0195136047.pdf

13.4 DIRECT-CURRENT MACHINES 607

RT 1 = 1. 56 + 0. 78 + 0. 39 + 0. 39 = 3. 12 

At starting, the counter emf is zero; the armature starting current is then

Ist=

Vt

RT 1

=

230

3. 12

= 73 .7A

By the time the armature current drops to its rated value of 37 A, the counter emf is

E= 230 − 3 × 12 × 37 = 230 − 115. 44 = 215 .57 V

The counter emf, when the motor is delivering rated load at a rated speed of 500 r/min with the
series starting resistor completely cut out, is

230 − 0. 39 × 37 = 230 − 14. 43 = 215 .57 V

The speed corresponding to the counter emf of 114.56 V is then given by

N=

114. 56

215. 57

× 500 = 265 .7 r/min

Step 2:The 1.56-step is cut out, leaving a total resistance of 3. 12 − 1. 56 = 1. 56 in the
armature circuit. The initial motor speed at this step is 265.7 r/min, which means that the counter
emf is still 114.56 V, if the effect of armature reaction is neglected. Accordingly, the resistance
drop in the armature circuit is still 115.44 V, so that

IaRT 2 = 115. 44 or Ia=

115. 44

1. 56

=74A

That is, if the inductance of the armature is neglected, the initial current is 74 A. With the final
current at 37 A, the counter emf is

E= 230 − 37 × 1. 56 = 230 − 57. 72 = 172 .28 V

corresponding to which the speed is

N=

172. 28

215. 57

× 500 = 399 .6 rpm

Step 3:The total resistance included in the armature circuit is 0.78at the beginning of this
step. The initial motor speed is 399.6 r/min, and the counter emf is 172.28 V. The initial armature
current is then (230−172.28)/0.78=57.72/0.78=74 A. With the final current at 37 A, the
counter emf is

E= 230 − 37 × 0. 78 = 230 − 28. 86 = 201 .14 V

corresponding to which the speed is

N=

201. 14

215. 57

× 500 = 466 .5 r/min

Thus, we have the results shown in Table E13.4.4.

TABLE E13.4.4

Step Current (A) Speed (r/min)
Number Initial Final Initial Final

1 74 37 0 266

2 74 37 266 400

3 74 37 400 467

4 74 37 467 500