EEET2263 – Tutorial #5, DC Motors -SOLUTION

EEET2263 – Tutorial #5, DC Motors - SOLUTION

1.

* A separately excited dc generator turning at 1400 rpm produces an armature voltage of 127V under no load conditions. The armature resistance is 2. If the machine then delivers a current of 12 A into a load, calculate (a) The terminal voltage ANS: Va  E a  I a Ra  127  12  2  103 A (b)

The power dissipated in the armature winding ANS: Parm  I a2 Ra  144  2  288 W

(c)

2.

3.

The braking torque exerted by the generator on the prime mover. P 1524  10.4 Nm ANS: Pmech  Ea I a  127 12  1524 W Tmech  mech  2 1400 60 

* A separately excited dc generator produces a no load voltage of 127V. What happens to the output voltage if (a) the speed is increased by 20%? ANS: Va  127 1.2  152.4 V (b)

the direction of rotation is reversed? ANS: Va  127 V (sign reverses)

(c)

the exciting current in the field winding is increased by 10%? ANS: Va  127 1.1  139.7 V (assuming linear magnetic field relationship with If)

(d)

the polarity of the field winding is reversed? ANS: Va  127 V (sign reverses)

A 240kW, 500V, 1750 rpm separately excited dc generator has an overall efficiency of 94%. The shunt field winding has a resistance of 60, and the rated field winding current is 5A. The resistive loss in the armature winding is 5.52kW at full rated current. Calculate (a) the rated armature current. P 240000 ANS: I a ( rated )  a   480 A 500 Va (b)

the total losses in the machine at full rated current. ANS: Ploss  Pa  Pf  5520  60  52  7020 W

(c)

the resistive losses in the armature when the machine is operating at half rated current. 5520 ANS: Pa  Ra I a2   1380 W (losses reduce to a quarter for half current) 4 (d) the torque required from the prime mover to drive the generator at 1750 rpm at half rated current. ANS: Pmech  Po  Pa  Pf  120  1.380  1.500  122.9 kW (assume supplies field losses despite being separately excited. Then

generator

still

Tmech 

4.

5.

Pmech





122880  670.5 Nm 2 1750 60

* A 230V shunt dc motor has a nominal armature current of 60A. If the armature resistance is 0.15, calculate (a) the internal backemf across the armature ANS: Ea  Va  Ra I a  230  60  0.15  221 V (b)

the power supplied to the armature ANS: Parm  Va I a  230  60  13.8 kW

(c)

the mechanical output power developed by the motor ANS: Pmech  Parm  Ra I a2  13800  0.15  60 2  13.26 kW

For problem 4, calculate (a) the initial starting current if the motor is directly connected across the 230V supply before it starts moving V ANS: I a  a  230  1533 A 0.15 Ra (b)

the value of a resistor that would need to be connected in series with the armature to limit the initial current to 115A. V  230  2  Rext  2  0.15  1.85  ANS: Rtot  a 115 115

6.

A separately excited dc motor turns at 1200 rpm under no load when the armature is connected to a 115V dc source. Calculate the armature voltage required to make the motor run (a) at 1500 rpm under no load 1500 ANS: Va  115   143.8 V 1200 (b) at 100 rpm under no load 100 ANS: Va  115   9.58 V 1200

7.

*A 180kW, 230V, 435 rpm dc shunt motor has a nominal full-load current of 860A. Calculate (a) the total losses and efficiency at full load and rated speed P  180  91.0 % ANS: Pin  Va I a  230  860  197.8 kW   mech Pin 197.8 (b)

the approximate shunt field exciting current if the shunt field causes 20% of the total losses (assume the field winding is directly connected to the 230V supply) ANS: Ploss  197800  180000  17.8 kW  Pf  3560 W  I f  3560  15.5 A 230 (c) the value of the armature resistance and the backemf, if 50% of the total losses at full-load are attributable to the armature resistance ANS: Pa  8900 W  Ra  8900  0.012   Ea  230  0.012  860  219.7 V 860 2

(d)

the approximate exciting current to make the motor run at 1000 rpm under no load ANS: I  15.5  435 I f 1000 rpm   f 435rpm  435 rpm    6.74A 1000 rpm  1000

(e)

the motor efficiency if it is operating at half rated current at 1000 rpm. ANS: Ea (1000 rpm )  Va  I a Ra  230  430  0.012  224.8V  Pe (1000 rpm )  224.8  430  96.66 kW

Pmechloss( 435rpm )  17800  0.3  5340 W (losses not due to field and armature) 1000  12276 W (assume mechanical losses are 435 linearly proportional to speed) 435 I f (1000 rpm )  15.5   6.74 A  Pf (1000rpm)  230  6.74  1550 W 1100 (assume external resistor in series with field to limit current) P P 96660  12276 Hence   e (1000 rpm ) mechloss (1000 rpm )   84 % Pe (1000 rpm )  Pa  Pf 96660  430 2  0.012  1550

 Pmechloss (1000rpm)  5340 

D. G. Holmes and B. McGrath, March 2012