Double Squirrel Cage Motor
The main disadvantage of a squirrel-cage motor is its poor starting torque, because of its low rotor resistance. The starting torque could be increased by having a cage of high resistance, but then the motor will have poor efficiency under normal running conditions (because there will be more rotor Cu losses). The difficulty with a cage motor is that its cage is permanently short-circuited, so no external resistance can be introduced temporarily in its rotor circuit during starting period. Many efforts have been made to build a squirrel-cage motor which should have a high starting torque without sacrificing its electrical efficiency, under normal running conditions. The result is a motor, due to Boucheort, which has two independent cages on the same rotor, one inside the other. A punching for such a double cage rotor is shown in Fig. 35.26.
The outer cage consists of bars of a high-resistance metal, whereas the inner cage has low-resistance copper bars.
Hence, outer cage has high resistance and low ratio of reactance-to-resistance, whereas the inner cage has low resistance but, being situated deep in the rotor, has a large ratio of reactance-to-resistance. Hence, the outer cage develops maximum As said earlier, at starting and at large slip values, frequency of induced e.m.f in the rotor is high. So the reactance of the inner cage (= 2p f L) and therefore, its impedance are both high. Hence, very little current flows in it. Most of the starting current is confined to outer cage, despite its high resistance. Hence, the motor develops a high starting torque due to high-resistance outer cage. Double squirrel-cage motor is shown in Fig. 35.27.
As the speed increases, the frequency of the rotor e.m.f. decreases, so that the reactance and hence the impedance of inner cage decreases and becomes very small, under normal running conditions. Most of the current then flows through it and hence it develops the greater part of the motor torque.
In fact, when speed is normal, frequency of rotor e.m.f. is so small that the reactance of both cages is practically negligible. The current is carried by two cages in parallel, giving a low combined resistance.
Hence, it has been made possible to construct a single machine, which has a good starting torque with reasonable starting current and which maintains high efficiency and good speed regulation, under normal operating conditions.
The torque-speed characteristic of a double cage motor may be approximately taken to be the sum of two motors, one having a high-resistance rotor and the other a low-resistance one (Fig. 35.28).
Such motors are particularly useful where frequent starting under heavy loads is required.
Equivalent Circuit
The two rotor cages can be considered in parallel, if it is assumed that both cages completely link the main flux. The equivalent circuit for one phase of the rotor, as referred to stator, is shown in Fig.
If the magnetising current is neglected, then the figure is simplified to that shown in Fig.
Hence, R0¢/s and Ri¢/s are resistances of outer and inner rotors as referred to stator respectively and X0¢ and X i¢ their reactances
Total impedance as referred to primary is given by
Example 35.25. A double-cage rotor has two independent cages. Ignoring mutual coupling between cages, estimate the torque in synchronous watts per phase (i) at standstill and at 5 per cent slip, given that the equivalent standstill impedance of the inner cage is (0.05 + j 0.4) ohm per phase and of the outer cage (0.5 + j 0.1) ohm per phase and that the rotor equivalent induced e.m.f. per phase is 100 V at standstill.
Solution. The equivalent circuit of the double-cage rotor is shown in Fig. 35.33.
(i) At standstill, s = 1
The combined impedance of the two cages is
Example 35.27. The resistance and reactance (equivalent) values of a double-cage induction motor for stator, outer and inner cage are 0.25, 1.0 and 0.15 ohm resistance and 3.5, zero and 3.0 ohm reactance respectively. Find the starting torque if the phase voltage is 250 V and the synchronous speed is 1000 r.p.m. (I.E.E. London)
Solution. The equivalent circuit is shown in Fig. 35.34 where magnetising current has been neglected. At starting, s = 1
Tutorial Problems 35.4
1. Calculate the steps in a 5-section rotor starter of a 3-phase induction motor for which the starting current should not exceed the full-load current, the full-load slip is 0.018 and the rotor resistance is 0.015 W per phase
r1 = 0.46 W ; r2 = 0.206 W ; r3 = 0.092 W ; r4 = 0.042 W ; r5 = 0.0185 W
(Electrical Machinery-III, Kerala Univ. Apr. 1976)
2. The full-load slip of a 3-phase double-cage induction motor is 6% and the two cages have impedances of (3.5 + j 1.5) W and (0.6 + j 7.0) W respectively. Neglecting stator impedances and magnetising current, calculate the starting torque in terms of full-load torque.
[79%]
3. In a double-cage induction motor, if the outer cage has an impedance at standstill of (2 + j 2) ohm and the inner cage an impedance of (0.5 + j 5) W, determine the slip at which the two cages develop equal torques.
[17.7%]
4. The two independent cages of a rotor have the respective standstill impedance of (3 + j 1) ohm and (1 + j 4) ohm. What proportion of the total torque is due to the outer cage (a) at starting and (b) at a fractional slip of 0.05 ?
[(a) 83.6% (b) 25.8%] (Principle of Elect. Engg.I, Jadavpur Univ. 1975)
5. An induction motor has a double cage rotor with equivalent impedance at standstill of (1.0 + j 1.0) and (0.2 + j 4.0) ohm. Find the relative value of torque given by each cage at a slip of 5%.
[(a) 40.1 (b) 0.4 : 1] (Electrical Machines-I, Gwalior Univ. Nov. 1977)
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