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Induction Motor •Why induction motor (IM)? –Robust; No brushes. No contacts on rotor shaft –High Power/Weight ratio compared to Dc motor –Lower Cost/Power –Easy to manufacture –Almost maintenance-free, except for bearing and other mechanical parts •Disadvantages –Essentially a “fixed-speed” machine –Speed is determined by the supply frequency –To vary its speed need a variable frequency supply

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Construction Stator

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Construction Squirrel Cage Rotor

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Construction

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**Performance of Three-Phase Induction Motor**

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Example 5.1 A 3-phase, 460 V, 100 hp, 60 Hz, four‑pole induction machine delivers rated output power at a slip of Determine the: (a) Synchronous speed and motor speed. (b) Speed of the rotating air gap field. (c) Frequency of the rotor circuit. (d) Slip rpm. (e) Speed of the rotor field relative to the rotor structure. Stator structure. Stator rotating field. (f) Rotor induced voltage at the operating speed, if the stator‑to‑rotor turns ratio is 1 : 0.5. Solution: (b) 1800 (same as synchronous speed)

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**Equivalent Circuit of the Induction Motor**

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**Equivalent Circuit of the Induction Motor**

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**IEEE‑Recommended Equivalent Circuit**

Theveninequivalent circuit

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If , then,

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**Tests To Determine The Equivalent Circuit**

No‑load test

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Locked‑rotor test

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**Blocked‑rotor equivalent circuit for improved value for**

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Example 4.2 A no‑load test conducted on a 30 hp, 835 r/min, 440 V, 3‑phase, 60 Hz squirrel‑cage induction motor yielded the following results: No‑load voltage (line‑to‑line): 440 V No‑load current: 14 A No‑load power: 1470 W Resistance measured between two terminals: 0.5 The locked‑rotor test, conducted at reduced voltage, gave the following results: Locked‑rotor voltage (line‑to‑line): 163 V Locked‑rotor power: 7200 W Locked‑rotor current: 60 A Determine the equivalent circuit of the motor. Solution: Assuming the stator windings are connected in way, the resistance per phase is: From the no‑load test:

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**From the blocked‑rotor test**

The blocked‑rotor reactance is:

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Example 5.3 The following test results are obtained from a three-phase 60 hp, 2200 V, six‑pole, 60 Hz squirrel‑cage induction motor. (1) No‑load test: Supply frequency = 60 Hz, Line voltage = 2200 V Line current = 4.5 A, Input power = 1600 W (2) Blocked‑rotor test: Frequency = 15 Hz, Line voltage = 270 V Line current = 25 A, Input power = 9000 W (3) Average DC resistance per stator phase: 2.8 (a) Determine the no‑load rotational loss. (b) Determine the parameters of the IEEE‑recommended equivalent circuit (c) Determine the parameters (Vth, Rth, Xth) for the Thevenin equivalent circuit of Fig.5.16.

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**(a) No-Load equivalent Circuit (b) Locked rotor equivalent circuit**

= .

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impedance at 15 Hz is: The blocked‑rotor reactance at 15 Hz is Its value at 60 Hz is at 60 Hz

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(c)

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**PERFORMANCE CHARACTERISTICS**

Where or

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At low values of slip,

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**At larger values of slip,**

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**Torque‑speed profile at different voltages.**

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Maximum Torque Then

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. Torque speed characteristics for varying

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If is small (hence is negligibly small) Then Then

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If is small (hence is negligibly small)

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Efficiency Power flow in an induction motor. The power loss in the stator winding is:

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Ideal Efficiency

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= N.m

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(c) (i) (c) (ii)

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Note that for parts (a) and (b) it is not necessary to use Thevenin equivalent circuit. Calculation can be based on the equivalent circuit of Fig.5.15 as follows:

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Solution: From no load test: From blocked rotor test:

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Input impedance Input power:

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Stator CU losses: Air gap power Rotor CU losses Mechanical power developed: From no load test:

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**Note that the equivalent circuit parameters are not given**

Note that the equivalent circuit parameters are not given. Therefore equivalent circuit parameters cannot be used directly for computation.)a) The synchronous speed is

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Solution:

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Solution:

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