EEE8082: Post-Exam Assignment on Electrical Power

This post-exam assignment covers topics such as CT connections, stabilizing resistor, fault current calculation, overcurrent relays, distance protection schemes, and more. It includes a detailed solution to a power system problem and explanations of permissive over-reaching, permissive-under reaching, and blocking communicating distance protection schemes. Course code EEE8082.

02.10.2018 EEE8082: Post-Exam Assignment
ELECTRICAL POWER
By
(Name)
(Course)
(Professor’s Name)
(Institution)
(State)
(Date)
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02.10.2018 EEE8082: Post-Exam Assignment
Post-Exam Assignment
Total marks = 30
Attempt all three parts.
PART 1
A busbar has four lines and its twelve 2000/5A CTs are connected in parallel with a
high impedance relay to provide earth fault protection.
The connection resistance from each group of three CTs to the relay is 0.3 Ω and the
winding resistance of each CT is 0.2 Ω. The relay has a setting voltage of 25 Volts and
the corresponding relay current is 25 mA. The excitation current of each CT at the
setting voltage is 80 mA.
i) Sketch the 3 phase connections of the CTs and the relay.
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02.10.2018 EEE8082: Post-Exam Assignment
ii) Calculate the value of a suitable stabilising resistor to ensure stability for a
through fault of 24000 A.
RS= Us
Is = 25
0.025 =1 K Ω
iii) On the sketch drawn, using the convention P1, P2 and S1, S2 indicate CT
polarities and show the secondary current flows for a single phase fault on the
busbar. Assume that all circuits contribute to the fault.
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02.10.2018 EEE8082: Post-Exam Assignment
iv) Calculate the minimum value of the fault current to cause the relay to operate
If =Us
Zf = 25
0.3+0.2+1000 =24.65 mA
v) v) How can the relay be made less sensitive?
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02.10.2018 EEE8082: Post-Exam Assignment
PART 2
Figure 1
For the network shown in Figure 1, the fault current for a phase-to-phase fault at Bus A
is 5000 A.
For the 33 kV, 1000 MVA fault level source, assume that the source positive and
negative sequence impedances are equal.
For the 15 MVA, 33/11 kV transformer, assume that the positive and negative
sequence impedances are equal.
Circuit impedance between Bus A and Bus B: Z1= 0.25 +j1.0 Ω
Circuit impedance between Bus B and Bus C: Z1= 0.2 +j2.5 Ω
5
R2
!
R3
!
R4
!
R1
!
A700
Load
380A
Load
130A
No other
protection
devices
Load
A115
300/5 200/5
100/5
600/5
Z1= 0.25 +j1.0 Ω Z1= 0.2 +j2.5 Ω Load
kV33
1000
MVA
R0
!
15 MVA
33/11 kV Bus A Bus B Bus C

02.10.2018 EEE8082: Post-Exam Assignment
Relay R0 is an instantaneous over current relay, determine the CT ratio of the CT
connected to relay R0 and appropriate settings for this relay.
Relays R1, R2, R3 and R4 are overcurrent relays with an IEC Standard Inverse
characteristic with a 5A nominal current rating.
Bearing in mind that overcurrent coordination should be determined at the maximum
fault current, determine appropriate current and time multiplier settings for these
relays.
The coordination margin between each relay should be a minimum of 0.25 seconds.
Provide current settings in multiples of relay nominal current.
Relay R4 should operate in 0.15 seconds for a close in fault.
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02.10.2018 EEE8082: Post-Exam Assignment
Solution:
Figure 1: Equivalent circuit of the power system
Let MVA-base be 25MVA and the voltage base at the 6.6 KV bus be 6.6 KV.
X newpu=
Xoldpu∗( KVold
KVnew )
2
∗Snew
Sold
Zpu = Zact
Zbase
Zbase= KV base
2
Sbase
X d 1 pu= j 0.15∗25
10 = j0.375 pu
X d 1 pu= j 0.125∗25
10 = j0.3125 pu
Xtx 1 pu = j 0.1∗25
10 = j 0.25 pu
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02.10.2018 EEE8082: Post-Exam Assignment
Xtx 2= j 0.08∗25
5 = j 0.4 pu
Z1 =?
Zbase=332
25 =43.56 Ω
Z1 =30∗( 0.27+ j0.36 )
43.56 = 45
242 + j 30
121 pu
Z2 =?
Zbase=6.62
25 =1.7424 Ω
Z2 =3∗( 0.135+ j0.08 )
1.7424 = 225
968 + j 50
363 pu
Pre-fault voltage at bus 6.6 KV:
V f = 6.6
6.6 =1 pu
X eq= ( X1 /¿ X2+ Xtx 1 + X tx 2 +Z1 )¿ /Z2
X1 /¿ X2 = j 0.375∗ j0.3125
j0.375+ j 0.3125 = j 15
88 pu
( X 1 /¿ X2+ Xtx 1 + Xtx 2 +Z1 ) = j 15
88 + j0.25+ j 0.4+ 45
242 + j 30
121 = 45
242 + j 5171
4840 pu
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02.10.2018 EEE8082: Post-Exam Assignment
X eq= ( 45
242 + j 5171
4840 )∗( 225
968 + j 50
363 )
( 45
242 + j 5171
4840 )+( 225
968 + j 50
363 )=0.229511<39.9083 pu
I f = V f
Xeq
= 1
0.229511< 39.9083 =4.357082←39.9083 pu
I 1 pu= ( 225
968 + j 50
363 )∗( 4.357082←39.9083 )
225
968 + j 50
363 + 45
242 + j 5171
4840
=0.922126←80.1267 pu
I base= MVAbase
√ 3∗KV base
= 25
√ 3∗11 =1.312159 kA
I1=I1 pu∗I base= ( 0.922126←80.1267 )∗1.312159 kA=1.20997593← 80.1267 kA
V 11 KV =V f −I f ∗( X tx 1+ Xtx 2 +Z1 )
V 11 KVpu=1− ( 0.922126←80.1267 )∗( j 0.25+ j 0.4+ 45
242 + j 30
121 )=0.157181<9.87329 pu
V 11 KV =V 11 KVpu∗V base=11∗0.157181< 9.87329=1.728991<9.87329 kV
PART 3
a) Describe permissive over-reaching, permissive-under reaching and blocking
communicating distance protection schemes.
Permissive signal must be detected from the remote end for the communication
aided trip. Absence of communication channel disables the accelerated tripping.
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02.10.2018 EEE8082: Post-Exam Assignment
On the hand Similar to POTT but permissive signal sent by under-reaching Z1
elements. At the receiving end, Z2 elements qualify the permissive signal. No
problems with current reversal since Z1 doesn’t overreach.
b) What would be the effect on a communication failure in each of these schemes?
Permissive overreaching signal:
1. Desensitization due to infeed.
2. Current reversal.
Permissive-under reaching
1. Coordinating time at fault inception is difficult.
2. External fault clearing failure.
3. Stop preference over start.
c) How can mal-operation of distance protection due to power swings be
prevented?
To guarantee steadfastness, most of the period shedding of the load is executed
regardless of the nature of the power swipe. On the other hand, in the course of
steady power swing, relay process ought to be blocked and in the course of
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02.10.2018 EEE8082: Post-Exam Assignment
unbalanced power swing relay nimble action ought to be activated. The
processes are stress-free to execute for uneven faults, as the negative and zero
series constituents do not occur in the course of power swing, which can be
utilized as fault uncovering criterion. Nevertheless, it is much more problematic
to detect balanced fault in the course of stable power swing, which may
interrupt the operation of relay.
d) Explain the effects of intermediate current in feed and out feed on the reach of a
distance protection relay.
The most important ruling of the in feed influence inquiry distresses the fact
that in feed can lead to important over-reach phenomena, in addition of the
well-known under-reach phenomena. Moreover, it is perceived that the strength
and form of the in feed effect differ in a definite way, reliant on the circulation
feeder conductors’ size, the DG link point and the DG penetration level. One
more critical verdict concerns the collective influence of fault resistance and K0
in the course of ground faults in above radial circulation feeders, which can lead
to concentrated overreach phenomena, even short of DG or reverse currents
feeding the fault. What is more, line load affects the impedance determination
of a distance relay significantly, mainly concerning the concurrent effect of
fault resistance.
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02.10.2018 EEE8082: Post-Exam Assignment
e) Describe what is meant by unit and non-unit protection.
Unit guard are capable of detecting and responding to error taking place only
contained within its zone protection. Have absolute discrimination. Its zone of guard
is defined. While in contrast non unit shield consist of the following schemes: time
graded over current shield, current corrupted over current shielded and distance or
impedance shield.
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02.10.2018 EEE8082: Post-Exam Assignment
Reference
Conference Board of Canada. (2004). Electricity restructuring: opening power
markets. Ottawa, Conference Board of Canada.
Grigsby, L. L. (2012). Power system stability and control. Boca Raton, Taylor &
Francis. Available from; http://www.crcnetbase.com/isbn/9781439883204. Date of
Access; 8th September, 2018
Musirin, I., & Sulaiman, S. I. (2015). Recent trends in power engineering: selected,
peer reviewed papers from the 2015 9th International Power Engineering and
Optimization Conference (PEOCO 2015), March 18-19, 2015, Melaka, Malaysia.
Pfaffikon, Trans Tech Publications. Available from;
http://search.ebscohost.com/login.aspx?
direct=true&scope=site&db=nlebk&db=nlabk&AN=1060570. Date of Access; 8th
September, 2018
Nagsarkar, T. K., & Sukhija, M. S. (2016). Power System Analysis: Power System
Analysis. New Delhi, Oxford University Press India. Available from;
http://app.knovel.com/hotlink/toc/id:kpPSAE0003/power-system-analysis. Date of
Access; 14th October, 2018
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02.10.2018 EEE8082: Post-Exam Assignment
World Bank. (2013). Vietnam Power Sector Generation Options. DC, Washington.
Available from; http://hdl.handle.net/10986/12860. Date of Access: 14th October 2018
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