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### Transcript of Siemens - Machine Protection Setting Exercises

• Machine Protection -Setting Exercises

• Exercise 1: Single line diagram

7UM62.

-T150 MVA, YNd11

110 52.5% / 11 kVuT(1) = 8 %

3 110 kV, 50 Hz

side 1iL1,2,3

uL1,2,3

G3~

-G146.6 MVA

11kV 7.5%50Hz

300/1A, 20VA5P20, Rct=1.2

3000/1A, 20VA5P20, Rct=12

300/1A, 20VA5P20, Rct=1.2

side 2iL1,2,3

IEE2

UE

IEE1

Excit.7XT71

TD1 (REF)TD2 (REF)TD3 (

• Exercise 1: Neutral transformer circuits

400/5A1FS515VA

P1

RL

P2

1B1

1A4

1A3

1A1

7XT34

1B4 4A1

4A3

7XT33

20 HzBandpass

20 HzGenerator

S1S2

7UM62.

R13 UER14

J8 IEE1J7

max. 3A (20Hz)

Protection cubicle

.. mA

20Hz500V 3kV/ 11

1A2

330

660

660

Burden < 0.5

• Exercise 1: Required protection elements

-- Threshold supervision (for Decoupling)64G-1 90% Stator Earth Fault U0> (calculated)64G-2 100% Stator Earth Fault (20Hz principle)64R Rotor Earth Fault (1-3Hz principle)87 Differential Protection46 Unbalanced Load (negative sequence)40 Under excitation49 Thermal Overload (Stator)24 Overexcitation (V/Hz)21 Impedance Protection78 1) Out of Step (loss of synchronism)

1) Option

• Exercise 1: Device configuration (partly)

for Decoupling

for Decoupling

• Exercise 1: Power System Data 1 (1/4)

1)

1) Neutral transformer is high resistive

(CT) ratio)(U ratioTransf.) Neutral (ratioSEF RFactor 0275 divider2 =

5.04400/55/2

500V311000V/SEF RFactor 0275

2

=

=

• Exercise 1: Power System Data 1 (2/4)

• Exercise 1: Power System Data 1 (3/4)

)/ratio(UUU

UU

Factor UE 0224dividerNTsec

NTprim

Esec

VTprim==

31.7V/(5/2) 500

3V/ 11000Factor UE 0224 ==

Refer to Setting Options for the UE Input and their Impacton the Protection Functions (refer to slide No. 10)

• Exercise 1: Power System Data 1 (4/4)

• Setting Options for the UE Input and their Impacton the Protection Functions

• Exercise 2: Generator Electrical Data (1/2)

• Exercise 2: Generator Electrical Data (2/2)

• Exercise 2.1: Calculation of load resistor and neutral transf. (1/3)

CK = 10 nFUHV = 110 kVULV = 11 kVprotected zone (stator) = 90% K = (100%-90%)/100% = 0.1

500V 3kV/ 11

90% (K=0.1)

CG

1L1

1L2

1L3

CL CTr

CK

RL

UHV

7UM62.

UE

ULV

equivalent circuit

~

CK

UE0RL primCE

CE = CG + CL + CTr neglectedRL prim

• Exercise 2.1: Calculation of load resistor and neutral transf. (2/3)

Formula symbols and definitions used:UE0 Displacement voltage on HV side of unit transformerFe Earthing factor, here: solid earthed Fe = 0.8ICprim Interference current on neutral transformer primary sideICsec Interference current on neutral transformer secondary sideCK Total capacitance (3x phase capacitance) between HV and LV side of

unit transformer (coupling capacitance)f Rated frequencyTRNT Transformation ratio of the neutral transformerUNTPrim Primary rated voltage of neutral transformerUNTSec Secondary rated voltage of neutral transformerRL Load resistorK Protected zone factorFS Safety factor FS = 2SNT(20s)Required output of neutral transformer when burdened by RL for 20 sIRLmax Current of load resistor R at 100 % UE

• Exercise 2.1: Calculation of load resistor and neutral transf. (3/3)

[ ]

L

NTsecRLmax

L

2NTsec

NT(20s)

sec C

NTsec

s

L

NTprim Csec C

NTsec

NTprimNT

KE0prim C

HVeE0

RUI

VAR

U =S

IU

FK

= R

TR I =IUU

TR

Cfpi2 U= I3/UFU

=

=

=

[ ]A 40.65

12.3V 500I

kVA 20.3VA 12.3V500

=S

12.3A 2.03V 500

20.1

= R

A 2.037.12A 0.16 =I

12.7V 500

3V/ 11000TR

A 0.16V

sA 101050s2pi 50800V= I

kV 50.83kV/ 1100.8U

RLmax

22

NT(20s)

L

sec C

NT

91-prim C

E0

==

=

=

=

==

=

==

TRRUI

:terminalsgenerator at fault Earth

NTL

NTsecprim E

= ok 10A A 3.212.7 12.3V 500I prim E

• Exercise 2.2: Decoupling - Example with Threshold supervision (1/3)

-dP (-50%)

(3IN, p.u.)

&

CFCMV2

• Exercise 2.2: Decoupling - Example with Threshold supervision (2/3)

• Exercise 2.2: Decoupling - Example with Threshold supervision (3/3)

Settings: primary values

• Exercise 2.3: Settings for 90% Stator Earthfault - U0>calculated (1/2)

UL1

UL3

UL2

90%

L1

L2

L3

Settings (primary value):5002: U0prim> = 100% - 90% = 10.0 %5003: T = 0.30 sec

Fuse Failure Monitor (FFM) to be enabledto block 90% SEF Element via CFC

• Exercise 2.3: Settings for 90% Stator Earthfault - U0>calculated (2/2)

Settings: primary values

• Exercise 2.4: Settings for unbalanced load (1/3)

(from Manufacturer)

• Exercise 2.4: Settings for unbalanced load (2/3)

Settings in primary values:1702: I2prim> = I2perm prim / IN Machine = 8.0 %1704: Kprim = (I2/IN)2t = 20 s1705: tCooldown = Kprim /(I2perm prim /IN Machine)2 = 20s/0.082 = 3125 s1706: I2prim>> = 60.0%1707: T I2>> = 3.00 sec

Conversion to secondary values:1702: I2sec> = I2prim> IN Machine/IN CT = 8.0% 2446A/3000A = 6.5 %1704: Ksec = Kprim (IN Machine/IN CT)2 = 20s(2446A/3000A)2 = 20s0.664 = 13.30 s1706: I2sec>> = I2prim>> IN Machine/IN CT = 60.0% 2446A/3000A 49 %

• Exercise 2.4: Settings for unbalanced load (3/3)

Settings: primary values

• Exercise 2.5: Settings for under excitation protection (1/4)Generator capability diagram

• Exercise 2.5: Settings for under excitation protection (2/4)

Settings in primary values (from Capability Diagram) 3002: 1/xd1prim = 0.58tan(1) = 0.7/0.2 = 3.5 , arctan(1) = 1.2923003: 1 743005: 1/xd2prim = 0.443006: 2 = 90Conversion to secondary values:

UNMACH = 11kV , UN VTprim = 11 kVINMACH = 46600kVA/(311kV) = 2446A IN CTprim = 3000A

CTprim N

VTprim N

NMACH

NMACH

dMachdsec IU

UI

x

1x

1=

0.815A 3000V11k

kV 11A 2446IU

UI

CTprim N

VTprim N

NMACH

NMACH=

= 3002: 1/xd1sec = 0.580.815 0.473005: 1/xd2sec = 0.440.815 0.36

• Exercise 2.5: Settings for under excitation protection (3/4)

generatorshaft

7XR6004 7UM62.

K13 +

TD1

K14Excit.

K15 +TD2

K16

20 k

7XT71

27

25

1921

1517

A3

A11

A6

G3~

20 k

20 kB11

B18

B1420 k

control

meas.

Example:LiYCY 4x1.5

Protection cubicle

3PP1326

5001

4 9 k

2

500

3

7UM62.

K17 +

TD3

K18

10F250V

1)

u

0 EXC

kU0.5Exc U:3013