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

[email protected]

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:

푍 = 푉

푃 − 푗 ∗ 푄

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:

퐼 = 푃 − 푗 ∗ 푄푉푡푟푎푛푝표푠푒

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 푆 = 푃 + 푗 ∗ 푄

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.

푉푝푉푠 =

푁푝푁푠

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.

푆푏푎푠푒 = 100푀푉퐴 푉푏푎푠푒 = 30푘푉

퐼푏푎푠푒 = 푆푏푎푠푒푉푏푎푠푒 =

100푀푉퐴30푘푉 = 3.33푘퐴

푍푏푎푠푒 = 푉푏푎푠푒퐼푏푎푠푒 =

30푘푉3.33푘퐴 = 9훺

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


Ω = 0.137 + j 0.089 pu

Imaginary part of Transformer’s impedances: Transformer 1 Xt1 = . ∗ 푍푡1.표푙푑 =

∗ 14% = 푗0.117pu

Transformer 2 Xt2 = . ∗ 푍푡2.표푙푑 =

∗ 11% = 푗0.22pu


∗ 11% = 푗0.22pu


∗ 6% = 푗0.3pu

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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푳풐풂풅푩

푍푙표푎푑퐵 = 30

58 − 푗12 = 14.88 + 푗3.07훺

푍푙표푎푑퐵푝푢 = 14.88 + 푗3.07훺

9훺 = 1.65+ 푗0.341푝푢 푳풐풂풅푪

푍푙표푎푑퐶 = 30

6 − 푗2 = 135 + 푗45훺

푍푙표푎푑퐶푝푢 = 135 + 푗45훺

9훺 = 15+ 푗5푝푢

푳풐풂풅푫

푍푙표푎푑퐷 = 30

24 − 푗5 = 35.94 + 푗7.48훺

푍푙표푎푑퐷푝푢 = 35.94 + 푗7.48훺

9훺 = 3.99+ 푗0.831푝푢

c.

Figure 1.1 Load Model, network scheme including the impedances of the loads

Real part of Transformer’s impedances:

Transformer 1

푅푡1푛푒푤푝푢 = 푃푘푆,표푙푑 ∗

푆푛푒푤푆,표푙푑 =

300푘푊120푀푉퐴 ∗

100푀푉퐴120푀푉퐴 = 0.00208푝푢

푻풓풂풏풔풇풐풓풎풆풓ퟐ

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200푘푊50푀푉퐴 ∗

100푀푉퐴50푀푉퐴 = 0.008푝푢

푻풓풂풏풔풇풐풓풎풆풓ퟑ




100푀푉퐴50푀푉퐴 = 0.008푝푢

NB : Pk = Short circuit Power (Korsleting Power) Real and Imaginary Impedances of Transformers: 푍푡1푝푢 = 0.00208 + 푗0.117푝푢

푍푡2푝푢 = 0.008+ 푗0.22푝푢

푍푡3푝푢 = 0.008+ 푗0.22푝푢

풁풏풐풓풕풐풏 = 풁풕ퟏ풑풖 + (풁풕ퟐ풑풖//풁풕ퟑ풑풖)

= 0.00208+ 푗0.117푝푢+ ((0.008+ 푗0.22푝푢)//(0.008+ 푗0.22푝푢))

= 0.11789푝푢+ ((0.2288푝푢)//(0.2288푝푢))

= 0.00208+ 푗0.117+ 0.004+ 푗0.11

= 0.00608+ 푗0.227푝푢

= 0.2270888.5푝푢

퐼퐵푝푢 = 푈푝푢

푍푛표푟푡표푛 = 30

300.22708 < 88.5 = 0.118– 푗4.402푝푢 = 4.404 < −88.5푝푢

Admittance (Y) 푌푐1 = 푌푐푎푏푙푒_1 = =

. .= 6.172 − 푗4.01푝푢

푌푐2 = 푌푐푎푏푙푒_2 = 1푍푐2 =

10.091 + 푗0.059 = 7.737 − 푗5.016푝푢


10.137 + 푗0.089 = 5.133 − 푗3.335푝푢

푌푒푞 = 푌푛표푟푡표푛 =1

푍푛표푟푡표푛 = 1

0.00608+ 푗0.227푝푢 = 0.118– 푗4.402푝푢

푌퐿푏 = 푌푙표푎푑_퐵 = 1

푍푙표푎푑_퐵푝푢 =1

1.65+ 푗0.341 = 0.581 − 푗0.12푝푢

푌퐿푐 = 푌푙표푎푑_퐶 = 1

푍푙표푎푑_퐶푝푢 =1

15 + 푗5 = 0.06 − 푗0.02푝푢

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푌퐿푑 = 푌푙표푎푑_퐷 = 1

푍푙표푎푑_퐷푝푢 = 1

3.99 + 푗0.831 = 0.24 − 푗0.05푝푢

퐼푏 = 푉푏(푌1+ 푌2 + 푌퐵) − 푉푐(푌2) − 푉푑(푌1) 퐼푐 = 푉푏(−푌2) + 푉푐(푌2 + 푌3+ 푌퐶) − 푉푑(푌3) 퐼푑 = 푉푏(−푌1) − 푉푐(푌3) + 푉푑(푌1+ 푌3 + 푌퐷)

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 푉 = 0.9210 − 0.1733푖0.9084 − 0.1750푖0.8989 − 0.1770푖

Transferring back to kV :

푉푏 = 0.9210 − 0.1733푖푝푢 = 0.9372− 10.6푝푢 푉푏 = 0.9372− 10.6푝푢 ∗ 30푘푉 = 28.116– 10.6푘푉

푉푐 = 0.9084 − 0.1750푖푝푢 = 0.925− 10.8푝푢

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푉푐 = 0.925− 10.8푝푢 ∗ 30푘푉 = 27.75− 10.8푘푉

푉푑 = 0.8989 − 0.1770푖푝푢 = 0.916 − 11.04푝푢 푉푑 = 0.916− 11.04푝푢 ∗ 30푘푉 = 27.48− 11.04푘푉

e. Also calculate the cable current in cable 1 in p.u. and transform it back to amperes.

푉푏 = 0.9210 − 0.1733푖푝푢 푉푑 = 0.8989 − 0.1770푖 푌1 = 6.172− 푗4.01푝푢 퐼푐푎푏푙푒_1 = (푉푏 − 푉푑) ∗ 푌1 퐼푐푎푏푙푒_1 = 0.1512 − 0.0658푖푝푢

퐼푐푎푏푙푒 = 0.1512 − 0.0658푖푝푢 ∗ 퐼 = (0.1512 − 0.0658푖) ∗ 3.33푘퐴 = 0.5035 − 푗0.2191푘퐴

=0.5491 − 23.52푘퐴

√3= 0.3170 − 23.52푘퐴

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.

푆푠푒푛푑 = 푉푏 × 푡푟푎푛푠푝표푠푒표푓퐼푐푎푏푙푒_1

= (28.116– 10.6푘푉) ∗ (0.549123.52푘퐴)

= 15.43812.92푀푉퐴

= 15.04716+ 푗3.45179푀푉퐴

푆푟푒푐푒푖푣푒 = 푉푑 × 푡푟푎푛푠푝표푠푒표푓퐼푐푎푏푙푒_1

= (27.48− 11.04푘푉) ∗ (0.549123.52푘퐴)

= 15.286912.48푀푉퐴

= 14.92569+ 푗3.30348푀푉퐴

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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– 10.6푘푉 28.207− 10.705푘푉 99.677385 Voltage at Node C 27.75− 10.8푘푉 27.847− 10.972푘푉 99.651668 Voltage at Node D 27.48− 11.04푘푉 27.579− 11.210푘푉 99.6410312

I at Cable 1 0.3170 − 23.52푘퐴 0.318 kA 99.6855345 P Sending 15.04716푀푊 15.200 MW 98.9944736

P Receiving 14.92569푀푊 14.889 MW 99.7541822 Q Sending 3.45179푀푉퐴푅 3.124 MVAR 90.537676

Q Receiving 3.30348푀푉퐴푅 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:

푋 = 1

2휋푓푐 = 3352.33Ω

The consumed reactive power can be written as follows:

푄 = |푉푟푚푠|

푋 = 268,470푉퐴푅

The total reactive power through the cable would then be inductive reactive subtracted with capacitive reactive element.

푄 = 푄 −푄

푄 = 3,451,790푉퐴푅 − 268,470푉퐴푅 = 3,183,320푉퐴푅

To be Compared Q sending Values Old Updated

VISION 3.124 MVAR 3.124 MVAR MatLab Calculation 3.45179푀푉퐴푅 3,183,320푉퐴푅

% 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. 푃 = 퐼 ∗ 푅

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 out from Trafo 2 941 A 760 A 181 A Current into Trafo 3 257 A 208 A 49 A Current out from Trafo 3 943 A 762 A 181 A Current at Cable 1 384 A 169 A 215 A Current at Cable 2 271 A 159 A 112 A Current at Cable 3 140 A 47 A 93 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 푃 = 퐼 ∗ 푅. 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:

푆 ( ) = 푆 ( ) + 푆 ( )

푆 ( ) = −0.24 − 0.05푗 + 0.1767− 0.0279 ∗ 푗

푆 ( ) = −0.0633 − 0.0779푖

푆 ( ) = 0.100450.90˚

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=1.00 p.u. and VD=1.00 p.u. Note that the embedded generator consumes reactive power!

푆푏푎푠푒 = 100푀푉퐴 푉푏푎푠푒 = 30푘푉

퐼 = 푆

√3 ∗ 푉= 1924.5퐴

푍푏푎푠푒 = 푉푏푎푠푒퐼푏푎푠푒 =

30푘푉1924.5 = 15.6훺

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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


Ω = 0.091 + j 0.059 pu


Ω = 0.137 + j 0.089 pu

Admittance (Y) 푌푐1 = 푌푐푎푏푙푒_1 = =

. .= 6.172 − 푗4.01푝푢


10.091 + 푗0.059 = 7.737 − 푗5.016푝푢


10.137 + 푗0.089 = 5.133 − 푗3.335푝푢

Imaginary part of Transformer’s impedances: Transformer 1 Xt1 = . ∗ 푍푡1.표푙푑 =

∗ 14% = 푗0.117pu


∗ 11% = 푗0.22pu


∗ 11% = 푗0.22pu


∗ 6% = 푗0.3pu

Real part of transformer’s Impedances: Transformer 1



300푘푊120푀푉퐴 ∗

100푀푉퐴120푀푉퐴 = 0.00208푝푢

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푻풓풂풏풔풇풐풓풎풆풓ퟐ




100푀푉퐴50푀푉퐴 = 0.008푝푢

푻풓풂풏풔풇풐풓풎풆풓ퟑ




100푀푉퐴50푀푉퐴 = 0.008푝푢

NB : Pk = Short circuit Power (Korsleting Power) Real and Imaginary Impedances of Transformers: 푍푡1푝푢 = 0.00208 + 푗0.117푝푢

푍푡2푝푢 = 0.008+ 푗0.22푝푢

푍푡3푝푢 = 0.008+ 푗0.22푝푢

푌 = (푌푐1 + 푌푐2) − 푌푐2 − 푌푐1−푌푐2(푌푐2 + 푌푐3) − 푌푐3−푌푐1 − 푌푐3(푌푐1 + 푌푐3)

Ybb = (푌푐1 + 푌푐2) Ycc = (푌푐2 + 푌푐3) Ydd =(푌푐1 + 푌푐3)

푉푘 =푆푘 ∗

푉푘 ∗.푌푘푘 −1푌푘푘 푌푘푛.푉푛

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 :

푉푏푝푢 = 0.931 − 9.851pu = 0.91727− j ∗ 0.15928 푉푐푝푢 = 0.919 − 9.721pu = 0.90833− j ∗ 0.15817 푉푑푝푢 = 0.922 − 9.878pu = 0.90580− j ∗ 0.15517

푆푐푝푢 =푆표푙푑푆푏푎푠푒 =

푃 + 푗푄100푀푉퐴 =

6푀푉퐴 + 푗2푀푉퐴100푀푉퐴 = 0.06 + 푗0.02

푃푐푝푢 = 0.06 푄푐푝푢 = 0.02

푆푑푝푢 =푆표푙푑푆푏푎푠푒 =


24푀푉퐴+ 푗5푀푉퐴100푀푉퐴 = 0.24 + 푗0.05

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푃푑푝푢 = 0.24 푄푑푝푢 = 0.05

푆푏푝푢 =푆표푙푑푆푏푎푠푒 =


58푀푉퐴+ 푗12푀푉퐴100푀푉퐴 = 0.58 + 푗0.12

푃푏푝푢 = 0.24 푄푏푝푢 = 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 % 푉푐 = 2.8538푒 + 004 = 28.538푘푉푓푟표푚푀푎푡푙푎푏 푉푐 = 27.648푘푉푓푟표푚푉퐼푆퐼푂푁

푉푑 = 2.8196푒 + 004 = 28.196푘푉푓푟표푚푀푎푡푙푎푏 푉푑 = 27.565푘푉푓푟표푚푉퐼푆퐼푂푁

b) Calculate the current in cable 1, in per units. Transform the answers back to amperes.

퐼 = 푆

√3 ∗ 푉= 1924.5퐴

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퐼 = ( ) ( )

( )∗ 퐼 = 172.5624퐴 from MatLab

퐼푐1 = 169퐴푓푟표푚푉퐼푆퐼푂푁

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; ibase = sbase/(sqrt(3)*Vbase); 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 % 푠푠푒푛푑2 = 6.1187푀푊 + 5.6762푀푉퐴푅 ∗ 푖 (Sending complex power in cable 1) 푝푓푠푒푛푑 = 0.7331 (Sending power factor ) 푠푠푟푒푐2 = 6.0271푀푊 + 5.6167푀푉퐴푅 ∗ 푖 (Receiving complex power in cable 1) 푝푓푟푒푐 = 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 28.538푘푉 27.648푘푉 96.8813511 Voltage at Node D 28.196푘푉 27.565푘푉 97.7620939

I at Cable 1 172.5624퐴 169 A 98.2558139 P Sending 6.1187푀푊 5.903 MW 92.02443943

P Receiving 6.0271푀푊 5.817 MW 93.3849063 Q Sending 5.6762푀푉퐴푅 5.445 MVAR 95.9268524

Q Receiving 5.6167푀푉퐴푅 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:

푆 = 푆 − 푆

푆 = 8,964,700VA− 933,700VA

푆 = 8,031,000푉퐴

Base on value done by calculation, the power factor sent is: 0.7331

푃 = 푆 ∗ 푝푓

The updated real power is found to be: 5,887,526.1 W

The updated reactive power can be determined:

휃 = 푐표푠 푝푓 = 42.85°

푄 = 푃 ∗ 푡푎푛휃 = 5,461,460.17푉퐴푅

To be Compared Apparent Power Sending Values Old Updated

VISION 8.031 MVA 8.031 MVA MatLab Calculation 8.9647 MVA 8,031,000푉퐴


89.5847044 99.987345

To be Compared Real Power Sending Values Old Updated

VISION 5.903 MW 5.903 MW MatLab Calculation 6.1187푀푊 5,887,526.1 W


92.02443943 99.7378638

Page 22 of 29

To be Compared Reactive Power Sending Values Old Updated

VISION 5.445 MVAR 5.445 MVAR MatLab Calculation 5.6762푀푉퐴푅 5,461,460.17푉퐴푅


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

Page 23 of 29

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

Page 24 of 29

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.

퐅퐚퐮퐥퐭퐋퐞퐯퐞퐥퐌퐕퐀퐚퐧퐝퐒퐡퐨퐫퐭퐂퐢퐫퐜퐮퐢퐭퐂퐮퐫퐫퐞퐧퐭퐂퐨퐦퐩퐚퐫퐢퐬퐨퐧 퐁퐮퐬 WithWindTurbines WithoutWindTurbines 퐁퐮퐬퐁 549.9MVA, Ik = 10.58kA 495.7MVA, Ik = 9.54kA 퐁퐮퐬퐂 443.0MVA, Ik = 8.53kA 400.3MVA, Ik = 7.70kA 퐁퐮퐬퐃 444.1MVA, Ik = 8.55kA 388.7MVA, Ik = 7.48kA

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

Page 25 of 29

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.

Page 26 of 29

Figure 3.5 Impedances in appropriate circuit for the 3phase fault on bus C.

sbase= 100000000; Vbase = 30000; ibase = sbase/(sqrt(3)*Vbase); %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 zc2=1/Yc2; %cable 2 impedances zc3=1/Yc3; %cable 3 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)

Page 27 of 29

Bus C Short Circuit Current and Fault Level Comparison To be Compared VISION MatLab

퐈퐤 7.70kA 7.5614kA 퐅퐚퐮퐥퐭퐋퐞퐯퐞퐥퐌퐕퐀 400.3MVA 357.18MVA

Figure 3.6 VISION Simulation Results

Part three:

퐅퐚퐮퐥퐭퐋퐞퐯퐞퐥퐌퐕퐀퐚퐧퐝퐒퐡퐨퐫퐭퐂퐢퐫퐜퐮퐢퐭퐂퐮퐫퐫퐞퐧퐭퐂퐨퐦퐩퐚퐫퐢퐬퐨퐧

퐁퐮퐬 WithWindTurbines WithoutWindTurbines 퐁퐮퐬퐁 549.9MVA, Ik = 10.58kA 495.7MVA, Ik = 9.54kA 퐁퐮퐬퐂 443.0MVA, Ik = 8.53kA 400.3MVA, Ik = 7.70kA 퐁퐮퐬퐃 444.1MVA, Ik = 8.55kA 388.7MVA, Ik = 7.48kA

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

Page 28 of 29

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) 퐈퐤 7.70kA 7.5614kA 98.2

퐅퐚퐮퐥퐭퐋퐞퐯퐞퐥퐌퐕퐀 400.3MVA 357.18MVA 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).

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)

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:

Page 29 of 29

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=(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 400.3MVA 400.3MVA MatLab Calculation 357.18MVA 392.90MVA


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)

퐈퐤 7.70kA 7.5614kA 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.

Date post:	19-Feb-2017
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