Series motor
A series motor is identical in construction to a shunt motor except for the field. The field is connected in series with the armature and must, therefore, carry the full armature current (Fig. 5.10a). This series field is composed of a few turns of wire having a cross section sufficiently large to carry the current.
Although the construction is similar, the properties of a series motor are completely different from those of a shunt motor. In a shunt motor, the flux F per pole is constant at all loads because the shunt field is connected to the line. But in a series motor the flux per pole depends upon the armature current and, hence, upon the load. When the current is large, the flux is large and vice versa. Despite these differences, the same basic principles and equations apply to both machines.
Although the construction is similar, the properties of a series motor are completely different from those of a shunt motor. In a shunt motor, the flux F per pole is constant at all loads because the shunt field is connected to the line. But in a series motor the flux per pole depends upon the armature current and, hence, upon the load. When the current is large, the flux is large and vice versa. Despite these differences, the same basic principles and equations apply to both machines.
Figure 5.10a. Series motor connection diagram b. Schematic diagram of a series motor
When a series motor operates at full-load, the flux per pole is the same as that of a shunt motor of identical power and speed. However, when the senes motor starts up, the armature current is higher than normal, with the result that the flux per pole is also greater than normal It follows that the starting torque of a senes motor is considerably greater than that of a shunt motor.This can be seen by comparing the T versus / curves of Figs 5. 8 and 5.11.
On the other hand, if the motor operates at less than full-load, the armature current and the flux per pole are smaller than normal. The weaker field causes the speed to rise in the same way as it would for a shunt motor with a weak shunt field. For example, if the load current of a senes motor drops to half its normal value, the flux diminishes by half and so the speed doubles. Obviously, if the load is small, the speed may nse to dangerously high values. For this reason we never permit a senes motor to operate at no-load It tends to run away, and the resulting centnfugal forces could tear the windings out of the armature and destroy the machine
Series motor speed control
When a senes motor carries a load, its speed may have to be adjusted slightly. Thus, the speed can be increased by placing a low resistance in parallel with the series field. The field current is then smaller than before, which produces a drop in flux and an increase in speed.
Figure 5.11 Typical speed-torque and current-torque characteristic of a series motor.
Conversely, the speed may be lowered by connecting an external resistor in senes with the armature and the field The total IR drop across the resistor and field reduces the armature supply voltage, and so the speed must fall.
Typical torque-speed and torque-current characteristics are shown in Fig. 5.11. They are quite different from the shunt motor characteristics given in Fig. 5.8b.
Example 5-5
A 15 hp, 240 V, 1780 r/min dc senes motor has a full-load rated current of 54 A. Its operating characteristics are given by the per-unit curves of Fig. 5.11.
Calculate
a. The current and speed when the load torque is
24 N×m
b. The efficiency under these conditions
Solution
a. We first establish the base power, base speed, and base current of the motor. They correspond to the full-load ratings as follows:
PB = 15 hp = 15 X 746 = 11 190 W
nB = 1780 r/min
/B= 54 A
nB = 1780 r/min
/B= 54 A
The base torque is, therefore,
A load torque of 24 N-m corresponds to a per-unit torque of
T(pu) = 24/60 = 0.4
Referring to Fig. 5.11, a torque of 0.4 pu is attained at a speed of 1.4 pu. Thus, the speed is
n = n(pu) X nB = I.4 X 1780
= 2492 r/min
= 2492 r/min
From the T vs / curve, a torque of 0.4 pu requires a current of 0.6 pu. Consequently, the load current is
/ = /(pu) X /B = 0.6 X 54 = 32.4 A
b. To calculate the efficiency, we have to know Po and Pi.
Pi = EI = 240 X 32.4 = 7776 W
Po = nT/9.55 = 2492 X 24/9.55
= 6263 W
h = Po/Pi = 6263/7776 = 0.805 or 80.5%
Po = nT/9.55 = 2492 X 24/9.55
= 6263 W
h = Po/Pi = 6263/7776 = 0.805 or 80.5%
< /div >
No comments:
Post a Comment