advanced Power system engineering analysis .pdf

power system engineering is a very crucial discipline of study in electrical engineering.

1. A 60 Hz, 4 pole turbo-generator rated 100 MVA, 13.8 kV has
an inertia constant of 10 MJ/MVA.

(i)
(ii)

(iii)

Find the stored energy in the rotor at synchronous
speed.

If the input to the generator is suddenly raised to 60 MW
for an electrical load of 50 MW, find the rotor
acceleration.

If the rotor acceleration calculated in part (ii) is
maintained for 12 cycles, find the change in torque angle
and rotor speed in rpm at the end of this period.

2. i) Shred energy = Bil
= (loo mua) (12m3/m%a)
= 1000 MT

= 447

Roter acieleradin (x)= fn = Mena
m

M= a Su.

px
1690

< 108 du des /

LE Ware

3. v gel we

= (2 Sec

_ b:277 0704 fx
meh Ted.

vpn = 1-37 pr

4. rx 60 eae reals

= Anno x2 rub weds

y

= One box 2 bO TP”
4
dr

190 rpm

% we Wst As
olt

So + 18

5. ous generator, capable of developing 500mW y

power angle of 8 © By how much can the inpu
ıcreased suddenly without loss of stability? Assume that Pmax

constant

6. ‘The transient stability of the Synchronous generator connected to the bus is analyzed by equal area criteria, A machine can go through several kinds of disturbance like the

sudden loss of generation, sudden change in demand, fault in the system. EAC gives the basic standing of Operating range to ensure the safe operation of Synchronous
generator.

7. Eau Aven

8. 2. A 50Hz synchronous generator capable of supplying
600MW of power is connected to a larger power system
and is delivering SOMW when a three-phase fault occurs at

its terminals, determine (a) the time in which the fault must
be cleared if the maximum power angle is to be -85°
assume H=5MJ/MVA on a 100MVA base (b) the critical

9. ON

10. A synchronous generator is connected to an infinite bus and supplying 0.45 pu of its
maximum power capacity. A fault occurs and the reactance between the generator and

the line becomes four times its value before the fault. The maximum power that can

be delivered after the fault is cleared is 70% of the original maximum value. Determine
the critical clearing angle.

12. Two generators, rated at 200 MW and 300 MW, are supplying power to a network. Both the

eir individual full rated capacity and the system frequency is

The load on the system decreases by 150 MW and the frequency rises by 0.5 Hz.

Compute the droop (speed regulation) R of each gene:

Assume that the load is decreased on each generator in proportion to their individual rating.

13. (1st optiin is correct.)

Initially system is operating at 50HZ frequency with 50% of their respective rated load.
Let

P1 = 200 MW (operating at 100MW)

P2 = 350 MW (operating at 175 MW)

Now with decrease in 1SOMW load frequency increases to 50.5 HZ
Also load decrease is directly proportional to individual rating.
Hence decrease in

P1 = 150x(2/5) = 60 MW

Hence new load of P1 = 100-60 = 40MW

Decrease in P2 = 150 - 60 = 90MW

new load of P2 = 175 - 90 = 85 MW

for load 1 decrease in 60 MW increases frequency by 0.5 HZ

Hence droop R = 0.5/60 = -0.008
For load 2 decrease in 90 MW Increases frequency by 0.5 HZ

Hence droop R = 0.5/90 = -0.006 equivalent

14. 10 MKcal/MWh at 25% of rating
9 MKcal/MWh at 40% of rating
8 MKcal/MWh at 100% of rating
and the cost of the fuel is Rs 2 per MKcal. Find
(a) C(Pg) in the form of Eq. (7.4);

(b) the fuel input rate (heat rate) and fuel cost when 25%
100% loaded;

(c) the incremental cost in Rs./MWh;

(d) the approximate cost and the cost
tion in Rs/h to deliver 101 MW.

, 50%, and

using the quadratic approxima-

15. expression for the heat rate AP) which leads to the

Eq. (7.5) as
HIP
Substituting the measured values of
(01123) + 4025
(7/40) + 6" +40
(a//100) + #4 c' 100
Solving for the three unknown co
Thus
HP)
Multiplying by Pg, we get the fuel
per MKcal (Rs 2) we get fuel cost €
HP)
CP
25 MW and
FOS)
cas)
Similarly at 40% and 100% rating,
F(40)
140) =
FC00)
1100)

A1 25% rating, Pc

The incremental cost is
IC

8 Ge. 100 MW)
10=

Approximate cost at 101 MW is
(100) + 1C(100) x AP
AP¿= 101 - 100

At 100% rati

where,

Pr CP
HP) at different rati
10

MKcalMWh

cients, a’. b and e, we 1000/9. # = 40/9 and c' = 2/45

CHILI PG) + 4.44 + 0.0444 P¿ MKeal/ MWh

input rate F(P¿) in MKcal and multiplying by the cost of fuel
(Pas
WM

4.44 P + 0.0444 PL, MKcallh

+ 8.8807, + 0.0889P", Reh.

250,00 MKcalih
250.00 x 2 = 500.00 Rs

360.00 MKcal/h

720.00 Rs'h

1000.00 MKeat/h

2000.00 Rs/h

R/MWKk

ACW, VAP; = 8.889 + 0.1778P,

8.889 + 0.1778 x 100 = 26.667 Rs/MWh

2000 + 26.667 x Reh

IMW

1 = 2026.67

Cost, using the quadratic approximation, is

eo)

222.22 + 8.889101)

202675 Rs/h

0.0889(101)

16. O A two bus power system is shown in Figure 1. Incremental Fuel Costs of
the two generators are given as (IC): = (0.35 Por + 41) Rs/MW-hr;
(IC) = (0.35 Pos + 41) Rs/MW-hr and the Loss expression is
P, = 0.001(Pcz - 70). Determine the optimal scheduling and power loss

17. (2 E HOP, +4 990)

= 3% 4O O18 op, ES
09-0018, 2-14 Po + 824.9 =o

mo

Lata xa

18. 202 1925

1.3245

Lost

2.26R

1 +4} Boh wy
035x192 5 414)
= 108.2
os de

de
a ah

DL Me

15 31M

7.108.375 $/Mtr te