) c. Constant Impedance (Ohms)
a. Constant Impedance or Admittance: 0% of constant P and Q
The load (P+jQ) is treated as constant impedance load. The impedance is calculated
by using the following equation:
=
− ∗
P = Load Active Power (kW)
Q = Load Reactive Power (kVAR)
V = Load Voltage
Z = Load equivalent constant Impedance
When a load of a rated “P” (kW), “Q” (kVAR) and “V” (Voltage) is entered, then it
calculates the equivalent impedance “Z” and defines the load as a constant shunt
impedance of “Z” (Ohms)
At constant impedance, the Z load is determined at nominal node voltage using Pload
and Qload. The following then applies in a load flow calculation:
if |U| increases : Iload increases
if |U| decreases: Iload decreases
Constant current: 50% constant P and Q, 50% constant admittance
The load (P+jQ) is treated as constant current load. The current is calculated by using
the following equation:
=
− ∗
P = Load Active Power (kW)
Q = Load Reactive Power (kVAR)
V = Load Voltage
I = Load equivalent constant current
When a load of rated “P” (kW), “Q” (kVAR) and “V” (Voltage) is entered, then it
calculates the equivalent current “I” and defines the load as a constant draw of “I”
(Amps).

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Constant power: 100% constant P and Q because = + ∗
At constant power, the rated power will always remain constant, independently of the
calculated node voltage |U|. The following then applies in a load flow calculation:
if |U| increases : Iload decreases
if |U| decreases: Iload increases
Note that in HV network models it is custom to use constant P and Q model for all
general loads. Note that in distribution network models the load behaviour tends to be
that of constant current or constant admittance. Also, the Load model is usually
determined before making the system and it’s based on the area where the system will
be built.
b. To regulate the secondary voltage level separated coils of winding are connected to
the tap changer at one of the transformer site, usually at the primary. Operation of the
tap changer modifies the total number of active coils of primary winding. For a fixed
number of coils at secondary winding, this action causes change of transformer
voltage ratio. If the supply voltage increases or load current decreases there will be an
increase in supply voltage which is not desirable. For example, the tap position in the
primary winding will rise towards positive direction i.e. +2.5%, and hence decreases
the primary windings. This will increase the turns ratio (Ns/Np) further decreases the
secondary voltage and vice versa.
=
Increasing tap-changer, decrease primary windings Np, decrease secondary voltage
and vice versa.
Part Two:
Choose a base power of 100 MVA.
a. Determine the base impedance, Zb at 30 kV and calculate the per unit values of the
cable impedances. Also determine the per unit impedances of the transformers and the
Source. Take no tap-changes into account.
= 100
= 30
= =
100
30
= 3.33
= =
30
3.33
= 9
Per unit values of the impedances:
Cable 1
Zc1 for 1km = 0.205 + j 0.134

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Zc1 = Length * ( 0.205 + j 0.134 ) = 5km * ( 0.205 + j 0.134 ) = 1.025 + j 0.67 Ω
Zc1pu = =
. . Ω
Ω
= 0.114 + j 0.074 pu
Cable 2
Zc2 for 1km = 0.205 + j 0.134
Zc2 = Length * ( 0.205 + j 0.134 ) = 4 km * ( 0.205 + j 0.134 ) = 0.82 + j 0.536 Ω
Zc2pu = =
. . Ω
Ω
= 0.091 + j 0.059 pu
Cable 3
Zc3 for 1km = 0.205 + j 0.134
Zc3 = Length * ( 0.205 + j 0.134 ) = 6 km * ( 0.205 + j 0.134 ) = 1.23 + j 0.804 Ω
Zc3pu = =
. . Ω
Ω
= 0.137 + j 0.089 pu
Imaginary part of Transformer’s impedances:
Transformer 1
Xt1 =
.
∗ 1. =

∗ 14% = 0.117 pu
Transformer 2
Xt2 =
.
∗ 2. =

∗ 11% = 0.22 pu
Transformer 3
Xt3 =
.
∗ 3. =

∗ 11% = 0.22 pu
Transformer 4
Xt4 =
.
∗ 4. =

∗ 6% = 0.3 pu
b. Derive a formula to express the impedance per phase of a load Zin the line to line
voltage VLL of the bus where it is connected to and the three phase power P and Q of
the given load. Use this formula to calculate the load impedances. Transform the
values to per unit. You can use these values for the load impedances in the handmade
calculation.
S =
Z =
Z =

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=
30
58 − 12
= 14.88 + 3.07
=
14.88 + 3.07
9
= 1.65 + 0.341

=
30
6 − 2
= 135 + 45
=
135 + 45
9
= 15 + 5

=
30
24 − 5
= 35.94 + 7.48
=
35.94 + 7.48
9
= 3.99 + 0.831
c.
Figure 1.1 Load Model, network scheme including the impedances of the loads
Real part of Transformer’s impedances:
Transformer 1
1 =
,
∗
,
=
300
120
∗
100
120
= 0.00208

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2 =
,
∗
,
=
200
50
∗
100
50
= 0.008

3 =
,
∗
,
=
200
50
∗
100
50
= 0.008
NB : Pk = Short circuit Power (Korsleting Power)
Real and Imaginary Impedances of Transformers:
1 = 0.00208 + 0.117
2 = 0.008 + 0.22
3 = 0.008 + 0.22
= + ( // )
= 0.00208 + 0.117 + (( 0.008 + 0.22 ) // (0.008 + 0.22 ))
= 0.117  89 + (( 0.22 88 ) // ( 0.22 88 ))
= 0.00208 + 0.117 + 0.004 + 0.11
= 0.00608 + 0.227
= 0.22708  88.5
= =
30
30
0.22708 < 88.5
= 0.118 – 4.402 = 4.404 < −88.5
Admittance (Y)
1 = _1 = =
. .
= 6.172 − 4.01
2 = _2 =
1
2
=
1
0.091 + 0.059
= 7.737 − 5.016
3 = _3 =
1
3
=
1
0.137 + 0.089
= 5.133 − 3.335
= =
1
=
1
0.00608 + 0.227
= 0.118 – 4.402
= _ =
1
_
=
1
1.65 + 0.341
= 0.581 − 0.12
= _ =
1
_
=
1
15 + 5
= 0.06 − 0.02

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= _ =
1
_
=
1
3.99 + 0.831
= 0.24 − 0.05
= ( 1 + 2 + ) − ( 2 ) − ( 1 )
= ( − 2 ) + ( 2 + 3 + ) − ( 3 )
= ( − 1 ) − ( 3 ) + ( 1 + 3 + )
d. Calculate the bus voltages in p.u. of the busses B, C and D and transform them back
to kV. You can use a powerful calculator with the function simult( ) or mathlab.
Matlab Code :
% Voltages calculation Vb, Vc, and Vd %
clear
%Cable Admittances
Yc1 = 6.172 - 4.01 * j ; % Y cable_1 pu %
Yc2 = 7.737 - 5.016 * j ; % Y cable_2 pu %
Yc3 = 5.133 - 3.335 * j ; % Y cable_3 pu %
%Load Admittances
YLb = 0.581 - 0.12 * j; % Y load_Bpu %
YLc = 0.06 - 0.02 * j; % Y load_Cpu %
YLd = 0.24 - 0.05 * j; % Y load_Dpu %
%Equivalent Admittances
Yeq = 0.118 - 4.402 * j; % Y Norton Admittance%
%Current Per Units
Ib = 0.118 - 4.402 * j;
Ic=0;
Id=0;
% Ib = Vb ( YLb + Yc2 + Yc1 + Yeq ) - Vc ( Yc2 ) - Vd ( Yc1 )
% Ic = Vb ( -Yc2 ) + Vc( YLc + Yc2 + Yc3 ) -Vd( Yc3 )
% Id = Vb ( -Yc1 ) - Vc( Yc3 ) + Vd( Yc3 + Yc1 + YLd )
Ytot1 = YLb + Yc2 + Yc1 + Yeq ; %Total admittances [1,1] in Matrices
Ytot2 = YLc + Yc2 + Yc3; %Total admittances [2,2] in Matrices
Ytot3 = Yc3 + Yc1 + YLd; %Total admittances [3,3] in Matrices
Y = [ Ytot1 -Yc2 -Yc1
-Yc2 Ytot2 -Yc3
-Yc1 -Yc3 Ytot3 ];
I = [Ib Ic Id];
V = I / Y
= 0.9210 − 0.1733 0.9084 − 0.1750 0.8989 − 0.1770
Transferring back to kV :
= 0.9210 − 0.1733 = 0.9372  − 10.6
= 0.9372  − 10.6 ∗ 30 = 28.116 – 10.6
= 0.9084 − 0.1750 = 0.925  − 10.8

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= 0.925  − 10.8 ∗ 30 = 27.75  − 10.8
= 0.8989 − 0.1770 = 0.916  − 11.04
= 0.916  − 11.04 ∗ 30 = 27.48  − 11.04
e. Also calculate the cable current in cable 1 in p.u. and transform it back to amperes.
= 0.9210 − 0.1733
= 0.8989 − 0.1770
1 = 6.172 − 4.01
_1 = ( − ) ∗ 1
_1 = 0.1512 − 0.0658
= 0.1512 − 0.0658 ∗
= (0.1512 − 0.0658 ) ∗ 3.33
= 0.5035 − 0.2191
=
0.5491  − 23.52
√3
= 0.3170  − 23.52
f. Calculate the sending end powers PS and QS on bus B and receiving end powers PR
and QR on bus D for cable 1.
= × _1
= (28.116 – 10.6 ) ∗ ( 0.5491  23.52 )
= 15.438 12.92
= 15.04716 + 3.45179
= × _1
= (27.48  − 11.04 ) ∗ (0.5491  23.52 )
= 15.2869  12.48
= 14.92569 + 3.30348

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Part Three:
Make a comparison table with all the values found by VISION and the calculated values by
hand.
Write a conclusion in which you evaluated you findings, explain the differences that you
observe.
Table 1.3 Values Comparison Table
To be Compared Calculation VISION % Error
Deviation Check
(Accuracy)
Z Cable 1 1.025 + j 0.67 Ω 1.025 + j 0.67 Ω 100
Z Cable 2 0.82 + j 0.536 Ω 0.82 + j 0.536 Ω 100
Z Cable 3 1.23 + j 0.804 Ω 1.23 + j 0.804 Ω 100
Voltage at Node B 28.116 – 10.6 28.207  − 10.705 99.677385
Voltage at Node C 27.75  − 10.8 27.847  − 10.972 99.651668
Voltage at Node D 27.48  − 11.04 27.579  − 11.210 99.6410312
I at Cable 1 0.3170  − 23.52 0.318 kA 99.6855345
P Sending 15.04716 15.200 MW 98.9944736
P Receiving 14.92569 14.889 MW 99.7541822
Q Sending 3.45179 3.124 MVAR 90.537676
Q Receiving 3.30348 3.152 MVAR 95.415325
Conclusion:
In theoretical way or manually calculating have assumptions that perhaps it is using an ideal
materials or components with no losses or less losses in constant temperature, constant
pressure in the environment etc. For example, if we are using voltmeter to measure the
voltage, ideally it has finite resistance but practically it has some resistance, so do Ammeter
ideally has zero resistance but in fact it has some resistance. Thus, it can causes a fluctuate
values between simulation and calculation. In power distribution, it is possible to have some
resistance, inductance or interference in cable, buses or load which can cause the current,
voltage or power unstable. But, in this case, values between calculation and simulation are
comparable with pretty much high of the error deviation check (accuracy) which could be
seen from the table.
Also, from the table it shows that voltages, current and sending Power found by VISION are
slightly higher than from calculation, but it is opposite direction if we look at the receiving
Power, sending and receiving Reactive Power (Q), the values from VISION are slightly
lower than calculation values. A possible conclusion could be the resistance or capacitance

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values of each transformer and each cable which are not taken into account when doing
calculation. This resulted in the error in reactive power on the sent end of the cable.
Cable Capacitance C (uF)
1 0.95
The Reactive power has more different values between calculation and simulation. This also
can be seen from error deviation check or the accuracy which has the lowest percentage than
the other elements. To diminish the error, the reactive power can be compensated by
following procedure:
First we have to calculate the capacitive reactance:
=
1
2
= 3352.33 Ω
The consumed reactive power can be written as follows:
=
| |
= 268,470
The total reactive power through the cable would then be inductive reactive subtracted with
capacitive reactive element.
= −
= 3,451,790 − 268,470 = 3,183,320
To be Compared Q sending Values
Old Updated
VISION 3.124 MVAR 3.124 MVAR
MatLab Calculation 3.45179 3,183,320
% Error Deviation Check
(Accuracy)
90.537676 98.1365367
It can be concluded that, previously the capacitive reactance was being neglected and this
cause the error or discrepancies which can make the values between VISION and MatLab
slightly different.
2. Power flow with loads of the constant P and Q type
Part One:
Now use the constant power load representation for all loads.
Perform two power flows. One with disconnected wind turbines and the other with
connected wind turbines. Print out the results. Again the voltage on bus B must be close to
the 30 kV and the other voltages and currents within acceptable limits.

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Figure 2.1 Power Flow Values with Wind Turbines Disconnected
Figure 2.2 Power Flow Values with Wind Turbines Connected
a) Compare the results of the case with the disconnected wind turbines and constant P and Q
loads with the constant Z case from the first instruction. What are the differences?

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Figure 2.2.a Result of case with constant impedance (left) and constant power (right)
Constant impedance load is simply a load that presents unchanging impedance, like a
resistor. A constant power load varies it's impedance on change of input voltage to keep
the power constant. It can be concluded that constant power needs to produce more
demanding current due to instability of impedance or load. = ∗
b) What is the change in currents flowing through the cables with and without wind turbines
and constant P and Q in both cases?
Table 2.1 Change of Values with and without wind turbines
Current Components Without Wind
Turbines
With Wind
Turbines
Different (Wo-W)
Current Source 142 A 114 A 28 A
Current out from Swing
Bus
142 A 114 A 28 A
Current out from Trafo 1 515 A 416 A 99 A
Current into Trafo 2 258 A 208 A 50 A
Current into Trafo 3 257 A 208 A 49 A
Current at Cable 1 384 A 169 A 215 A
Without turbines the power flowing to the loads is solely reliant on the original source
feeding this power to all loads; this results in higher cable currents as shown in the previous
table. The wind turbines can be seen as a separate generator being fed into the system which
overall reduces the power requirements on the cables because the wind turbines also provide
power to the load.

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c) What change in losses in the cables do you see? (Losses are found in Results).
Fig 2.3 Results with connected wind turbines Fig 2.4 Results with disconnected wind
turbines
Table 2.2 Change of Losses Table
Components Losses Without Wind
Turbines
With Wind
Turbines
Different (Wo – W)
Real Power in Cable 1 452 kW 86.5 kW 365.5 kW
Real Power in Cable 2 180 kW 61.3 kW 118.7 kW
Real Power in Cable 3 71.4 kW 7.2 kW 64.2 kW
Conclusion:
In advance we know that = ∗ . From the table and simulation results above, the
real power and current through each cable in the case without wind turbines is much
higher than with wind turbines. When current increases the power loss is also increase.
Why current is increasing without wind turbines? First, the power which is coming out is
kept to be constant regardless of changing current. Wind turbines behaves as generator, it
is also delivering current to the system, if it is connected, current will comes through
cable 1, cable 2, cable 3 and also from wind turbines, but if it is disconnected, current is
only coming from cable 1, cable 2 and cable 3. That’s why to keep power out constant,
cables will need to carry more current. But if wind turbines are connected, cables just
deliver less current because wind turbines are also delivering current to the system. Back
to the formula above, more current equals more losses in the materials.
Injected load powers per-unit
Given in the following table are the per-unit values of the dejecting loads.

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Load (S) Power injected (S) PU Rectangular (S) PU Polar
C 6+2j 0.06+0.02j 0.06 18.4˚
D 24+5j 0.24+0.05j 0.24 11.77˚
Injected power from wind turbines per-unit
The injected power coming from the wind turbines is: 17.666 MW - 2.793 MVAR. The
resulting per-unit power is found to be: 0.1767-0.0279 * j PU.
Load D and the wind turbines have an influence on the power on node D. As a result the net
power in per-unit must be found as follows:
( ) = ( ) + ( )
( ) = −0.24 − 0.05 + 0.1767 − 0.0279 ∗
( ) = −0.0633 − 0.0779
( ) = 0.1004 50.90˚
Part two:
This can be prepared at home and at school!
Here we will perform the power flow with the wind turbines connected to the 30 kV system
by hand. The power flow theory is treated in chapter 8 of [1]. We need a swingbus. We take
the feeding bus B of the 30 kV system as the swingbus (source). So there are three busses for
the handmade calculations, the swingbus B with a given voltage, and the load busses C and
D. Take the voltage of bus B, VB from the outcome of the power flow calculation in
VISION in per units. It must be nearly 1 p.u. with the angle found. The delivered power by
the wind turbines, found in the first part VISION simulation, will be used as injected
power for the handmade calculations. At first you have to choose a Base Power. Take Sb=100
MVA again. Transform the power of the embedded generator (wind turbines) and the loads to
per units. Compose the bus admittance matrix (3x3) of the system using the node-voltage
method from network theory. VB (in p.u.) from the power flow will be the voltage of the
swingbus. Starting values of VC=1.00 p.u. and VD=1.00 p.u. Note that the embedded
generator consumes reactive power!
= 100
= 30
=
√3 ∗
= 1924.5
= =
30
1924.5
= 15.6

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Per unit values of the impedances:
Cable 1
Zc1 for 1km = 0.205 + j 0.134
Zc1 = Length * ( 0.205 + j 0.134 ) = 5km * ( 0.205 + j 0.134 ) = 1.025 + j 0.67 Ω
Zc1pu = =
. . Ω
Ω
= 0.114 + j 0.074 pu
Cable 2
Zc2 for 1km = 0.205 + j 0.134
Zc2 = Length * ( 0.205 + j 0.134 ) = 4 km * ( 0.205 + j 0.134 ) = 0.82 + j 0.536 Ω
Zc2pu = =
. . Ω
Ω
= 0.091 + j 0.059 pu
Cable 3
Zc3 for 1km = 0.205 + j 0.134
Zc3 = Length * ( 0.205 + j 0.134 ) = 6 km * ( 0.205 + j 0.134 ) = 1.23 + j 0.804 Ω
Zc3pu = =
. . Ω
Ω
= 0.137 + j 0.089 pu
Admittance (Y)
1 = _1 = =
. .
= 6.172 − 4.01
2 = _2 =
1
2
=
1
0.091 + 0.059
= 7.737 − 5.016
3 = _3 =
1
3
=
1
0.137 + 0.089
= 5.133 − 3.335
Imaginary part of Transformer’s impedances:
Transformer 1
Xt1 =
.
∗ 1. =

∗ 14% = 0.117 pu
Transformer 2
Xt2 =
.
∗ 2. =

∗ 11% = 0.22 pu
Transformer 3
Xt3 =
.
∗ 3. =

∗ 11% = 0.22 pu
Transformer 4
Xt4 =
.
∗ 4. =

∗ 6% = 0.3 pu
Real part of transformer’s Impedances:
Transformer 1
1 =
,
∗
,
=
300
120
∗
100
120
= 0.00208

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2 =
,
∗
,
=
200
50
∗
100
50
= 0.008

3 =
,
∗
,
=
200
50
∗
100
50
= 0.008
NB : Pk = Short circuit Power (Korsleting Power)
Real and Imaginary Impedances of Transformers:
1 = 0.00208 + 0.117
2 = 0.008 + 0.22
3 = 0.008 + 0.22
=
( 1 + 2) − 2 − 1
− 2 ( 2 + 3) − 3
− 1 − 3 ( 1 + 3)

Ybb = ( 1 + 2)
Ycc = ( 2 + 3)
Ydd = ( 1 + 3)
=
∗
∗.
−
1
.
a) Now perform a necessary number of iterations by hand to calculate VC and VD. You
have to use a high accuracy in order to get good results because further on we use voltage
differences of voltages of nearly the same value. If your calculator is not powerful enough
you better use Matlab or Sci-lab. Transform the answers back to volts. Include the
calculations (including the result of every step of the iteration process).
Per unit calculation of the loads :
= 0.931  − 9.851 pu = 0.91727 − j ∗ 0.15928
= 0.919  − 9.721 pu = 0.90833 − j ∗ 0.15817
= 0.922  − 9.878 pu = 0.90580 − j ∗ 0.15517
= =
+
100
=
6 + 2
100
= 0.06 + 0.02
= 0.06
= 0.02
= =
+
100
=
24 + 5
100
= 0.24 + 0.05

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= 0.24
= 0.05
= =
+
100
=
58 + 12
100
= 0.58 + 0.12
= 0.24
= 0.05
% Iterations of Voltages calculation Vb, Vc, and Vd %
clear
Vbase = 30000;
Pc = 0.06; % PU real power in Load C %
Qc = 0.02; % PU reactive power in Load C %
Pd = 0.24; % PU real power in Load D %
Qd = 0.05; % PU reactive power in Load D %
% Voltages at Nodes in PU %
Vbpu = 0.91727-0.15928 * j ; % Voltage NOde B pu%
Vcpu = 1; % Voltage NOde C pu%
Vdpu = 1; %Vd PU Voltage%
%Cable Admittances
Yc1 = 6.172 - 4.01 * j ; % Y cable_1 pu %
Yc2 = 7.737 - 5.016 * j ; % Y cable_2 pu %
Yc3 = 5.133 - 3.335 * j ; % Y cable_3 pu %
Ycc = Yc2 + Yc3; %total admittances in Node C%
Ydd = Yc1 + Yc3; %total admittances in Node C%
N = 1;
% Node D Iterations %
for N = 1:1000
Vdpu = 1/Ydd * (((Pd - Qd * 1j)/(0.90580+0.15517 * j))-(Yc1 * Vbpu)-(Yc3 *
Vcpu));
disp(Vdpu);
N = N + 1 ;
End
Vd = abs(Vdpu)* Vbase % Actual voltage at node D %
= 2.8538 + 004 = 28.538
= 27.648
= 2.8196 + 004 = 28.196
= 27.565
b) Calculate the current in cable 1, in per units. Transform the answers back to amperes.
=
√3 ∗
= 1924.5

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=
( ) ( )
( )
∗ = 172.5624 from MatLab
1 = 169
sbase= 100000000;
Vbase = 30000;
ibase = sbase/(sqrt(3)*Vbase);
Vbpu = 0.91727-0.15928 * j ; % Voltage NOde B pu%
Vcpu = 0.90883-0.15817 * j; % Voltage NOde C pu%
Vdpu = 0.90580-0.15517 * j; %Vd PU Voltage%
zc1=0.114 + 0.074 * j ; %Impedances at Cable 1 PU %
Icable_1 = ( Vbpu - Vdpu ) / zc1 ;
I1 = Icable_1* ibase;
I=abs(I1)
c) Determine the sending and receiving end complex powers only for cable 1. Transform the
answers back to MVA's.
sbase= 100000000;
Vbase = 30000;
Vb = 0.91727-0.15928 * j ; % Voltage NOde B pu%
Vc = 0.90883-0.15817 * j; % Voltage NOde C pu%
Vd = 0.90580-0.15517 * j; %Vd PU Voltage%
zc1=0.114 + 0.074 * j ; %Impedances at Cable 1 PU %
Icable_1 = ( Vbpu - Vdpu ) / zc1 ;
I1 = Icable_1* ibase;
I=abs(I1)
Iconj = conj(Icable_1); %Conjugate form of I cable 1%
ssend = Vb * Iconj; %reactive power sent at cable 1 PU %
ssend2 = ssend*sbase % transferring back to MVA %
ss=abs(ssend) %absoulte value of S send %
pfsend=real(ssend)/ss %Power factor of sending power %
srec = Vd * Iconj; %reactive power sent at cable 1 PU %
ssrec2 = srec*sbase % transferring back to MVAR %
sr=abs(srec) %absoulte value of S send %
pfrec=real(srec)/sr %Power factor of sending power %
2 = 6.1187 + 5.6762 ∗ (Sending complex power in cable 1)
= 0.7331 (Sending power factor )
2 = 6.0271 + 5.6167 ∗ (Receiving complex power in cable 1)
= 0.7316 (Receiving power factor )

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Fig 2.4 Simulation Result from VISION
Part three:
a) Compare the results of the handmade calculation with the voltages, current and sending
and receiving end powers of cable 1, found in VISION.
To be Compared Calculation VISION % Error
Deviation Check
(Accuracy)
Z Cable 1 1.025 + j 0.67 Ω 1.025 + j 0.67 Ω 100
Z Cable 2 0.82 + j 0.536 Ω 0.82 + j 0.536 Ω 100
Z Cable 3 1.23 + j 0.804 Ω 1.23 + j 0.804 Ω 100
Voltage at Node C 28.538 27.648 96.8813511
Voltage at Node D 28.196 27.565 97.7620939
I at Cable 1 172.5624 169 A 98.2558139
P Sending 6.1187 5.903 MW 92.02443943
P Receiving 6.0271 5.817 MW 93.3849063
Q Sending 5.6762 5.445 MVAR 95.9268524
Q Receiving 5.6167 5.618 MVAR 99.97686
S Sending 8.9647 MVA 8.031 MVA 89.5847044
S Receiving 8.238522519 MVA 8.087 MVA 98.1608047
b) Draw conclusions from the comparison, and explain sources of differences in your report.
It can be concluded that the results from both handmade calculation and VISION
Simulation are little bit different. The causes of differences could be by the software that
takes some elements or parameters from the components into account. This could be also
by some little errors in hand calculation.
From the table, it is showing that voltage node C, voltage node D, cable 1 current, real
power (P), apparent power (S) and the reactive power (Q) values from calculation are
slightly higher than values found by VISION.

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Cable Capacitance C (uF)
1 0.95
2 0.76
3 1.14
Possible conclusion could be the capacitance in each cable which is not included in the
calculation.
It also possible to conclude another conclusion that based on table the value of current
through cable 1 from calculation is higher than from VISION. It shown 172.5624 A is from
calculation and value from VISION is found to be 169 A. This error discrepancy is around
+3.5624 A causes in a change of the apparent power (S) through the line about +933,700 VA.
This roughly shown in the comparison table.
Solution to compensate this could be done in the following way:
= −
= 8,964,700 VA − 933,700 VA
= 8,031,000
Base on value done by calculation, the power factor sent is: 0.7331
= ∗
The updated real power is found to be: 5,887,526.1 W
The updated reactive power can be determined:
= = 42.85°
= ∗ = 5,461,460.17
To be Compared Apparent Power Sending Values
Old Updated
VISION 8.031 MVA 8.031 MVA
MatLab Calculation 8.9647 MVA 8,031,000
(Accuracy)
89.5847044 99.987345
To be Compared Real Power Sending Values
Old Updated
VISION 5.903 MW 5.903 MW
MatLab Calculation 6.1187 5,887,526.1 W
(Accuracy)
92.02443943 99.7378638

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To be Compared Reactive Power Sending Values
Old Updated
VISION 5.445 MVAR 5.445 MVAR
MatLab Calculation 5.6762 5,461,460.17
(Accuracy)
95.9268524 99.6986122
It can be concluded that the current error margin gives the effect to the updated power
sending. After the compensation, it increases the deviance check of the accuracy.
3. Short circuit calculations
Part one:
We use the same network to study a three-phase fault on the load bus C. We will use the
Fault Analysis mode of VISION. Here at first a power flow is run in order to get the pre-
fault voltages. The transformers stay controlled. You will get about the same bus voltages as
before. They are given now on a phase to neutral base. Print them out only once for the
case with connected wind turbines.
Use Views in Tools to make also visible the branch currents on the scheme. Describe the case
on the scheme in a Frame. Don't forget your name.
Figure 3.1 Pre-fault Voltages

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a) Select bus C for the calculation of a symmetrical fault on this bus. Print out the results.
Figure 3.2 calculation of a symmetrical fault in Bus C with wind turbine
b) Disconnect the wind turbines and perform the same simulation. Print out the pre-fault
voltages of this case.
Figure 3.3 pre-fault voltages without wind turbines

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Figure 3.4 calculation of a symmetrical fault in Bus C without wind turbines
c) Print the results of the short circuit calculation and give your opinion about the difference
in fault level with and without the wind turbines.

With Wind Turbines Without Wind Turbines
549.9 MVA,Ik = 10.58 kA 495.7 MVA, Ik = 9.54 kA
443.0 MVA, Ik = 8.53 kA 400.3 MVA, Ik = 7.70 kA
444.1 MVA, Ik = 8.55 kA 388.7 MVA, Ik = 7.48 kA
In a power system, the maximum fault current (or fault MVA ) that can flow into a zero
impedance fault is necessary to be known for example, for switch gear solution. This can

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either be the balanced three phase value or the value at an asymmetrical condition. The
Fault Level defines the value for the symmetrical condition. The fault level is usually
expressed in MVA (or corresponding per-unit value), with the maximum fault current
value being converted using the nominal voltage rating.
Part two:
This can be prepared at home and at school!
Here we perform hand made calculations of the fault current on load bus C. For
simplification we do not take into account the contribution of the wind turbines.
The calculation of the three phase symmetrical fault is done in the single line equivalent. The
loads are considered passive so they have no contribution to the fault current! We have to
take into account all the network impedances, including the ones of the transformers and the
source in the fault current path.
a) Draw the appropriate circuit for the 3phase fault on bus C. Indicate the impedances in per
unit and calculate the fault current. Transform the per unit value back to amperes. Include
all calculations in the report.

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Figure 3.5 Impedances in appropriate circuit for the 3phase fault on bus C.
sbase= 100000000;
Vbase = 30000;
%Cable Admittances
Yc1 = 6.172 - 4.01 * j ; % Y cable_1 pu %
Yc2 = 7.737 - 5.016 * j ; % Y cable_2 pu %
Yc3 = 5.133 - 3.335 * j ; % Y cable_3 pu %
%Impedances
zc1=1/Yc1; %cable 1 impedances
%Trafo Impedances
zt1 = 0.00208 + j*0.117 ;
zt2 = 0.008 + j*0.22 ;
zt3 = 0.008 + j*0.22 ;
%Parallel/Series Impedances
zt23= (zt2*zt3)/(zt2+zt3);
zc123=((zc1+zc3)*zc2/(zc1+zc2+zc3));
%Node C Fault Level Calculation
ztot=zt1 + zt23 + zc123; %total impedance
ikpu=1/ztot; %korsleting I per unit
ikc=ikpu * ibase; %convert back to Amps
ikcreal=abs(ikc); %%absolute Ik
ikcreal2=ikcreal * 1.1 %maximum system voltage based on IEC 60909
fl=ikpu * sbase %Fault Level in MVA
flabs=abs(fl)

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Bus C Short Circuit Current and Fault Level Comparison
To be Compared VISION MatLab
7.70 kA 7.5614 kA
400.3 MVA 357.18 MVA
Figure 3.6 VISION Simulation Results
Part three:

With Wind Turbines Without Wind Turbines
549.9 MVA,Ik = 10.58 kA 495.7 MVA, Ik = 9.54 kA
443.0 MVA, Ik = 8.53 kA 400.3 MVA, Ik = 7.70 kA
444.1 MVA, Ik = 8.55 kA 388.7 MVA, Ik = 7.48 kA
Sources of Short Circuit Fault Currents
It is very important to consider all sources of fault current and to know the impedance
characteristics of the fault current sources when calculating the magnitudes of short
circuit fault currents. Following equipment feeds fault current into a short circuit:
 Generators

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 Synchronous Motors
 Induction Motors
 Electric Utility Systems
Generators
Generators are driven by turbines, water wheels, diesel engines or other types of prime
movers. When a short circuit occurs on the system powered by a generator, the generator
continues to produce voltage at the generator terminals as the field excitation is
maintained and the prime mover drives the generator at normal speed. The generated
voltage causes a large magnitude fault current flow from the generator to the short
circuit. The flow of fault current is limited only by the generator impedance and the
impedance of circuit between the generator and short circuit. In case of a short circuit at
the generator terminals, the fault current is limited by generator impedance only. That’s
way short circuit current in the case with connected wind turbines is bigger than the case
with disconnected wind turbines.
Bus C Short Circuit Current and Fault Level Comparison
To be Compared VISION MatLab % Error Deviation Check
(Accuracy)
7.70 kA 7.5614 kA 98.2
400.3 MVA 357.18 MVA 89.2280789
Observation and Conclusion:
Table above shows that Korsleting current value on bus C found by VISION are slightly
higher than values done by MatLab calculation. But the fault level value found by VISION
get more different if we compare the different in the Ik between calculation and simulation.
Possible reason why it is happen could be in matlab calculation. In the last code above, an
error has occurred when compensating the voltage factor of the maximum system voltage.
Note that, based on IEC 60909, voltage factor which accounts for the maximum system
voltage (1.05 for voltages <1kV, 1.1 for voltages >1kV).
fl=ikpu * sbase %Fault Level in MVA
flabs=abs(fl)
The old matlab code above written that Fault Level, fl=ikpu * sbase. Since that ‘ikpu’
has not been compensated yet with the voltage factor of the maximum system voltage, it
might cause the error for the final value of Fault Level MVA. To minimalism the error, the
MatLab code will be changed to the code below:

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fl=(ikcreal2/ibase) * sbase %Fault Level in MVA
flabs=abs(fl)
The new formula for Fault level is now written like fl=(ikcreal2/ibase) * sbase. Now,
we use ‘ikcreal2’ in the new Fault Level formula, since ‘ikcreal2’ has been compensated with
the voltage factor of the maximum system voltage. From that, the error is decreased as shown
in the table below.
Table 3.3 Table new Fault Level MVA
To be Compared Fault Level MVA Values
Old Updated
VISION 400.3 MVA 400.3 MVA
MatLab Calculation 357.18 MVA 392.90 MVA
(Accuracy)
89.2280789 98.1513864
This improvement minimalism the discrepancy between old and updated values of Fault
Level MVA.
To be Compared VISION MatLab % Error Deviation Check
(Accuracy)
7.70 kA 7.5614 kA 98.2
For the short circuit current, the possible reason why it has few different could be that the
capacitance of the cable which not included while doing calculation or another possible
evidence is refer to the transformer per unit impedances which could be slightly not precisely.

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