Generator Relay Protection Setting Calculation

(1)

Generator relay protection setting

calculation instruction

Instruction: This setting calculation instruction is only

Instruction: This setting calculation instruction is only forfor reference,reference,

protection device running setting is confirmed by user.

XJ Electricity Co., Ltd

2010.11

(2)

Catalog

1. Generator Differential Protection 87G ...3

2. Overall Differential 87ALL ...4

3. Generator Inter-turn Fault 95G ...6

4.Generator Inadvertent Energize protection Stator Earth Fault Protection of startup condition 99V ... ... ... 7

5. Generator Over-voltage 59G ...7

6. Reverse Power Protection 32G ; Low Forward Power Protection 37G ...8

7. Generator Stator Overload 49G ...9

8. Generator Negative-Sequence Over-current 46G ...10

9. Generator Pole Slipping 78G ... 11

10. Generator loss of excitation protection 40G ...13

11. Generator Under & Over Frequency 81G ...14

12. Generator Over Fluxing 24G ...15

13. Generator Under-voltage 27G ...16

14. Stator Earth Fault 100% 64G1 ...16

17. Field Winding Earth Fault( 64F ...18

18. Generator 95% Voltage Check 59GB ...18

19. Back Up Impedance 21G ...18

(3)

1. Generator Differential Protection 87G

1.1 Basic parameter

CT ratio n_TA 8000/5

rated primary current I gn1 7060A

rated secondary current I gn2 _n A

I I TA gn gn 4.41 1600 7060 1 2

=

1.2 Setting calculation

1 Min operation current op.0

Setting by the max unbalanced current under the condition of avoid the normal generator rated load.

2 2

2 0

. rel 2 0.03 gn 1.5 2 0.03 gn 0.09 gn

op K I I I

I

=

×

=

×

=

Suggest to select I _op_.₀

=

0.2I _gn₂

2 Min brake current I _res_.₀

2 0

. gn res I

I

=

3 Ratio brake coefficientS

External three-phase short-circuit the maximum short-circuit current:

69 . 32 1600 8 . 13 3 08 . 0 100000 8 . 13 3 , , ) 3 ( max .

=

×

=

×

=

TA d b n X S I k A I I k res 32.69 ) 3 ( max .

=

.max

=

The max unbalanced current of differential protection when the generator is outer short circuit

A I K K K I k cc er ap unb 2 0.1 0.5 32.69 3.27 ) 3 ( max .

=

×

.max

=

×

=

Therein ap is the non-periodic branch coefficient select 1.5 2.0 cc is the

same type coefficient of transformer select 0.5 er is the error coefficient of

transformer ratio select 0.1

Put the condition of the differential protection won’t error operate under the max

outer short circuit current the secondary current value I op.max of correspondent max

(4)

A 9 . 4 27 . 3 5 . 1 I K

I_op_._max

=

_rel _unb_._max

=

×

=

Therein rel is the reliable coefficient, select 1.3 1.5.

The ratio brake coefficient S is 12 . 0 53 . 3 69 . 32 324 . 1 9 . 4 I I I I S 0 . res (3) max 0 . op max . op

₌

−

=

−

=

Suggest to select S=0.3 4 Sensitivity check

Sensitivity check principle : Generator terminal side of two-phase metallic short-circuit occurs when generator is parallel off:

A n X S I TA d b k 28.3 1600 8 . 13 3 08 . 0 100000 866 . 0 8 . 13 3 866 . 0 , , min .

=

×

=

×

=

A I I _res k 14.15 2 min .

=

A I I S I

I _op

=

_op_.₀

+

( _res

−

_res_.₀)

=

1.324

+

0.3

×

(14.15

−

3.53)

=

4.51

2 27 . 6 51 . 4 3 . 28 min .

=

>

=

op k sen I I K

5 Output model : Trip All CBs

2. Overall Differential 87ALL

2. 1 Basic parameter

Name M.T. HV Side Generator terminal A.T. HV Side

Rated Voltage kV 230 13.8 13.8

Rated Current A 452.4 7539.6 7539.6

CT ratio 1000/1 8000/5 8000/5

Second side Current A 0.45 4.71 4.71

2.2 Setting Calculation

1 Min operation current I op.0

Setting by the max unbalanced current under the condition of avoid the normal main transformer rated load.

b b I I 0.24 ) 05 . 0 05 . 0 06 . 0 ( 5 . 1 I ) m U f ( K I_op_.₀

=

_rel _i₍_n₎

+

∆

+

∆

_b

=

×

+

=

Suggest to select I_op_.₀

=

0.5 I _b_b

Therein

b

(5)

secondary side.

rel

K

is the reliable coefficient K rel

=

1.3~ 1.5 )

n ( i

f

is the transformer ratio error of the current transformer under rated current.

U

∆ _{is the error caused due to the regulation of transformer tapping (relative to the}

percent of the rated voltage).

m

∆ _{is the error caused due to the incomplete matching of transformer ratios of TA and}

TAA, ∆m takes 0.05 in general.

2 Min brake current I res.0

Suggest to select I _res_.₀

=

I _b

3 Ratio brake coefficient S

The calculation of three-phase main transformer high-voltage busbar metallic short-circuit occurs current when generator is parallel off:

A n S X X I TA b T d k 1.674 1000 230 3 100000 07 . 0 08 . 0 1 230 3 1 ,, ) 3 ( max .

=

×

+

=

×

+

=

External three-phase short-circuit the maximum short-circuit current: (3) max . i aper st max . unb (K K f U m)I I

=

+

∆

+

∆

_k A 4185 . 0 674 . 1 ) 05 . 0 05 . 0 1 . 0 5 . 1 1 (

×

+

×

=

st

K _{is the same type coefficient of TA} K _st =1.0

aper

K

is non-periodic coefficient of TA K aper =1.5~2.0 _{5P or 10P type TA}

or K aper

=

1.0 _{TP type TA}

Setting of ratio braking coefficient S

31 . 0 36 . 0 674 . 1 225 . 0 4185 . 0 5 . 1 I -I I -I K S res.0 res.max op.0 unb.max rel

₌

−

×

=

Suggest to select: S

=

0.4 4 Sensitivity check

Sensitivity check principle : Main transformer high-voltage side of two-phase metallic short-circuit occurs when generator is parallel off:

A n X X S I TA T d b k 1.45 1000 230 3 ) 07 . 0 08 . 0 ( 100000 866 . 0 230 3 ) ( 866 . 0 , , min .

₊

_×

=

×

=

×

+

×

=

(6)

A I

I _res_._min

=

_k_._min

=

1.45

A I

I S I

I _op

=

_op_.₀

+

( _res

−

_res_.₀)

=

0.225

+

0.4

×

(1.45

−

0.36)

=

0.661

5 . 1 19 . 2 661 . 0 45 . 1 min .

=

>

=

op k sen I I K

5 Setting of second harmonic braking coefficient

Suggest to select: 15%

6 Difference current quick brake

Select: I _op

=

5I _b

7 Difference current quick brake sensitivity check

Sensitivity check principle : Main transformer low-voltage side of two-phase metallic short-circuit occurs:

A n X X X S I TA T S d b k 5.08 1000 230 3 ) 07 . 0 022 . 0 //( 08 . 0 100000 866 . 0 230 3 ) //( 866 . 0 , ,

+

×

=

×

=

×

+

×

=

2 . 1 26 . 2 25 . 2 08 . 5

₌

_>

=

op k sen I I K

8 Output model: Trip all CBs

3. Generator Inter-turn Fault 95G

3.1 Basic parameter

Rated secondary current I _gn₂ A n I I TA gn gn 4.41 1600 7060 1 2

=

PT ratio n_TV 3 1 . 0 3 1 . 0 3 8 . 13 3.2 Setting Calculation 1 Zero-sequence Voltage

Setting by the max unbalanced zero-sequence voltage under the condition of avoid the normal generator rated load.

V U _op

=

2

2)Fault branch negative-sequence direction elements

A I _gn i

=

3% 2

=

3%

×

4.41

=

0.13 ε V U _gn u

=

1% 2

=

1%

×

110/ 3

=

0.63 ε

(7)

VA U I n n P gn gn TA TV gn p 0.1% 3 0.1% 3 4.41 100 0.76 % 1 . 0 2 2

×

=

×

=

×

=

×

=

ε 3) Time delay t

=

0.1s

4 Output model: Trip All CBs

4.Generator Inadvertent Energize protection Stator Earth Fault Protection of

startup condition 99V

4.1 Basic parameter

Rated secondary current I _gn₂ 4.41A

Rated secondary voltage U _gn₂ 110V

1 Low impedance element setting

( )

Ω

=

×

=

×

=

38.4 41 . 4 3 . 0 3 110 8 . 0 3 . 0 3 8 . 0 2 2 gn gn set I U Z

( )

Ω

=

×

=

0.85 _set 0.85 38.4 32.6 set Z R

2 Over current element setting

( )

A I

I _op

=

0.3 _gn₂

=

0.3

×

4.41

=

1.32

3 Stator Earth Fault Protection of startup condition operation voltage

V U _op

=

10

Time delay t

=

2s

4 Output model

Generator Inadvertent Energize protection: Trip 220kV CB & excitation CB Stator Earth Fault Protection of startup condition: Trip excitation CBs

5. Generator Over-voltage 59G

5.1 Basic parameter

Rated secondary voltage U _gn₂ ₁₁₀_V

5.2 Setting Calculation 1 Operation voltage

I: U _op

=

1.05U _gn₂

=

1.05

×

110

=

115.5V

II: U _op

=

1.1U _gn₂

=

1.1

×

110

=

121V

(8)

2 Time delay Select

I: t

=

30s

II: t

=

10s

III: t

=

0.5s

6. Reverse Power Protection 32G ; Low Forward Power Protection 37G

6.1 Basic parameter

Generator rated power P_n 135WM

Generator rated secondary power P_n₂ 672.6W

6.2 Setting Calculation 1 Min operation power

W K _rel

op

=

+

)

=

0.5

×

(3%

+

1

−

98.6%)

×

672.6

=

14.8

Suggest to select _op

=

10W

Therein K rel is the reliable coefficient select 0.5~0.8

is the min loss when steam turbine is in reverse power operation generally

select 2 ~4 of the rated power

is the min loss when steam turbine is in reverse power operation generally select

η Pgn is the rated capacity of generator.

2 Time delay

Reverse Power Protection 32G t 1

=

5s(32G) ) 32 ( 60 2 s G t

=

Low Forward Power Protection 37G

3 Output model

Reverse Power Protection 32G :

Time delay t 1: Alarm

Time delay t 2: Trip All CBs Low Forward Power Protection 37G : Trip All CBs

) 37 (

1s G

(9)

7. Generator Stator Overload 49G

7.1 Basic parameter

Rated secondary current I _gn₂ ₄_.₄₁_A

allowed heat time constant of stator winding K

_37.5

7.2 Setting calculation 1 Time specified overload

Stator winding time specified over-load can be set by the condition of the long term allowed loading current can reliable return.

A

I

K

gn r rel op

4 .

41

5 .

14

9 .

0

05 .

1

2

=

×

=

×

=

Therein K rel _{is the reliable coefficient} _{select 1.05} r

K is the return coefficient select0.9

Time specified overload time delay: t

=

9s

2 Reverse time specified overload

Reverse time specified over-current can be set by the over-load ability allowed by stator winding it should be determined by the over-load ability allowed by the stator winding of the motor manufacturer. The relation of allowed duration time is :

) 1 ( I K t ₂ *

−

+

α

=

Therein K is the allowed heat time constant of stator winding it should be based on the parameter provided by the motor manufacturer

* is the per-unit value based on stator rated current

α is the heat radiation constant and related to the stator winding temperature

rising and temperature margin generally select 0.01~0.02.

3 Reverse time specified startup current

Reverse time specified startup current op.min should be set by the condition of

matched with time specified over-load protection

A

K

_C _op

op.min

=

0

=

1 .

05 ×

5 .

14 =

5 .

4

Therein op is the set value of time specified startup current K C 0 _{is the matching}

coefficient select 1.05

Reverse time specified delay lower limit: t _min

=

120s

Reverse time specified delay upper limit current op.max can be set by the condition of three phase metal short circuit at generator side

(10)

A

X

I

d gn op

32 .

7

135 .

0

41 .

4

" 2 max .

=

(

)

s I K t op 7 . 0 01 . 0 1 41 . 4 1 . 55 5 . 37 ) 1 ( 2 2 max* . max

=

+

−

⎟

⎠

⎞

⎜

⎝

⎛

=

+

−

=

α 4 Output model

Time specified overload: Alarm

Reverse time specified overload: Programming Trip

8. Generator Negative-Sequence Over-current 46G

8.1 Basic parameter

Rated secondary current I _gn₂ 4.41 A

The per-unit value of generator long term allowed

negative-sequence current I ₂_∞

. . 08 .

0 PU

time constant of withstanding negative-sequence current

ability of rotor surface A

10

8.2 Setting calculation 1 Time specified overload

The negative-sequence time specified over-load should be set on the condition of

under the generator long term allowed negative-sequence current 2∞ can reliable

return

A

K

I

K

r gn rel op

0 .

39

95 .

0

41 .

4

08 .

0

05 .

1

2 2

=

×

=

∞

Therein K rel _{is the reliable coefficient} _{select 1.05} K is the return coefficientr

select0.85 0.95

∞

2 is the per-unit value of generator long term allowed negative-sequence current

Time specified overload time delay: t

=

5s

2 Reverse time specified overload

The reverse time specified negative-sequence over-load was confirmed by the allowed negative-sequence over-current ability of the generator rotor surface. The relation mode between the generator short time withstanding negative-sequence over-current multiple and allowed duration time is

2 2 2 * 2

−

∞

=

I

A

t

(11)

Therein is the per-unit value of generator negative-sequence current

∞

2

I _{is the per-unit value of generator long term allowed negative-sequence}

current

A is the time constant of withstanding negative-sequence current ability of rotor

surface

Reverse time specified startup current

The reverse time specified startup current _op_._min generally be set by the

operation current in correspondent with delay1000s set value of reverse time

specified delay lower limit

A

I

A

I

_gn op

0 .

08

0 .

56 1000

10

41 .

4 1000

2 2 2 2 min .

=

+

∞

=

×

+

=

Reverse time specified delay upper limit current _op_._max should be set by the condition of main transformer HV side two phase metal short circuit

A

X

I

t G d gn op

8 .

38

125 .

0

2

141 .

0

135 .

0

41 .

4

2

2 " 2 max .

=

₊

=

₊

_×

=

s I I A t

2 .

76

08 .

0

41 .

4

38 .

8

10

2 2 2 2 2 * 2 min

=

−

⎟

⎠

⎞

⎜

⎝

⎛

=

−

=

∞ 3 Output model

Time specified overload: Alarm

Reverse time specified overload: Programming Trip

9. Generator Pole Slipping 78G

9.1 Basic parameter

Generator neutral CT ratio 8000/5=1600 Generator terminal CT ratio 13.8/0.11=125.4 9.2 Setting calculation

1 Reduced generator transformer system reactance etc. to the named unit (ohm)value with the generator side voltage is 13.8kV.

Generator

=

( )

Ω

×

=

0.267 7060 3 13800 237 . 0 ' d X Main transformer

=

×

=

×

=

0.132

( )

Ω

180 8 . 13 125 . 0 2 2 n gn k t S U X X System

(

÷

)

=

( )

Ω

=

0.031 8 . 13 220 8 . 7 2 s X

(12)

st

X setting of system relation impedance

( )

Ω

=

+

=

+

=

_t _s 0.132 0.031 0.163 st X X X

2 Reduce X _d' X _t& X _st to the secondary side value of generator side TV TA.

) ( 4 . 3 4 . 125 1600 267 . 0 ' 2 '

₌

_×

₌

_×

₌

_Ω

TV TA d d n n X X ) ( 68 . 1 4 . 125 1600 132 . 0 2

=

×

=

×

=

Ω

TV TA t t n n X X ) ( 08 . 2 4 . 125 1600 163 . 0 2

=

×

=

×

=

Ω

TV TA st st n n X X

3 Setting lens principal axis obliquity

Select system impedance angle :

0

85

=

Ψ

_z

4 Setting of operation power angle _set

) ( 71 . 0 8 . 0 7060 15 . 1 3 13800 9 . 0 15 . 1 3 9 . 0 min . 1

×

=

Ω

×

=

×

=

COS ϕ I U R n n 0 0 0 0 min . ₁₈₀ ₂ ₆₈_.₅ ₄₃ 267 . 0 163 . 0 71 . 0 54 . 1 2 180 54 . 1 2 180

=

−

×

=

+

×

−

=

′

+

−

°

=

_arctg X X R arctg d st L set

Suggest to select δ _set

=

1200

5 Tripping blocking current setting

This protection use generator neutral CT, CT ratio is 8000/5 Main transformer high side circuit breaker rated breaking current is 40kA

=

×

=

8 . 13 230 1600 40000 5 . 0 set I 208A

Suggest to select I _set

=

180 A

6 Slipper times setting

Outer zone Slipper times setting:N =4

Inner zone, Slipper times setting:N =1 #1Generator 2 #2 Generator 3 #3 Generator 4 #4 Generator

7 Reactance line position Z _C

) ( 377 . 1 53 . 1 9 . 0 9 . 0 ₂

=

×

=

Ω

=

_t C X Z 8 Startup current

(13)

) ( 3 . 5 41 . 4 2 . 1 2 . 1 I ₂ A I _st

=

_n

=

×

=

9 Time delay Outer zone :0.5s Inner zone :0.1s 10 Output model

Outer zone :Alarm Inner zone :Trip all CBs

10. Generator loss of excitation protection 40G

10.1 Basic parameter

Reduced generator transformer system reactance etc. to the named unit (ohm)value with the generator side voltage is 13.8kV.

Generator

=

( )

Ω

×

=

₂_.₃₃ 7060 3 13800 065 . 2 d X

( )

Ω

=

×

=

×

=

×

=

0.132

( )

Ω

180 8 . 13 125 . 0 2 2 n gn k t S U X X System

(

÷

)

=

( )

Ω

=

0.031 8 . 13 220 8 . 7 2 s X st

X setting of system relation impedance

( )

Ω

=

+

=

+

=

_t _s ₀_.₁₃₂ ₀_.₀₃₁ ₀_.₁₆₃ st X X X

As X _d

=

X _q Salient pole power equal 0W 1 Set of static and stable boundary impedance

) ( 7 . 29 4 . 125 1600 33 . 2 1

=

×

=

×

=

Ω

TV TA d B n n X Z ) ( 1 . 2 4 . 125 1600 163 . 0 1

=

×

=

×

=

Ω

TV TA st A n n X Z

2 Set of stable asynchronous impedance

) ( 7 . 29 4 . 125 1600 33 . 2 2

=

×

=

×

=

Ω

TV TA d B n n X Z

(14)

) ( 7 . 1 4 . 125 1600 267 . 0 5 . 0 5 . 0 ' 2

=

×

=

×

=

Ω

TV TA d A n n X Z

3 Set of generator terminal low voltage

) ( 88 110 8 . 0 8 . 0 U ₂ V U _st

=

_n

=

×

=

4 Time delay Select t1=1.5S t2=1.5S t3=3S 5 Output model

loss of excitation zone 1: Alarm loss of excitation zone 2: Trip all CBs loss of excitation zone 3: exit

11. Generator Under & Over Frequency 81G

Generator under & over frequency capability tables provided by the equipment manufacturing factor:

Allow run-time Allow run-time

Frequency Hz accumulated time (min) Each time(s) Frequency Hz accumulated time (min) Each time(s) 5 . 51 0 . 51

<

F

≤

30 30 47.5

<

F

≤

48 60 60 51 5 . 50

<

F

≤

180 180 5 . 47 47

<

F

≤

10 20 5 . 50 5 . 48

<

F

≤

Run Continuously 47 5 . 46

<

F

≤

2 5 5 . 48 48

<

F

≤

300 300 11.2 Setting calculation 1 Under frequency zone I

under-frequency zone I frequency setting f ₁_._set

=

48.5Hz

under-frequency zone I accumulated time

∑

_t₁_._set

=

18000s

under-frequency zone I time delay t ₁_._set

=

300s

2 Under frequency zone II

under-frequency zone II frequency setting f ₂_._set

=

48Hz

(15)

under-frequency zone II time delay t ₂_._set

=

60s

3 Under frequency zone III

under-frequency zone III frequency setting f ₃_._set

=

47.5Hz

under-frequency zone III accumulated time

∑

_t₃_._set

=

600s

under-frequency zone III time delay t ₃_._set

=

20s

4 Under frequency zone IV

under-frequency zone IV frequency setting f ₄_._set

=

47Hz

under-frequency zone IV time delay t ₄_._set

=

20s

5 Over-frequency

over-frequency setting f _set

=

51Hz

over-frequency time delay t _set

=

30s

6 Generator terminal low voltage setting U _set

=

0.8U _n

=

88V

7 Output model: Programming Trip

12. Generator Over Fluxing 24G

Generator over excitation capability tables provided by the equipment

manufacturing factor: Stator voltage/Frequency 1.25 1.19 1.15 1.12 1.10 1.09 1.08 1.07 1.05 Time(sec) 5 7.5 10 15 20 30 45 60 ∞ 12.2 Setting calculation

1 Time specified over fluxing

Over fluxing times: N ₀

=

1.06, time delay: t ₀

=

5s

Select rated voltage as reference voltage 110V 2 Reverse time specified over fluxing

Over fluxing times: N ₁

=

1.07 time delay:t ₁

=

60s

Over fluxing times: N ₂

=

1.08 time delay:t ₂

=

45s

Over fluxing times: N ₃

=

1.09 time delay:t ₃

=

30s

(16)

Over fluxing times: N ₅

=

1.12 time delay:t ₅

=

15s

Over fluxing times: N ₆

=

1.15 time delay:t ₆

=

10s

Over fluxing times: N ₇

=

1.19 time delay:t ₇

=

7.5s

Over fluxing times: N ₈

=

1.25 time delay:t ₈

=

5s

13. Generator Under-voltage 27G

Generator terminal PT ratio

3 11 . 0 3 11 . 0 3 8 . 13 13.2 Setting calculation V U U _set

=

0.6 _n

=

66 Time delay: 0.5s 13.3 Output model: Alarm

14. Stator Earth Fault 100% 64G1

Neutral earthing transformer ratio 13.8/0.3kV

Neutral earthing transformer secondary resistance ( R_n) 1.3

Ω

Neutral earthing transformer primary resistance ( R_N) 2750.8

Ω

Neutral earthing transformer secondary CT ratio(n_TA₀) 100/1

Generator 3 relative capacitance (3C ₀) 1.392µ F

Generator capacitive reactance ( X _C₀) 6860

Ω

14.2 Setting calculation 1) earthing resistance

The setting of earthing resistance low set value should be set by the principle of one point earthing in the distance within 20% of generator neutral, the earthing fault

point current 3I0 safe earthing current Is(1A). Means

S C N g gn I jX R R U I

≤

−

+

×

=

) //( 3 3 3 3 2 . 0 3 0 0

(17)

S C N set g gn I jX R R U

=

−

+

×

) //( 3 3 3 3 2 . 0 0 . 1 ) 6860 //( 8 . 2750 3 3 3 13800 3 2 . 0 .

=

−

×

+

×

j R_g_set

Ω

=

k R_g_._set 5.2 Suggest to select

Earthing resistance high value: R_hg_._set

=

10k

Ω

Earthing resistance low value: R_lg_._set

=

5k

Ω

Time delay:

high value time delay:3s low value time delay:0.5s

A n K R U I TA I n n set 0.5 100 1 . 1 3 . 1 300 2 . 0 0 2 . 50

_×

=

×

=

×

=

α

Therein α can generally be selected as 20

I

K is the reliable coefficient generally select 1.1.

2) Output model

high value : Alarm low value : Trip all CBs

15.1Basic parameter

Generator terminal PT ratio

3 1 . 0 3 1 . 0 3 8 . 13

Generator neutral PT ratio 0.1 3

8 . 13

Setting value by the protection device according to measured data Time delay : 5s

15.3 Output model: Alarm

The fundamental wave zero-sequence voltage has two sections protection including high value and low set value.

(18)

unbalance fundamental wave zero-sequence voltage of neutral single phase voltage transformer when normal operation or three phase voltage transformer open triangle winding at generator side. The setting value applied in the project should avoid the max zero-sequence unbalance voltage that transmitted to the generator side when system HV side and plant transformer LV side earthing short circuit.

The setting of high set value zero-sequence voltage is 20 30V.

Suggest to select:U _op_._H

=

20V high value U _op_._l

=

10V low value According to the measured maximum value of zero sequence voltage imbalance adjustment

Time delay:

High value: 0.1s Low value: 0.3s 16.2 Output model

High value: Trip all CBs Low value: Trip all CBs

17. Field Winding Earth Fault( 64F

Non-electrical time delay is setting by the owner

18. Generator 95% Voltage Check 59GB

rated secondary voltage U _gn₂ ₁₁₀_V

18.2 Setting Calculation 1 Operation voltage

V U

U _op

=

0.95 _gn₂

=

0.95

×

110

=

104

2 output model: Alarm

19. Back Up Impedance 21G

Based on actual engineering experience, this protection is valueless ,we suggest don’t use it. 19.1 Basic parameter

1 Reduced generator transformer to the named unit (ohm)value with the generator side voltage is 13.8kV. Generator

=

( )

Ω

×

=

×

=

×

=

0.132

( )

Ω

180 8 . 13 125 . 0 2 2 n gn k t S U X X

(19)

) ( 19 . 3 4 . 125 1600 25 . 0 ' 2 '

₌

_×

₌

_×

₌

_Ω

TV TA d d n n X X ) ( 68 . 1 4 . 125 1600 132 . 0 2

=

×

=

×

=

Ω

TV TA t t n n X X 3 R_set₁

=

X _d'2

×

1.2

=

3.83(

Ω

) ) ( 01 . 2 2 . 1 2 2

=

t

×

=

Ω

set X R 20. TV Fuse-failure 160G 21.1 Basic Parameter

rated secondary current I _gn₂ 4.41 A

21.2 Setting Calculation