Setting by the max unbalanced current under the condition of avoid the normal generator rated load. I op.0 = K rel × 2 × 0.03 I gn 2 = 1.5 × 2 × 0.03 I gn 2 = 0.09 I gn 2
Suggest to select I op.0 = 0.2 I gn 2 2
Min brake current I res.0
I res.0 = I gn 2 3
Ratio brake coefficient S External three-phase short-circuit the maximum short-circuit current: ) I k ( 3. max =
S b X d ,, × 3 × 13.8 × nTA
=
100000 0.08 × 3 × 13.8 × 1600
= 32.69
) I res. max = I k (3. max = 32.69 A
The max unbalanced current of differential protection when the generator is outer short circuit ) I unb. max = K ap × K er × K cc × I k ( 3. max = 2 × 0.1 × 0.5 × 32.69 = 3.27 A
Therein
ap
is the non-periodic branch coefficient
same type coefficient of transformer
select 0.5
er
select 1.5
2.0
cc
is the
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 operation current
the secondary current value I op .max of correspondent max
I op.max = K rel I unb.max = 1.5 × 3.27 = 4.9 A Therein
is the reliable coefficient, select 1.3
rel
1.5.
The ratio brake coefficient S is S=
I op.max − I op.0 I (3) max − I res.0
4.9 − 1.324
=
32.69 − 3.53
= 0.12
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: I k . min = I res =
0.866 × S b X d ,, × 3 × 13.8 × nTA
I k . min
2
=
0.866 × 100000 0.08 3 × 13.8 × 1600
= 28.3 A
= 14.15 A
I op = I op.0 + S ( I res − I res.0 ) = 1.324 + 0.3 × (14.15 − 3.53) = 4.51 A K sen = 5
I k . min I op
=
28.3 4.51
= 6.27 > 2
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
1000/1
8000/5
8000/5
0.45
4.71
4.71
CT ratio Second side Current
A
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.
I op.0 = K rel (f i ( n ) + ∆U + ∆m)I b = 1.5 × (0.06 + 0.05 + 0.05) I b = 0.24 I b Suggest to select
I op.0 = 0.5 I b b
Therein
I b
is the value of rated current of transformer norm side converted to the TA
secondary side. K rel f i( n )
K rel = 1.3 ~ 1.5
is the reliable coefficient
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
Suggest to select 3
I res.0
I res .0 = I b
Ratio brake coefficient S
The calculation of three-phase main transformer high-voltage busbar metallic short-circuit occurs current when generator is parallel off: ) I k (3. max =
1 X + X T ,, d
S b
×
=
3 × 230 × nTA
1 0.08 + 0.07
×
100000 3 × 230 ×1000
= 1.674 A
External three-phase short-circuit the maximum short-circuit current:
I unb . max = ( K st K aper f i + ∆ U + ∆ m ) I (3) k . max
Sensitivity check Sensitivity check principle : Main transformer high-voltage side of two-phase metallic short-circuit occurs when generator is parallel off:
4
I k . min =
0.866 × S b ( X d ,, + X T ) × 3 × 230 × nTA
=
0.866 × 100000 (0.08 + 0.07) × 3 × 230 × 1000
= 1.45 A
I res. min = I k . min = 1.45 A I op = I op.0 + S ( I res − I res.0 ) = 0.225 + 0.4 × (1.45 − 0.36) = 0.661 A K sen =
5
I k . min I op
=
1.45 0.661
= 2.19 > 1.5
Setting of second harmonic braking coefficient Suggest to select: 15%
6
Difference current quick brake Select: I op
= 5 I b
Difference current quick brake sensitivity check Sensitivity check principle : Main transformer low-voltage side of two-phase metallic short-circuit occurs:
7
I k =
0.866 × S b X //( X S + X T ) × 3 × 230 × nTA ,, d
Setting by the max unbalanced zero-sequence voltage under the condition of avoid the normal generator rated load. U op = 2V 2) Fault branch negative-sequence direction elements
ε i = 3% I gn 2 = 3% × 4.41 = 0.13 A ε u = 1%U gn 2 = 1% × 110 / 3 = 0.63V
ε p =
0.1% Pgn
t = 0.1s
3) Time delay 4
= 0.1% × 3 × I gn 2 × U gn 2 = 0.1% × 3 × 4.41 × 100 = 0.76VA
Stator Earth Fault Protection of startup condition operation voltage
U op = 10V Time delay 4
t = 2 s
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 gn 2
110V
5.2 Setting Calculation 1
Operation voltage
I:
U op = 1.05U gn 2 = 1.05 × 110 = 115 .5V
II:
U op = 1.1U gn 2 = 1.1 × 110 = 121V
III:
U op = 1.2U gn 2 = 1.2 × 110 = 132V
2
Time delay
Select I: II: III: 3
t = 30s t = 10 s t = 0.5s
Output model: Trip all CBs
6. Reverse Power Protection
32G
; Low Forward Power Protection
37G
6.1 Basic parameter Generator rated power
Pn
135WM Pn 2
Generator rated secondary power
672.6W
6.2 Setting Calculation 1
Min operation power op
= K rel
+
Suggest to select
) = 0.5 × (3% + 1 − 98.6%) × 672 .6 = 14 .8W op
= 10W
K rel is the reliable coefficient
Therein
select 0.5~0.8
is the min loss when steam turbine is in reverse power operation select 2
~4
of the rated power
is the min loss when steam turbine is in reverse power operation η
2
Pgn is the rated capacity of generator.
Time delay
Reverse Power Protection
t 1 = 5s(32G )
32G
t 2 = 60s(32G ) Low Forward Power Protection
37G
t = 1s (37G ) 3
generally
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
generally select
7. Generator Stator Overload
49G
7.1 Basic parameter Rated secondary current
I gn 2
4.41 A
allowed heat time constant of stator winding
37.5
K
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. op
=
K rel K r
Therein
× I gn 2 =
K rel
1.05 0.9
× 4.41 = 5.14 A
is the reliable coefficient
select 1.05
K r is the return coefficient select 0.9 Time specified overload time delay: t = 9 s 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 : t =
K I * − ( 1 + α ) 2
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 op . min
Therein
= K C 0 op
op
= 1.05 × 5.14 = 5.4 A
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
three phase metal short circuit at generator side
can be set by
the condition of
op. max
t max =
4
=
I gn 2
=
X d "
4.41 0.135
K 2 op. max*
I
− (1 + α )
= 32.7 A =
37.5 2
⎛ 55.1 ⎞ ⎜ ⎟ − (1 + 0.01) ⎝ 4.41 ⎠
= 0.7 s
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 gn 2
4.41 A
The per-unit value of generator long term allowed 0.08 P.U . negative-sequence current I 2∞ 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 op
= K rel
Therein
K rel
I
2 ∞ gn 2
K r
= 1.05 ×
0.08 × 4.41 0.95
is the reliable coefficient
= 0.39 A select 1.05
K r is the return coefficient
select 0.85 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
t =
A 2 2 I 2* − I 2 ∞
is the per-unit
Therein I 2∞
is the per-unit
value of generator negative-sequence current
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 operation current in correspondent with specified delay lower limit op . min
A
= I gn 2
1000
op . min
delay1000s
10
+ I 22∞ = 4.41 ×
1000
Reverse time specified delay upper limit
current
generally be set by the set value of reverse time
+ 0.08 2 = 0.56 A op . max
should be set by the
condition of main transformer HV side two phase metal short circuit op. max
t min =
3
=
I gn 2 X + X G 2 + 2 X t " d
A I − I 2 2*
2 2∞
=
=
4.41 0.135 + 0.141 + 2 × 0.125
10 2
⎛ 8.38 ⎞ 2 ⎜ ⎟ − 0.08 ⎝ 4.41 ⎠
= 8.38 A
= 2.76s
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 X d
= 0.237 ×
Main transformer X t
System
X s =
13800 3 × 7060
= X k ×
7.8
(220 ÷ 13.8)
2
U gn S n
= 0.267(Ω )
2
= 0.125 ×
= 0.031(Ω )
13.82 180
= 0.132(Ω )
X st setting of system relation impedance X st = X t + X s = 0.132 + 0.031 = 0.163(Ω ) 2
'
Reduce X d
X d 2 = X d × '
'
X t 2 = X t ×
nTA
= 0.267 ×
nTV
nTA nTV
X st 2 = X st × 3
X t & X st to the secondary side value of generator side TV TA.
= 0.132 ×
nTA nTV
1600 125.4
1600
= 1.68(Ω)
125.4
= 0.163 ×
= 3.4(Ω)
1600 125.4
= 2.08(Ω)
Setting lens principal axis obliquity
0 Ψ = 85 z Select system impedance angle :
4
Setting of operation power angle
R1. min =
0.9U n 3 × 1.15 I n
set = 180° − 2arctg
COS ϕ =
set
0.9 × 13800 3 × 1.15 × 7060
1.54 R L. min X st + X d ′
× 0.8 = 0.71(Ω)
= 1800 − 2arctg
1.54 × 0.71 0.163 + 0.267
= 1800 − 2 × 68.50 = 430
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
I set = 0.5 ×
40000 1600
×
230 13.8
= 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 Generator 7
4
#4 Generator
Reactance line position Z C
Z C = 0.9 X t 2 = 0.9 × 1.53 = 1.377(Ω) 8
Startup current
#1Generator
2
#2 Generator
3
#3
I st = 1.2 I n 2 = 1.2 × 4.41 = 5.3( A) 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 Generator neutral CT ratio 8000/5=1600 Generator terminal CT ratio 13.8/0.11=125.4 10.2 Setting calculation Reduced generator transformer system reactance etc. to the named unit (ohm)value with the generator side voltage is 13.8kV. Generator X d
= 2.065 ×
X d = 0.237 ×
System
3 × 7060
X t = X k ×
7.8
X s =
= 2.33(Ω)
3 × 7060
13800
'
Main transformer
13800
(220 ÷ 13.8)2
= 0.267(Ω)
U gn
2
S n
= 0.125 ×
13.82
= 0.031(Ω )
X st setting of system relation impedance X st = X t + X s = 0.132 + 0.031 = 0.163(Ω ) As X d 1
= X q
Salient pole power equal 0W
Set of static and stable boundary impedance
Z 1 B = X d ×
nTA nTV
Z 1 A = X st × 2
nTA nTV
= 2.33 ×
1600 125.4
= 0.163 ×
= 29.7(Ω)
1600 125.4
= 2.1(Ω)
Set of stable asynchronous impedance
Z 2 B = X d ×
nTA nTV
= 2.33 ×
1600 125.4
= 29.7(Ω)
180
= 0.132(Ω)
Z 2 A = 0.5 X d × '
3
nTA
= 0.5 × 0.267 ×
nTV
1600 125.4
= 1 .7 ( Ω )
Set of generator terminal low voltage
U st = 0.8U n 2 = 0.8 × 110 = 88(V ) 4
Time delay Select t1=1.5S
5
t2=1.5S
t3=3S
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
11.1 Basic parameter
Generator under & over frequency capability tables provided by the equipment manufacturing factor: Allow run-time Frequency
Hz
accumulated
Each
time (min)
time(s)
30
30
51.0 < F ≤ 51.5 50.5 < F ≤ 51
48.5 < F ≤ 50.5 48 < F ≤ 48.5
Allow run-time Frequency Hz
47.5 < F ≤ 48
180
time (min)
time(s)
60
60
180
10
20
2
5
46.5 < F ≤ 47 Run Continuously
300
300
Under frequency zone I under-frequency zone I frequency setting
f 1. set = 48.5Hz
under-frequency zone I accumulated time
∑ t 1.set = 18000s
under-frequency zone I time delay 2
Each
47 < F ≤ 47.5
11.2 Setting calculation 1
accumulated
t 1.set = 300s
Under frequency zone II under-frequency zone II frequency setting
f 2.set = 48Hz
under-frequency zone II accumulated time
∑ t 2.set = 3600s
t 2. set = 60s
under-frequency zone II time delay 3
Under frequency zone III under-frequency zone III frequency setting
f 3.set = 47.5Hz
under-frequency zone III accumulated time
∑ t 3.set = 600s
t 3. set = 20s
under-frequency zone III time delay 4
Under frequency zone IV under-frequency zone IV frequency setting
t 4. set = 20s
under-frequency zone IV time delay 5
f 4. set = 47 Hz
Over-frequency over-frequency setting
f set = 51Hz
over-frequency time delay
t set = 30s
6
Generator terminal low voltage setting
7
Output model: Programming Trip
12. Generator Over Fluxing
U set = 0.8U n = 88V
24G
12.1 Basic parameter
Generator over excitation manufacturing factor: Stator
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 3 I 0 =
0.2 × 3U gn
3
3 Rg + 3 R N //(− jX C 0 )
≤ I S
0.2 × 3U gn
3
3 Rg .set + 3 R N //(− jX C 0 ) 0.2 × 3 × 13800
= I S
3
3 Rg . set + 3 × 2750.8 //( − j 6860 )
=1
R g .set = 5.2k Ω Suggest to select Earthing resistance high value: Rhg . set = 10k Ω Earthing resistance low value: Rlg .set
= 5k Ω
Time delay: high value time delay:3s low value time delay:0.5s
I 50. set =
α × U 2 n Rn × K I × nTA 0
Therein
α
=
0.2 × 300 1.3 × 1.1× 100
= 0.5 A
can generally be selected as 20
K I
is the reliable coefficient
generally select 1.1. 2) Output model high value : Alarm low value : Trip all CBs
15. Stator Earth Fault 100%
64G2
15.1Basic parameter Generator terminal PT ratio
Generator neutral PT ratio
13.8
0.1
0.1
3
3
3 13.8 3
0.1
15.2 Setting calculation Setting value by the protection device according to measured data Time delay : 5s 15.3 Output model: Alarm
16. Stator Earth Fault 95%
64G3
16.1 Setting calculation
The fundamental wave zero-sequence voltage has two sections protection including high value and low set value. Principle ,the low setting value zero-sequence voltage should be set by the max
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
U op.l = 10V
high value
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( Non-electrical
64F
time delay is setting by the owner
18. Generator 95% Voltage Check
59GB
18.1 Basic parameter rated secondary voltage
U gn 2
110V
18.2 Setting Calculation 1
Operation voltage
U op = 0.95U gn 2 = 0.95 × 110 = 104V 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 Generator neutral CT ratio 8000/5=1600 Generator terminal CT ratio 13.8/0.11=125.4 19.2 Setting calculation 1
Reduced generator
transformer to the named unit (ohm)value with the generator side
voltage is 13.8kV. Generator
X d = 0.221 × '
Main transformer
2
'
Reduce X d
13800 3 × 7060
X t = X k ×
U gn S n
= 0.25(Ω )
2
= 0.125 ×
13.82 180
= 0.132(Ω)
X t to the secondary side value of generator side TV TA.
X d 2 = X d × '
'
X t 2 = X t ×
3
nTA
= 0.25 ×
nTV
nTA nTV
= 0.132 ×
1600 125.4 1600
125.4
= 3.19(Ω)
= 1.68(Ω)
Rset 1 = X d 2 × 1.2 = 3.83(Ω) '
Rset 2 = X t 2 × 1.2 = 2.01(Ω) 20. TV Fuse-failure
