SNS COLLEGE OF ENGINEERING, COIMBATORE-641 107.
5.
Write
an
equation
to
find
unbalanced
vectors
from
symmetrical components.
DEPARTMENT OF ELECTICAL AND ELECTRONICS ENGINEERING
Internal Assessment Exam-III, April 2015
VI Semester ,B.E.(EEE)
EE2351 POWER SYSTEM ANALYSIS
Time: 3Hrs
Date: 21.04.2015
Max marks: 100
Part- A
Answer all Questions
1.
(10*2=20)
6.
What is meant by negative sequence impedance?
What is meant by symmetrical components?
The impedance of a circuit when negative sequence currents are
The key idea of symmetrical component analysis is
present, the impedance is called the impedance to negative
to decompose the unbalanced system into three
sequence current.
Za2 = 1/3 (Za +a2 Zb +aZc) negative sequence
sequence of balanced networks. The networks are
then coupled only at the point of the unbalance (i.e.,
7.
Draw the zero sequence network of an alternator having its
neutral is grounded through a reactance.
the fault)
The three sequence networks are known as the
– positive sequence (this is the one we’ve been using)
– negative sequence
– zero sequence
2.
Define sequence operator.
An sequence operator 'a' is introduced, which when operates
upon a phasor rotates it by +120° without changing the
magnitude of the phasor upon which it operates.
3.
Define positive sequence components.
A balanced three-phase system with the same phase sequence
4.
8.
Draw the zero sequence network of a transformer for the
as the original sequence.
following configurations:
Write any four properties of sequence operator.
(i)
∆-Υ
∆-∆
(ii)
Part- B
Answer all Questions
(5*16=80)
11. a) For a 3 bus network shown below obtain Zbus by building
algorithm.
(OR)
11. b) Determine Zbus matrix for the system whose reactance
diagram is shown below where impedances are given in pu.
Υ–Υ
(iii)
12. a) The following figure shows the single line diagram of a
simple power system by omitting the resistance and shunt
admittances. Draw the positive, negative and zero sequence
network .
(OR)
13. b) Determine the positive, negative and zero sequence
networks for the system shown below. Assume zero
9.
What
is
meant
by
unsymmetrical
fault?
List
its
sequence reactances for the generators and synchronous
classifications.
motors as 0.06pu. Current limiting reactors of 2.5Ω are
Faults in which the balanced state of the network is disturbed are
connected in the neutral of the generator and motor no.2.
called unsymmetrical or unbalanced faults.
The zero sequence reactance of the transmission line is

Single-line-to-ground (1LG) fault

Line-to-line (LL) fault

Double-line-to-ground (2LG) fault
10. Draw the Thevenin’s equivalent of sequence network for line
to line fault.
j300Ω.
13. a) Derive the relationship for fault current in terms of
symmetrical components when there is a line to line fault
between phase b and c and also draw the Thevenin’s
equivalent circuit.
13.b) A 50 MVA, 11KV, 3Φ alternator has a direct sub transient
reactances of 0.1 pu. The negative and zero sequence reactances are
0.15 and 0.25 pu respectively. The neutral of the generator is solidly
earthed. Determine the sub transient current in the generator and the
line to line voltages for sub transient conditions when a single line
to ground fault occurs at the generator terminals with the generator
operating unloaded at rated voltage.
14.a) Draw the Thevenin’s equivalent circuit of symmetrical
components by deriving relations between symmetrical component
voltages and currents when a double line to ground fault occurs in a
3Φ alternator.
Double Line-To-Ground Fault
If the fault involves phases b, c, and ground, the “terminal” relationship at
the point of the fault is:
V b = 0, V c = 0, Ia = 0
(OR)
Then, using the sequence transformation:
Single-Phase to Ground Fault
V1=V2=V0=Va3
Consider the single-phase fault at bus k, as shown in Figure
Combining the inverse transformation:
Ia = I1 + I2 + I0 = 0
These describe a situation in which all three sequence networks are
connected in parallel, as shown above.
(OR)
14.b) A 30 MVA, 11Kv, 3Φ synchronous generator has a direct axis
sub transient reactance of 0.25 pu. The negative and zero sequence
reactances are 0.35 and 0.1 pu respectively. The neutral of the
generator is solidly grounded. Find the sub transient currents and
the line to line voltage at the fault when a line to line fault occurs at
the terminals of the generator. Assume that the generator is
unloaded and operating at rated terminal voltage when the fault
occurs.
15.a) Draw the Thevenin’s equivalent circuit of symmetrical
components by deriving relations between symmetrical component
voltages and currents when a single line to ground fault occurs in a
3Φ power system.
Adding the three rows yields
15.b) An unloaded, solidly grounded 10 MVA, 11 KV generator has
positive, negative and zero sequence impedances are j1.2Ω, j0.9 Ω
and j 0.04 Ω respectively. A double line to ground fault occurs at the
terminals of the generator. Calculate the current in the faulted line
and voltage at the healthy line.
Having the key fault equation, then the resulting bus voltages and branch
currents throughout the network during the fault can be determined.
Recall from (1P_3) that the 012 fault current components are
F
I ka
F
F
F
I k 0  I k1  I k 2 
3
All 012 network voltages can then be found from the 012 Thevenin
equation
e
V jF012  V jPr012
 Z j 012 , k 012 I kF012
Then, 012 fault currents in branches can be found using Ohm’s Law and
the corresponding positive, negative, and zero sequence branch
impedances. Afterward, 012 voltages and currents can be converted to
abc. See later sections on how to include transformer phase shifts when
converting 012 to abc.
Note that if
z k 0, k 0  z k1, k 2
, a single-phase fault will have a higher
value than does a three-phase fault.
(OR)