Advanced Electrical Machines and Drives Assignment Solution ENEE20002

This assignment solution addresses problems related to advanced electrical machines and drives. The first part analyzes a separately excited DC motor, calculating and plotting motor speed versus armature voltage under no-load and full-load torque conditions using MATLAB. The second part focuses on an inverter connected to the utility network through a transformer, deriving the control transfer function for the d and q axis loops and calculating the control loop bandwidth and damping. Finally, the solution examines a balanced, positive-sequence, Y-connected voltage source applied to a balanced Y-connected resistive load, calculating voltages and currents at the load terminals. The solution provides detailed calculations and MATLAB code for each problem, offering a comprehensive understanding of the concepts.

Electrical Machines and Drives
Solution 1:
Separately Excited DC Motor
Parameters :
Voltage = 500 V , 100 hp, 2500 r / min
Field Resistance = Rf = 109 Ohm
Rated field voltage = Vf0 = 300 V
Armature Resistance = Ra = 0.084 Ohm
Geometric constant = kf = 0.694 V / ( A rad / s )
Case 1 : Motor operating under no load
Vf = If Rf
Armature circuit :
Va = Ia Ra + Eg = Ia Ra + Kv . w . If
Back emf = Eg = Kv . w . If

W = Vs – Va / kv * If
If = Vf / Rf
Torque = k * Ia
Td = kt * If * ia
W = 500 – Va / ( 0.694 * 300 / 109 )
At constant field voltage = 300 V
Plot of motor speed versus the armature voltage ( as the armature voltage is varied from 250
V to 500 V ) using MATLAB software :
MATLAB Code :
V = 250 : 1 : 500 ;
w = ( 500 – V ) / ( 0.694 * 300 / 109 ) ;
plot ( V , w ) ;
xlabel ( ' Armature voltage ' ) ;
ylabel ( ' Motor Speed ' ) ;

Figure 1
Case 2 : Motor operating under rated full load torque
Plot of motor speed versus the armature voltage ( as the armature voltage is varied from 250
V to 500 V ) using MATLAB software :
MATLAB Code :
Va = 250 : 1 : 500 ;
W1 = ( 500 – Va ) / ( 0.694 * 300 / 109 ) ;
plot ( Va , w1 ) ;
xlabel ( ' Armature voltage ' ) ;
ylabel ( ' Motor Speed ' ) ;

Figure 2
Solution 2 :
Inverter  Transformer Utility network

Transformer can be modelled by a per phase series inductance of 0.8 mH and resistance of
0.4 Ohm
(i) Derivation of control transfer function for the ‘d’ and ‘q’ axis loops
Solution :
Z = R + w L
R = 0.4
L = 0.0008
Z = 0.4 + w 0.0008
(ii) If a q loop control with gains of kp = 14 and ki = 900 are used, then
Solution :
Control loop bandwidth = ?
1 + Kp ( s + ki / kp ) / s2
= s2 + kp ( s + ki / kp)
= s2 + 14 ( s + 900 / 14 )
= s2 + 14 s + 900
Damping for this loop = ?
W = 30
2 e w = 14
E = 14 / 2 w = 14 / 2 * 30 = 0 . 233333 = Damping coefficient

Bandwidth = 30 √ 1 – (2 * 0.233*0.233) + √(2-(4*0.233*0.233)+(4*0.233^4))
= 30 √0.8914 + 1.34 = 44.83 rad/s
Solution 3 :
Balanced Positive Sequence, Y connected voltage source
It is applied to a balanced Y connected resistive load of 30 Ohm.
Eab = 450 angle ( 0 degree ) V = 450 V
Line impedance between the source and the load = ZL = 1 angle ( 82 degree ) Ohm = 0.139
+ j 0.99 for each phase
Eca = 450 angle ( - 240 degree) = -225 + j 389.71 V
Ebc = 450 angle ( - 120 degree ) = -225 - j 389.71V
450 = √ 3 x Vp
Voltages ( at the load terminals in per unit ) = ?
Ean = 450 / √ 3 = 260.12 V
Ebn = 450 / √ 3 angle ( -120 degree ) = 260.12 angle ( -120 degree ) = -130.06 – j 225.27 V
Ecn = 450 / √ 3 angle ( - 240 degree ) = 260.12 angle ( -240 degree ) = -130.06 + j 225.27 V
Line / Load currents ( both values same )= ?

Ia = Ean / Zy = 450 / 30 x √ 3 = 8.67 A
Ib = Ebn / Zy = 450 / 30 x √ 3 angle ( -120 degree ) = 8.67 angle ( -120 degree ) = -4.335 – j
7.51 A
Ic = Ecn / Zy = 450 / 30 x √ 3 angle ( - 240 degree ) = 8.67 angle ( -240 degree ) = -4.335 + j
7.51 A