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Georgia Institute of Technology
School of Electrical and Computer Engineering
ECE6323 Power System Protection - Spring 2024
Homework Assignment # 4 Solution
Problem P1 (10 points): Consider an 800 A circuit, 25.0 kV, 2 miles long, as it is illustrated in Figure
P1. The equivalent source impedance (on a 100 MVA basis) is:
z 1 =z 2 =j 0. 15 pu, z 0 =j 0. 18 pu
The CT is rated 1200:5A. The impedance of the circuit is:
z 1 =z 2 =j 0. 70 ohms/mile, z 0 =j 2. 10 ohms/mile
A time overcurrent relay is located at the indicated location. Assume that the relay is a numerical relay
that has instantaneous overcurrent protection as well as time-overcurrent protection with the following
selections:
Moderately Inverse: d (^ ) d
r
t t
I
t 0. 114
(^0 0). 02
Very Inverse: d (^ ) d
r
t t
I
t 0. 491
Extremely Inverse: d (^ ) d
r
t t
I
t 0. 1217
Where tdis the time dial that can be set from 0.1 to 15.0 (any numerical value in this interval, it does not
have to be discrete values).
The relay settings are: pickup current=6A, time dial=1.0. Compute the time to trip for:
- A line to line fault at the middle of the distribution line.
- A single line to ground fault at the middle of the distribution line.
- A line to line fault at the end of the distribution line.
- A single line to ground fault at the end of the distribution line.
Solution:
100 / 3 MVA
2.3094 kA
25 / 3 kV
I base= =
6.25 ohms
base base base
V
Z
I
- A line-to-line fault at the middle of the distribution line
z 1 = z 2 = j 0.7ohms/mile =j 0.112.. /p u mile
z 0 = j 2.1ohms/mile =j 0.336.. /p u mile
1
I j p u j j j j
I 2 =j1.9084.. p u
I (^) a = 0 p.u. ; I (^) b = −3.3054 p.u. ; Ic=3.3054 p.u.
Convert to Ampere:
100 / 3 3.3054 7.6335 kA 25
I b = − pu = −
Current in secondary of CT:
7.6335kA / 240 = 31.8 A
r 5. PickUp
I
I
I
I
The resulting time to trip is:
0
0
0
1.6324 s (Moderately Inverse)
1.2149 s (Very Inverse)
1.1627 s (Extremely Inverse)
t
t
t
- Line to line fault at the end of the distribution line:
1
I j p u j j j j
I 2 =j1.3369.. p u
I (^) a = 0 p.u. ; Ib = −2.3156 p.u. ; Ic=2.3156 p.u.
Convert to Ampere:
2.3156 5.3476 kA 25
I b = − pu = −j
Current in secondary of CT:
5.3476 / 240 kA = 22.2817 A
r 3. PickUp
I
I
I
I
The resulting time to trip is:
0
0
0
2.0510 s (Moderately Inverse)
2.0241 s (Very Inverse)
2.3264 s (Extremely Inverse)
t
t
t
- A single line to ground fault at the end of the distribution line
0
0.6250 p.u. 0.15 0.15 0.18 0.224 0.224 0.
I j j j j j j j
I (^) F= 3 I 0 = −j1.8750 p.u.
Convert to Ampere:
1.8750 4.3301 kA 25
I (^) F= − pu = −
4.3301 kA / 240 = 18.0422 A
r 3. PickUp
I
I
I
I
The resulting time to trip is:
0
0
0
2.4272 s (Moderately Inverse)
2.9294 s (Very Inverse)
3.6282 s (Extremely Inverse)
t
t
t
Problem P2 (10 points) : Consider the 75 MVA, 230 kV /13.8 kV , z =j 0.118puthree-phase,
delta-wye connected variable tap step-up transformer indicated in Figure P2. This transformer is protected
with a differential relay. The delta side is the 13.8 kV side. The current transformer ratios arek and k 1 2
for the 230 kV and 13.8 kV sides, respectively.
Figure P 2
- Compute the transformation ratio of each of the three power transformers of Figure P2.
- Compute the operating current in the operating coil of the differential relay, assuming steady state,
3 - phase balanced operating conditions. The operating current should be provided as a function of the current transformer ratio on the 13.8 kV side, the current transformer ratio on the 230 kV side, and the current of phase A on the 13.8 kV side of the transformer.
k = =
Then, calculate the operating current:
I op = − kA = A
- When the transformer ratio is 230kV/14.49kV., the operating current in the differential relay is:
First, calculate the transformation ratio:
k = =
Then, calculate the operating current:
I op = − kA = A
Question 2:
During a specific fault, the distance relay at one of the line ends recorded the following values:
Determine (a) the type of fault and (b) the location of the fault.
Solution:
Question 1:
- The compensation factor is:
j j j m e j
For numerical relays use:
j m e
−
Zone 1:
83.70 83. 1
j j z e e
Relay setting for zone 1: 5.38 Ohms, 83.7 degrees, compensation factor = 2. 53 𝑒−𝑗^11.^4
° , 3 cycle delay.
Zone 2:
83.70 83. 2
j j z e e
Relay setting for zone 2: 8.96 Ohms, 83.7 degrees, compensation factor = 2. 53 𝑒−𝑗^11.^4
° , 12 cycle delay.
Zone 3:
83.70 83. 3
j j z e e
Relay setting for zone 3: 17.92 Ohms, 83.7 degrees, compensation factor = 2. 53 𝑒−𝑗^11.^4
° , 30 cycle delay.
Question 2:
The type of fault is Line-to-Ground fault in Phase C.
Location of the fault:
Calculate the zero-sequence current from phase currents:
0 0.^
j I = e kA
Calculate the fault impedance:
0
c^ j f c
V
Z e I mI
Calculate the fault distance:
1
j f j
Z (^) e l miles e z
The fault distance is 42. 85 miles from the distance relay of the measurement (Zone 2).
