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# 3-Phase Induction Motor Rotor Frequency, EMF, Current and Power Factor

## Rotor Current Frequency

The frequency of current and voltage in the stator of a 3-phase induction motor must be same as the supply frequency and is given by,

$$\mathrm{𝑓 =\frac{𝑁_{𝑆}𝑃}{120}… (1)}$$

But, *the frequency of the current and EMF in the rotor circuit* of the 3-phase induction motor is variable and depends upon the difference between the synchronous speed *(N _{S})* and the rotor speed

*(N*, i.e., on the slip. Thus, the rotor frequency is given by,

_{r})$$\mathrm{𝑓_{𝑟} =\frac{(𝑁_{𝑆} − 𝑁_{𝑟} )𝑃}{120}… (2)}$$

Now, from the equations (1) and (2), we get,

$$\mathrm{\frac{𝑓_{𝑟}}{𝑓}=\frac{𝑁_{𝑆} − 𝑁_{𝑟}}{𝑁_{𝑆}}}$$

$$\mathrm{∵ \:Slip,\:𝑠 =\frac{𝑁_{𝑆} − 𝑁_{𝑟}}{𝑁_{𝑆}}}$$

$$\mathrm{∴ 𝑓_{𝑟} = 𝑠𝑓 … (3)}$$

i. e. , Rotor Current Frequency = Per unit slip × Supply frequency When the rotor is stationary, i.e., N_{r} = 0, then,

$$\mathrm{𝑠 =\frac{𝑁_{𝑆} − 𝑁_{𝑟}}{𝑁_{𝑆}}= 1}$$

$$\mathrm{∴ \: 𝑓_{𝑟} = 𝑓}$$

Hence, when the rotor is stationary, the frequency *(f _{r})* of the rotor current is the same as that of the supply frequency

*(f)*.

When the rotor picks up speed, the relative speed between the rotating magnetic field and the rotor decreases. As a result of this, the slip (s) and hence the rotor current frequency decreases.

At synchronous speed, i.e., *N _{r} = N_{S}*,

$$\mathrm{𝑠 =\frac{𝑁_{𝑆} − 𝑁_{𝑟}}{𝑁_{𝑆}}= 0}$$

$$\mathrm{∴ 𝑓_{𝑟} = 0}$$

## Rotor EMF

When the rotor is stationary, the 3-phase induction motor behaves as a 3-phase transformer with secondary winding short circuited. Thus, the per phase induced EMF in the rotor (or secondary) is given by,

$$\mathrm{Rotor\:EMF/Phase,\:𝐸_{2} = 𝐸_{1} × \frac{𝑁_{2}}{𝑁_{1}}= 𝐾\:𝐸_{1} … (3)}$$

Where,

**E**_{1}= Per phase stator voltage.**N**_{1}= Number of turns in stator winding per phase.**N**= Number of turns in rotor winding per phase._{2}

When the rotor is running at slip ‘s’, then the relative speed between the rotating magnetic field of the stator and the rotor is *(N _{S} – N_{r})*. Therefore, the rotor EMF is directly proportional to the

*(N*or slip (s), i.e.

_{S}– N_{r})$$\mathrm{Rotor\:EMF/Phase\:at\:slip\:𝑠,\:{𝐸^{′}_{2}} = 𝑠\:𝐾\:𝐸_{1} … (4)}$$

## Rotor Current and Power Factor

Consider a 3-phase induction motor at any slip value ‘s’ as shown in the figure below.

Here, the rotor is assumed to be wound rotor and is connected in star. Therefore,

$$\mathrm{Rotor\:EMF\:Phase,𝐸^{′}_{2} = 𝑠\:𝐸_{2 }\:\:\:(let\:𝐾 = 1)}$$

$$\mathrm{Rotor\:reactance/Phase,{𝑋^{′}_{2}} = 𝑠\:𝑋_{2}}$$

Where, X_{2} is the rotor reactance per phase at standstill condition.

The resistance of the rotor circuit is R_{2} per phase and is independent of the frequency and hence does not depend upon slip. Similarly, the resistance (R_{1}) and reactance (X_{1}) of the stator winding do not depend upon slip.

Since the 3-phase induction represents a balanced 3-phase load, then we need to analyse one phase only and the conditions in the other two phases being similar.

### Case 1 – When the rotor is stationary

When the rotor is stationary, the motor is at standstill (slip, s = 1)

$$\mathrm{Rotor\:current/phase ,\:𝐼_{2} =\frac{𝐸_{2}}{𝑍_{2}}=\frac{𝐸_{2}}{\sqrt{{𝑅_{2}}^{2} + {𝑋_{2}}^{2}}}… (5)}$$

$$\mathrm{Rotor\:Power\:Factor,\:cos \:φ_{2}=\frac{R_{2}}{𝑍_{2}}=\frac{R_{2}}{\sqrt{{𝑅_{2}}^{2} + {𝑋_{2}}^{2}}}… (6)}$$

### Case 2 – When the motor is running at slip ‘s’

$$\mathrm{Rotor\:current/phase,\:{𝐼^{′}_{2}}=\frac{𝑠\:𝐸_{2}}{𝑍^{′}_{2}}=\frac{𝑠\:𝐸_{2}}{\sqrt{{R_{2}}^{2} +(SX_{2})^{2}}}}…… (7)$$

$$\mathrm{Rotor\:Power\:Factor,\:cos\:{φ^{′}_{2}} =\frac{R_{2}}{𝑍^{′}_{2}}=\frac{R_{2}}{\sqrt{{𝑅_{2} }^{2}+(SX_{2})^{2}}}}… (8)$$

## Numerical Example 1

A 3-phase, 50 Hz induction motor has 8 poles and operates with a slip of 4 % at a certain load. Determine the frequency of the rotor current.

## Solution

$$\mathrm{Rotor\:current\:frequency, 𝑓_{𝑟} = 𝑠\:𝑓 = 0.04 × 50 = 2\:Hz}$$

## Numerical Example 2

A 3-phase, 4-pole induction motor is connected to a 50 Hz supply. The voltage induced in the rotor bar conductors is 5 V when the rotor is at standstill. Calculate the voltage and frequency induced in the rotor conductors at 500 RPM.

## Solution

$$\mathrm{Synchronous\:speed,\:𝑁_{𝑆} =\frac{120\:𝑓}{𝑃}=\frac{120 × 50}{4}= 1500\:RPM}$$

$$\mathrm{Slip,\:𝑠 =\frac{𝑁_{𝑆} − 𝑁_{𝑟}}{𝑁_{𝑆}}=\frac{1500 − 500}{1500}= 0.67}$$

Therefore, corresponding to the slip ‘s’,

$$\mathrm{Induced\:voltage\:in\:rotor,\:𝑉^{′}_{2} = 𝑠\:𝑉_{2} = 0.67 × 5 = 3.35\:V}$$

$$\mathrm{Rotor\:frequency,\:𝑓_{𝑟} = 𝑠\:𝑓 = 0.67 × 50 = 33.5\:Hz}$$

## Numerical Example 3

An 8-pole, 3-phase, 50 Hz induction motor is running at full load with a slip of 5 %. The rotor is star connected and its per phase resistance and standstill reactance are 0.35 Ω and 2 Ω respectively. The EMF between slip rings is 150 V. Determine the rotor current per phase and rotor power factor. Assuming the slip rings are short circuited.

## Solution

The per phase rotor EMF at standstill is,

$$\mathrm{𝐸_{2} =\frac{150}{\sqrt{3}}= 86.61\:V}$$

$$\mathrm{Rotor\:EMF/phase\:at\:full\:load,\:𝐸^{′}_{2} = 𝑠\:𝐸_{2} = 0.05 × 86.61 = 4.33\:V}$$

$$\mathrm{Rotor\:reactance/phase\:at\:full\:load,\: 𝑋^{′}_{2} = 𝑠\:𝑋_{2} = 0.05 × 2 = 0.1\:Ω}$$

$$\mathrm{Rotor\:impedance/phase\:at\:full\:load, \:𝑍^{′}_{2}=\sqrt{{0.35}^{2} + {0.1}^{2}}= 0.364\:Ω}$$

Therefore,

$$\mathrm{Rotor\:current/phase =\frac{𝐸^{′}_{2}}{𝑍^{′}_{2}}=\frac{4.33}{0.364}= 11.895\:A}$$

$$\mathrm{Rotor\:Power\:Factor =\frac{𝑅_{2}}{𝑍^{′}_{2}}=\frac{0.35}{0.364}= 0.96\:(lagging)}$$

- Related Questions & Answers
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