Siemens - Machine Protection Setting Exercises

Machine Protection -Setting Exercises

Exercise 1: Single line diagram

7UM62.

-T150 MVA, YNd11

110 ±5·2.5% / 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=1.2Ω

3000/1A, 20VA5P20, Rct=12Ω

300/1A, 20VA5P20, Rct=1.2Ω

side 2iL1,2,3

IEE2

UE

IEE1

Excit.

7XT71TD1 (REF)

TD2 (REF)

TD3 (<Excit.)

Ucontr.Umeas.

7XR6004

3PP1326

~7XT34 7XT3320Hz Gen.

3

kV 0.1

3

kV 0.1

3

kV 11

sensitive current input only!!

500V

3kV/ 11

Q7

Q8

3000/1A, 20VA5P20, Rct=12Ω

400/5A 5/2

SEF (20Hz)SEF (20Hz)

Exercise 1: Neutral transformer circuits

400/5A

1FS5

15VAP1

RL

P2

1B1

1A4

1A3

1A1

7XT34

1B4 4A1

4A3

7XT33

20 HzBandpass

20 HzGenerator

S1S2

7UM62.

R13 UE

R14

J8 IEE1

J7

max. 3A (20Hz)

Protection cubicle

.. mA

20Hz500V

3kV/ 11

1A2

330Ω

660Ω

660Ω

Burden < 0.5Ω

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 Protection

46 Unbalanced Load (negative sequence)

40 Under excitation

49 Thermal Overload (Stator)

24 Overexcitation (V/Hz)

21 Impedance Protection

78 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 ⋅=

5.04400/5

5/2

500V

311000V/SEF RFactor 0275

2

=⋅

=


)/ratio(UU

U

U

UFactor UE 0224

dividerNTsec

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)

Setting Options for the UE Input and their Impact

on 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 kV

protected 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 << 1/(ω·CK)

IC prim

IE prim


Formula symbols and definitions used:

UE0 Displacement voltage on HV side of unit transformer

Fe Earthing factor, here: solid earthed Fe = 0.8

ICprim Interference current on neutral transformer primary side

ICsec Interference current on neutral transformer secondary side

CK Total capacitance (3x phase capacitance) between HV and LV side ofunit transformer (coupling capacitance)

f Rated frequency

TRNT Transformation ratio of the neutral transformer

UNTPrim Primary rated voltage of neutral transformer

UNTSec Secondary rated voltage of neutral transformer

RL Load resistor

K Protected zone factor

FS Safety factor FS = 2

SNT(20s)Required output of neutral transformer when burdened by RL for 20 s

IRLmax Current of load resistor R at 100 % UE


[ ]

L

NTsecRLmax

L

2

NTsecNT(20s)

sec C

NTsec

s

L

NTprim Csec C

NTsec

NTprim

NT

KE0prim C

HVeE0

R

UI

VAR

U =S

I

U

F

K = R

TR I =I

U

UTR

Cfπ2 U= I

3/UFU

=

⋅

⋅

=

⋅⋅⋅⋅

⋅=

[ ]

A 40.65Ω 12.3

V 500I

kVA 20.3VAΩ 12.3

V500 =S

Ω 12.3A 2.03

V 500

2

0.1 = R

A 2.037.12A 0.16 =I

12.7V 500

3V/ 11000TR

A 0.16V

sA 101050s2π 50800V= I

kV 50.83kV/ 1100.8U

RLmax

22

NT(20s)

L

sec C

NT

91-

prim C

E0

==

=

=⋅

=⋅

==

=⋅⋅⋅⋅

=⋅=

−

TRR

UI

:terminalsgenerator at fault Earth

NTL

NTsecprim E

⋅= ok 10A A 3.2

12.7Ω 12.3

V 500I prim E ⇒<=

⋅=

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

-dP (-50%)

<I2 (<10%)

I>> (3·IN, p.u.)

&

CFC

MV2<, 8503, 8504

MV4<, 8507, 8508

0113

07961

07963

01808

External Trip 104526

8602: t = 0.15s

S Q

R Q0T

It can be assumed that the Generator will run out of step in case a three-phase

short circuit close to the power station will last for (example) more than 150ms.This situation can be described by the following AND logic.

If the fault is not cleared immediately the unit will be decoupled from the net after 150 ms.


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)


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

(from Manufacturer)


Settings in primary values:

1702: I2prim> = I2perm prim / IN Machine = 8.0 %

1704: Kprim = (I2/IN)2·t = 20 s

1705: tCooldown = Kprim /(I2perm prim /IN Machine)2 = 20s/0.082 = 3125 s

1706: 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 = 20s·0.664 = 13.30 s

1706: I2sec>> = I2prim>> · IN Machine/IN CT = 60.0% · 2446A/3000A ≈ 49 %

Exercise 2.5: Settings for under excitation protection (1/4)

Generator capability diagram


Settings in primary values (from Capability Diagram)

3002: 1/xd1prim = 0.58

tan(α1) = 0.7/0.2 = 3.5 , arctan(α1) = 1.292

3003: α1 ≈ 74°

3005: 1/xd2prim = 0.44

3006: α2 = 90°


UNMACH = 11kV , UN VTprim = 11 kV

INMACH = 46600kVA/(√3·11kV) = 2446A IN CTprim = 3000A

CTprim N

VTprim N

NMACH

NMACH

dMachdsec I

U

U

I

x

1

x

1⋅⋅=

0.815A 3000V11k

kV 11A 2446

I

U

U

I

CTprim N

VTprim N

NMACH

NMACH =⋅

⋅=⋅ 3002: 1/xd1sec = 0.58·0.815 ≈ 0.47

3005: 1/xd2sec = 0.44·0.815 ≈ 0.36


generatorshaft

7XR6004 7UM62.

K13

+

TD1

K14Excit.

K15 +TD2

K16

20 kΩ

7XT71

27

25

19

21

15

17

A

3

A1

1

A6

G3~

20 kΩ

20 kΩB11

B1

8

B1

420 kΩ

contro

l

meas.

Example:

LiYCY 4x1.5

Protection cubicle

3PP1326

500Ω1

49 kΩ

2

500Ω

3

7UM62.

K17

+

TD3

K18

10µF250V

1)

u

0 EXC

k

U0.5Exc U:3013 ⋅<

UEXC0 = 45Vku (voltage divider) = (0.5 kΩ+ 0.5 kΩ + 9 kΩ)/ 0.5 kΩ = 20

3013 U Exc = 0.5·45 V /20 ≈ 1.13 V


Settings: primary

values

Exercise 2.6: Settings for (stator-) thermal overload (1/5)

(from Manufacturer)



k-factor: without additional information's the voltage deviation can be taken into

account. From Generator electrical data: voltage deviation (-) = 7.5%

for nominal load and -7.5% voltage the current will increase to 1.075 p.u.

1602: k-Factor (prim.) = 1.07

From generator electrical data: ILoad = 1.3·In t(trip) = 60s at IPreload = 1·In

( )255s

0.2357

60s

1.2658ln

60s

1I1.07

I1.3

I1.07

I1

I1.07

I1.3

ln

60s

1Ik

I

Ik

I

Ik

I

ln

tτ

2

n

n

2

n

n

2

n

n

2

n

Load

2

n

Preload

2

n

Load

≈==

−

⋅

⋅

⋅

⋅−

⋅

⋅=

−

⋅

⋅−

⋅

=

1603: thermal time constant = 255 sec

thermal alarm stage: setting must be higher than 1/k2 = 1/1.072 = 0.873

1604: thermal alarm stage = 90 %

1610A: Current Overload Alarm Setpoint = 107%



1612A: kt-Factor when Motor Stops = 1.0 (xxxx)

The thermal Overload should not trip for example before Over current protection1615A: Maximum Current for Thermal Replica = 250%

1616A: Emergency Time = 100 sec (xxxx)

From Generator electrical data: winding temp. rise Stator = 61 K1605: Temperature Rise at Rated Sec. Curr. = 61°C


1602: k-Factor (sec.) = k-Factor (prim.) · IN Machine/IN CT = 1.07· 2446A/3000A = 0.87

1610A: Current Overload Alarm Setpoint (Sec.) = 1.07· IN Machine· IN CT sec/IN CT prim

= 2446A·1A/3000A = 0.87 A

1605: Temp. Rise (Sec.) = Temp. Rise (Prim.) · (IN CT/IN Machine)2

= 61°C · (3000A/2446A)2 = 92°C


1 2 3 41

10

100

1000

10000

Trip. characteristic for τ = 255s, k =1.07 and preload = 1·In

Trip. characteristic for τ = 255s, k =1.07 and preload = 0

Trip. characteristic from generator electrical data

I/In

t [s]

Exercise 2.7: Settings for generator over excitation U/f (1/2)

(from Manufacturer)

Exercise 2.7: Settings for generator over excitation U/f (2/2)

Exercise 2.8: Settings for impedance protection (1/4)

2*)

G3~

Z< , 7UM6

X1 ≈ 0.70·X1 Transformer

T1,T1B

T2

X1 ≥≥≥≥ X1 Transformer

TEND

Z1(R1,X1) Z2

(R2,X2)Z1b(R1b,X1b)

1*)

t

Z

1*) To be coordinated with net protection

2*) Setting can (must) be higher than the real Z up to the HV C.B.

BI. (activating Z1b)


Im (Z)

Re (Z)R1= Z1

X1=Z1

X2=Z2

X1b=Z1b

R2=Z2R1b= Z1b

sec 3.00T :here

protectionnet with dcoordinate be toT :3312

sec 0.50T2 :here

protectionnet with dcoordinate be toT2 :3311

Ω 0.0 Z:here

protectionnet with dcoordinate be to Z

ZS

U

100%

u Z2:3310

T1T1b :3309

2k:here

kS

U

100%

u Z1b:3308

sec 0.10T1 :3307

70%) k:(heremer transfor

intoreach zoneProtectionk

S

U

100%

u

100%

k Z1:3306

END

END

Net

Net

Net

Trf N

2

Trf.LV NT(prim)

x

x

Trf N

2

Trf.LV NT(prim)

r

r

Trf N

2

Trf.LV NTr(prim)

=

→

=

→

=

→

+⋅=

=

=

⋅⋅=

=

=

=

⋅⋅=


sec 3.00T :3312

sec 0.50 T2 :3311

Ω 0.1940ΩMVA 50

kV 11

100%

8% Z2:3310

0.10secT1b :3309

Ω 0.3872MVA 50

kV 11

100%

8% Z1b:3308

sec 0.10T1 :3307

Ω 0.136MVA 50

kV 11

100%

8%

100%

70% Z1:3306

END

2

(prim)

2

(prim)

2

(prim)

=

=

=+⋅=

=

=⋅⋅=

=

=⋅⋅=



Ω 5.29Ω 0.194V11000V/100

3000A/1AZ2

k

k Z2:3310

Ω 10.55Ω 0.387V11000V/100

3000A/1AZ1b

k

k Z1b:3308

Ω 3.71Ω 0.136V11000V/100

3000A/1AZ1

k

k Z1:3306

(prim)

VT

CT(sec)

(prim)

VT

CT(sec)

(prim)

VT

CT(sec)

=⋅=⋅=

=⋅=⋅=

=⋅=⋅=

Exercise 2.9: Settings for out of step protection (1/3)

°=

+⋅=

−

=

==

→

=

=

⋅

⋅⋅≈

⋅⋅

=≈

90 P :3508

)Z(Z0.289 Z:3504

Z Z:3507

k

10Ω Z

/UU

/II

k

kk with :here

!!net with thedcoordinate be to Z

5%)8k:(here

ormerintoTransf 1 sticCharacteri ofreach k

S100%

Uu

%100

k Z:3506

xI3

UX Z:3505

c(prim)(prim)ba(prim)

c(prim)d(prim)

sec

d(prim)

VTsnVTpn

CTsnCTpn

VT

CTsec

d(prim)

r(OoS)

r(OoS)

TrN,

2

Tr(LV)N,Tr(OoS)

c(prim)

'

d

GenN,

GenN,'

db(prim)

ϕ

Za

Zb

Im (Z)

Re (Z)

Zd-Zc

Zc

Zd

φP

Char. 2

Char. 1


°=

=+⋅=

=−=

=====

=⋅

⋅⋅≈

=⋅⋅

≈

90 P :3508

Ω 0.210Ω) 0.165Ω (0.5610.289 Z:3504

Ω 0.202Ω 0.165Ω 0.367Z- Z:3507

Ω 0.36727.27

Ω 10 Z 27.27

V V/100 11000

A A/1 3000

k

kk

Ω 0.165MVA 50100%

kV118%

100%

85% Z:3506

Ω 0.5610.2162446A3

11000V Z:3505

a(prim)

c(prim)d(prim)

d(prim)

VT

CTsec

22

c(prim)

b(prim)

ϕ



( )Ω 5.73Ω 0.21027.27Zk Z:3504

Ω 5.51Ω 202.027.27Z- ZkZ- Z:3507

Ω 10 Z

Ω 4.50Ω 0.16527.27Zk Z:3506

Ω 15.30Ω 0.56127.27Zk Z:3505

above) (from 27.27k

a(prim)seca(sec)

c(prim)d(prim)secc(sec)d(sec)

d(sec)

c(prim)secc(sec)

b(prim)secb(sec)

sec

=⋅=⋅=

=⋅=⋅=

=

=⋅=⋅=

=⋅=⋅=

=

Siemens - Machine Protection Setting Exercises

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Transcript of Siemens - Machine Protection Setting Exercises