Machine Protection Setting Exercises

Machine Protection Setting Exercises

Exercise 1: Single line diagram 3∼ 110 kV, 50 Hz

7UM62 . side 1 iL1,2,3

300/1A, 20VA 5P20, Rct=1.2 -T1 50 MVA, YNd11 110 ±5·2.5% ±5·2.5% / 11 kV uT(1) = 8 %

Q7

300/1A, 20VA 5P20, Rct=1.2 11 kV 3

0.1 kV 3

sensitive current input only!! Q8

0.1 kV

uL1,2, 3

3

3000/1A, 20VA 5P20, Rct=12 -G1 46.6 MVA 11kV ±7.5% 50Hz

7XR6004

G 3~

7XT71

Excit.

11 kV/ 3 500V

7XT34 400/5A

7XT33 20Hz Gen.

~ 5/2

Ucontr . Umeas .

TD1

(REF)

TD2

(REF)

TD3

(
side 2 iL1,2,3

3PP1326

3000/1A, 20VA 5P20, Rct=12

IEE2

UE IEE1

SEF (20Hz) SEF

Exercise 1: Neutral transformer transformer circuits

Protection cubicle

max. 3A (20Hz)

20 Hz Bandpass

20 Hz Generator

7XT34

7XT33

1B1 1A4

1B4

4A1

660 660 

.. mA

11 kV/ 3

660 660 

RL

500V

P2 S2

400/5A 1FS5 15VA

1A1

1A2 P1

20Hz

330 330  1A3 4A3

S1

7UM62. Burden < 0.5

R13

UE

R14 J8 J7

IEE1

Exercise 1: Required protection elements --

Threshold supervision (for Decoupling)

64G 64 G-1

90% Stato ator Ea Earth Fa Fault U0 U0> (ca (callculat lated)

64G 64 G-2

100% Stat Stato or Ear Eartth Fau Faullt (20 (20H Hz pr princ inciple) le)

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 Exercis e 1: Device configuration (partly)

for Decoupling

for Decoupling

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

1)

0275 Factor R SEF = (ratio Neutral Transf.) 2 ⋅

 11000V/ 3   0275 Factor R SEF =   500V   

2

1) Neutral transformer is high resistive

⋅

5/2 400/5

ratio (U divider )

=

ratio (CT) 5.04



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

0224 Factor UE

=

0224 Factor UE

=

U VTprim U Esec

U NTprim

=

U NTsec /ratio(U divider )

11000 V/ 3 500 V/(5/2)

=

31.7


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)

90% (K=0.1)

ULV

CK

UHV

11 kV/ 3

1L1

500V 1L2 1L3

RL

CG

UE

CL

IE prim

CK = 10 nF UHV = 110 kV ULV = 11 kV

equivalent circuit CK

7UM62.

CE

RL prim

IC prim

~

CTr

UE0

CE = CG + CL + CTr  neglected RL prim << 1/(ω·CK)

protected zone (stator) = 90%  K = (100%-90%)/100% = 0.1

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 transformer

Fe

Earthi Earthing ng factor, factor, here: 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 of unit transformer (coupling capacitance)

f

Rated frequency

TRNT

Transformation ratio of the neutral transformer

UNTPrim Primary Primary rated voltage voltage of of neutral neutral transformer 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

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

U E0

=

Fe ⋅ U HV / 3

I C prim = U E0 ⋅ 2 ⋅ π ⋅ f ⋅ C K TR

NT

=

U NTsec K

0.8 ⋅ 110 kV/ 3

I C prim = 50800V

U NTprim TR

I C sec = I C prim ⋅ TR RL =

=

U E0

⋅

Fs

NT

U NTsec

NT

I C sec

11000 V/ 3

0.1 2

S NT(20s) = I RLmax

=

RL

[VA ]

U NTsec RL

S NT(20s) = I RLmax

=

⋅

500 V 2.03 A

2

U NTsec

⋅ 2 π ⋅ 50s

500 V = 0.16 A ⋅ 12 . 7

RL =

I C sec

=

500 2 V 2 12.3  500 V

=

12.3

=



=

50.8 kV -1

⋅ 10 ⋅ 10

−9

As

=

V

= 12.7

2.03 A

= 12.3 

[VA ] = 20.3

kVA

40.65 A

Earth fault at generator terminals : I E prim

=

U NTsec R L ⋅ TR NT

I E prim

=

500 V 12.3  ⋅12.7

=

3.2 A

< 10A

⇒ ok

0.16 A

Exercise 2.2: Decoupling Decoupling - Example with Threshold Threshold supervision supervision (1/3) (1/3) It can be assumed that the Generator will run out of step in case a three-phase short circuit circuit close to the power power station will will last for (example) (example) more than 150ms. 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. CFC MV2<, 8503, 8504

-dP -dP (-50 (-50%) %)

07961 T

07963

0113

I>> (3·IN, p.u.)

Q

R

Q

8602: t = 0.15s

&

MV4<, 8507, 8508

0

S

01808

04526 External Trip 1

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

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

Settings: primary values

Exercise Exercise 2.3 2.3:: Settings Settings for 90% Stator Stator Earthfaul Earthfaultt - U0>calcul U0>calculated ated (1/2) (1/2) L1 L2

UL3

L3 90%

Settings (primary value): 5002: U0prim> = 100% - 90% = 10.0 %

UL1

5003: T = 0.30 sec

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

UL2

Exercise Exercise 2. 2.3: 3: Setting Settings s for 90% 90% Stator Stator Earthfau Earthfault lt - U0>calcu U0>calculated lated (2/2)


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

(from Manufacturer)

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

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 8.0% % · 2446A/3 2446A/3000A 000A = 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% 60.0% · 2446A/3 2446A/3000 000A A ≈ 49 %

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


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

Exercise Exercis e 2.5: Settings for under excitation protection (2/4) 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°

Conversion to secondary values: 1

1

=

x dsec

x dMach

I NMACH

⋅

U NMACH

⋅

I NMACH U NMACH

U N VTprim I N CTprim

=

⋅

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

U N VTprim I N CTprim

2446 A ⋅ 11 kV 11k V ⋅ 3000 A

=

0.815

3002: 1/xd1sec = 0.58·0.815 ≈ 0.47 3005: 1/xd2sec = 0.44·0.815 ≈ 0.36

Exercise 2.5: Settings for under excitation protection (3/4) Protection cubicle 7XT71

7XR6004 20 k k

A 3

G 3~

A6

TD1

27 19 21

1)

Excit.

7UM62.

contro l

K13 + K14 TD2

A1 1 B11

generator shaft

B1 8

20 k k

20 k k 20 k k

15 17 B1 4

25 Example: LiYCY LiYCY 4x1.5 4x1.5

7UM62.

3PP1326 1

500 500 

TD3 2

500 500  4

3013 : U Exc < 0.5 ⋅

9 k k

3

U EXC 0 k u

UEXC0 = 45V ku (voltag (voltage e divide divider) r) = (0.5 (0.5 k+ 0.5 k + 9 k)/ 0.5 k = 20 3013 U Exc = 0.5· 0.5·45 45 V /20 ≈ 1.13 V

meas.

K15 + K16

10µF 250V

K17 + K18

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


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

(from Manufacturer)

Exercise 2.6: Settings for (stator-) thermal overload (2/5) Settings in primary values:

1602 16 02::

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. k-F k-Factor (pri prim.) = 1.07 1.07 From generator electrical data: ILoad = 1.3·In  t(trip) = 60s at IPreload = 1·In τ=

t 2

 I Load   I Preload    −   ⋅ ⋅ k I k I n  n   ln  2  I Load    − 1 ⋅ k I  n 

2

60s

=

2

 1.3 ⋅ I n   1⋅ I n    −   ⋅ ⋅ 1.07 I 1.07 I n  n   ln  2  1.3 ⋅ I n    − 1 ⋅ 1.07 I n  

2

=

60s ln(1.2658)

=

1603:

the herrmal time co const nstant = 255 255 sec

1604:

thermal alarm alarm stage: stage: setting setting must be higher than than 1/k2 = 1/1.072 = 0.873 thermal alarm stage = 90 %

1610A: 1610A: Curren Currentt Over Overloa load d Alar Alarm m Setpo Setpoint int = 107% 107%

60s 0.2357

≈

255s

Exercise 2.6: Settings for (stator-) thermal overload

(3/5)

Settings in primary values: 1612A: 1612A: kt-Facto kt-Factorr when Motor Motor Stops Stops = 1.0

(xxxx)

The thermal Overload should not trip for example before Over current protection 1615A: Maximum Current for Thermal Thermal Replica Replica = 250% 250% 1616A: 1616A: Emergenc Emergency y Time Time = 100 sec sec

(xxxx)

From Generator electrical data: winding temp. rise Stator = 61 K 1605: Temper 1605: Temperat atur ure e Ris Rise e at R Rate ated d Sec. Sec. Cur Curr. r. = 61°C

Conversion to secondary values: 1602 16 02::

k-Fa k-Fact ctor or (sec (sec.) .) = k-F k-Fac acto torr (pr (priim. m.)) · IN Machine /IN CT = 1.0 1.07· 7· 2446A/3 2446A/3000A 000A = 0.8 0.87 7

1610A 1610A:: Cu Curr rrent ent Over Overlo load ad Ala Alarm rm Setp Setpoi oint nt (Sec (Sec.) .) = 1.07· 1.07· IN Machine· IN CT sec /IN CT prim = 2446A·1A/3000A = 0.87 A 1605 16 05::

Temp Temp.. Rise Rise (Sec (Sec.) .) = Tem emp. p. Ris Rise (Prim Prim.) .) · (IN CT /IN Machine)2 = 61°C 61°C · (3000A/2446A) (3000A/2446A)2 = 92°C


(4/5)

10000 t [s]

Trip. characteristic for τ = 255s, k =1.07 and preload = 1·In Trip. characteristic for τ = 255s, k =1.07 and preload = 0

1000

Trip. characteristic from generator electrical data

100

10

1 1

2

3

I/In

4


(5/5)


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 impedance protection (1/4) t TEND 1*) T2

2*)

T1,T1B Z1 (R1,X1)

X1 ≈ 0.70·X1 Transformer X1 ≥ X1 Transformer

Z Z2 (R2,X2)

G 3~

Z< , 7UM6

BI. (activating (activati ng Z1b)

1*) To be coordinated with net protection 2*) Setting can (must) be higher than the real Z up to the HV C.B.

Z1b (R1b,X1b)

Exercise 2.8: Settings for impedance impedance protection (2/4)

3306 : Z1(prim) Im (Z)

k r

⋅

⋅

U 2N Trf.LV

100% 100%

S N Trf

Protectionzone reach into = 70%)

3307 : T1 = 0.10 sec

X2=Z2

R1= Z1

=

uT

transformer (here : k r

X1b=Z1b

X1=Z1

=

k r

R2=Z2 R1b= Z1b Re (Z)

=

3308 : Z1b(prim)

2

uT

⋅

U N Trf.LV

100%

here : k x

=

S N Trf

⋅ k x

2

3309 : T1b = T1 3310 : Z2(prim) Z Net

→

uT

=

2

⋅

U N Trf.LV

100%

S N Trf

+ Z Net

to be coordinated with net protection

here : Z Net

= 0.0 

3311 : T2 → to be coordinated with net protection here : T2 = 0.50 sec 3312 : TEND

→

to be coordinated with net protection

here : TEND

= 3.00 sec

Exercise 2.8: Settings for impedance protection (3/4) Settings in primary values: 70%

=

3306 : Z1(prim)

8%

⋅

⋅

11 kV 2

=

100% 100% 50 MVA 3307 : T1 = 0.10 sec =

3308 : Z1b (prim)

8%

⋅

11 kV 2

100% 50 MVA 3309 : T1b = 0.10sec 3310 : Z2 (prim)

=

8%

⋅

11 kV 2

100% 50 MVA 3311 : T2 = 0.50 sec 3312 : TEND

⋅2 =

0.136 

0.387 

+ 0 =

0.194 

= 3.00 sec

Conversion to secondary values: 3306 : Z1(sec)

k VT

3308 : Z1b (sec) 3310 : Z2

k CT

=

=

=

⋅ Z1(prim) =

k CT k VT

k CT

3000A/1A 11000V/100 V

⋅ Z1b (prim) =

⋅ Z2 (prim) =

⋅ 0.136  =

3000A/1A 11000V/100 V

3000A/1A

3.71 

⋅ 0.387  = 10.55 

⋅ 0.194  = 5.29 

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


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

Im (Z)

3505 : Zb(prim)

≈

Zd Char. 2

Zd-Zc Zc φP

Za

Re (Z)

3506 : Zc(prim)

≈

k r(OoS)

=

U N,Gen

=

3 ⋅ I N,Gen ⋅

100% 100% ⋅S N,Tr reach of Characteristic 1 intoTransf ormer = 85%)

→ to be coordinated

here : with k sec Zd(prim) Zb

'

⋅ xd

k r(OoS) u T ⋅ U 2N,Tr(LV)

(here : k r(OoS) Zd(prim)

Char. 1

X d'

=

=

k CT k VT

=

with the net !!

I CTpn /I CTsn U VTpn /U VTsn

10 k sec

3507 : Zd(prim) − Zc(prim) 3504 : Za(prim)

=

3508 : ϕ P = 90°

0.289 ⋅ (Zb (prim)

+ Zc(prim) )

Exercise 2.9: Settings for out of step protection (2/3)

Settings in primary values: 3505 : Z b(prim) 3506 : Z c(prim) k sec

=

≈

11000V 3 ⋅ 2446A

0.561 

8% ⋅ 11 2 kV 2

85%

⋅ = 0.165 100% 100% ⋅ 50 MVA k CT 3000 A/1 A = = 27.27 k VT 11000 V/100 V

≈

3507 : Z d(prim) - Z c(prim) 3504 : Z a(prim)

⋅ 0.216 =

=

0.367

 − 0.165  =

0.289 ⋅ (0.561  + 0.165

=

) =



Z d(prim)

0.202



0.210



=

10 

=

27.27

3508 : ϕ P = 90 °

Conversion to secondary values: k sec

=

27.27 (from above)

3505 : Z b(sec) 3506 : Z c(sec) Z d(sec)

= =

k sec ⋅ Z b(prim)

=

27.27 ⋅ 0.561 

k sec ⋅ Z c(prim)

=

27.27 ⋅ 0.165

(Z

d(prim)

 =

4.50



= 10 

3507 : Z d(sec) - Z c(sec) 3504 : Z a(sec)

= 15.30 

=

=

k sec

k sec ⋅ Z a(prim)

⋅

=

- Z c(prim)

27.27 ⋅ 0.210

) = 27.27 ⋅ 0.202  = 5.51   =

5.73



0.367



Exercise 2.9: Settings for out of step protection

(3/3)


Machine Protection Setting Exercises

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