BMS COLLEGE OF ENGINEERING, BANGALORE-19

(Autonomous College under VTU)

CONTINUOUS INTERNAL EVALUATION FIRST TEST 2023
TH
Section: 5 SEM YEAR BACTH:2021
Department: ELECTRICAL AND ELECTRONICS ENGINEERING Semester: V
Course Title: Course Code: 22EE5PCPSA Date: 04.11.2024 Time:12.45PM-10.15PM
POWER SYSTEMS ANALYSIS
Signature of FIC with Date: Dr.C.lakshminarayana. 04.11.2024
SL QUESTION M CO PO
NO
PART A
1.a Importance of online line symbols of power systems 02 1
1.b Show that the per unit impendence of two winding transformer will remain same 03 1
referred to primary as well as secondry.
PART B
2.a 1. What is the need for base value? 07 1 1
2. Write the equation for converting the P.U. impedance expressed in one base
to another.
3. What are the approximations made in reactance diagram?

2.b Draw the per unit reactance diagram for the system shown in figure below. Choose 08 1 2
Common a base MVA of 300 MVA,20KV on generator 03(G3) as shown in figure
below:

Generator G1=200MVA, 20KV, Xd=15%

Generator G2=300MVA, 18KV, Xd=20%
Generator G3=300MVA, 20KV, Xd=20%
TransformerT1=300MVA, 220 Y/22KV, Xd=10%
TransformerT2=Three single phase unit of 100MVA,130Y/25KV, Xd=10%
TransformerT2=300MVA, 220 Y/22KV, Xd=10%
PART C
3.a Analyses and develop an expreesion for the doubling effect of the short circuit 10 1 2
during symmetrical fault on a transmission line including waveforms.
3.b Draw the per unit reactance diagram for the system shown in figure below. Choose 10 1 2
a base of 22KV, 100 MVA in the generator circuit. The Three phase load of 57
MVA,0.6 pf lagging at 10.45 KV. Line 1&2 have reactance of 48.4 & 65.3 ohm
respectively.
 Ite S V Xpu Ite S V Xpu
m MVA KV % m MVA KV %
G 90 22 18 T1 50 22/220 10

T2 40 220/11 6 T3 40 22/110 6.4

T4 40 110/11 8 M 66.5 66.5 18.5

OR
4.a Analyses and develop expreesion for the maximum mometry current twice the 10 1 2
maximum value of symmetrical short circuit current of the short circuit during
symmetrical fault on a transmission line including waveforms.
4.b A radial power system network is shown in fig. a three phase balanced fault 10 1 2
occurs at F. Determine the fault current and the line voltage at 11.8 KV bus under
fault condition.
SOLUTION CI1: POWER SYSTEMS ANALYSIS: 22EE5PCPSA- ABRAR
PART A
1a The symbols are standardized and universally recognized, making it easier for engineers, 02
technicians, and electricians to understand and interpret electrical plans.
Why is symbol study important?
 To document your work.
 To efficiently communicate and share with others.
 To facilitate construction.
 To explain what is happening in the circuit.
 To reduce the memory load while working.
Not studying electrical/electronic symbols is like trying to read without studying the alphabet.
The electrical/electronic symbols are the map that gives you the required directions to do what
you want to do.
1b 𝐿𝑒𝑡 (𝑀𝑉𝐴)𝐵 = 𝑅𝑎𝑡𝑒𝑑 𝑀𝑉𝐴 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 03
(𝐾𝑉1)𝐵 = 𝐵𝑎𝑠𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
(𝐾𝑉2)𝐵 = 𝐵𝑎𝑠𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑐𝑜𝑛𝑑𝑟𝑦 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
(𝑧)𝑒𝑞1 = 𝐼𝑚𝑝𝑒𝑛𝑑𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑟𝑒𝑓𝑒𝑟𝑒𝑑 𝑡𝑜 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑠𝑖𝑑𝑒
(𝑧)𝑒𝑞2 = 𝐼𝑚𝑝𝑒𝑛𝑑𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑟𝑒𝑓𝑒𝑟𝑒𝑑 𝑡𝑜 𝑠𝑒𝑐𝑜𝑛𝑑𝑟𝑦 𝑠𝑖𝑑𝑒
(𝑀𝑉𝐴)𝐵 (𝑀𝑉𝐴)𝐵
(𝑍𝑒𝑞1)𝐵 = 𝑍𝑒𝑞1(Ω) × 2 . . (1) & (𝑍𝑒𝑞1)𝐵 = 𝑍𝑒𝑞2(Ω) × … (2)
(𝐾𝑉1)𝐵 (𝐾𝑉2)2𝐵
(𝐾𝑉2)2𝐵
𝑊ℎ𝑒𝑟𝑒 𝑍𝑒𝑞2(Ω) = 𝑍𝑒𝑞1(Ω) × … (3); 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛(3)𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2), 𝑤𝑒𝑔𝑒𝑡
(𝐾𝑉1)2𝐵
(𝐾𝑉2)2𝐵 (𝑀𝑉𝐴)𝐵 (𝑀𝑉𝐴)𝐵
(𝑍𝑒𝑞2(Ω)𝑃𝑈 = 𝑍𝑒𝑞(Ω) × 2 × 2 = 𝑍𝑒𝑞(Ω) × = (𝑍𝑒𝑞1(Ω)𝑃𝑈
(𝐾𝑉1)𝐵 (𝐾𝑉1)𝐵 (𝐾𝑉1)2𝐵
Comparing above equations we conclude that the per-unit impedance is the same regardless of the side
from which it is viewed. Thus the simple equivalent circuit for the two-winding transformer in
which(𝑍𝑒𝑞2(Ω)𝑃𝑈 = (𝑍𝑒𝑞1(Ω)𝑃𝑈
PART B
2a 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑒𝑒𝑑 𝑓𝑜𝑟 𝑏𝑎𝑠𝑒 𝑣𝑎𝑙𝑢𝑒? (𝟎𝟐𝑴): The components or various sections of power 07
system may operate at different voltage and power levels. It will be convenient for analysis of
power system if the voltage, power, current and impedance ratings of components of power
system are expressed with a common value called base value. Hence for analysis purpose a base
value is chosen for voltage, power, current and impedance.

𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝. 𝑢. 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑖𝑛 𝑜𝑛𝑒 𝑏𝑎𝑠𝑒 𝑡𝑜 𝑎𝑛𝑜𝑡ℎ𝑒𝑟.
2
(𝑀𝑉𝐴) (𝐾𝑉)
(𝑍)𝑝𝑢,𝑁𝑒𝑤 = 𝑍𝑜𝑙𝑑 ∗ ⌈ (𝑀𝑉𝐴)𝑏,𝑁𝑒𝑤 ⌉ ∗ {(𝐾𝑉) 𝑏,𝑂𝑙𝑑 } … . (𝟎𝟑𝑴)
𝑏,𝑂𝑙𝑑 𝑏,𝑁𝑒𝑤
What are the approximations made in reactance diagram? (𝟎𝟐𝑴)
 The neutral reactance are neglected.
 The shunt branches in equivalent circuit of induction motor are neglected.
 The resistances are neglected.
 All static loads and induction motor are neglected.
 The capacitances of the transmission lines are neglected. → (𝟎𝟐 + 𝟎𝟑 + 𝟎𝟐 = 𝟎𝟕𝐌)
2b 2 08
(𝑀𝑉𝐴)𝑏,𝑁𝑒𝑤 (𝐾𝑉)𝑏,𝑂𝑙𝑑
𝐹𝑜𝑟𝑚𝑢𝑙𝑎: (𝑋)𝑝𝑢,𝑁𝑒𝑤 = 𝑋𝑜𝑙𝑑 ∗ ⌈ ⌉∗{ }
(𝑀𝑉𝐴)𝑏,𝑂𝑙𝑑 (𝐾𝑉)𝑏,𝑁𝑒𝑤
𝐻𝑇 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑎𝑡𝑖𝑛𝑔
(𝐾𝑉)𝐵 𝑜𝑛 𝐻𝑇 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = (𝐾𝑉)𝐵 𝑜𝑛 𝐿𝑇 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 × ;
𝐿𝑇 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑎𝑡𝑖𝑛𝑔
𝐿𝑇 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑎𝑡𝑖𝑛𝑔
(𝐾𝑉)𝐵 𝑜𝑛 𝐿𝑇 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = (𝐾𝑉)𝐵 𝑜𝑛 𝐻𝑇 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 × (𝟎𝟐𝑴)
𝐻𝑇 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑎𝑡𝑖𝑛𝑔
220 22
New KV base: Transmission line: 20 ∗ 22
= 200𝐾𝑉; 𝐺1,𝑛𝑒𝑤 = 200 ∗ 220 = 20𝐾𝑉;
25
𝐺2,𝑛𝑒𝑤 = 200 ∗ = 225𝐾𝑉 ; (𝟎𝟐𝑴)
√3×130
 300 20 2 300 18 2
(𝑋)𝑔1, = 0.15 ∗ ⌈ ⌉ ∗ { } = 0.225 𝑝𝑢; (𝑋) 𝑔2, = 0.2 ∗ ⌈ ⌉∗{ } = 0.1312 𝑝𝑢;
200 20 200 22.22
2 2
300 20 300 220
(𝑋)𝑔3, = 0.2 ∗ ⌈ ⌉ ∗ { } = 0.2 𝑝𝑢; (𝑋) 𝑇1, = 0.1 ∗ ⌈ ⌉∗{ } = 0.121 𝑝𝑢;
200 20 200 200
2
300 25 300 22 2
(𝑋) 𝑇2, = 0.1 ∗ ⌈ ⌉∗{ } = 0.1266 𝑝𝑢; (𝑋) 𝑇3, = 0.1 ∗ ⌈ ⌉ ∗ { } = 0.121 𝑝𝑢;
200 22.25 200 20
300 300
(𝑋) 𝑇𝐿 75 Ω, = 75 ∗ = 0.5625 𝑝𝑢; (𝑋) 𝑇𝐿 50 Ω, = 50 ∗ = 0.375 𝑝𝑢; (𝟎𝟐𝑴)
(200)2 (200)2

(𝟎𝟐𝑴)
→ (𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 = 𝟎𝟖𝐌)
PART C
3.a Transients on Transmission Line: The following assumptions are made: 10
 The supply is a constant voltage source,
 The short circuit occurs when the line is unloaded and
 The line capacitance is negligible.

Figure 1. Short Circuit Transients on an Unloaded Line. →02M

→03M
V  Vm sin( wt   )  iR  L dtdi , i  im sin( wt     )

Vm  2 V , im  2  i, z  ( R)2  (wL)2 ,  tan 1 ( wL

R )

I s  Vzm sin( wt     ); IR  L dtdi  RL  dtdi  0 ;

(  )
; 
t
CF  C1e L
R
 C1   I s (0)   Vzm sin(    )  Vzm sin(    ) ;
 ( t )  ( t )
it   I sim e  Vzm sin(    )e
 ( RL ) t
i  I s  it  2 V
sin( wt     )  sin(    )  e
Vm
z z

i mm  2 V
z
sin(    )  2 V
z
; R  negligible ,   90 0 ;
i mm  2 V
z
 cos( )  2 V
z
, when,   0, at , (i mm ) max
→03M
imm  2  2V
z  Doubling , Effect →it is observed that the maximum momentary
current is twice the maximum value of symmetrical short circuit current. This is referred as
the doubling effect of the short circuit current during the symmetrical fault on a transmission
line.
Short Circuit Current Oscillogram: The symmetrical short circuit current can be divided
into three zones: the initial sub transient period, the middle transient period and finally the
steady state period. The corresponding reactance’s, Xd,” Xd’ and Xd respectively, are
offered by the synchronous machine during these time periods.

→02M
→02+03+03+02=10M

3b 2 10
(𝑀𝑉𝐴)𝑏,𝑁𝑒𝑤 (𝐾𝑉)𝑏,𝑂𝑙𝑑
𝐹𝑜𝑟𝑚𝑢𝑙𝑎: (𝑋)𝑝𝑢,𝑁𝑒𝑤 = 𝑋𝑜𝑙𝑑 ∗ ⌈ ⌉∗{ } ;
(𝑀𝑉𝐴)𝑏,𝑂𝑙𝑑 (𝐾𝑉)𝑏,𝑁𝑒𝑤
𝐻𝑇 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑎𝑡𝑖𝑛𝑔
(𝐾𝑉)𝐵 𝑜𝑛 𝐻𝑇 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = (𝐾𝑉)𝐵 𝑜𝑛 𝐿𝑇 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 × ;
𝐿𝑇 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑎𝑡𝑖𝑛𝑔
𝐿𝑇 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑎𝑡𝑖𝑛𝑔
(𝐾𝑉)𝐵 𝑜𝑛 𝐿𝑇 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = (𝐾𝑉)𝐵 𝑜𝑛 𝐻𝑇 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 × → (𝟎𝟏𝐌)
𝐻𝑇 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑎𝑡𝑖𝑛𝑔

New KV base:
220 110
𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑛 𝑙𝑖𝑛𝑒 − 1 = 22 × = 220𝐾𝑉; 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑛 𝑙𝑖𝑛𝑒 − 2 = 22 × = 110𝐾𝑉
22 22
11 11
𝑀𝑜𝑡𝑜𝑟 = 220 × = 11𝐾𝑉; 𝑇4 = 110 × = 11𝐾𝑉 → (𝟎𝟐𝐌)
220 110
100 22 2 100 22 2
(𝑋𝑔 ) = 𝑗0.18 ∗ ⌈ ⌉ ∗ { } = 𝑗0.2𝑝𝑢; (𝑋𝑇1 )𝑝𝑢 = 𝑗0.1 ∗ ⌈ ⌉ ∗ { } = 𝑗0.2𝑝𝑢
𝑝𝑢 90 22 50 22
100 100
(𝑋𝑇𝐿1 )𝑝𝑢 = 𝑗48.4 ∗ = 𝑗01 𝑝𝑢; (𝑋𝑇𝐿2 )𝑝𝑢 = 𝑗65.3 ∗ ⌈110⌉2 = 𝑗0.54𝑝𝑢
⌈220⌉2
100 220 2 100 22 2
(𝑋𝑇2 )𝑝𝑢 = 𝑗0.08 ∗ ⌈ ⌉∗{ } = 𝑗0.15𝑝𝑢; (𝑋𝑇3 )𝑝𝑢 = 𝑗0.064 ∗ ⌈ ⌉ ∗ { } = 𝑗0.16𝑝𝑢
40 220 40 22
2
100 110
(𝑋𝑇4 )𝑝𝑢 = 𝑗0.08 ∗ ⌈ ⌉∗{ } = 𝑗0.2𝑝𝑢
40 110
100 10.45 2
(𝑋𝑚 )𝑝𝑢 = 𝑗0.185 ∗ ⌈ ⌉∗{ } = 𝑗0.25𝑝𝑢
66.5 11
3∅ 𝑙𝑜𝑎𝑑: cos(∅) = 0.6; ∅ = cos −1 (0.6) = 53.13°
 ⌈𝑉𝐿𝐿 ⌉2 ⌈10.45⌉2
(𝑍)𝐴𝑐𝑡𝑢𝑎𝑙 = ∗ = = (1.1495 + 𝑗1.5326);
(𝑆𝑙,3∅ ) 57 × 106 ∟53.13°
⌈11⌉2
(𝑍)𝐵 = = 1.21𝛺
100
(1.1495 + 𝑗1.5326)
(𝑍𝑙 )𝑝𝑢 = ⌈ ⌉ = 0.95 + 𝑗1.266𝑝𝑢 → (𝟎𝟑𝐌)
1.21

→ (𝟎𝟐𝐌)
→ (𝟎𝟏 + 𝟎𝟐 + 𝟎𝟑 + 𝟎𝟐 + 𝟎𝟐 = 𝟏𝟎𝐌)
OR
4a 3.a 10
4b Let Base MVA-12MVA; Base voltage =11.8KV 10
Xg1=j0.12 pu; Xg2=j0.12 pu; XT1=j0.12 pu; XT2=j0.08 *(12/3)-j0.32 pu
Base voltage for line-1 is 33KV & Base voltage for line-2 is 6.6KV→02M
(33)2 (6.6)2
𝑍𝐵,𝐿𝑖𝑛𝑒−1 = = 90.75 𝑜ℎ𝑚 & 𝑍𝐵,𝐿𝑖𝑛𝑒−2 = = 3.63 𝑜ℎ𝑚
12 12
(9.45 + 𝑗12.6)
𝑍 𝐿𝑖𝑛𝑒−1 = = (0.104 + 𝑗0.139)𝑝𝑢 &
90.75
(0.54+𝑗0.40)
𝑍 𝐿𝑖𝑛𝑒−2 = = (0.148 + 𝑗0.11)𝑝𝑢 →02M
3.63

→02M
12 × 1000 1∠0°
𝐼𝐵 = = 1049.7𝐴; ∴ 𝐼𝐹 = = 1.256∠ − 71.5° 𝑝𝑢
√3 × 6.6 (0.252 + 𝑗0.755)
∴ 𝐼𝐹 = 1.256∠ − 71.5° × 1049.7; ∴ 𝐼𝐹 = 1318.4∠ − 71.5° 𝐴→02M
Total impedance between ‘F’ and 11.8KV bus=(0.252 + 𝑗0.689)𝑝𝑢
Voltage at 11.8 KV bus = 1.256∠ − 71.5° × (0.252 + 𝑗0.689) = 0.921∠ − 1.6°
Voltage at 11.8 KV bus = 0.921∠ − 1.6° × 118 𝐾𝑉 = 10.89∠ − 1.6° 𝐾𝑉 → 𝟎𝟐𝐌
→(02+02+02+02+02=10M)
 BMS COLLEGE OF ENGINEERING, BANGALORE-19
(Autonomous College under VTU)
CONTINUOUS INTERNAL EVALUATION SECOND TEST 2024
TH
Section: 5 SEM YEAR BACTH: 2021
Department: ELECTRICAL AND ELECTRONICS ENGINEERING Semester: FIFTH
Course Title: POWER SYSTEMS Course code:22EE5PCPSA Date: 10.12.2024 Time:02.15PM-03.30PM
ANALYSIS
Signature of FIC with Date: Dr.C.lakshminarayana. 10.12.2024
ANSWER THE FOLLOWING
Q QUESTION M
PART A
1 a List out the types of faults in power system. 02M
b Significance of ‘a’ operator in Symmetrical components. 03M
PART B
2 a Develop an expression for positive negative and zero sequence network and also Draw 07M
the different configuration of transformer for zero sequence networks
b A 50MVA, 11KV, synchronous generator has a sub transient reactance of 20%. The 08M
generator supplies two motors over a transmission line with transformers at both ends as
shown in fig. The motors have rated inputs of 30 and 15 MVA, both 10KV, with 25% sub
transient reactance. The three phase transformers are both rated 60MVA, 10.8/121KV, with
leakage reactance of 10% each. Assume zero sequence reactance for the generator and
motors of 6% each. Current limiting reactors of 2.5 ohms each are connected in the neutral
of the generator and motor number 2. The zero sequence reactance of the transmission line
is 300 ohms. The series reactance of the line is 100 ohms. Draw the positive, negative and
zero sequence networks. Assume base MVA =50MVA and base KV =11 KV of Generator.

(CO2 PO2)

PART C
a Analyses and develop expression for phase shift symmetrical components in Y-∆ Transformer 10M
3 bank with positive, negative sequences voltage and current vector diagrams. (CO2 PO2)
b Analyses and develop an expression for fault current, line current for double line to ground 10M
fault on unloaded generator through impedance. Draw the inter connection of sequence
network. (CO2 PO2)
OR
a Analyses and develop an expression for fault current, line current for single line to ground fault 10M
4 on unloaded generator through impedance. Draw the inter connection of sequence network.
(CO2 PO2)
b Analyses and develop the relation between: 10M
i). Phase voltages in terms of symmetrical components (05M)
ii).Prove that a balanced set of three phase voltages will have only positive sequence
components of voltages only. (05M). (CO2 PO2)
SCHEME OF SECOND TEST POWER SYSTEMS ANALYSIS Course Code:22EE5PCPSA
PART A
1 a Shunt faults: SLG faults; LL faults; LLG faults & 3 LLL faults 02M
Series faults: One conductor open type faults & Two conductor open type faults→ 𝟎𝟐𝑴

b Importance of ‘a’ operator in Symmetrical components: 03M

𝑎 = 1∠120° = 𝑐𝑜𝑠120° + 𝑗𝑠𝑖𝑛120° = −0.5 + 𝑗0.866
𝑎2 = 1∠ − 120° = 1∠ − 240° = 𝑐𝑜𝑠240° + 𝑗𝑠𝑖𝑛240° = −0.5 − 𝑗0.866
𝑎 + 𝑎2 + 𝑎3 = 0 𝑓𝑜𝑟 𝑏𝑎𝑙𝑛𝑐𝑒𝑑 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ;
𝑎3 = 1; 𝑎4 = 𝑎; (𝑎 − 𝑎2 ) = 𝑗√3; (𝑎2 − 𝑎) = −𝑗√3; 𝑎∗ = 𝑎2 → 𝟎𝟑𝑴

PART B
2 a 07M

𝑉𝑎0 = −𝑍0 𝐼𝑎0; 𝑉𝑎1 = 𝑉𝑓 − 𝑍1 𝐼𝑎1; 𝑉𝑎2 = −𝑍2 𝐼𝑎2;

Where Z0, Z1 and the Z2 are the total equivalent impedance of the zero, positive and
negative sequence network up to the fault point.

b (MVA)B=50MVA,( KV)B=11KV on generator 08M

121 10.8
New KV base of: 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑙𝑖𝑛𝑒 = 11 × = 123.2 𝐾𝑉; 𝑀𝑜𝑡𝑜𝑟 = 123.2 × = 11𝐾𝑉
10.8 121
50 11 2 50 10.8 2
(𝑋𝑔 )𝑝𝑢 = 𝑗0.2 ∗ ⌈ ⌉ ∗ { } = 𝑗0.2𝑝𝑢; (𝑋𝑇1 )𝑝𝑢 = (𝑋𝑇2 )𝑝𝑢 = 𝑗0.10 ∗ ⌈ ⌉ ∗ { }
50 11 60 11
= 𝑗0.0805𝑝𝑢
 50
+𝑣𝑒& − 𝑣𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑙𝑖𝑛𝑒 ∶ (𝑋𝑇𝐿 )𝑝𝑢 = 𝑗100 ∗
⌈123.2⌉2
= 𝑗0.33𝑝𝑢;
50
𝑍𝑒𝑟𝑜 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑙𝑖𝑛𝑒 (𝑋𝑇𝐿 )𝑝𝑢 = 𝑗300 ∗ ⌈123.2⌉2 = 𝑗0.99𝑝𝑢
50 10 2
+𝑣𝑒& − 𝑣𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑚𝑜𝑡𝑜𝑟1 ∶ (𝑋𝑚1 )𝑝𝑢 = 𝑗0.25 ∗ ⌈ ⌉ ∗ { } = 𝑗0.345𝑝𝑢
30 11
50 10 2
𝑍𝑒𝑟𝑜 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒𝑜𝑓 𝑚𝑜𝑡𝑜𝑟1: (𝑋𝑚1 )𝑝𝑢 = 𝑗0.06 ∗ ⌈ ⌉ ∗ { } = 𝑗0.082𝑝𝑢
30 11
50 10 2
+𝑣𝑒& − 𝑣𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑚𝑜𝑡𝑜𝑟2 ∶ (𝑋𝑚2 )𝑝𝑢 = 𝑗0.25 ∗ ⌈ ⌉ ∗ { } = 𝑗0.69𝑝𝑢
15 11
50 10 2
𝑍𝑒𝑟𝑜 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒𝑜𝑓 𝑚𝑜𝑡𝑜𝑟1: (𝑋𝑚2 )𝑝𝑢 = 𝑗0.06 ∗ ⌈ ⌉ ∗ { } = 𝑗0.164𝑝𝑢
15 11
50
𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑜𝑟𝑠 (𝑋𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 )𝑝𝑢 = 𝑗12.5 ∗ = 𝑗1.033𝑝𝑢 = 3 ∗ 1.033 = 3.10
⌈11⌉2
→ 𝟎𝟑𝐌
The positive-, negative- and- zero sequence networks are shown in Figs. 4c(b), 4c (c) and 4c (d)
respectively:

→ 𝟎𝟐𝑴

→ 𝟎𝟑𝑴

(→ 𝟎𝟑 + 𝟎𝟐 + 𝟎𝟑 = 𝟎𝟖𝑴)

PART C
a Phase shift of Y-∆ trans former Bank: +ve and -ve sequence voltages and currents 10M
3 undergo a phase angle change in passing a star-delta transformer is known as phase
shift.
1. The HT side terminals are marked as H1, H2 and H3 and the corresponding LT side
terminals are marked X1, X2 and X3.
2. The phases in the HT side are marked in uppercase letters as A, B, and C. Thus for the
sequence abc, A is connected to H1, B to H2 and C to H3. Similarly, the phases in the LT
side are marked in lowercase letters as a, b and c.
3. The standard for designating the terminals H1 and X1 on transformer banks requires
that the positive-sequence voltage drop from H1 to neutral lead the positive sequence
voltage drop from X1 to neutral by 300 regardless of the type of connection in the HT→
𝟎𝟒𝑴
 → 𝟎𝟑𝑴

→ 𝟎𝟑𝑴
→ (𝟎𝟒 + 𝟎𝟑 + 𝟎𝟑 = 𝟏𝟎𝑴
b 10M

1. 𝑇𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠: 𝐼𝑎 = 0 … (1); 𝑉𝑏 = (𝐼𝑏 + 𝐼𝑐 )𝑍𝑓 … (2); 𝑉𝑐 = (𝐼𝑏 + 𝐼𝑐 )𝑍𝑓 … (3) → 𝟎𝟐𝑴
2. 𝑠𝑦𝑚𝑚𝑡𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑟𝑒𝑙𝑡𝑖𝑜𝑛𝑠 → 𝟎𝟐𝑴:
1 1 𝑉𝑎 − 𝑉𝑏
𝑉𝑎1 = [𝑉𝑎 + 𝑎𝑉𝑏 + 𝑎2 𝑉𝑐 ] = [𝑉𝑎 + (𝑎 + 𝑎2 )𝑉𝑏 ] =
3 3 3
1 2
1 2
𝑉𝑎 − 𝑉𝑏
𝑉𝑎2 = [𝑉𝑎 + 𝑎 𝑉𝑏 + 𝑎𝑉𝑐 ] = [𝑉𝑎 + (𝑎 + 𝑎 )𝑉𝑏 ] = ; 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑉𝑎1 = 𝑉𝑎2 . . (4)
3 3 3
1
𝑉𝑎0 − 𝑉𝑎2 = [3𝑉𝑏 ] = 𝑉𝑏 = (𝐼𝑏 + 𝐼𝑐 )𝑍𝑓 = 3𝐼𝑎0 𝑍𝑓 … (5)
3
𝑇ℎ𝑢𝑠 𝑉𝑎0 = 𝑉𝑎2 + 3𝐼𝑎0 𝑍𝑓 … (5) ; 𝐼𝑎0 + 𝐼𝑎1 + 𝐼𝑎2 = 0 … (6);
3. 𝐼𝑛𝑡𝑒𝑟𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑦𝑚𝑚𝑡𝑟𝑖𝑐𝑎𝑙 𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠(𝐼𝑆𝑁): → 𝟎𝟐𝑴
From the equations (5) &(6) it indicates that interconnection of symmetrical networks would be
connected in parallel
4. 𝑆𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑒𝑠: → 𝟎𝟐𝑴: 𝑉𝑎1 = 𝑉𝑎2 = 𝑉𝑎0 = 𝐸𝑎 − 𝐼𝑎1 𝑍1 . . (6);
𝐸𝑎 𝐸𝑎 𝑍0 +𝑍𝑓
𝐼𝑎1 = 𝑍2 (3𝑍𝑓 +𝑍0 ) = 𝑍2 (3𝑍𝑓 +𝑍0 ) … (7); 𝐼𝑎2 = −𝐼𝑎1 [ ] … . (8);
𝑍1 + 𝑍1 + 𝑍0 +𝑍1+ 3𝑍𝑓
𝑍0 +𝑍1+ 3𝑍𝑓 𝑍0 +𝑍1+ 3𝑍𝑓
 𝑍2
𝐼𝑎0 = −𝐼𝑎1 [ ] … . (8);
𝑍0 +𝑍1+ 3𝑍𝑓
5. 𝐹𝑎𝑢𝑙𝑡 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 → 𝟎𝟐𝑴: 𝐼𝑓 = 𝐼𝑏 + 𝐼𝑐 = (𝐼𝑎0 + 𝑎2 𝐼𝑎1 + 𝑎𝐼𝑎2 ) + (𝐼𝑎0 + 𝑎𝐼𝑎1 + 𝑎2 𝐼𝑎2 )
𝐼𝑓 = 2𝐼𝑎0 + (𝑎2 + 𝑎)𝐼𝑎1 + (𝑎2 + 𝑎)𝐼𝑎2
𝐼𝑓 = 𝐼𝑏 + 𝐼𝑐 = 2𝐼𝑎0 − 𝐼𝑎1 − 𝐼𝑎2 = 2𝐼𝑎0 − (𝐼𝑎1 + 𝐼𝑎2 ); 𝑏𝑢𝑡𝐼𝑎1 + 𝐼𝑎2 = −𝐼𝑎0 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑓𝑖𝑔𝑢𝑟𝑒
𝐼𝑓 = 𝐼𝑏 + 𝐼𝑐 = 2𝐼𝑎0 − (−𝐼𝑎0 ) = 3𝐼𝑎0;
𝑍2
𝐼𝑓 = −3𝐼𝑎1 𝑍 . . (10); → (𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 = 𝟏𝟎𝐌
2 +𝑍0

OR
a 10M
4

1. 𝑇𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠: 𝑉𝑎 = 𝐼𝑎 𝑍𝑓 … (1); 𝐼𝑏 = 0 … (2); 𝐼𝑐 = 0 … . (3) → (𝟎𝟐𝐌)

1 𝐼𝑎
2. 𝑠𝑦𝑚𝑚𝑡𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑟𝑒𝑙𝑡𝑖𝑜𝑛𝑠: 𝐼𝑎0 = [𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 ] = ; 𝐼𝑎1 = 𝐼𝑎2 = 𝐼𝑎0 . . (4)
3 3
𝐹𝑟𝑜𝑚 𝑇𝐶 ∶ 𝑉𝑎 = 𝐼𝑎 𝑍𝑓 ; 𝑉𝑎0 + 𝑉𝑎1 + 𝑉𝑎2 = 𝐼𝑎 𝑍𝑓 = 3𝐼𝑎0 𝑍𝑓 . . (5) → (𝟎𝟐𝐌)
3. 𝐼𝑛𝑡𝑒𝑟𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑦𝑚𝑚𝑡𝑟𝑖𝑐𝑎𝑙 𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠(𝐼𝑆𝑁): → (𝟎𝟐𝐌)
𝐹𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (4)&(5); 𝐼𝑆𝑁 𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑠𝑒𝑟𝑖𝑒𝑠 𝑐𝑜𝑛𝑛𝑒𝑡𝑖𝑜𝑛
𝐸𝑎
4. 𝑆𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑒𝑠: 𝐼𝑎0 = 𝐼𝑎1 = 𝐼𝑎2 = … (6)
[𝑍1 + 𝑍2 + 𝑍0 + 3𝑍𝑓 ]
𝑍2 + 𝑍0 + 3𝑍𝑓
𝑉𝑎1 = 𝐸𝑎 − 𝐼𝑎1 𝑍1 = 𝐸𝑎 [ ] … (7)
[𝑍1 + 𝑍2 + 𝑍0 + 3𝑍𝑓 ]
𝐸𝑎 𝑍2
𝑉𝑎2 = −𝐼𝑎2 𝑍2 = − [ ] … (8);
[𝑍1 + 𝑍2 + 𝑍0 + 3𝑍𝑓 ]
𝐸𝑎 𝑍0
𝑉𝑎0 = −𝐼𝑎0 𝑍0 = − [ ] … (9) → (𝟎𝟐𝐌)
[𝑍1 + 𝑍2 + 𝑍0 + 3𝑍𝑓 ]
𝐸𝑎
5. 𝐹𝑎𝑢𝑙𝑡 𝑐𝑢𝑟𝑟𝑒𝑛𝑡: 𝐼𝑓 = 𝐼𝑎 = 3𝐼𝑎0 = . . (10); 𝑖𝑓 𝑍𝑓 = 0, 𝑡ℎ𝑒𝑛𝐼𝑓 = 0
[𝑍1 + 𝑍2 + 𝑍0 + 3𝑍𝑓 ]
→ (𝟎𝟐𝐌)
→ (𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 = 𝟏𝟎𝐌)
b Analyses and develop the relation between: 10M
i). Phase voltages in terms of symmetrical components (05M)
𝑉𝑏1 𝑙𝑒𝑎𝑑𝑠 𝑉𝑎1 𝑏𝑦 240° 𝑎𝑛𝑑 𝑉𝑏1 𝑙𝑒𝑎𝑑𝑠 𝑉𝑐1 𝑏𝑦 120°
𝑉𝑏1 = 𝑉𝑎1 ∠240° = 𝑎2 𝑉𝑎1 & 𝑉𝑐1 = 𝑉𝑎1 ∠120° = 𝑎𝑉𝑎1 ;
𝑠𝑖𝑚𝑖𝑙𝑙𝑒𝑟𝑙𝑦 𝑉𝑏2 = 𝑉𝑎2 ∠240° = 𝑎2 𝑉𝑎2 & 𝑉𝑐2 = 𝑉𝑎2 ∠120° = 𝑎𝑉𝑎2
1 1
𝑉𝑎1 = [𝑉𝑎 + 𝑎𝑉𝑏 + 𝑎2 𝑉𝑐 ]& 𝑉𝑎2 = [𝑉𝑎 + 𝑎2 𝑉𝑏 + 𝑎𝑉𝑐 ] &𝑉𝑎𝑜 = 𝑉𝑏𝑜 = 𝑉𝑐𝑜
3 3
𝑉𝑎 = 𝑉𝑎𝑜 + 𝑉𝑎1 + 𝑉𝑎2 . . (1); 𝑉𝑏 = 𝑉𝑏𝑜 + 𝑉𝑏1 + 𝑉𝑏2 . . (2); 𝑉𝑎 = 𝑉𝑐𝑜 + 𝑉𝑐1 + 𝑉𝑐2 . . (3);
The equations 1-3, can be expreesed interms of ‘a,
𝑉𝑎 = 𝑉𝑎𝑜 + 𝑉𝑎1 + 𝑉𝑎2 . . (3); 𝑉𝑏 = 𝑉𝑎𝑜 + 𝑎2 𝑉𝑎1 + 𝑎𝑉𝑎2 . . (4); 𝑉𝑎 = 𝑉𝑎𝑜 + 𝑎𝑉𝑎1 + 𝑎2 𝑉𝑎2 . . (5);
In matrix form
𝑉𝑎 1 1 1 𝑉𝑎𝑜
[𝑉𝑏 ] = [1 𝑎2 𝑎 ] [𝑉𝑎1 ]
𝑉𝑐 1 𝑎 𝑎2 𝑉𝑎2
ii).Prove that a balanced set of three phase voltages will have only positive sequence
components of voltages only. (05M).
𝑉𝑎 = 𝑉𝑎 ; 𝑉𝑏 = 𝑎2 𝑉𝑎 ; 𝑉𝑐 = 𝑎𝑉𝑎 ; we have
𝑉𝑎𝑜 𝑉𝑎 𝑉 + 𝑎2 𝑉𝑎 + 𝑎𝑉𝑎
1 1 12 1 𝑉𝑎 1 1 12 1 1 𝑎
[𝑉𝑎1 ] = [1 𝑎 𝑎 ] [𝑉𝑏 ] = [1 𝑎 𝑎 ] [𝑎 𝑉𝑎 ] = [𝑉𝑎 + 𝑎3 𝑉𝑎 + 𝑎3 𝑉𝑎 ]
2
3 3 3
𝑉𝑎2 1 𝑎 𝑎 2 𝑉𝑐 1 𝑎 𝑎 2 𝑎 𝑉𝑎 𝑉𝑎 + 𝑎 4 𝑉𝑎 + 𝑎 2 𝑉𝑎
𝑉𝑎𝑜 𝑉 + 𝑎 𝑉𝑎 + 𝑎 2 𝑉𝑎 𝑉𝑎 (1 + 𝑎 + 𝑎2 )
1 𝑎 1 1 0 3𝑉𝑎
𝑉𝑎1 = [𝑉𝑎 + 𝑎3 𝑉𝑎 + 𝑎3 𝑉𝑎 ] = [ 3𝑉𝑎 ] = [3𝑉𝑎 ] = = 𝑉𝑎
𝑉𝑎2 3 𝑉𝑎 + 𝑎𝑉𝑎 + 𝑎2 𝑉𝑎 3
𝑉𝑎 (1 + 𝑎 + 𝑎2 )
3
0
3

→ (𝟎𝟓 + 𝟎𝟓 = 𝟏𝟎𝑴)
 BMS COLLEGE OF ENGINEERING, BANGALORE-19
(Autonomous College under VTU)
CONTINUOUS INTERNAL EVALUATION THIRD TEST 2025
Section: 5TH SEM YEAR BACTH: 2022
Department: ELECTRICAL AND ELECTRONICS ENGINEERING Semester: FIFTH
Course Title: POWER Course code: Date: 08.01.2025 Time:09.45PM-11.00 AM
SYSTEMS ANALYSIS 22EE5PCPSA
Signature of FIC with Date: Dr.C.lakshminarayana. 08.01.2025
ANSWER THE FOLLOWING
Q QUESTION M
PART A
1 a Define dynamic stability 02M
b What are the assumptions made in stability studies 03M
PART B
2 a Analyses and develop the relation for Power-angle equation of a salient pole Synchronous 07M
machine..(CO3, PO2)
b Evaluate the SSSL of a system consisting of a generator of equivalent reactance 0.5pu 08M
connected to an infinite bus through a series reactance of 1.0pu. The terminal voltage of the
generator is held at 1.2 pu and voltage of the infinite bus is 1.0pu. (CO3 PO2)
PART C
a Analyses and develop expression for Swing Equation and swing Curve of trasient stability. 10M
3
(CO3 PO2)
b Analyses and develop an expression for application of equal area criterion with the case of 10M
Sudden loss of one of the parallel lines. (CO3 PO2)
OR
a i). Analyses and develop expression for critical clearing angle and critical clearing time. 10M
4
(06M)
ii). A 50Hz generator is delivering 50% of the power that it is capable of delivering trough
a transmission line to an infinite bus. a fault occurs that increase the reactance between the
generators & the infinite bus to 500% of the value before the fault. When the fault is isolated
the maximum power that can be delivered is 75% of the original maximum value. Evaluate
the critical clearing angle for the condition described. (04M) (CO3 PO2)
b A loss free alternator supplies 50MW to an infinite bus, the SSSL being 100MW. Evaluate 10M
if the alternator will remain stable if the input to the prime mover of the alternator is abruptly
increased by 40 MW. (CO3 PO2)
 THIRD CIE Solution for POWER SYSTEM ANALYSIS : 22EE5PCPSA
Q QUESTION M
PART A
1 a It is the ability of a power system to remain in synchronism after the initial swing (transient stability 02
period) until the system has settled down to the new steady state equilibrium condition .→ 𝟎𝟐𝑴
b Assumptions made in stability studies. 03
(i). Machines represents by classical model
(ii). The losses in the system are neglected (all resistance are neglected)
(iii). The voltage behind transient reactance is assumed to remain constant.
(iv). Controllers are not considered ( Shunt and series capacitor )
(v). Effect of damper winding is neglected.→ 𝟎𝟑𝑴
PART B
2 a A salient pole synchronous machine has a number of projecting (salient) poles. Hence air gap is no- 07
uniform along the rotor periphery. It is least along the axis of the main poles (called the direct axis)
and maximum along the axis of the inter-polar region (called the quadrature axis). Hence, the flux
linkages is non-uniform . Correspondingly, the machine offers a direct axis reactance (Xd) and
quadrature axis reactance (Xq) for flow of armature current. A circuit model of the machine cannot
be easily drawn. However, the phasor diagram of machine neglecting its armature resistance is
shown in Fig.6

Iq
Vcosδ E
0
δ
῀ Vsinδ (jXq)Iq

G Infinite bus E
V (jXd)Id
Id Ia Fig.6. Phasor diagram

𝐿𝑒𝑡 𝐸∠𝛿 = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒𝑚𝑓 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑛𝑐ℎ𝑟𝑜𝑛𝑜𝑢𝑠 𝑚𝑎𝑐ℎ𝑖𝑛𝑒

𝛿 = 𝐿𝑜𝑎𝑑 𝑎𝑛𝑔𝑙𝑒 𝑜𝑟 𝑇𝑜𝑟𝑞𝑢𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑟 𝑝𝑜𝑤𝑒𝑟 𝑎𝑛𝑔𝑙𝑒
𝑋𝑑 = 𝐷𝑖𝑟𝑐𝑒𝑡 𝑎𝑥𝑖𝑠 𝑠𝑦𝑛𝑐ℎ𝑟𝑜𝑛𝑜𝑢𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒
𝐸𝑡 ∠𝜃 = 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑐ℎ𝑖𝑛𝑒
𝑋𝑞 = 𝑄𝑢𝑎𝑑𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑥𝑖𝑠 𝑠𝑦𝑛𝑐ℎ𝑟𝑜𝑛𝑜𝑢𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒
𝑉∠0° = 𝐵𝑢𝑠 𝑏𝑎𝑟 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 (𝑇𝑎𝑘𝑒𝑛 𝑎𝑠 𝑅𝑒𝑓𝑒𝑛𝑐𝑒 )
I = 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑑 𝑎𝑡 𝑙𝑎𝑔𝑔𝑖𝑛𝑔 𝑃𝑓 𝑜𝑓 𝑐𝑜𝑠Ф

Refering to the diagarm, we can write the expression for power developed as :
𝑃 = |𝑉|𝑐𝑜𝑠𝛿. |𝐼𝑞 | + |𝑉|𝑠𝑖𝑛 𝛿. |𝐼𝑑 | (1)

Also from the Fig.6, we observe that

|𝑉|𝑠𝑖𝑛𝛿
|𝐼𝑞 . 𝑋𝑞 | = |𝑉|𝑐𝑜𝑠𝛿 𝑂𝑅 𝐼𝑞 = (2)
𝑋𝑞
|𝐸 − 𝑉𝑐𝑜𝑠𝛿|
|𝐼𝑑 . 𝑋𝑑 | = |𝐸 − 𝑉𝑐𝑜𝑠𝛿| 𝑂𝑅 𝐼𝑑 = (3)
𝑋𝑑

Using equations (2)&(3) in equation(1), we get

|𝑉|𝑠𝑖𝑛𝛿 |𝐸 − 𝑉𝑐𝑜𝑠𝛿|
𝑃 = |𝑉|𝑐𝑜𝑠𝛿. + |𝑉|𝑠𝑖𝑛 𝛿.
𝑋𝑞 𝑋𝑑
𝑠𝑖𝑛2𝛿 |𝑉|. |𝐸|𝑠𝑖𝑛 𝛿 𝑠𝑖𝑛2𝛿
𝑃 = |𝑉|2 . + − |𝑉|2 .
2𝑋𝑞 𝑋𝑑 2𝑋𝑑
𝑠𝑖𝑛2𝛿 1 1 𝑠𝑖𝑛2𝛿
𝑃 = |𝑉|2 . [ − ] + |𝑉|. |𝐸|.
2 𝑋𝑞 𝑋𝑑 𝑋𝑑
|𝑉|. |𝐸| 2
|𝑉| . 𝑠𝑖𝑛2𝛿 𝑋𝑑 − 𝑋𝑞
𝑃= . 𝑠𝑖𝑛𝛿 + .[ ]
𝑋𝑑 2 𝑋𝑑 𝑋𝑞
 |𝑉|. |𝐸| |𝑉|2 . (𝑋𝑑 − 𝑋𝑞 )
𝑇ℎ𝑢𝑠 , 𝑃 = . 𝑠𝑖𝑛𝛿 + . 𝑠𝑖𝑛2𝛿 (4)
𝑋𝑑 2(𝑋𝑑 𝑋𝑞 )

As evident from equation.(4), there is a funadamental and a second harmonic component of power.
The first term is the same as far round-rotor machine with XS=Xd. This constitutes the major part of
power transfer.The second termis quit small(10-20%) compared to the first term and known as
relutance power.
The power angle curve of the machine is shown in Fig.7. It is noticed that the maxmimum power
output(SSSL) occurs at δ < 90° (about 70° ). This value of ‘δ’ at which the power flow is maximum can
be computed by equating the synchorising power co-efficient i.e.dp/dδ to zero.

b 𝐸−𝑉 1.2∠𝛿−1∠0° 08
𝐼= =
𝑗(𝑋𝑠 ) 1∠90°
1.2∠𝜃 − 1∠0°
𝐸 = 𝐸𝑡 + 𝑗𝐼. 𝑋𝑠 = 1.2∠𝛿 + 𝑗 [ ] 0.5 = 1.8∠𝜃 − 0.5
1∠90°
𝐸 = 1.8 cos(𝜃 − 0.5) + 𝑗1.8𝑠𝑖𝑛𝜃 (1)
SSSL is reached when generated emf ‘E’ has an angle of 90° ,i.e., δ=90° . This imposes the condition
that real part of equation.1 is zero. Thus
1.8 cos(𝜃 − 0.5) = 0 𝑜𝑟𝜃 = 73.87°
𝑁𝑜𝑤𝐸𝑡 = 1.2∠73.87° = 0.333 + 𝑗1.152
𝐸 − 𝑉 1.2∠73.87° − 1∠0° 0.333 + 𝑗1.152 − 1∠0°
𝐼= = =
𝑗(𝑋𝑠 ) 1∠90° 1∠90°
𝐼 = (1.152 + 𝑗0.667)
𝐸 = (0.333 + 𝑗1.152). 𝑗0.5(1.152 + 𝑗0.667)
𝐸 = −0.002 + 𝑗1.728 = 1.728∠90° , thus the SSSL is given by,
|𝐸| − |𝑉| 1𝑥1.728
𝑃𝑚 = 𝑆𝑆𝑆𝐿 = = = 1.152 𝑝𝑢 → 𝟎𝟐𝑴
(𝑋𝑠 + 𝑋𝑒 ) (0.5 + 1)
→ (𝟎 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 = 𝟎𝟖𝑴
PART C
a Swing Equation:The load angle or the torque angle ‘δ’ depends upon the loading of the machine. 10
3 Larger the loading, larger is the value torque angle. if some lod is added or removed from the shaft of
the synchronous machine; the rotor will decelerate or acclerate respectively with respect to the
sychronously rotating stator field and relative motion begins. It is said that the rotor is swinging with
respect to the stator field. The equation describing the relative motion of the rotor (load angle ‘δ’)
with respect to the stator field as function of time is called as swing equation → 𝟎𝟏𝑴
 Pe

Te

PS GENRATOR

ω
TS

Fig.6

ws Rotor field

δ
0 w0 Reference Rotating Axis
θ
Fig.7

Assuming that winding and frication losses to be negligiable, the accelarting torque on the rotor is
given by:𝑇𝑎 = 𝑇𝑠 − 𝑇𝑒 (1)
Multiplying by ‘w’ on both sides, we get:𝑤𝑇𝑎 = 𝑤𝑇𝑠 − 𝑤𝑇𝑒 → 𝟎𝟐𝑴
But; wTa=Pa= Accelerating power; wTs=Ps= mechnanical power
wTe=Pe= Electrical power output assuming that power losses is negliable.
∴ 𝑃𝑎 = 𝑃𝑠 − 𝑃𝑒 (2)
Under steady state conditions, Ps=Pe , so that Pa =0.
When Ps.Pe balance is disturbed, the machine undergoes dyanamic goverened by
𝑑2 𝜃 𝑑2 𝜃
𝑃𝑎 = 𝑇𝑎 𝑤 = 𝐼𝛼. 𝑤 = 𝑀. 𝑑𝑡 2 (3), 𝑤ℎ𝑒𝑟𝑒 𝛼 = 𝑑𝑡 2 = 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑡𝑜𝑟 ,
Since the angular position ‘θ’ of the rotor is continully varying with time,it is more convient tomeasuire
the angular position and velocity with respect to a synchronously rotating axis(Fig.7 )
From the Fig.7, it can be inferred that:𝛿 = 𝜃 − 𝑤0 𝑡 (4); Where,
W0=Angular velocity of the reference rotating axis;δ=Rotor angular displecement with respect to the
stator field
𝑑𝛿 𝑑𝜃 𝑑2 𝑡 𝑑2 𝜃
Taking time derivatives of Equation (4), 𝑑𝑡 = 𝑑𝑡
− 𝑤0 𝑎𝑛𝑑 𝑑𝑡 2
= 𝑑𝑡 2
(5)
𝑑2 𝛿
Combining equations (2),(3) and (5) we get, 𝑀 𝑑𝑡 2
= 𝑃𝑎 = 𝑃𝑠 − 𝑃𝑒 (6) → 𝟎𝟐𝑴
This equation is called as the swing equation of the synchronous machine. When the machine is
connected to infinite bus bars, then;
|𝐸||𝑉| 𝑑2 𝛿
𝑃𝑒 = − 𝑋 𝑠𝑖𝑛𝛿 = 𝑃𝑚 𝑠𝑖𝑛𝛿 𝑜𝑟𝑀 𝑑𝑡 2 = 𝑃𝑠 − 𝑃𝑚 𝑠𝑖𝑛𝛿 (7) → 𝟎𝟐𝑴
Swing Curve:The solution to swing equation gives the relation between rotor angle ‘δ’ as a function
of time ‘t’. The plot of ‘δ’ versus ‘t’ is called as swing curve. The exact solution of the swing equation
is however a very tedious task. Normally, step-by-step method or any other numerical solution
techniques like Euler’s method, Runge-Kutta’s method ae used for solving the swing equation. The
swing curve is used to determine the stability of the system. In case ‘δ’ increases indefinitely, it
indicates instability. Where as if it reaches a maximum will be damped out with time. A sample swing
curve is below. For the stability of the system ,dδ/dt=0 (8)
 The system will be un-stable if dδ/dt >0 for a sufficient long time (normally more than one second)
For the stabity of the system ,dδ/dt=0…(8); The system will be un-stable if dδ/dt >0 for a sufficent
long time (noramally more than one second) → 𝟎𝟑𝑴 → (𝟎𝟏 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟑 = 𝟏𝟎𝑴
b Sudden loss of one of the parallel lines: 10

Let us consider a single machine connected to an infinite bus through two parallel lines as shown in
Fig.4. The circuit model of the system is given in Fig.5.
The transient stability of the system when one of the lines is suddenly switched OFF from the system,
while operating under steady load conditions is now being considered.
Case-1: Before switching OFF, the power angle equation is,
|𝐸||𝑉|
𝑃𝑒1 = . 𝑠𝑖𝑛𝛿 = 𝑃𝑚1 𝑠𝑖𝑛𝛿 𝐶𝑢𝑟𝑣𝑒 1 → 𝟎𝟐𝑴
𝑋𝑠 +(𝑋1 ||𝑋2 )

Case-2: On switching OFF line-2, the power angle equation is,

|𝐸||𝑉|
𝑃𝑒2 = . 𝑠𝑖𝑛𝛿 = 𝑃𝑚2 𝑠𝑖𝑛𝛿 𝐶𝑢𝑟𝑣𝑒 2 → 𝟎𝟐𝑴
(𝑋𝑠 +𝑋1 )

→ 𝟎𝟐𝑴

The Fig.6 shows the two curves where in 𝑃𝑚2<𝑃𝑚1 as(𝑋𝑠 + 𝑋1 ) > (𝑋𝑠 + (𝑋1 ||𝑋2 )). As soon as line-2
was switched OFF, the original operating point ‘a’ on cruve-1 is shifted to a point ‘b’ on cruve-2.
Accelerating energy corresponding to area A1 is put into rotor followed by decelerating energy. If an
area A2 equal to A1 is found above the Ps line, the system will be stable, and finally at ‘c’ corresponding
to a new rotor angle𝛿1 > 𝛿0 . → 𝟎𝟐𝑴
For the limiting case of stability, 𝛿2 has a maximum value 𝛿𝑚 is given by
𝛿2 = 𝛿𝑚 = 180° − 𝛿1 ; Which is the same condition as the previous example. → 𝟎𝟐𝑴
→ (𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 = 𝟏𝟎𝑴
 OR
4 a Critical Clearing Angle and Critical Clearing Time: 10
In the previous case, if Ps is increased, then δ1 increase, area A1 increases and to find A2=A1, δ2 is
increased till it has a value δm, the maximum allowable limit for stability. Then the system said to be
critically stable. The angle δ1 is then called as the critical angle (δcc). The time corresponding to this is
called critical clearing time (tcc). It is possible to estimate critical clearing time using the swing curve.
This time is very much essentials in designing the protective circuit breakers for the system.
The case of critical stability of the system is shown in Fig.15.

Applying EAC to the above case, we get; A1=A2 → 𝟎𝟐𝑴

𝛿 𝛿 𝑃 𝑃
∫𝛿 (𝑃𝑠 − 𝑃𝑚2 𝑠𝑖𝑛𝛿)𝑑𝛿 = ∫𝛿 (𝑃𝑚3 𝑠𝑖𝑛𝛿 − 𝑃𝑠 )𝑑𝛿 ;Where 𝛿0 = 𝑠𝑖𝑛−1 (𝑃 𝑠 ); 𝛿𝑚 = 𝜋 − 𝑠𝑖𝑛−1 (𝑃 𝑠 )
𝑐𝑐 𝑚
0 𝑐𝑐 𝑚1 𝑚3
𝛿𝑐𝑐 𝛿𝑚
Integrating, we get ; {(𝑃𝑠 𝛿 + 𝑃𝑚2 . 𝑐𝑜𝑠𝛿)} = {(𝑃𝑚3 . 𝑐𝑜𝑠𝛿 − 𝑃𝑠 𝛿)}
𝛿0 𝛿𝑐𝑐
𝑶𝑹 𝑃𝑠 (𝛿𝑐𝑐 − 𝛿0 ) + 𝑃𝑚2 (cos𝛿𝑐𝑐 − cos 𝛿0 ) + 𝑃𝑠 (𝛿𝑚 − 𝛿𝑐𝑐 ) + 𝑃𝑚3 (cos𝛿𝑚 − cos 𝛿𝑐𝑐 ) = 0

𝑃𝑠 (𝛿𝑚 −𝛿0 )−𝑃𝑚2 cos𝛿0 +𝑃𝑚3 cos𝛿𝑚

𝑶𝑹 𝑐𝑜𝑠𝛿𝑐𝑐 = (1) → 𝟎𝟐𝑴
(𝑃𝑚3 −𝑃𝑚2 )
The angles in the above equation are in radians. If the angles are in degrees, the equation modifies
as below;
𝜋
180 ° 𝑃𝑠 (𝛿𝑚 − 𝛿0 ) − 𝑃𝑚2 cos𝛿0 + 𝑃𝑚3 cos𝛿𝑚
𝑐𝑜𝑠𝛿𝑐𝑐 = (2) → 𝟎𝟐𝑴
(𝑃𝑚3 − 𝑃𝑚2 )
II).Solution:Let Pm be the maximum power that can be delivered by the generator, from the given
data, we find that
1
𝑃𝑚1 = 𝑃𝑚 ; 𝑃𝑚1 = . 𝑃 = 0.2𝑃𝑚 = 0.75 𝑃𝑚
500 𝑚
100
𝐴𝑙𝑠𝑜 𝑃𝑠′ = 0.5 𝑃𝑚 𝑖𝑠 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟; 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑃𝑠 = 𝑃𝑚1 𝑠𝑖𝑛𝛿0
𝑃𝑠 0.5𝑃𝑚
∴ 𝛿0 = 𝑠𝑖𝑛−1 ( ) = 𝑠𝑖𝑛−1 ( ) = 30° → 𝟎𝟐𝑴
𝑃𝑚1 𝑃𝑚
𝑃 0.5𝑃𝑚
∴ 𝛿𝑚 = 180° − 𝑠𝑖𝑛−1 ( 𝑠 ) = 180° − 𝑠𝑖𝑛−1 ( ) = 138.2°; we have;
𝑃𝑚3 0.75𝑃𝑚
𝜋
180 ° 𝑃𝑠 (𝛿𝑚 − 𝛿0 ) − 𝑃𝑚2 cos𝛿0 + 𝑃𝑚3 cos𝛿𝑚
𝑐𝑜𝑠𝛿𝑐𝑐 =
(𝑃𝑚3 − 𝑃𝑚2 )
𝜋 ° ° ° °
180 ° . 0.5𝑃𝑚 (138.2 − 30 ) − 0.2𝑃𝑚 cos30 ) + 0.75cos(138. 2 )
𝑐𝑜𝑠𝛿𝑐𝑐 = = 0.385
0.75𝑃𝑚 − 0.2𝑃𝑚 )
∴ 𝛿𝑐𝑐 = 67.35° This is desired results→ 𝟎𝟐𝑴

b 10

Solution:The power angle curve of the alternator depicting the problem is shown in Fig.7
Let ‘δ0’ be operating power angle when load is 50MW. → 𝟎𝟐𝑴
we have 50 = 100𝑠𝑖𝑛𝛿 = 1.69𝑠𝑖𝑛𝛿; ∴ 𝛿0 = 30° → 𝟎𝟐𝑴
At point ‘b’, when load is 50+40=90MW, the operating angle is given as,
90
∴ 𝛿1 = 𝑠𝑖𝑛−1 (100) = 64° ; At point ‘c’, the operating angle ∴ 𝛿2 = 180° − 64° = 116°
→ 𝟎𝟐𝑴
From the power angle diagram, it can be observed that,
64°
𝜋
𝐴1 = ∫ (90 − 100 𝑠𝑖𝑛𝛿)𝑑𝛿 = 90(64° − 30° ) °
+ 100(𝑐𝑜𝑠64° − 𝑐𝑜𝑠30° ) = 10.6
30 ° 180
→ 𝟎𝟐𝑴
116°
𝜋
𝐴2 = ∫ (100 𝑠𝑖𝑛𝛿 − 90)𝑑𝛿 = −100(𝑐𝑜𝑠116° − 𝑐𝑜𝑠64° ) − 90(116° − 64° ) = 5.99
64 ° 180°
→ 𝟎𝟐𝑴
It can observed that from the results that the accelerating area A1 is greater than the decelerating
area A2. Hence the machine will fall out of synchronisation i.e, it will lose its stability when the input
is suddenly increased by 40 MW.→ (𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 + 𝟎𝟐 = 𝟏𝟎𝑴
Prerequisites: Transmission and Distribution, Electrical Energy Systems Course Description: This course
covers various techniques for analysis of different types of faults occurring in the Power System. Methods of
evaluating Power System Stability are also discussed.

Course Title POWER SYSTEMS

Course Code 22EE5PCPS Credits 3 L-T-P 2-1-0
CIE 50 Marks (100% weightage) SEE 100 Marks (50% weightage)
Prerequisites: Transmission and Distribution, Electrical Energy Systems
Course objectives:
The purpose of the course is to facilitate the learners to:
 Develop the mathematical model of given power system components
 Analyses the behavior of power systems under faulted conditions
 Study the stability studies of power systems under trainset conditions
UNIT-1 08Hrs
Representation of Power System Components: Circuit models of Transmission line, Synchronous
machines, Transformer and load. Single line diagram, impedance and reactance diagram. Per unit system,
per unit impedance and reactance diagrams.
Teaching-Learning Process Chalk and Board, Power Point Presentation.
UNIT-2 08Hrs
Symmetrical 3 - Phase Faults: Transients on a transmission line, Short-Circuit currents and the
reactance of synchronous machines on no load and on load, Short circuit of a loaded synchronous
machine, Short circuit current computation through Thevenin’s theorem.
Teaching-Learning Process Chalk and Board, Power Point Presentation.
UNIT-3 08Hrs
Symmetrical components - Resolution of unbalanced phasors into their symmetrical components,
Analysis of unbalanced load against balanced Three-phase supply, Analysis of balanced and unbalanced
loads against unbalanced 3 phase supply, Phase shift of symmetrical components in star-delta
transformer bank, Power in terms of symmetrical components. Sequence impedances and networks of
power system elements (alternator, transformer and transmission line) Sequence networks of power
systems.
Teaching-Learning Process Chalk and Board, Power Point Presentation.
UNIT-4 08Hrs
Unsymmetrical faults - Unsymmetrical faults on an unloaded alternator with and without fault
impedance. Unsymmetrical faults on a power system with and without fault impedance. Open conductor
faults in power system.
Teaching-Learning Process Chalk and Board, Power Point Presentation.

UNIT-5 08Hrs
Stability Analysis- Importance of stability analysis in power system planning and operation,
classification of power system stability, Rotor dynamics and the swing equation. Power angle equation,
Steady State Stability, synchronizing power coefficients, Transient Stability, Equal area criterion for
transient stability evaluation and its applications.
Teaching-Learning Process Chalk and Board, Power Point Presentation.

COURSE OUTCOME:
CO1 Model and analyses power systems using complex mathematical transformations under short
circuit and unbalanced conditions
CO2 Analyses different unsymmetrical faults on unloaded alternator and on complex power systems
using symmetrical component transformations
CO3 Apply mathematical techniques to evaluate system stability.
CO4 Ability to engage in independent study to make an effective presentation and submit report on
applications of power systems concepts in various domains
 CO-PO mapping
PO PO PO PO PO PO PO PO PO PO1 PO1 PO1 Tota
1 2 3 4 5 6 7 8 9 0 1 2 l
CO1 3
CO2 3
CO3 3
CO4 2 2 2

UNIT CHOICE:UNIT II and UNIT-V

Text books:
1. Elements of Power System Analysis- WD Stevenson, McGraw Hill Publications, 2nd
Edition,1994
2. Modern Power System Analysis-Nagrath & D P Kothari, Tata McGraw Hill
Publications, 3rd Edition, 2003
Reference books:
1. Elements of Power System Analysis- WD Stevenson, McGraw Hill Publications, 2nd
Edition,1994
2. Power System Analysis-Hadi Sadat, Tata McGraw Hill Publications, 3rd edition, 2002
3. Computer Techniques and Models in Power Systems- Uma Rao
4. Computer aided Power System Analysis- G L Kusic, CRC Press, 2nd edition, 2008
E-learning:
1. NPTEL Course titled: Computer Aided Power System Analysis.
Link: http://nptel.ac.in/courses/108107028/
S. USN NAME THEORY

QUIZ 1/AAT-05

QUIZ 2/AAT-05
NO.

ATTENDANC
MARKS -50

ATTENDED
TEST 3 - 20
TEST 1-20

TEST 2-20

% OF
HELD

E
1 1BM22EE001 ADARSH KASHYAP AB 13 18 4 5 40

2 1BM22EE003 AITHA NITYA AB 16 19 5 5 45

3 1BM22EE004 AJAY KUMAR A M 12 14 20 4 5 43

4 1BM22EE005 AKASH P 15 17 AB 4 5 41
VAKKALAD
5 1BM22EE006 AKSHAY BAHUBALI 11 16 20 4 5 45
PATIL
6 1BM22EE007 AKSHAY GOWDA H AB 17 19 4 5 45
K
7 1BM22EE008 ANAGHA SURESH AB 16 20 4 5 45
8 1BM22EE010 ANURAG VINAY 18 AB 19 5 5 47
NAIR
9 1BM22EE011 B NARASIMHA 19 20 AB 4 5 48
BHAT
10 1BM22EE012 BHOOMIKA H G AB 17 19 4 5 45

11 1BM22EE013 CHANDRAKALA AB 16 19 4 5 44

12 1BM22EE014 DEEPIKA G 17 AB 20 5 5 47

13 1BM22EE015 DHANUSH KUMAR AB 18 20 4 5 47

S
14 1BM22EE017 DINAKAR AB 19 20 4 5 48

15 1BM22EE018 GOPIKA P PRASAD AB 16 16 4 5 41

16 1BM22EE019 HARIKA L T AB 16 20 4 5 45

17 1BM22EE020 HARISH K 16 20 AB 4 5 45

18 1BM22EE021 HARSHITH D S AB 17 18 4 5 44

19 1BM22EE022 K G MONIKA 15 18 AB 4 5 42

20 1BM22EE023 K UDHYANA 17 18 AB 5 5 45
21 1BM22EE024 KAANISHKA N 02 15 17 4 5 41
22 1BM22EE025 LAXMIPUTRA 13 11 18 4 5 40
SHEGAJI
23 1BM22EE026 LOHITH R AB 13 15 4 5 37

24 1 BM22EE027 MAHAMMAD AB 16 16 4 5 41
SUHAIL
25 1BM22EE028 MIZBA ANJUM 15 0 AB 5 5 25

26 1BM22EE029 MOHAMMED 14 20 AB 5 5 44
ABRAR
27 1BM22EE030 MOHAMMED AB 17 16 4 5 45
OWAIS ABDUS
SAMAD
28 1BM22EE032 NEEL MANNUEL AB 7 13 4 5 29
SEBASTIAN
29 1BM22EE033 PAPPU SHAH AB 7 20 4 5 36

30 1BM22EE034 PAVANKUMAR B K AB 20 20 4 5 49

31 1BM22EE035 PRAJAPATI AB 20 14 4 5 43
KRISHNAKUMAR
MAHESH
32 1BM22EE036 PRATYUSH AB 13 20 5 5 43
PANDEY
33 1BM22EE037 PREETHAM P H AB 18 12 4 5 39

34 1BM22EE038 RADHA BIRADAR 15 14 AB 5 5 39

35 1BM22EE039 RAGHU KURUVA 13 16 17 4 5 42

36 1BM22EE040 SAKETH REDDY AB 10 20 4 5 39

37 1BM22EE042 SAKSHI SHARMA AB 20 20 5 5 50

38 1BM22EE043 SAMARTH AB 18 18 4 5 45
SADASHIV
GAVAROJI
39 1BM22EE044 SANANDAN R AB 13 18 4 5 40
40 1BM22EE045 SANCHITHA RAO 18 14 AB 5 5 42

41 1BM22EE046 SANIYA AHAMAD AB 20 20 4 5 49

42 1BM22EE047 SARFARAZ 11 13 AB 4 5 33
HUSSAIN
43 1BM22EE048 SARVVESH.R AB 11 20 4 5 40

44 1BM22EE050 SHREYAS.B AB 19 19 5 5 48

45 1BM22EE051 SHRIKANTH M 16 20 AB 5 5 46
KANNUR
46 1BM22EE052 SHWETA AB 16 20 5 5 46
SHIVANAND
AMBALI
47 1BM22EE053 SKANDA KUMAR AB 15 20 5 5 45
RAMAKANTH RAO
48 1BM22EE054 SKANDA R AB 20 18 5 5 48

49 1BM22EE055 SOURABH MAHESH AB 12 16 4 5 37

MAVINAKATTI
50 1BM22EE056 SRI GOWRI P G 20 15 AB 5 5 45

51 1BM22EE057 SUMANTH 20 19 AB 5 5 49
PURUSHOTTAM A
52 1BM22EE058 SUPRITHA B S AB 20 19 5 5 49
53 1BM22EE059 SUPRIYA B J 12 18 19 4 5 46

54 1BM22EE061 SYAD ALI AB 16 19 4 5 44

55 1BM22EE062 TEJASWINI M B 15 17 20 5 5 47

56 1BM22EE063 VAISHALI 18 20 AB 4 5 47

57 1BM22EE064 VIDYA K S 14 20 20 4 5 49

58 1BM22EE066 VISHAAL SRIDHAR AB 20 19 4 5 48

59 1BM22EE067 VISHNUVARDHAN AB 20 19 5 5 49
SV
60 1BM22EE068 PAWAN JOSHI 19 15 AB 4 5 43

61 1BM22EE069 SOORAJ 17 17 AB 4 5 43
SHREESHAIL
SANGANTTI
62 1BM23EE400 CHANDANA H AB 17 20 5 5 47

63 1BM23EE401 CHIRAG H T 18 19 AB 5 5 47

64 1BM23EE402 PRADEEP 13 20 AB 5 5 43
NAGGAPA BRAKAR
65 1BM23EE403 SAI CHARAN K 18 15 AB 5 5 43

66 1BM23EE404 SANJANA M 20 18 AB 5 5 48

67 1BM23EE405 YASHAS SINGH N 19 20 AB 5 5 49

RE REGSTRATED

1 1BM21EE002 A HUSSIAN BASHA 15 17 AB 4 5 41