Factors Affecting Slip !!

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Load, of course. However for a same load if the rotor it is a wound-rotor induction motor you can control the speed [then the slip] by varying the exterior connected resistance.

If it is a squirrel cage rotor-it depends on how the bar and the slot are built [as dip bar, double cage and so on] the goal is to change the resistance-which it depends on skin effect mainly.

'... factors/parameters ( design part) on which the slip of an induction motor depend...'

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-Is the induction motor of the type female, or simply male operating in drag?

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-Is the slip supported via hip or shoulder?

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-How much has the induction motor been drinking?

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-What level of seriousness does one expect in replies to a post titled 'Factors affecting slip !!' (specifically the single-spaced, double-exclamation) that asks a question that begins as 'What all are...'?

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The non-design part is another story, entirely.

Dear Mr.Solar Eagle,

Good Humour. Laughed Loudly.

DHAYANANDHAN.S

Dear Mr. Sibin,

The equation for slip of Induction Motor is that the TORQUE is PROPORTIONAL To Slip/Resistance in the Rotor Circuit. i.e. T = K x (S/R) where T is Torque, K is constant, S is Slip, and R is Resistance in the Rotor, decided during Design Stage.

Note: This applies to only to SLIP-RING MOTOR and NOT Squirrel Cage Motor.

DHAYANANDHAN.S

By design:

Architecture and precision of the rotor cage and stator. (air gap) Iron volume and thickness of the sheets.

R2 & X2 Sibin! Aren't they?

Load and supply voltage.

I would assume slip varies linearly with load from zero at no load to manufacturer's figure (expressed as full-load speed) at design load. But somebody might disagree the details.

For voltage, there was a similar discussion back in October 2009, below is copy of a post then.

I came up with a formula relating motor speed to voltage, assuming, among other things, that back-EMF = applied volts at synchronous speed, and varies directly with speed.

ω_V2 = (V₂² _* ω_s - V₁² _* (ω_s - ω_V1)) / V₂²

^{where V}1 ^{= voltage 1, V}2^{= voltage 2, ω_V1 = speed at voltage 1, ω}V2^{= speed at voltage 2, ω}s ^{= synchronous speed.}

Agrees pretty well with published motor data.

As the speed increases slightly with increasing voltage, load power also increases, so the current falls a bit less than calculated from voltage ratios. In principle, the load could be modified to maintain constant power, but in practice it varies as speed³ for a squared-torque load, directly with speed for constant torque load.