DEDAN KIMATHI UNIVERSITY OF TECHNOLOGY
DEPARTMENT OF ELECTRICAL AND ELECTRICAL ENGINEERING
MASTER OF SCIENCE IN ELECTRICAL AND ELECTRONIC ENGINEERING
EEE6156; POWER SYSTEMS STABILITY AND CONTROL
CONTINUOUS ASSESSMENT TEST I
SUBMITTED BY: VINCENT KIPROTICH, E229-01-0087/2023
1. Draw the per-unit impedance diagram of the system shown in Fig below.
Assumed base values are 100MVA and 100kV.
Solution:
[10 Marks]
Per unit impedance diagram:
2. A synchronous generator of reactance 1.20 pu is connected to an infinite bus
(V = 1.0 p.u) through transformers and a line of total reactance of 0.60pu. The
generator no load voltage is 1.20 pu and its inertia constant is H = 4MW- s/MVA.
The resistance and machine damping may be assumed negligible. The system
frequency is 50Hz. Calculate the frequency of natural oscillations if the
generator is loaded to
[10 Marks]
Solution:
The total natural frequency of oscillation is given by:
1
 dPe
 2
H 1*4



 0.0255
n   d M  Where M 

f 50




EV
Pe 
sin 
X
dPe EV
1.2*1

cos  
cos   0.6667 cos 
d
X
1.2  0.6
(i) 50% loading,
sin   0.5
  30
1
 0.6667 cos 30  2
n  
  4.7584rad / sec
0.0255


4.7584
fn  n 
 0.7573Hz
2
2
(ii) 80 % of its maximum power limit
Solution:
sin   0.8
  53.13
1
 0.6667 cos 53.13  2
n  
  3.9607rad / sec
0.0255


3.9607
fn  n 
 0.6304 Hz
2
2
3. A 50Hz, synchronous generator capable of supplying 400MW of power is
connected to a large power system and is delivering 80MW when a threephase fault occurs at its terminals. Determine, the time in which the fault must
be cleared if the maximum power angle is to be 85 o. Assume H = 7 MJ/MVA
on a 100 MVA base. Also, calculate the critical clearing angle. [10 Marks]
Solution:
At the point of fault occurrence:
  1  85 
85  
 1.4835rad
180
pe  pmax sin 
pmax  400MW ,
pe  80MW
   0 , implying that;
pe  pm
  0
pe
80
 sin  0 
 0.2
pmax
400
11.54 
 0.2014rad
180
 c  cos 1 (cos 1  (1   0 ) sin  0 )  cos 1 (cos1.4835  (1.4835  0.2014)sin 0.2014)  1.4880rad
 0  sin 1 0.2  11.54 
The clearing time is given by;
On a 100MVA base, pm =0.8
tc 
2 H ( c   0 )
2  7(1.4880  0.2014)

 0.3786s
 fPm
  50  0.8