POLYTECHNIC OF NAMIBIA
SCHOOL OF ENGINEERING
http://www.polytechnic.edu.na
DEPARTMENT OF ELECTRICAL ENGINEERING
http://www.polytechnic.edu.na/EEE
ETP820S – ELECTRICAL PROTECTION 425
‘HOME’ ASSIGNMENT 2: MEMO
DATE GIVEN: 20/02/2013
DATE DUE: 25/02/2013
TOTAL MARKS: 90
QUESTION ONE
(25 MARKS)
Transformer Differential Protection: Selection and Connection of CTs
Consider a 30 MVA, 11.5/69 kV, Yd1 transformer as shown in the single-line diagram in Figure
Q1.1. Determine the transformation ratio and connections of the CTs required in order to set the
differential relays. CTs with ratios in steps of 50/5 up to 250/5, and in steps of 100/5 thereafter,
should be used. Use relays with a variable-percentage characteristic. The available current taps are:
5.0-5.0, 5.0-5.5, 5.0-6.0, 5.0-6.6, 5.0-7.3, 5.0-8.0, 5.0-9.0, and 5.0-10.0 A. Give the complete
schematic of the three-phase connections.
Figure Q1.1. Single-line diagram for Question 1.
Solutions:
Figure Q3.1 shows the complete schematic of the three-phase connections. The currents in the
windings and in the lines are drawn and show that the restraint currents on the star and delta sides
of the relay are in phase.
For a through-put of 30 MVA the load currents are
I load (11.5 kV )
30 106
3 11.5 103
1506.13 A … (2.0)
I sec. (69 kV )
30 106
3 69 103
251.0 A … (2.0)
In order to increase the sensitivity, CT ratios are normally chosen very close to the rated currents.
Therefore, at 11.5 kV the CT ratio selected is 1500/5 and at 69 kV the CT ratio selected is 250/5.
… (3.0)
1
The above ratios should be checked to see if they are compatible with the taps that are available on
the relay. … (1.0)
With the two ratios selected above, the currents in the secondary windings of the CTs for nominal
conditions are:
ICT-Pri. (11.5 kV) = 1506.13 × (5/1500) = 5.02 A …(2.0) ICT-Sec.(69 kV) = 251 × (5/250) = 5.02 A…(2.0)
Figure Q3.2(a) shows the transformer windings. In order to eliminate the phase shift from the
primary to the secondary, the CTs on the delta side of the transformer are connected in star and
those on the star side in delta. …(2.0)
Figure Q3.2(b) shows the inside of the three-phase differential relay of Figure Q3.2(a). The complete
winding and tap arrangements for only one of the three phases has been shown.
The available CT ratios are as follows (see description in the problem):
2
50/5
100/5
600/5
1100/5
1600/5
60/5
200/5
700/5
1200/5
1700/5
70/5
300/5
800/5
1300/5
1800/5
80/5
400/5
900/5
1400/5
1900/5
90/5
500/5
1000/5
1500/5
2000/5
(NB: the correct connections and flow of currents should be given)
Figure Q3.2(a). Three-phase connection diagram for the one-line network of Figure Q3.1.
Available relay current taps
1
5.0 – 5.0 A
(see description in the problem):
2
5.0 – 5.5 A
3
5.0 – 6.0 A
4
5.0 – 6.0 A
5
5.0 – 7.3 A
6
5.0 – 8.0 A
7
5.0 – 9.0 A
8
5.0 – 10.0 A
Figure Q3.2(b). Inside the differential relay of Figure Q3.2(a), showing the windings and taps for one
phase only. The other two phases are configured similarly.
3
Transformer and relay currents:
I Ra
Ia
CT
Ic
CT
5.02 0
CT
5.02
240
5.02
j 0.0 ( 2.51
j 4.35) 7.53
j 4.35 8.696
30 A
… (0.5)
IA
L
IA
ph
144.9 0
I Ra
ph
144.9
251
YCT
IC
30
240
5
250
144.9
5.02
j 0.0 ( 72.45
j125.5)
217.35
j125.5 251
5.0
5.0
8.696
30 A
30 A
30 A …(0.5)
The chosen relay taps are as follows:
Delta CT side: 5.0 – 5.0 A
… (0.5)
Star CT side: 5.0 – 9.0 A … (0.5)
I1
I2
8.696
I Ra
YCT
30
9
5
5.02
30
9.036
30 A
Therefore, the differential current for the chosen CT ratios and relay taps is
I diff
I2
9.036
0.2954
I1
30 8.696
30
j 0.168 0.33983
7.8254 j 4.518 (7.53
29.6 A 0.34 -30 A
j 4.35)
In order for the relay not to operate on an external fault or on load current, the slope of the relay
characteristic, K, must be chosen such that I diff
I1 I 2
K or I diff
2
KI rest .
4
QUESTION TWO
(22 MARKS)
Inverse, Time-Delay Over-current Relays: Setting Rules
Referring to Figure Q2.1, determine the CT ratio, pickup and time dial settings for the relays at
breakers 2 and 1 (Rb and Rab, respectively) assuming that no coordination of relay 2 with any other
relay is required. Assume that the maximum load is 95 A. The minimum and maximum faults are
shown in Figure Q2.1. Use Table Q2.1 to select your CT ratios. Assume relays with the following
taps 4.0, 6.0, 8.0, 10.0 and 12.0
Figure Q2.1.
Table Q2.1. Standard current transformer multi-ratios
(MR represents multi-ratio CTs)
600 : 5 MR 1200 : 5 MR 2000 : 5 MR 3000 : 5 MR
50 : 5
100 : 5
300 : 5
300 : 5
100 : 5
200 : 5
400 : 5
500 : 5
150 : 5
300 : 5
500 : 5
800 : 5
200 : 5
400 : 5
800 : 5
1000 : 5
250 : 5
500 : 5
1100 : 5
1200 : 5
300 : 5
600 : 5
1200 : 5
1500 : 5
400 : 5
800 : 5
1500 : 5
2000 : 5
450 : 5
900 : 5
1600 : 5
2200 : 5
500 : 5
1000 : 5
2000 : 5
2500 : 5
600 : 5
1200 : 5
3000 : 5
5
Solutions:
Settings for Relay Rb:
Select a CT ratio to give 5.0 A secondary current for maximum load, i.e. 95/5 = 19 : 1.
(2.0)
Since this is not a standard CT ratio, from Table Q2.1, the nearest CT ratio of 20 : 1 or 100 : 5 is
selected. (2.0)
Select the 10.0 A relay tap, giving a primary current relay pick-up of 10.0 × 20 = 200 A. (2.0)
For this configuration no coordination is required, so one can set the time delay at the lowest dial
setting (fastest time) of ½. (See Figure Q2.2). (2.0)
Settings for Relay Rab:
The relay protecting the next line segment closest to the source, Rab, must protect its own line and, if
possible, back up the relay protecting the next line. (2.0)
The pick-up should therefore be for the same primary current as the down-stream relay and the time
setting must coordinate with it. (2.0)
This time setting should be made with maximum current conditions, i.e. assuming a three-phase
fault, with maximum generation behind the relay. (1.0)
In a radial system, all relays for which co-ordination is required must be examined for operation at
this same primary current. When co-ordination is achieved at maximum current, the shape of the
inverse curves, provided they are all of the same family of inverseness, will ensure co-ordination at
all lesser current values.
The pick-up setting of Rb is 10.0 A and the time dial setting is ½.
Theoretically, to ensure that Rab backs up Rb it should be set for the same pick-up current, i.e. it sees
the same faults, but is set at a slower (higher) time dial. (2.0)
The operating time of Rb is determined from Figure Q2.2 at the maximum fault current at bus B
(1500 A) divided by its pick-up setting (20 × 10 A = 200 A) or 1500/200 = 7.5 × pu and the ½ time
dial. (2.0)
This is seen to be 0.25 s. Add 0.3 s co-ordinating time and Rab operating time should be 0.55 s.
(3.0)
If reference is made again to Figure Q2.2, at the same maximum fault current of 1500 A at bus B
and pu of 20 × 10 A, the multiple of the tap setting is 7.5 × pu and the operating time is 0.55 s.
Interpolating between the time dial setting curves of 1 and 2 gives a setting of 1.5. (The actual time
dial should be determined by test.) (2.0)
Summary:
Relay
Ra
Rab
CT Ratio
100:5
100:5
Relay Tap Setting (TS)
10.0
10.0
Time-dial Setting (TDS)
½
1½
6
Figure Q2.2. Time-delay over-current relay operating characteristic
7
QUESTION THREE
(28 MARKS)
Unit Protection of a Transformer Feeder
Figure Q3.1 shows unit protection applied to a transformer feeder. The feeder is assumed to be a
100 m length of cable, such as might be found in some industrial plants or where a short distance
separates the 33 kV and 11 kV substations. While 11 kV cable capacitance will exist, it can be
regarded as negligible for the purposes of this problem.
Figure Q3.1. Unit protection of a transformer feeder
The delta/star transformer connection requires phase shift correction of CT secondary currents
across the transformer, and in this case software equivalents of interposing CTs are used.
(a) You are required to replace the software equivalents of the inter-posing CTs (ICTs) with actual
ICTs and to replace the digital communication channel with pilot telephone wires and a three-phase
differential relay. Determine the ratio correction factors (or ICT turns ratios) to achieve balance in the
differential relay and indicate the vector groups of the ICTs required on both sides of the differential
relay.
(b) Draw the three-phase equivalent of the installation.
Solution:
(a) Since the LV side quantities lag the HV side quantities by 30° (from the transformer vector
group), it is necessary to correct this phase shift by using ICT settings that produce a 30° phase
shift. (2.0)
There are two obvious possibilities:
(i) HV side: Yd1 and LV side: Yy0 and
(ii) HV side: Yy0 and LV side: Yd11 (2.0)
Only the second combination is satisfactory, since only this one provides the necessary zerosequence current trap to avoid mal-operation of the protection scheme for earth-faults on the LV
side of the transformer outside the protected zone. (4.0)
8
Ratio correction must also be applied, in order to ensure that the relays see currents from the
primary and secondary sides of the transformer feeder that are well balanced under full-load
conditions. This is not always inherently the case, due to selection of the main CT ratios.
The transformer (inverse) turns ratio at nominal tap is 11/33 = 0.3333 (it is a step-down transformer).
Required (inverse) turns ratio according to the CT ratios used =
400
1
1250
1
0.32
The currents are as follows:
High-voltage side:
Low-voltage side:
Rated primary line current:
Rated secondary line current:
20 106
3 11 103
6
20 10
3 33 103
Ip
350 0 A (2.0)
CT secondary current:
I CT
p
1
350 0.875 A (1.0)
400
Is
1050
30 A (2.0)
CT secondary current:
I CT
s
1
1050 0.84 A (1.0)
1250
Figure Q3.1 is re-drawn in Figure Q3.2 to show all the currents associated with the installation.
Figure Q3.2. Unit protection of a transformer feeder with current flows.
The spill current that will arise due to the incompatibility of the CT ratios used with the power
transformer turns ratio may cause relay mal-operation. This has to be eliminated by using the ICTs
connected between the line CTs and the differential relay.
For this particular relay, the correction factors are chosen such that the full-load current seen by the
relay is 1A. (2.0)
The appropriate correction factors are: HV: 400/350 = 1.14 and LV: 1250/1050 = 1.19 (2.0)
9
The ICT turns ratios are
N1
N
= 1.14 on the HV side and 1 = 1.19 3 on the LV side.
N2
N2
i.e. On HV side compensation factor = 1.14 and on the LV side compensation factor = 1.19.
(2.0)
NB:
With the above compensation factors, the relay currents are:
(i) HV side: IR = ICT-sec. × comp. Factor = 0.875 A × 1.14 = 0.9975 A
(ii) LV side: IR = ICT-sec. × comp. Factor = 0.84 A × 1.19 = 0.9996 A
(iii) Therefore, current mis-match (differential current) = 0.9996 – 0.9975 = 0.0021 A
(iv) In both cases
I CT sec . ( ICTPr i. )
I R ( ICTsec . )
N2
N1
(b)
(a) Main transformer and CT connections
(b) ICT, Relay and Pilot Wire connections.
Figure Q3.3. Three-phase schematic connection diagram of the one-line diagram of Figure Q3.2.
10
QUESTION FOUR
(15 MARKS)
GENERATOR STATOR WINDING PROTECTION AGAINST SHORT-CIRCUITS
It is the standardised practice of manufacturers to recommend differential protection for generators
rated l000 kVA or higher and most of such generators are protected by differential relays. Above
10,000 kVA, it is almost universally the practice to use differential relays. Percentage-differential
relaying is the best for the purpose and it should be used wherever it can be justified economically.
(a) Draw the arrangement of Current-Transformers (CTs) and Percentage-Differential relays for
the protection of a Delta-connected Synchronous Generator. Show the location of the circuitbreakers as well and label completely.
(7.0 marks)
Solution:
Figure Q4.1. Percentage-differential relaying for a delta-connected synchronous generator.
11
(b) Assume an external line-to-line (b-c) fault and analyse the fault-current by simply drawing the
flow of current in the various branches of the installation. Indicate whether or not, the relays will
operate).
(5.0 marks)
Solution:
The relays do not operate since there is no current flowing in the operating coils.
… (1.0)
Figure Q4.2. Percentage-differential relaying for a delta-connected synchronous generator with an
external fault.
12
(c) Assume an internal line-to-line (b-c) fault and analyse the fault-current by simply drawing the
flow of current in the various branches of the installation. Indicate whether or not, the relays will
operate).
(3.0 marks)
Solution:
The relays for phases ‘b’ and ‘c’ will operate since there is current flowing in the operating coils.(1.0)
Figure Q4.3. Percentage-differential relaying for a delta-connected synchronous generator with an
internal fault.
13