Tanta University Electrical Power and Machines Department
Faculty of Engineering
Fourth Year- Second Term
2020-2021
Power System Protection
Section #1
Sheet #1
Eng. Eatmad Nahas
Differential Protection
1- Principle of operation
2- Characteristics
3- Errors
4- Stability and Sensitivity
5- How to increase stability
Differential Protection
1- Principle of operation
a- Healthy
Any Equipment I2
I1
G
M
XFMR
B.B
i1 i2
T.L
R IR=i2 - i1
Differential Protection
1- Principle of operation
b- Internal fault
Any Equipment I2
I1
G
M
XFMR
B.B
i1 i2
T.L
R IR=i2 + i1
Differential Protection
1- Principle of operation
c- External fault
Any Equipment I2
I1
G
M
XFMR
B.B
i1 i2
T.L
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
Any Equipment
G
M Iomin
XFMR IPu
B.B ISet
T.L No operation
R Status
Differential Protection
3- Errors (That is why Iomin not equal to zero)
CT errors
Leads
Relay constrains
Differential Protection
4- Stability and Sensitivity
Stability : Max throughout fault current at which the relay will not operate
Sensitivity : Min internal fault current at which the relay will operate
Any Equipment
G
M
XFMR
B.B
T.L
R
Differential Protection
4- Stability and Sensitivity
RCT Rlead Rlead RCT
I1/N1 Xm Relay Xm
I2/N2
Differential Protection
4- Stability and Sensitivity RCT Rlead Rlead RCT
To increase stability: I1/N1 Xm Relay Xm I2/N2
1- Increase Iomin
2- Stabilizing resistance
3- Biasing C/Cs
Differential Protection
4- Stability and Sensitivity
RCT Rlead Rlead RCT
To increase stability:
2- Stabilizing resistance I1/N1 Xm Relay Xm I2/N2
Stability calculation
Differential Protection
Stability calculation
1- External fault occur and one of the RCT Rlead Rlead RCT
CTs is fully saturated and the other CT
is ideal Rs
𝐼𝐹𝑒𝑥𝑡 IFext/N1 Xm
= 𝐼𝑜𝑚𝑖𝑛 + 𝐼𝑚
𝑁1 VR Relay Iomin I2/N2
Vomin
𝑉𝑜𝑚𝑖𝑛 = 𝐼𝑜𝑚𝑖𝑛 ∗ 𝑅𝑟𝑒𝑙𝑎𝑦 Im
𝑉𝑅 = 𝐼𝑜𝑚𝑖𝑛 ∗ 𝑅𝑟 + 𝑅𝑠
(𝐼𝐹𝑒𝑥𝑡. 𝑁1)∗ 𝑅𝑙 +𝑅𝐶𝑇 −𝑉𝑜𝑚𝑖𝑛
𝑅𝑠 =
𝐼𝑜𝑚𝑖𝑛
Differential Protection
Sensitivity calculation
1- An internal fault occur and using the RCT Rlead Rlead RCT
worst condition for energization
(Feed from one end) Rs
Xm Xm
IFint/N1 VR Relay Iomin
Apply K.C.L at node A: Vomin I2/N2
Im1 Im2
𝐼𝐹𝑖𝑛𝑡
= 𝐼𝑜𝑚𝑖𝑛 + 𝐼𝑚1 + 𝐼𝑚2 A
𝑁
If Sensitivity is less than the required, It must be either replace the relay
or using the biased characteristics
Differential Protection
Biased Relay
or
High impedance relay
or
IR
Percentage differential Relay
Trip
Iomin
No operation
I Fext.
Differential Protection
Biased Relay
or
High impedance relay
𝑰𝒐 = 𝒊𝟏 − 𝒊𝟐 or
Percentage differential Relay
Trip
Setting are:
Iomin
Iomin
No operation
𝑰𝑹=
𝒊𝟏 + 𝒊𝟐 Slope
𝟐
Differential Protection
Biased Relay or High impedance relay or Percentage differential Relay
𝒊𝟏 + 𝒊𝟐 𝑰𝒐 = 𝒊𝟏 − 𝒊𝟐
𝑰𝑹=
𝟐
Trip
𝑰𝒐 = 𝒊𝟏 − 𝒊𝟐
Iomin
𝒊𝟏 + 𝒊𝟐
No operation 𝑰𝑹=
𝟐
Setting are:
𝐹𝑜𝛼 𝐼𝑜2 𝐹𝑟𝛼 𝐼𝑟2
𝐹𝑠𝛼 𝐾𝑠
Iomin
Slope
Sheet (1)
1- A generator, having normal current = 100 A is to be protected by current differential relays using stabilizing
resistor. The through fault required is up to 15 times full load current; assuming one end of the CT gets completely
saturated, CT ratio = 100/1. CT resistance = 0.046 ohm, and total lead burden = 1 ohm, determine the required
stabilizing resistor and the stability ratio. Given also the relay pickup current = 0.1 A, relay operating voltage = 12
volts, and the magnetizing current of CT = 200 mA for induced voltage of 15.6 volt.
𝑉𝑅 − 𝑉𝑜𝑚𝑖𝑛
𝑅𝑠 =
𝐼𝑜𝑚𝑖𝑛
𝐼𝐹𝑒𝑥𝑡
𝑉𝑅 = 𝑅𝑙 + 𝑅𝐶𝑇 = 15.6 𝑉
𝑁
𝑅𝑠 =36.9 Ω
𝐼𝐹𝑖𝑛𝑡
= 𝐼𝑜𝑚𝑖𝑛 + 𝐼𝑚1 + 𝐼𝑚2
𝑁
𝐼𝐹𝑖𝑛𝑡 = 50𝐴
𝐼𝐹𝑒𝑥𝑡 1500
𝑆𝑡𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑅𝑎𝑡𝑖𝑜 = = = 30.
𝐼𝐹𝑖𝑛𝑡 50
Sheet (1)
2- A differential protection circulating current is used to protect a generator having a full-load current of 600 A, CT
ratio 2000/5 A and the distance between the CTs at opposite ends of the machine is 200 yds. Under straight through
fault condition of 15 times full-load the CTs at one end have a voltage of 80% of that of other end. The relay having
an impedance of 100 ohm is connected across the physical midpoint of the pilots. Determine (1) at what distance
from the physical midpoint will zero voltage be located. (2) At what current the relay will have to be set to give a
stability factor of 3. Magnetizing current of CT = 100 mA.
𝑉 𝑋 V
= ∴ 𝑋 = 111 𝑦𝑑𝑠
0.8 𝑉 200−𝑋
𝐼𝐹𝑒𝑥𝑡 15∗600
𝑆𝑡𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑅𝑎𝑡𝑖𝑜 = = =3 ∴ 𝐼𝐹𝑖𝑛𝑡 = 3000 𝐴
𝐼𝐹𝑖𝑛𝑡 𝐼𝐹𝑖𝑛𝑡
𝐼𝐹𝑖𝑛𝑡 Relay
= 𝐼𝑜𝑚𝑖𝑛 + 𝐼𝑚1 + 𝐼𝑚2
𝑁
0.8V
𝐿𝑒𝑡 𝐼𝑚 = 100 𝑚𝐴
200- X X
∴ 𝐼𝑜𝑚𝑖𝑛 = 7.3 𝐴
Sheet (1)
3- In differential relay using an electromagnetic comparator, the relay pickup current is 0.4 A. When a through fault
current of 10 times full load passes through the generator winding, the secondary currents at the two ends of the CTs
are 50 A and 47.5 A respectively. Assuming the slope of the differential relay to be 5% , determine whether through
fault stability will be achieved. In case this is not achieved what modification will have to be done in the relay so as
to achieve stability, keeping the relay pickup current to be the same.
𝐼𝑜 = 𝑖1 − 𝑖2 𝑰𝒐 = 𝒊𝟏 − 𝒊𝟐
slope =5%
𝐼𝑜 = 50 − 47.5 = 2.5 𝐴
𝑖1 + 𝑖2 Trip
𝐼𝑅 =
2 Iomin
𝐼𝑅 = 48.75 𝐴 No operation
𝒊𝟏 + 𝒊𝟐
Draw the C/Cs firstly by, 𝑰𝑹=
𝟐
Iomin= 0.4 A
Slope =5 % 𝐼𝑜/𝐼𝑅 = 0.05
Sheet (1)
3- In differential relay using an electromagnetic comparator, the relay pickup current is 0.4 A. When a through fault
current of 10 times full load passes through the generator winding, the secondary currents at the two ends of the CTs
are 50 A and 47.5 A respectively. Assuming the slope of the differential relay to be 5% , determine whether through
fault stability will be achieved. In case this is not achieved what modification will have to be done in the relay so as
to achieve stability, keeping the relay pickup current to be the same.
𝐼𝑜 = 𝑖1 − 𝑖2 𝑰𝒐 = 𝒊𝟏 − 𝒊𝟐
slope =5%
𝐼𝑜 = 50 − 47.5 = 2.5 𝐴
𝑖1 + 𝑖2 Trip
𝐼𝑅 =
2 Iomin
𝐼𝑅 = 48.75 𝐴 No operation
Modification: 𝒊𝟏 + 𝒊𝟐
𝑰𝑹=
2.5 𝟐
𝑆𝑙𝑜𝑝𝑒 = ( ) ∗ 100 = 5.16%
48.75
𝑁𝑒𝑤 𝑠𝑙𝑜𝑝𝑒 > 5.16 %
Sheet (1)
4- Consider the power system shown in Figure 1 which represents a unit-connected generator prior to being
synchronized to the system and protected with an overcurrent relay connected as a differential relay. Determine the
full load current, select a CT ratio for the generator differential, calculate the relay operating currents for a three-
phase fault at F1 and F2 and set the relay. Assume there is no CT error and the relay has the CO-11 time–current
characteristics.
Sheet (1)
4- Consider the power system shown in Figure 1 which represents a unit-connected generator prior to being
synchronized to the system and protected with an overcurrent relay connected as a differential relay. Determine the
full load current, select a CT ratio for the generator differential, calculate the relay operating currents for a three-
phase fault at F1 and F2 and set the relay. Assume there is no CT error and the relay has the CO-11 time–current
characteristics.
Full load current
𝑆 975 ∗ 1000
𝐼𝑓.𝑙 = = = 25587 𝐴
3𝑉 3 ∗ 22
CT Ratio 30000/5
Three phase Fault
𝑉𝑡ℎ 1
𝐼𝐹 = = = 4.762 𝑝𝑢
𝑍𝑡ℎ 0.21
𝐼𝐹 =4.762*2557=121843 A
For F1:
𝐼1 = 𝐼𝐹 , 𝐼2 = 0
5
𝐼𝑅𝑒𝑙𝑎𝑦 = 𝑖1 − 𝑖2 = 121843 ∗ = 20.3 𝐴
30000
∴ The relay will operate
Sheet (1)
4- Consider the power system shown in Figure 1 which represents a unit-connected generator prior to being
synchronized to the system and protected with an overcurrent relay connected as a differential relay. Determine the
full load current, select a CT ratio for the generator differential, calculate the relay operating currents for a three-
phase fault at F1 and F2 and set the relay. Assume there is no CT error and the relay has the CO-11 time–current
characteristics.
For F2:
𝐼1 = 𝐼2 = 𝐼𝐹 ,
𝐼𝑅𝑒𝑙𝑎𝑦 = 𝑖1 − 𝑖2 = 0 𝐴
∴ The relay will not operate
Sheet (1)
5- Repeat problem 4 assuming that the line-side CT has an error of (a) 1%, (b) 5% and (c) 10% of its secondary
current. Set the overcurrent relay so it will not operate for an external fault.
6- Repeat problem 4 for a phase-to-phase fault at F1.
Sheet (1)
5- Repeat problem 4 assuming that the line-side CT has an error of (a) 1%, (b) 5% and (c) 10% of its secondary
current. Set the overcurrent relay so it will not operate for an external fault.
For F2:
𝐼1 = 𝐼2 = 𝐼𝐹 ,
𝑖1 = 20.3 𝐴
𝑖2 = 20.3 ∗ 0.99 = 20.097 𝐴
𝐼𝑅𝑒𝑙𝑎𝑦 = 𝑖1 − 𝑖2 = 0.203 𝐴
∴ The relay will operate
Solution:
1- Stabilizing resistance
2- Biased Relay
Sheet (1)
5- Figure 2 shows a percentage differential relay applied for the protection of a generator winding. The relay has a
0.1 A minimum pickup and a 10% slope. A high-resistance ground fault has occurred as shown near the grounded-
neutral end of the generator winding while it is carrying load with the currents flowing at each end of the generator
as shown. Assume that the CT ratios are as shown in the figure and they have no error. Will the relay operate to trip
the generator under this condition? Would the relay operate if the generator were carrying no load with its breaker
open? Draw the relay operating characteristic and the points that represent the operating and restraining currents in
the relay for the two conditions.
𝑫𝒖𝒓𝒊𝒏𝒈 𝒍𝒐𝒂𝒅 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏
𝐼1 = 300 𝐴, 𝐼2 = 280 𝐴
𝑖1 = 300 ∗ 5/400 = 3.75 𝐴
𝐼2 = 280 ∗ 5/400 = 3.5 𝐴
𝐼𝑜 = 𝑖1 − 𝑖2 = 0.25 𝐴
𝐼𝑅 = (𝑖1 + 𝑖2)/2 = 3.625 𝐴
𝑫𝒖𝒓𝒊𝒏𝒈 𝒏𝒐 𝒍𝒐𝒂𝒅 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏
𝑰𝒐 = 𝒊𝟏 − 𝒊𝟐
𝐼1 = 20 𝐴, 𝐼2 = 0 𝐴 slope =10%
𝑖1 = 20 ∗ 5/400 = 0.25 𝐴
Trip
𝐼2 = 0 𝐴
Iomin = 0.1 A
No operation
𝐼𝑜 = 𝑖1 − 𝑖2 = 0.25 𝐴
𝐼𝑅 = (𝑖1 + 𝑖2)/2 = 0.125 𝐴 𝒊𝟏 + 𝒊𝟐
𝑰𝑹=
𝟐
