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
=⋅
=
Exercise 1: Power System Data 1 (2/4)
Exercise 1: Power System Data 1 (3/4)
)/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)
Exercise 1: Power System Data 1 (4/4)
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
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 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
Exercise 2.1: Calculation of load resistor and neutral transf. (3/3)
[ ]
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.
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)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 2.4: 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 = 0.58
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:
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
Exercise 2.5: Settings for under excitation protection (3/4)
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
Exercise 2.5: Settings for under excitation protection (4/4)
Settings: primary
values
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:
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%
Exercise 2.6: Settings for (stator-) thermal overload (3/5)
Settings in primary values:
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
Conversion to secondary values:
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
Exercise 2.6: Settings for (stator-) thermal overload (4/5)
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 2.6: Settings for (stator-) thermal overload (5/5)
Settings: primary values
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)
Exercise 2.8: Settings for impedance protection (2/4)
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)
=
→
=
→
=
→
+⋅=
=
=
⋅⋅=
=
=
=
⋅⋅=
Exercise 2.8: Settings for impedance protection (3/4)
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)
=
=
=+⋅=
=
=⋅⋅=
=
=⋅⋅=
Conversion to secondary values:
Settings in primary values:
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 2.8: Settings for impedance protection (4/4)
Settings: primary values
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
Exercise 2.9: Settings for out of step protection (2/3)
°=
=+⋅=
=−=
=====
=⋅
⋅⋅≈
=⋅⋅
≈
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
Settings in primary values:
Conversion to secondary values:
( )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
=⋅=⋅=
=⋅=⋅=
=
=⋅=⋅=
=⋅=⋅=
=
Exercise 2.9: Settings for out of step protection (3/3)
Settings: primary values
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