ELEC
C4612 Po ower Sys stem Ana alysis Midsess sion SOL
LUTION (
(Session 1, 2016) n QUES
STION 1
For the system sho own in Figu 1, comp ure pute the matrix. T system has an off-nominal
The
tap chan nging transformer betw ween buses 1 and 2 wi a tap setting of 1.03 The line b ith 3. between buses 1 and 4 has half line charging sus sceptance o 0.25 pu at each end. There is a mutual of a
.
ng u d own. couplin of 0.15 pu between lines 2-3 and 4-3 as sho

2

t = 1.03

1

3

j0.45

j0.4
4

j0.15 j0.1 j0. 2 j0.3 j0.1
4

Figure 1 on Solutio etween line 2-3 and 4-3 es Mutual coupling be
=

=

0.4 0.15
5
0.15 0.1
0
3 − 2 3 − 4
3 − 2 − 5.7143 8.5714
=
8.5714
3 − 4 8.5714

Buildin block matrix ng Tempor rary values are,
= − 5.7143
=− 2
2.8571
= − 2.8571
= − 11.4286
= 8.5714
= 14
4.2857

= 8.57
714
= 14.2
2857
= − 22
2.8571

Tap changing transformer in line 1-2
1
=
= − 2.222
0.45
.
.
=
= − 2.1575,
=
.
Temporary values are,
= − 2.0947
= 2.1575
= 2.1575
= − 2.2222

(
.

.

)

= 0.0628,

=

.

( .
.

)

= − 0.0647

Line 1-4 with half line charging susceptance of j0.25 at each end
=
= − 3.3333,
= 0.25,
= 0.25
.

The
=−
=
=
=
=−
=
=
=−
=
=−

matrix is therefore,
3.3333 + 0.25 − 2.0947 = − 5.178
= 2.1575
=0
= 3.3333
2.2222 − 5.7143 = − 7.9365
= − 2.8571
= 8.5714
11.4286
= 14.2857
3.3333 + 0.25 − 22.8571 = − 25.9404

5.178
−2.1575
0
−3.3333
−2.1575 7.9365
2.8571
−8.5714
=−
0
2.8571
11.4286 −14.2857
−3.3333 −8.5714 −14.2857 25.9404

QUESTION 2
The matrix of a 3-bus system is given by,
0
13.8675 − 56.31 13.8675123.69
= 13.8675123.69 41.6025 − 56.31 27.7350123.69
0
27.7350123.69 27.7350 − 56.31
Obtain the initial Jacobian matrix of Newton-Raphson method. Bus 1 is a slack bus with V1 =
1.00. Take P2 + jQ2 = -5.96 + j1.46, P3= 6.02 and |V3| = 1.02. Assume a flat start with
V20 = 10 and V30 = 1.020.
Solution

∆
∆
∆

| |
=

∆
∆
| |
∆| |
| |
Page 2 of 4

= 1 (
=|

),
(

−

|

=|
→

= 1 (

(

|

→

(

|

→

=|

(

|

−

(

= −|
→

) + 2|

)+|

+

(

−

→

(

|

→

=|

−

(

|

−

+

)−|

=|

|

(

−

+

)+|

→

−

+

|

(

(

−

)+|

+

|

|

(

−

)

+

|

−|
|

(

−

= −13.8675 x 1 x
− 27.735 x 1.02 x

(

|

−

−

)

+

)

+

(123.69) = −23.3846

)
(123.69) = 15.6923

= −27.735 x 1 x 1.02 x
= −|

(−56.31)

(123.69) + 27.735 x 1 x 1.02 x

= 13.8675 x 1 x 1 x

| |

)

+

(123.69) = −15.6923
−

→

−

)

+

(

(

(

|

(123.69) = 23.5385

|

|

(123.69) = 35.0769

+|

|

) +|

= −|

= −|

)

+

(123.69) = −23.5385

+

= 27.735 x 1.02 x

→

)

)

+

= 0 + 27.735 x 1.02 x 1 x

| |

+

(123.69) = 22.7692

= −27.735 x 1.02 x 1 x
=|

|

(123.69) + 2 x 41.6025 x 1 x
−

|

−

−

(123.69) = −23.5385

+

+27.735 x 1.02 x
|

(

(

|

)

+

−

= 13.8675 x 1 x
=|

+|
|

)

(123.69) + 27.735 x 1 x 1.02 x

= −27.735 x 1 x 1.02 x

| |

|

) +|

+

= 13.8675 x 1 x 1 x
= −|

= 1.02 (

),

)+|

+

−

+

) − 2|

|

−|

(123.69) − 2 x 41.6025 x 1 x

|

(

−

+

)

(−56.31)

(123.69) = 34.1539

35.0769 −23.5385 22.7692
[ ] = −23.5385 23.5385 −15.6923
−23.3846 15.6923
34.1539
Page 3 of 4

QUESTION 3
(i)

Which among the following methods are generally used for the calculation of symmetrical faults?
a. Norton theorem
b. Thevnin’s theorem
c. Kirchhoff’s laws
d. Only (b) and (c)
e. All of these

(ii)

Which among the following reactance has a greater value?
a. Sub transient reactance
b. Transient reactance
c. Synchronous reactance
d. All of these
e. None of these

(iii) On which among the following factors does the magnitude of the fault current depend?
a. Total impedance upto the fault.
b. Voltage at the fault point
c. Both (a) and (b)
d. None of these
(iv) In power-system short-circuit studies, for calculating subtransient fault currents, transformers are represented by their (a) leakage reactance, transmission lines by their equivalent (b) series reactance, and synchronous machines by (c) constant voltage behind subtransient reactance.
(v)

The load currents in short-circuit calculations are neglected because, short-circuit currents are (a) many times larger than and (b) out of phase from load currents.

Page 4 of 4