Microsoft Word - DT705-Protection_Qns-Feb18
Dr. J Kearney Power Sys Analysis 2 DT705
1
Tut Qns
DT705 MEng.
Power System Analysis 2
Protection.
Ex 1
(a). On the Figure .. of the small power system circle the zones of protection.
(b). State which circuit
eakers trip for a fault at:
(i) L1
(ii) Bus 3
(iii) Generator G2
(iv) Transformer T4
(v) Bus 2
(vi) Load
(vii) M1
Ex 2 (a) An induction-disc overcu
ent relay has a characteristic as shown in Table 1 and
has a plug setting (PS) of 150% and a time multiplier of 0.2. The relay is supplied by a 200 A/5
A cu
ent transformer on a three-phase, 10 kV local feeder of reactance 3 Ω per phase. The
feeder is supplied by a 100 MVA, 38 kV/10 kV transformer of 0.15 p.u. reactance. A three-
phase solid earth occurs at the end of the 10 kV feeder.
Calculate:
(i) the fault cu
ent,
(ii) the time of operation of the relay.
Table 1
PSM XXXXXXXXXX XXXXXXXXXX
Time in Seconds XXXXXXXXXX XXXXXXXXXX
XXXXXXXXXXG2)
Generator
1
L1
3
B2 T1
B1
B3 B8
B4 B9
L2 L3
2
B5 B6
B7
B10
T3
B11
L4
B12
B13
(M1)
Motor
XXXXXXXXXXG1)
Generator
T2
T4
(Transformer)
Load
B1-B14
(Circuit
Breakers)
B14
Dr. J Kearney Power Sys Analysis 2 DT705
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Ans.
Actual relay time = XXXXXXXXXX)= 0.74 sec.
Ex. 3 A 50 Hz, 20 kV radial system is protected be a circuit
eaker A as shown in Figure..The
maximum load on the line is 4 MVA. The fault level at the 20 kV bus is 400 MVA and the positive
sequence reactance of the line is 3 Ω. Calculate the maximum load cu
ent and the fault cu
ent for a
three-phase to ground fault at Bus 2. The CT for the relay has a ratio of 200:5. If the relay setting is
such that the relay should trip when the cu
ent is at least twice the maximum load cu
ent and 1/5 of
the minimum fault cu
ent, determine the appropriate cu
ent tap setting. The characteristic of the relay
is given by:
k
I
t .
1
014.0
02.0
−
=
Where t is the response time, I is the tap setting multiple and k is the time dial constant setting.
Determine the constant k in the relay characteristics to ensure that the relay responds in 0.3 seconds.
Fig.
Ans CTS = 14.44, k=0.864
Ex 4 Differential relay protection for a single-phase transformer. Explain with aid of a sketch
the operation of a differential method of protection for a single phase two-winding
transformer. A single-phase two-winding, 10 MVA, 80kV/20kVtransformer has differential
elay protection. Select suitable CT ratios, from below;
Table 2 Cu
ent Transformer ratios
50:5 100:5 150:5 200:5 250:5 300:5 400:5 450:5 500:5 600:5 800:5 900:5 1000:5
Select k for a mismatch of 25%.
Ans
Similar in notes. CT1 = 150:5, CT2 =600:5, k=0.222
Ex 5
Load
20kV
kV CT 200/5 A
Line
Grid
A
Bus 1 Bus 2
Relay
Load
CT 200/5 A
0.15 pu
3 Ω/ph
CB
38/10 kV
Dr. J Kearney Power Sys Analysis 2 DT705
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Two-source system protection with directional and time-delay overcu
ent relays.
Explain how directional and time-delay overcu
ent relays can be used to protect the system in
the figure.Which relays should be coordinated for a fault (a) at P1, (b) at P2.
(c) Is the system also protected against bus faults?
XXXXXXXXXX1 XXXXXXXXXX2 XXXXXXXXXX3
XXXXXXXXXXB XXXXXXXXXXB1 XXXXXXXXXXB XXXXXXXXXXB23 B XXXXXXXXXXB3
XXXXXXXXXXP2 P1
XXXXXXXXXXLoad L1 Load L XXXXXXXXXXLoad L3
Figure
Ans.
In notes
Ex 6 A 40 MVA, 110 kV/38 kV delta/star transformer has its star point earthed through a
esistor which limits the cu
ent, due to an earth fault on one of the line terminals, to rated
value. Circulating-cu
ent protection is employed, with 5 A in the relay circuits at rated load.
(a) Determine the cu
ent transformer ratios required.
(b) Draw a circuit diagram showing the cu
ents an all windings when a phase-to-phase fault of
1000A occurs on the 38kV side beyond the protected zone.
(c) Find the maximum cu
ent setting of the relay such that it will trip if an earth fault occurs
halfway along one of the 38 kV windings.
Note: The diagram in (b) may also be used to obtain the solution for (c).
Dr. J Kearney Power Sys Analysis 2 DT705
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Ans Primary 200/5 CT’s , 40/1 ratio. Secondary 1000/5 CT’s ratio 200/1.
Ans
Relay R&T will operate with 0.75 A setting for fault ½ way along 38 kV winding.
Ex 7
A three-phase delta-star connected 30 MVA 38/10kV transformer is protected by a differential
elay. Calculate the relay cu
ent setting for faults drawing up to 200 per cent of the rated
cu
ent. The CT cu
ent ratio on the primary side is 500:5 and on the secondary side is 2000:5.
Ans I
`
1 = 4.56A, I
`
2 = 7.49A, Io (200%) = 5.862A
Ex 8 A single-phase two-winding, 5 MVA 20/8.66 kV transformer is protected by a differential relay.
Available relay tap settings are 5:5, 5:5.5, 5:6.6, 5:7.3, 5:8, 5:9 and 5:10. Giving tap ratios of 1.00,
1.10, 1.32, 1.46, 1.60,1.80 and XXXXXXXXXXCT ratios, are: 50:5, 100:5, 150:5, 200:5, 250:5, 300:5, 400:5,
450:5, 500:5, 600:5, 800:5, 900:5.
Select CT ratios and relay tap settings. Also determine the percentage mismatch for the selected tap
setting.
Relay Tap Settings 5:5 5:5.5 5:6.6 5:7.3 5:8 5:9 5:10
Relay Tap Ratios XXXXXXXXXX XXXXXXXXXX
CT Ratios 50:5 100:5 150:5 200:5 250:5 300:5 400:5 450:5 500:5 600:5 800:5 900:5
Ans Primary CT = 300:5, Sec= 600:5,Tap setting = 1.1 Mismatch =4.7 %
If2 s/c
Y
B
R
Oper. Coil
Restraining Coil
38kV 110kV
Direction of Fault Cu
ent
B Y R
Dr. J Kearney Power Sys Analysis 2 DT705
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G&S
Ex. 9
Table below gives the positive-sequence line impedances and the CT and VT ratios at B12 for
the 345 kV system as shown in the Fig above.
(a) Determine the settings Zr1, Zr2 and Zr3 for the B12 three-zone, directional impedance
elays connected as shown (Fig . below ) . Consider only solid, three-phase faults. (b)
Maximum cu
ent for line 1-2 during emergency loading conditions is 1500 A at a
power factor of 0.95 lagging. Verify that B12 does not trip during normal and
emergency conditions.
B12 B21
B13
B31 B32
B23
B24 B42
Zone 1
Zone 2
Zone 3
1 2
3
4
P1
P2 P3
Dr. J Kearney Power Sys Analysis 2 DT705
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Table
Line Positive-Sequence Impedance
Ω
1-2
2-3
2-4
1-3
8 + j50
8 + j50
5.3 + j33
4.3+ j27
Breaker CT ratio VT Ratio
B12 1500:1 3000:1
Ans Zr1 =4.05 ∠80.9
o Ω , Zr2 = =6.08 ∠80.9