Tanta University Electrical Power and Machines Department
Faculty of Engineering
Fourth Year- Second Term
2020-2021
Power System Protection
Section # 5
Sheet #3
Eng. Eatmad Nahas
Generator Protection
Internal External (System)
Stator Rotor
(DC System)
1- loss of Prim-mover
2- Over excitation
Phase earth
Fault Fault Electrical Mechanical 3- Unbalance current
4- Over voltage
OC SC • Bearing
• Symmetrical
Generator Protection
1- 87 Differential
2- 50/51 Overcurrent
3- 46 Unbalance/ negative sequence OCR
4- 51V voltage control OCR
5- 32 Power directional relay
6- 27 H under voltage relay
7- 59 G Over voltage relay
8- 81 O/U Over/Under frequency
9- 24 Over excitation
87 G - Differential Protection
1- Principle of operation
a- Healthy
I1 I2
i1
G i2
R IR=i2 - i1
Differential Protection
1- Principle of operation
b- Internal fault
I1 I2
G i2
i1
R IR=i2 + i1
Differential Protection
1- Principle of operation
c- External fault
I1 I2
i1
G i2
R IR=i2 - i1
Differential Protection
2- Characteristics
➢ Iomin = (0.1-0.2 ) INCT G-M-B.B- Cable
❑ IR> = Iomin Trip IR
➢ Iomin = (0.15-0.4 ) INCT XFMR
❑ IR< Iomin No op.
Trip
Iomin
G IPu
ISet
No operation
R Status
Sheet (3)
1. Three 21,875 kVA, 13.8 kV generators with Xd =13.9% are connected to individual buses, from which various loads are
supplied. These buses are connected to another bus through 0.25 Ω reactors as shown in Figure 1. The generators are all
ungrounded. In this system:
(a) Calculate a three-phase fault at the terminals of one of the generators.
(b) Choose a current transformer ratio for differential relays to protect the generators. If the generator differential relays have a
minimum pickup of 0.14 A, how many times pickup does the three-phase fault provide?
(c) Calculate a single-phase-to-ground fault at the terminals of one Of the generators.
(d) Will this ground fault operate the generator differential relays? If so, how many times pickup will the ground fault
provide?
Sheet (3)
(a) Calculate a three-phase fault at the terminals of one of the generators.
❑PU calculation:
𝑆𝑏 100 ∗ 1000
𝑆𝑏 = 100 𝑀𝑉𝐴 , 𝑉𝑏 = 13.8 𝑘𝑉 , 𝐼𝑏 = = = 4183𝐴
3𝑉𝑏 3 ∗ 13.8
𝑆𝑏,𝑛𝑒𝑤
𝑋𝑔 = 𝑋𝑔,𝑜𝑙𝑑 ∗ = 0.635 𝑝𝑢
𝑆𝑏,𝑜𝑙𝑑
𝑉𝑏2
Reactor impedance , 𝑍𝑏 = , 𝑋𝑟𝑒𝑎𝑐𝑡𝑜𝑟 = 0.1313 𝑝𝑢
𝑆𝑏
Three – Phase fault
+ ve seq. impedance
𝑉𝑡ℎ
𝐼𝑓 =
𝑍𝑡ℎ
Vg X1g X
Vg X1g X Xsystem
Vg X1g X Vs
𝑰𝒇
Three – Phase fault
𝑉𝑡ℎ + ve seq. impedance
𝐼𝑓 =
𝑍𝑡ℎ
𝑅𝑎 𝑅𝑏 X1g X
𝑅1 =
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐 X1g X Xsys
X1g X
𝑋𝑒𝑞1 ∗ 𝑋
𝑋1 =
𝑰𝒇
𝑋𝑒𝑞1 + 𝑋 + 𝑋1𝑔 Vth
𝑍𝑡ℎ = ((𝑋1 +𝑋𝑠𝑦𝑠 )/ /(𝑋2 )) + 𝑋3 )
Xeq1
X1
𝑍𝑡ℎ = j0.2307 pu X2 Xsys
X3
X1g X
𝐼𝑓 = 4.332 𝑝𝑢
𝑰𝒇
Vth
𝐼𝑓 = 18127 𝐴
Sheet (3)
(b) Choose a current transformer ratio for differential relays to protect the generators. If the generator differential
relays have a minimum pickup of 0.14 A, how many times pickup does the three-phase fault provide?
𝑆g 21.875 ∗ 1000
𝐼𝑓.𝑙 = = = 915𝐴
3𝑉g 3 ∗ 13.8
CTR 1000/5
For secondary side of CT
5
𝐼𝑓 = 18127 ∗ = 90.635 𝐴
1000
𝐼𝑓
= 647.39
𝐼𝑜𝑚𝑖𝑛
Sheet (3)
(c ) Calculate a single-phase-to-ground fault at the terminals of one Of the generators.
𝑍𝑡ℎ,1 = 𝑍𝑡ℎ,2 = 𝑍𝑡ℎ = j0.2307 pu
𝑍𝑡ℎ,0 = 𝑋 + 3𝑍𝑜 = 𝑗0.1313 + 3 ∗ (11 + 𝑗4.73)
Zero seq. impedance
isolated
X1g X
isolated
X1g X Xsys
X1g X
𝑰𝒇 3Zo
Sheet (3)
(c ) Calculate a single-phase-to-ground fault at the terminals of one Of the generators.
𝑍𝑡ℎ,1 = 𝑍𝑡ℎ,2 = 𝑍𝑡ℎ = j0.2307 pu, 𝑍𝑡ℎ,0
1
𝐼𝑓 = 3 𝐼𝑜 = 3 ∗ = 0.083𝑝𝑢
2 ∗ j0.2307 + 𝑍𝑡ℎ,0
𝐼𝑓𝑎𝑐𝑡 = 347 A
❑ For secondary side of CT:
5 𝐼𝑓
𝐼𝑓 = 347 ∗ = 1.736 𝐴 > 𝐼𝑜𝑚𝑖𝑛 , = 12
1000 𝐼𝑜𝑚𝑖𝑛
The relay will trip
Equivalent Circuit
𝒁𝒕𝒉,𝟏 𝒁𝒕𝒉,𝟐 𝒁𝒕𝒉,𝒐
Vg 𝑰𝒐
Sheet (3)
2. The unit generator shown in figure 2 has the following capacitance to ground values in microfarads per phase:
Generator surge capacitors 0.25
Generator to transformer leads 0.004
Power transformer low voltage windings 0.03
Station service transformer high voltage windings 0.004
Voltage transformer windings 0.0005
The ground resistor R has a 64.14 kW rating at 138 volts.
▪ Determine the fault current magnitude for a single line to ground fault between the generator and the power
transformer.
▪ Determine the three phase fault current magnitude for a fault between the generator and the power transformer.
▪ Choose a CT ratio for the generator differential protection .
▪ Compare the fault currents of parts a and b with the generator relay pick up value of 0.15 amp.
▪ How much voltage is available to operate an overvoltage relay 59G when connected across the grounding resistor?
▪ What is the multiple of pick up if 59G minimum operating value is 5.4 volt?
▪ How much current flows through the resistor?
▪ Select a CT and suggested over-current pick up values for the 50/51 relay.
Sheet (3)
Sheet (3)
❑PU calculation:
Choose : 𝑆𝑏 = 160 𝑀𝑉𝐴 , 𝑉𝑏1 = 18 𝑘𝑉 , 𝑉𝑏2 = 345 𝑘𝑉,
For Generator : 𝑋1𝑔 = 0.21 𝑝𝑢
For MT : 𝑋1𝑡𝑟 = 0.15 𝑝𝑢
𝑆𝑏,𝑛𝑒𝑤 𝑉𝑏,𝑜𝑙𝑑 2 160
For System : 𝑋1,2 = 𝑋𝑜𝑙𝑑 ∗ ∗ = 0.04 ∗ = 0.064 𝑝𝑢
𝑆𝑏,𝑜𝑙𝑑 𝑉𝑏,𝑛𝑒𝑤 2 100
160
𝑋𝑜 = 0.12 ∗ = 192 𝑝𝑢
100
Sheet (3)
❑Equivalent circuit 3-ph fault
𝑋𝑡ℎ = (𝑋𝑔1 //𝑋𝐶 )// (𝑋𝑡𝑟 + 𝑋𝑠𝑦𝑠 )
X1g Xtr Xsys
𝐶𝑒𝑞 = 𝐶 = 0.2885 μ 𝐹
1 𝑋𝐶
𝑋𝐶 = = −𝑗11033 Ω Eg E
𝑗ω𝐶
For PU:
2
𝑉𝑏1
𝑍𝑏 = = 2.025 Ω
𝑆𝑏
𝑋𝐶 = − j5448 pu
Note: We can neglect Xc W.r.t Xg
Sheet (3)
X1g Xtr Xsys
❑Equivalent circuit 3-ph fault
𝑋𝐶
Eg E
𝑋𝑡ℎ = (𝑋𝑔1 )// (𝑋𝑡𝑟 + 𝑋𝑠𝑦𝑠 ) = 0.106 𝑝𝑢
𝑉𝑡ℎ
𝐼𝑓 = = 9.433 𝑝𝑢
𝑋𝑡ℎ X1g Xtr Xsys
𝑆𝑏 160 ∗ 1000
𝐼𝑏 = = = 5132 𝐴
3𝑉𝑏 3 ∗ 18
Vth
𝐼𝑓 = 48410 𝐴
Fault current represent Generator and system contribution
Sheet (3)
❑Equivalent circuit 3-ph fault
X1g Xtr Xsys
𝑋𝐶
Eg E
For Generator contribution
𝑋𝑡𝑟 + 𝑋𝑠𝑦𝑠
𝐼𝑓 = 48410 ∗
𝑋𝑡𝑟 + 𝑋𝑠𝑦𝑠 + 𝑋𝑔1
Sheet (3)
Xgo Xtro
❑Equivalent circuit L-G fault
𝑍𝑡ℎ,1 = 𝑍𝑡ℎ,2 = 𝑍𝑡ℎ 𝑋𝐶
Xsyso
3Rg
Zero sequence impedance:
For grounding resistance :
𝑉2
𝑃=
𝑅
𝑅 = 0.296 Ω 𝑍𝑡ℎ𝑜 = 3𝑅𝑔 + Xgo // -jXc
By referring to primary side of Distribution transformer:
2 ∘
18000 𝑍𝑡ℎ𝑜 = 2245∠ − 24.33 𝑝𝑢
𝑅𝑔 = 𝑅 ∗ = 1663Ω
240
1663
PU value : 𝑅𝑔 = = 821 𝑝𝑢
2.025
Sheet (3)
❑Equivalent circuit L-G fault
Equivalent Circuit
𝒁𝒕𝒉,𝟏 𝒁𝒕𝒉,𝟐 𝒁𝒕𝒉,𝒐
𝑍𝑡ℎ,1 = 𝑍𝑡ℎ,2 = 𝑍𝑡ℎ
∘ Vg 𝑰𝒐
𝑍𝑡ℎ𝑜 = 2245∠ − 24.33 𝑝𝑢
𝑉𝑡ℎ 3
𝐼𝑓 = 3 ∗ =
𝑍𝑡ℎ 𝑍𝑡ℎ,1 + 𝑍𝑡ℎ,2 + 𝑍𝑡ℎ𝑜
∘
𝐼𝑓 = 1.336 ∗ 10`3 ∠24.33 pu
∘
𝐼𝑓 = 6.858∠24.33 A
Sheet (3)
• Choose a CT ratio for the generator differential protection .
• Compare the fault currents of parts a and b with the generator relay pick up value of 0.15 amp.
𝑆g
𝐼𝑓.𝑙 = = 5132 𝐴
3𝑉g
CTR 6000/5
For secondary side of CT
5
Correct operation as 87
𝐼𝑓−3𝑝ℎ = 48410 ∗
6000
>> 0.15 A G is mainly for phase
fault not ground faults
5
𝐼𝑓−𝑆𝑙−𝐺 = 6.858 ∗
6000
<< 0.15 A
Sheet (3)
▪ How much voltage is available to operate an overvoltage relay 59G when connected across the grounding resistor?
▪ What is the multiple of pick up if 59G minimum operating value is 5.4 volt?
▪ How much current flows through the resistor?
Xgo Xtro
𝐼𝑓−𝑆𝑙−𝐺 = 6.858∠24.33 ∘ = 6.249+j2.82 A
Xsyso
𝐼𝑓𝑅 = 6.249 𝐴 3Rg 𝑋𝐶
18000
𝐼𝑓𝑅 = 6.249 ∗ = 468 𝐴
240
𝑉𝑅 = 𝐼𝑓𝑅 ∗ 𝑅 = 138.6 𝑉
This is the voltage that seen by 59G (Vomin=5.4 V)
The relay 59G will trip (138.6/5.4=25.5)
Sheet (3) 6.24 A
▪ How much current flows through the resistor?
▪ Select a CT and suggested over-current pick up values for the 50/51 relay.
𝐼𝑓𝑅 59 G
R
18000
𝐼𝑓𝑅50/51 = 6.249 ∗ = 468 𝐴
240 468.6
18000
𝐷𝑇 = =75 50/51 G
240
CTR = 75/1………… 375 /5 6.24 A
CTR: 400/5
1.5 < 𝐼𝑝𝑢 < 2 𝐼𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒 20% 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒
20% * 5A=1A, 𝑰𝒑𝒖 =2 A
