Solved Assignment on Electric Drives and Power Systems

ASSIGNMENT
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Question 1
A four pole, 20 hp, 440 V motor is supplying its rated power to a centrifugal load at 60 Hz
frequency. Its rated speed is 1706 rpm. Calculate its speed, slip frequency, slip when it is
supplied by a 230 V, 50 Hz source.
Solution:
Speed:
n2 = f 2
f 1
∗n1= 50∗1706
60 =1421.7 rpm
Slip frequency:
ns= 120 f
p = 120∗50
4 =1500 rpm
f s=( ns−n2)
120 ∗p=1500−1421.7
120 ∗4=2.61 Hz
Slip:
s= ns−n2
ns
= 1500−1421.7
1500 =0.0522
Solution for question 1 in Matlab:
Nr=1706;
p=4;
f1=60;
f2=50;
% Speed of the motor
n2=f2/f1*Nr;
speed=n2
% slip frequency
ns=120*f2/p;
fslip=(ns-n2)*p/120
% slip
slip=(ns-n2)/ns
speed =
1.4217e+03
fslip =
2.6111
slip =
0.0522
Question 2
In the DC servo motor, a PI regulator is used in the speed loop to a transfer of the following
form:
Fω ( s ) = 1
1+ s ( 2 D
ωn )+ s2 ( 1
ωn
2 )
1
s
Where 𝐷=0.5 and 𝜔𝑛=314 𝑟ad/𝑠
i. Draw the bode plot of the closed loop transfer function if a gain 𝑘𝑝=60 is used for
the proportional regulator (use MATLAB).
Solution:

We will assume unity feedback i.e H(s)=1
T ( s ) = G ( s )
1+G ( s ) H ( s )
G ( s )= k p
s +s2
( 2 D
ωn )+ s3 ( 1
ωn
2 )
Therefore:
T ( s ) = k p
k p +s + s2
( 2 D
ωn )+ s3 ( 1
ωn
2 )
% i. Bode plot of the closed system with proportional regulator
kp=60;
D=0.5;
wn=314;
num=kp;
den=[1/(wn^2) 2*D/wn 1 kp];
sys=tf(num,den)
figure;
bode(sys)
sys =
60
-------------------------------------
1.014e-05 s^3 + 0.003185 s^2 + s + 60
Continuous-time transfer function.
-100
-80
-60
-40
-20
0
Magnitude (dB)
100 101 102 103 104
-270
-180
-90
0
Phase (deg)
Bode Diagram
Frequency (rad/s)
ii. What is the bandwidth of the closed loop system?
% ii. Bandwidth of the closed loop system
Bandwidth=bandwidth(sys)
Bandwidth =
80.2154