ECE529: Lecture 37 - uidaho.eduece.uidaho.edu/ee/power/ECE529/Lectures/L37/L37.pdf · Session 37 Page 1/31 Spring 2019 ECE529: Lecture 37 MW 1000kW MVA MW MVAR MW pu 1 Problem 1:

ECE 529: Utility Apps of Power Electronics

Session 37 Page 1/31 Spring 2019

ECE529: Lecture 37 MW 1000kW MVA MW MVAR MW pu 1

Problem 1: For the system below we want to increase the power transfer from Bus 1 to Bus 2 by adding a SVCat the midpoint.

A. Calculate the power transfer from bus 1 to bus 2 if the SVC is disconnected.

B. Suppose the SVC is connected through a 18:1 transformer with a leakage reactance of .005 on low voltageside, determine the MVAR requirements for the SVC to do the following:

(1) Increase P12=650MW with the TCR is not gated.(2) How much capacitance and inductance is needed to increase P12 to 633MW with the TCR operating with

= 75 deg.) Assume delta connection of the TCR and capacitors.(3) Modify your solution for a 12 pulse SVC (one TCR in and the other in Y)

C. Implement your solution for the 12 pulse SVC in ATP.

BUS 1

|V1|= 345kVLL

at 0o

BUS 2BUS M

j30 ohm j30 ohm

|V2|= 345kVLL

at -17.605o|Vm|= ??

at (1 - 2)/2

SVC



A. Calculate the power transfer from bus 1 to bus 2 if the SVC is disconnected.Sending and Receiving End Voltages

V1L 345kV V1pV1L

3 V1p 199.186 kV

V2L 345kV V2pV2L

3 V2p 199.186 kV

Sending and Receiving End Angles

θ1 0deg θ2 17.605deg

Line Reactance Xline 60ohm

P12V1L V2L sin θ1 θ2

Xline P12 599.991 MW

B. Suppose the SVC is connected through a 18:1 transformer with a leakage reactance of .005 on low voltage side,determine the MVAR requirements for the SVC to do the following: (1) Increase P12=650MW with the TCR is not gated.

XhalfXline

2 Xhalf 30 ohm

Lossless line, so the following relationships holds :

P12max 650MW

P1m P12max Pm2 P12max



θmθ2 θ1

2 θm 8.803 deg

Assuming that the angles at the ends of the line to not change with change in P12, start from:

P1mV1L VmL sin θ1 θm

Xhalf=

3 V1p Vmp sin θ1 θm

Xhalf=

Then

VmpP1m Xhalf

3 V1p sin θ1 θm Vmp 213.246 kV

VmL 3 Vmp VmL 369.353 kV

OverVoltVmL

V1L OverVolt 107.059 %

Qm1VmL

2

Xhalf

VmL V1L cos θm θ1

Xhalf

Qm1 349.861 MVAR

Qm2VmL

2

Xhalf

VmL V2L cos θm θ2

Xhalf

Qm2 349.861 MVAR

Qcompensator_max Qm1 Qm2

Qcompensator_max 699.721 MVAR This is the net capacitive injection to increase P12 to 650MW.



On the secondary of the transformer (18:1)

VsecmaxVmL

18 Vsecmax 20.52 kV line to line

Xxfmr_low .005ohm Xxfmr_low

2π 60 Hz0.0133 mH

Since the TCR is not gated, this leaves capacitor only. Assume it is connected in Delta, so mult by 3The capacative will also need to supply some reactive power to transformer leakage

Qcomp3 VmL

2

Xeff=

3 VmL2

Xxfmr_low Xc=

Solving for effective compensating reactance:

Xeff 3Vsecmax

2

Qcompensator_max Xeff 1.805 Ω

Xc1 Xeff Xxfmr_low Xc1 1.81 Ω

Ccomp1

2 π 60 Hz Xc1 Ccomp 1465.3203 μF

If we split the capacitance into two banks (one associated with each 6 pulse bridge). Capacitorsin parallel add, so we divide by two.

Ccomp

2732.66 μF



Now calculate how much inductance is needed for P12=633MW

P12 633MW

P1m P12 Pm2 P12

θmθ2 θ1

2

Assuming that the angles at the ends of the line to not change with change in P12

VmpP1m Xhalf

3 V1p sin θ1 θm

Vmp 207.669 kV

VmL 3 Vmp VmL 359.693 kV

OverVoltVmL

V1L OverVolt 104.259 %

Qm1VmL

2

Xhalf

VmL V1L cos θm θ1

Xhalf

Qm1 224.889 MVAR

Qm2VmL

2

Xhalf

VmL V2L cos θm θ2

Xhalf

Qm2 224.889 MVAR

Qcompensator Qm1 Qm2

Qcompensator 449.778 MVAR This is the net capacitive contribution.



On the secondary of the transformer (18:1)

VsecVmL

18 Vsec 19.983 kV

Xeff633 3Vsec

2

Qcompensator Xeff633 2.663 Ω Multiply by 3 for delta

Subtract off transformer, leaving parallel LC

XSVC Xeff633 Xxfmr_low XSVC 2.668 Ω

Now solve for Xtcr from: -jX_SVC= jXtcr * (-jXc) / (jXtcr - jXc)

Xtcr_σ 1000ohm

Given

j XSVCjXtcr_σ j Xc1

j Xtcr_σ j Xc1 0=

Xtcr_σ Find Xtcr_σ Xtcr_σ 5.629 Ω

This is X_tcr for =75 deg, now find actual value of XL and then L

σ633 75deg

XL Xtcr_σ

σ633 sin σ633

π

XL 0.615 Ω LtcrXL

2 π 60 Hz Ltcr 1.63 mH



Since there will be two TCR's in parallel, we need to double this value if they areboth delta connected

L12pulse 2 Ltcr L12pulse 3.261 mH

For the TCR that is Y connected we need to perform a to Y conversion (divide by 3)

LYL12pulse

3 LY 1.087 mH

Simulation parameters:

Since the transformer is connected delta - wye - wye the turns ratio is 31.177:1 (the sqrt 3 needs to be added)

3 18 31.177

The transformer reactances are added to the transformer secondaries. Since there are two transformers, the reactanceare doubled so the effective reactance meets the problem specifications:

Xxfmr_low

2π 60 Hz0.0133 mH 2

Xxfmr_low

2π 60 Hz 0.0265 mH

C. Implement your solution for the 12 pulse SVC in ATP. (see ATPDraw file for better view).



345kV @ 0

V1

30 ohm

BUS1

345kV@ -17.6

V2

30 ohmV2

VM

HV

I

V

18:1 xfmrHV

TCR1

TCR2

SA

T YY

1.087mH

I

V

P2B

N2B

V

732.6uF

V TCR1AT

TCR1B

T

TCR1CT

0.001ohm

0.001ohm Snubber

P2C

N2C

0.001ohm0.001ohm

1.087mH

I

P2A

I

N2A I

0.001ohm

0.001ohm

I

I

Snubber

1.087mHI

Snubber

3.261mH

TCRD1I

3.261mHTCRD1

I

N1ABN1BC

P1BC

0.001ohm

0.001ohm

P1AB

0.001ohm

TCRD1

0.001ohm

3.261mH

I

P1CA

N1CA

0.001ohm

0.001ohm

Snubber Snubber

Snubber

V TCR2AT

TCR2B

T

TCR2CT

Ground Ref

Ground Ref

Ground Sw

I

V1

T

T

T

T

T

T

732.6uF

345 kV System

TACS Curret Measurements

Small R fordamping

Small R fordamping

Power Circuit



Notes about the power circuit part of the ATPDraw file:1. Note that the wye connected TCR is grounded, and that it is connected to a wye grounded transformer.

a. This allows triplen harmonic currents to flow. b. The connection from this TCR to the power system must be delta to keep the triplen harmonics off the power system (this will be a

problem if the TCR's a not loaded equally (but this is a problem in general with TCR's).2. The main power transformer is a three winding transformer connected with the HV winding in delta, and the secondary and tertaries ea

in wye. 3. The capacitors in each 6 pulse SVC has a small resistance added in series to provide damping (it tends to ring with the TCR and line

inductances. The added resistance is 0.01 ohm per phase. Note that the size of the resistance makes a big difference in the performancesince the capacitive reactance is only 1.8ohm. When = 90 degrees, P12 should be 650MW. If this added resistance is 0, the systemachieves this. If R = 0.1 ohm, P12 = 660MW, of R = 0.01 ohm, P12 = 651.5MW. This has a bigger effect when the TCR is firing tooas will be noted below. In practice this is resolved with a closed loop controller, rather than the open loop control shown here.

4. I have added a wye grounded resistance of 10kOhm on each TCR circuit. This primary matters on the delta connected TCR, as itprovides a balanced ground reference point (this is more to satisfy ATP numerical behavior as anything else.

Controls

S21AB

PLUS1

TCR2A

S22AB

DELTAT0.5 T

T

G=21600

RMP2APLUS1

S24AB+

-S23AB

T

F2AB

PLUS1PLUS1ZERO

ALPHA RMP2A

TD2AB

PWIDTH

N2A

HALFCY

P2A

90 deg

ALPHA

90 deg

T

52.5 deg

1. Synchronization and firing circuit (shown above) is similar to that used in homework 22. The delta connected TCR gets synchronization reference from the line to line voltage

across each phase leg of the TCR3. The wye connected TCR gets synchronization reference from the line to ground voltage on

each phase.4. Initially operate with alpha = 90 degress (sigma=0), and then move to calculated alpha of

52.5 degrees at t = 100 ms.



Simulation ResultsThe simulations start out with = 90 deg (this corresponds to = 0 deg so the TCR's don't conduct) and then there is a step changeto = 52.5 deg (corresponds to = 75 deg). This covers the two operating conditions used for performing the calcualtions.

( f i l e p r o b 1 _ 1 2 p V 0 . p l 4 ; x - v a r t ) f a c t o r s :o f f s e t s :

10

t : A L P H A 1- 9 0

0 . 0 0 0 . 0 5 0 . 1 0 0 . 1 5 0 . 2 0 0 . 2 5[ s ]0

2 0

4 0

6 0

8 0

1 0 0Firing angle

90 deg

52.5deg

( f i l e p r o b 1 _ 1 2 p V 0 . p l 4 ; x - v a r t ) c : T C R D 2 B - X X 0 1 6 8 c : T C R D 1 B - X X 0 1 7 8 0 . 0 0 0 . 0 5 0 . 1 0 0 . 1 5 0 . 2 0 0 . 2 5[ s ]

- 9 0 0 0

- 6 0 0 0

- 3 0 0 0

0

3 0 0 0

6 0 0 0

9 0 0 0

[ A ]Single phase TCRcurrents (one phaseon each TCR)

The current starts outat zero while=90deg



(f ile p rob1_12pV0.p l4 ; x -v a r t ) c :TC R D 2B-XX0168 c :TC R D 1B-XX0178 0 .20 0.21 0.22 0.23 0.24 0.25[s ]

-9000

-6000

-3000

0

3000

6000

9000

[A ]

Single phase TCRcurrents (one phaseon each TCR) after=52.5 deg

Note:The phase shift between1.the currents due to thewye and delta connectedTCRsThe magnitude difference2.(this difference isSQRT(3)) again due tothe difference betweenthe wye and deltaconnections.

Line current between the two 6 pulse SVC's and the transformer winding (includes the capacitor current) for each SVCFirst both currents over the entire simualtion period. 1.During the initial part of the simulation the TCR is not gated and the capacitor currents dominate.2.Note that the currents are identical when the TCR's are not conducting3.

( f i l e p r o b 1 _ 1 2 p V 0 . p l 4 ; x - v a r t ) c : T C R D 1 A - T C R 1 A c : T C R D 2 A - T C R 2 A 0 . 0 0 0 . 0 5 0 . 1 0 0 . 1 5 0 . 2 0 0 . 2 5[ s ]

- 1 5

- 1 0

- 5

0

5

1 0

1 5

2 0

[ k A ]



(f ile prob1_12pV0.p l4; x -v ar t ) c :TC R D 1A-TC R 1A 0 .20 0.21 0.22 0.23 0.24 0.25[s ]

-15

-10

-5

0

5

10

15

[kA ]

Line current Drawn by delta connected TCR + capacitors Line current Drawn by wye connected TCR + capacitors

(f ile p rob1_12pV 0.p l4 ; x -v ar t ) c :TC R D 2A -TC R 2A 0 .20 0.21 0.22 0.23 0.24 0.25[s ]

-15

-10

-5

0

5

10

15

[kA ]

Harmonic currents from delta SVC Harmonic currents from SVC with wye connected TCRHarm. Amplitude Phase1 6558.7 -39.0563 0.99421 172.425 1079.8 165.827 44.576 86.7589 0.20633 29.40711 52.285 24.00813 7.383 -150.6615 0.061123 2.15117 11.034 64.92619 30.67 167.321 0.054103 -1.584123 0.77281 7.1004THD=16.505%

Harm. Amplitude Phase1 6594.9 -39.1083 2104.5 -116.495 1090.9 -14.3387 43.981 -93.179 264.3 10.51411 55.943 -135.3413 6.0736 -135.4815 85.812 137.7417 11.126 -114.8119 30.617 -13.33521 16.378 -95.67123 0.77066 21.461THD=36.21%

Note that the wye connected TCR has significant triplen harmonicThis is why the transformer winding connected to it is wye (to complete the circuit path1.The primary winding is delta to keep these off the power system2.



Note the magnitude and angles of the 5th and 7th harmonics especially

Phase relationship between the currents drawn by the TCR's plus capacitors (note the fundamental components are in phase)

( f i le p r o b 1 _ 1 2 p V 0 . p l4 ; x - v a r t ) c : T C R D 1 A - T C R 1 A c : T C R D 2 A - T C R 2 A 0 .2 0 0 .2 1 0 .2 2 0 .2 3 0 .2 4 0 .2 5[s ]

- 1 5

- 1 0

- 5

0

5

1 0

1 5

2 0

[k A ]

For comparison, here is the same waveforms without the capacitors in the circuit



(f ile p ro b 1 _ 1 2 p V 0 . p l4 ; x -v a r t ) c : TC R D 1 A -TC R 1 A c : TC R D 2 A -TC R 2 A 0 .20 0 .21 0 .22 0 .23 0 .24 0 .25[s ]

-8000

-6000

-4000

-2000

0

2000

4000

6000

8000[A ]

The two waveforms havenearly the ssame fundamentalcomponent magnitudes andangles.

( f i l e p r o b 1 _ 1 2 p V 0 . p l 4 ; x - v a r t ) c : V M A - H V A c : V M B - H V B c : V M C - H V C 0 . 0 0 0 . 0 5 0 . 1 0 0 . 1 5 0 . 2 0 0 . 2 5[ s ]

- 2 0 0 0

- 1 5 0 0

- 1 0 0 0

- 5 0 0

0

5 0 0

1 0 0 0

1 5 0 0

2 0 0 0[ A ]

Current on this line side of the transformer.Note the transient when the thyristorsstart firing. This shows much better damping if2.there is more resistance in the circuit(for example in the capacitors or theline inductances)This also improves if the firing angle3.changes in a ramp instead of a step

Harmonic content (after TCR's gated):Harm. Amplitude Phase



1 730.78 170.893 0.085471 51.0135 0.61514 134.357 0.023955 -140.349 0.012679 -168.8411 1.1243 85.66113 0.73386 68.37215 7.3747E-03 -174.1117 4.3742E-03 159.4519 4.8875E-03 -167.3821 5.3131E-03 -176.923 0.090596 162.51THD=0.20337%

(f ile prob1_12pV0.pl4 ; x -v ar t ) t : P3PH 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35[s ]

400

450

500

550

600

650

700

750

800*10 6

Note that power transfer without the TCR gated isslightly higher than calculated, and significantlyhigher than calculated after the TCR is gatedNotice the oscillation in the "steady-state" powertransfer after the TCR is gated.This will be muchlarger without the resistors.This error increases as the damping resistance inseries with the capacitor gets larger

651.52MW 634MW

Power transfer with resistance of 0.1 ohm in series with the capacitors:



(f ile prob1_12pV0.pl4; x -v ar t ) t : P3PH 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35[s ]

400

450

500

550

600

650

700

750

800*10 6

Notice improved damping, but increased error inpower transfer

660.3MW643MW

Since the calculations all neglected resistanec inthe circuit, they would need to be updated toinclude resistance. A closed loop control will alsoimprove things

Problem 2: Design 5th and 7th harmonic filters for a 132 kV system such that they supply 30MVAr capacitive (at 60Hz). Assume



the inductors have an X/R ratio of 20, the capacitors have no resistance. Specify L to the nearest 0.1mH and C to the nearest 0.1Farad. Connect capacitors in Delta.

If the filters are connected in parallel with a harmonic current source (load) and the impedance looking into the power system is Z= 1 + j12 (at 60Hz) what percentage of the harmonic current goes into the filter and what percentage goes to the powersystem?

Test your filters in ATP. Rather than modeling a full SVC, just implement harmonic current sources at the appropriate frequencies(implement one at a time).

VLL 132kV ω60 2 π 60 Hz VLL23

107.778 kV

Qcap 30MVAR

Rsystem 1ohm

Xsystem 12ohm LsystemXsystem2 π 60 Hz

Lsystem 31.831 mH

f5 5 60 Hz f5 3001s

f7 7 60 Hz f7 4201s

Fifth Harmonic FilterNeglect resistance since the X/R ratio is highUse Xc and C for Wye connected and then convert to delta later

QcapVLL( )2

XL XCwye= VLL( )2

ω60 L51

ω60 C5wye

= Derive an expression only in terms ofC5wye and known frequency terms



f51

2 π L5 C5wye=

L51

C5wye 2 π f5( )2= substitute into the Qcap equation

QcapVLL2

ω601

C5wye 2 π f5( )2

1ω60 C5wye

=

QcapVLL2

ω60 C5wye 2 π f5( )2

ω602 2 π f5( )2=

Express this in terms of the capacitance:

C5wyeQcap ω602 2 π f5( )2

VLL2ω60 2 π f5( )2

C5wye 4.384 μF

Convert to delta by dividing C5wye by 3 (this is the same as 3*Xc5wye = Xc5delta):

C5deltaC5wye

3 C5delta 1.461 μF

L51

C5wye 2 π f5( )2 L5 64.192 mH



Now express L and C within specified tolerances.

L5set 64.2mH C5delta_set 1.5μF

f5_o1

2 π L5set 3 C5delta_set f5_o 296.105

1s

effectively detuned to slightlybelow 300 Hz.

RL5ω60 L5set

20 RL5 1.21 Ω

Sfilter_5VLL2

RL5 j ω60 L5set1

ω60 3 C5delta_set

Sfilter_5 0.066 30.825i( ) MVA

Im Sfilter_5( ) 30.825 MVAR Close, would be exact without rounding.

Repeat for the 7th Harmonic Filter

QcapVLL2

ω60 C7wye 2 π f7( )2

ω602 2 π f7( )2=



Express this in terms of the capacitance:

C7wyeQcap ω602 2 π f7( )2

VLL2ω60 2 π f7( )2

C7wye 4.474 μF

Convert to delta by dividing C7wye by 3 (this is the same as 3*Xc7wye = Xc7delta):

C7deltaC7wye

3 C7delta 1.491 μF

L71

C7wye 2 π f7( )2 L7 32.096 mH

Now express L and C within specified tolerances.

L7set 32.1mH C7delta_set 1.5μF

f7_o1

2 π L7set 3 C7delta_set f7_o 418.756

1s

detuned to slightly below 420 Hz.

RL7ω60 L7set

20 RL7 0.605 Ω

Sfilter_7VLL2

RL7 j ω60 L7set1

ω60 3 C7delta_set



Sfilter_7 0.032 30.179i( ) MVA

Im Sfilter_7( ) 30.179 MVAR Close, would be exact without rounding.

Now do current divider relationships:

First, see what percentage of 5th harmonic current enters the 5th harmonic filter, the 7th harmonic filter, and the power system

Zsystem f( ) Rsystem j 2 π f Lsystem

Zfilter5 f( ) RL5 j 2 π f L5set1

2 π f 3 C5delta_set

Zfilter7 f( ) RL7 j 2 π f L7set1

2 π f 3 C7delta_set

Zsystem f5( ) 1 60i( ) Ω Ignoring skin effect on the resistance.....

Zfilter5 f5( ) 1.21 3.122i( ) Ω

Zfilter7 f5( ) 0.605 57.385i( ) Ω

1. If the 7th harmonic filter is neglected:

PercentFilter5th f5( )Zsystem f5( )

Zsystem f5( ) Zfilter5 f5( ) PercentFilter5th f5( ) 95.01 %

The 5th harmonic filter will take 95% of the 5th harmonic current, only 5% of the 5thharmonic current will enter the power system..

2. If the 7th harmonic filter is included:



Now need to consider the 7th harmonic filter in parallel with the system impedance (for the current divider with the 5thharmonic filter) and in parallel with the system impedance (for the current divider with the system).

ParaFilt7_and_System f( )1

Zsystem f( )1

Zfilter7 f( )

1

ParaFilt7_and_System f5( ) 29.335 Ω

ParaFilt7_and_Filt5 f( )1

Zfilter5 f( )1

Zfilter7 f( )

1

ParaFilt7_and_Filt5 f5( ) 3.163 Ω

ParaFilt5_and_System f( )1

Zsystem f( )1

Zfilter5 f( )

1


PercentFilter5th2 f5( )ParaFilt7_and_System f5( )

ParaFilt7_and_System f5( ) Zfilter5 f5( ) PercentFilter5th2 f5( ) 89.756 %

PercentSystem2 f5( )ParaFilt7_and_Filt5 f5( )

ParaFilt7_and_Filt5 f5( ) Zsystem f5( ) PercentSystem2 f5( ) 5.008 %





The system and the 7th harmonic filter each draw roughly equal amounts of 5th harmonic current.

Repeat for the 7th harmonic case:

Zsystem f7( ) 1 84i( ) Ω Ignoring skin effect on the resistance.....

Zfilter5 f7( ) 1.21 85.211i( ) Ω

Zfilter7 f7( ) 0.605 0.501i( ) Ω

1. If the 5th harmonic filter is neglected:

PercentFilter7th f7( )Zsystem f7( )

Zsystem f7( ) Zfilter7 f7( ) PercentFilter7th f7( ) 99.074 %

PercentSystem7th f7( )Zfilter7 f7( )

Zsystem f7( ) Zfilter7 f7( ) PercentSystem7th f7( ) 0.926 %

2. If the 5th harmonic current is included: Now need to consider the 5th harmonic filter in parallel with the system impedance (for the current divider with the 7thharmonic filter) and in parallel with the system impedance (for the current divider with the system).


ParaFilt7_and_Filt5 f7( )1

Zfilter5 f7( )1

Zfilter7 f7( )

1 ParaFilt7_and_Filt5 f7( ) 0.598 0.502i( ) Ω





PercentSystem2 f7( )ParaFilt7_and_Filt5 f7( )

ParaFilt7_and_Filt5 f7( ) Zsystem f7( ) PercentSystem2 f7( ) 0.921 %



As a check, the percentages should add up to 100%.

PercentFilter7th2 f7( ) PercentFilter5th2 f7( ) PercentSystem2 f7( ) 100 %

Note that the percentage to the 7th harmonic filter goes down slightly, since some of the current now enters the5th harmonic filter.

Test your filters in ATP. Rather than modeling a full SVC, just implement harmonic current sources at the appropriatefrequencies (implement one at a time).

ATP test system (t = 1 sec):



Harm AHarm BHarm C

IBUS

HARMI

5th or 7th

HARMA

T

FILT5

I

FILT7

I

SYSTI

1 ohm

31.831 mH

1.21 ohm

1.5uF1.5uF

HARMB

T

HARMC

T

0.605 ohm

64.2 mH 32.1 mH

126.456kV

V

TACS sources to create 10A_pk, 3 phase harmonic sources 1. Negative sequence for 5th 2. Positive sequence for 7th

Determine source voltage such that filter bus is at 1:0pu to verify reactive power:First calculate the current the filters should draw (based on the design parameters, not the desired results)

I1_filt5VLL

3 Zfilter5 60Hz( ) I1_filt5 134.823A arg I1_filt5 89.877 deg

I1_filt7VLL

3 Zfilter7 60Hz( ) I1_filt7 131.997A arg I1_filt7 89.94 deg

I60Hz I1_filt5 I1_filt7

Now calcuate at the source if the filters draw this current:

VsourceVLL

3I60Hz Zsystem 60Hz( ) Vsource 73.009 kV arg Vsource 0.213 deg



Vsource 2 103.251 kVAs a check: Fourier spectrum from simulation result:

Vsource_pk 103.251kV e j 89.797 deg 76.212kVVLL

3

1 pu Harm. Amplitude Phase (Deg)1 76212 0

Fifth Harmonic Current Source CaseThe 5th harmonic source is a balanced three phase (negative phase rotation) source with that has 10A peak.

Current through the 5th harmonic filter: 5th Harmonic Filter Current

(f ile Harm F ilt .pl4; x -v ar t) c :IBUSA -F ILT5A 0.00 0.02 0.04 0.06 0.08 0.10[s]

-250.0

-187.5

-125.0

-62.5

0.0

62.5

125.0

187.5

250.0[A ]

Harm. Amplitude Phase (deg)1 134.82 89.8513 0.16617 -176.675 7.0396 88.9857 0.080307 24.4459 0.043343 12.562THD=5.2595%

The fundamental current is very close to thecurrent calculated above:

I1_filt5 134.823A arg I1_filt5 89.877 deg

The percentage of the 7th harmonic current is:



7.0396A10A

2

99.555 %

Compared to calculated value of: PercentFilter5th2 f5( ) 89.756 %

Current through the 7th harmonic filter:7th Harmonic Filter Current (a little visibledistortion)

Harm. Amplitude Phase (deg)1 132.02 89.9413 0.096796 156.715 0.30847 -98.8097 0.060121 -101.119 0.017849 -78.287THD=0.94753%

(f ile HarmFilt.pl4; x-v ar t) c:IBUSA -FILT7A 0.00 0.02 0.04 0.06 0.08 0.10[s]

-200

-150

-100

-50

0

50

100

150

200[A]




0.30847A10A

2

4.362 %


Source CurrentSource current (a little visible distortion) Harm. Amplitude Phase (deg)



1 266.85 89.8793 0.25535 175.735 0.34449 -82.0937 0.065774 -20.8899 0.046507 -9.1146THD=0.33691%

(f ile HarmFilt.pl4; x-v ar t) c:SYSTA -IBUSA 0.00 0.02 0.04 0.06 0.08 0.10[s]

-400

-300

-200

-100

0

100

200

300

400[A]


I60Hz 266.82A arg I60Hz 89.908 deg


0.34449A10A

2

4.872 %

Compared to calculated value of: PercentSystem2 f5( ) 5.569 %

Notice that the percentages of the 5th harmonics currents don't compare very well to the theoretical results (and add to well over100%). Lets add them as phasors:

Subtract Isys due to measurementpolarityI5th 7.0396A ej 88.985 deg 0.30847A e j 98.809 deg 0.34449A e j 82.093 deg

I5th 7.075A arg I5th 89.756 deg simulation had I5_rms=7.042 at an angle of -90.23deg

2 I5th 10.005A So percentage of magnitudes might not be the best way....

Seventh Harmonic Current Source Case

Again, the 7th harmonic source has 10A peak, but now with a positive sequence rotation.



Current through the 7th harmonic filter:

Harm. Amplitude Phase (deg)1 131.99 89.9073 0.044533 -149.875 0.056157 -98.5987 7.0306 90.1649 0.020174 29.305THD=5.3699%

7th Harmonic Filter Current


-250

-160

-70

20

110

200

[A]




7.0306A10A

2

99.428 %


Current through the 5th harmonic filter:



5th Harmonic Filter Current (a little visibledistortion)

Harm. Amplitude Phase (deg)1 134.84 89.8443 0.049522 163.895 0.06254 33.3477 0.048741 16.4469 9.3974E-03 -68.074THD=0.42168%


-200

-150

-100

-50

0

50

100

150

200[A]




0.048741A10A

2

0.689 %


Source Current



Source current (a little visible distortion) Harm. Amplitude Phase (deg)1 266.81 89.9073 0.08622 179.615 0.048949 -26.1057 0.028382 -118.819 0.021815 14.557THD=0.15502%

(f ile HarmFilt.pl4; x-v ar t) c:SYSTA -IBUSA 0.00 0.02 0.04 0.06 0.08 0.10[s]

-400

-300

-200

-100

0

100

200

300

400[A]


I60Hz 266.82A arg I60Hz 89.908 deg


0.028382A10A

2

0.401 %

Compared to calculated value of: PercentSystem2 f7( ) 0.921 %

Documents

ECE529: Lecture 37 - uidaho.eduece.uidaho.edu/ee/power/ECE529/Lectures/L37/L37.pdf · Session 37 Page 1/31 Spring 2019 ECE529: Lecture 37 MW 1000kW MVA MW MVAR MW pu 1 Problem 1: