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Induction motor slip
- Thread starter Jk1996
- Start date
BrianPetersen
Mechanical
The correlation that you describe is correct but the way you are describing it suggests that you haven't gotten the cause and effect right.
electricpete
Electrical
To elaborate Brian's point you might roughly view that changes in torque originate at the load.
load torque increases -> motor slip increases -> motor current increases.
(we neglect the effect of any small change in motor speed upon the load)
This pattern holds up unless load torque increase to/beyond breakdown torque, at which point motor torque can't increase any further (in fact starts decreasing) but slip and current will continue to increase.
=====================================
(2B)+(2B)' ?
load torque increases -> motor slip increases -> motor current increases.
(we neglect the effect of any small change in motor speed upon the load)
This pattern holds up unless load torque increase to/beyond breakdown torque, at which point motor torque can't increase any further (in fact starts decreasing) but slip and current will continue to increase.
=====================================
(2B)+(2B)' ?
I'd add a few things to that:
load torque increases -> motor rotor slows, so motor slip increases -> higher slip produces an increase in voltage induced in the rotor -> rotor current increases -> motor torque increases to equal load torque -> motor reaches stable speed at new higher slip value, i.e., lower rotor speed, at the end of the transient. Simultaneously, as the rotor speed slows it reduces counter emf induced in stator windings -> stator current increases to new higher stable value at the end of the transient. Note that some rotor speed oscillations may occur but die out by the end of the transient until there is stable operation, assuming breakdown torque isn't reached as electricpete mentioned.
xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
load torque increases -> motor rotor slows, so motor slip increases -> higher slip produces an increase in voltage induced in the rotor -> rotor current increases -> motor torque increases to equal load torque -> motor reaches stable speed at new higher slip value, i.e., lower rotor speed, at the end of the transient. Simultaneously, as the rotor speed slows it reduces counter emf induced in stator windings -> stator current increases to new higher stable value at the end of the transient. Note that some rotor speed oscillations may occur but die out by the end of the transient until there is stable operation, assuming breakdown torque isn't reached as electricpete mentioned.
xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
Dear Mr. Jk1996 (Electrical)(OP)19 Dec 21 15:26
"...Is it true that the greater the slip in an induction motor the the greater the motor torque which causes an increase in motor current?..."
1. Both learned advice by Messrs. Brain and electripete are correct.
2. In general, most induction asynchronous motors having these characteristics. See 3. below.
2.1. Attention: this is NOT applicable at very low speed, say during starting.
3. (Pay particular attention to) [the region between say 0.5-1 syn-speed (= 0.5-0 slip)], on:
a) "torque vs speed" curve,
b) "current vs speed" curve.
Che Kuan Yau (Singapore)
"...Is it true that the greater the slip in an induction motor the the greater the motor torque which causes an increase in motor current?..."
1. Both learned advice by Messrs. Brain and electripete are correct.
2. In general, most induction asynchronous motors having these characteristics. See 3. below.
2.1. Attention: this is NOT applicable at very low speed, say during starting.
3. (Pay particular attention to) [the region between say 0.5-1 syn-speed (= 0.5-0 slip)], on:
a) "torque vs speed" curve,
b) "current vs speed" curve.
Che Kuan Yau (Singapore)
Re-reading your textbooks [assuming you've hung onto them: I did, and have never once regretted that decision] should go a long way to refreshing your memory and answering your questions.
Full disclosure, that's not completely true; I did let go of some of them, and before long repented of my sin and redeemed that error by finding copies elsewhere and restoring them to my collection . . .
Since then, I often scour the bins at flea markets, trunk sales, book and thrift stores and like venues for references, handbooks, and so on, and have stumbled upon some real gems; I heartily recommend this to others as well.
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
Full disclosure, that's not completely true; I did let go of some of them, and before long repented of my sin and redeemed that error by finding copies elsewhere and restoring them to my collection . . .
Since then, I often scour the bins at flea markets, trunk sales, book and thrift stores and like venues for references, handbooks, and so on, and have stumbled upon some real gems; I heartily recommend this to others as well.
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
To be very specific: higher slip in a squirrel cage induction machine does not equate to it being able to develop higher torque across all speeds.
It ALWAYS equates to higher torque at ZERO SPEED.
It MAY equate to higher torque at specific speed points below about 50% of synchronous speed.
It will USUALLY equate to LOWER torque as you approach closer to synchronous speed (50-80%, for example).
It will ALWAYS have slightly LOWER maximum breakdown torque.
Think of the typical speed-torque curve. A "standard" low slip (e.g. slip < 3%) design has the torque generally sloping from lower left to upper right as speed increases, representing a net increase in developed (and thus available) torque. A "mid-slip" design (e.g. 3% < slip < 5%) has a fairly horizontal slope: no real difference, although it is very slightly higher at the right compared to the left. A "high slip" design (e.g. slip > 5%) will generally slope from upper left to lower right.
Converting energy to motion for more than half a century
It ALWAYS equates to higher torque at ZERO SPEED.
It MAY equate to higher torque at specific speed points below about 50% of synchronous speed.
It will USUALLY equate to LOWER torque as you approach closer to synchronous speed (50-80%, for example).
It will ALWAYS have slightly LOWER maximum breakdown torque.
Think of the typical speed-torque curve. A "standard" low slip (e.g. slip < 3%) design has the torque generally sloping from lower left to upper right as speed increases, representing a net increase in developed (and thus available) torque. A "mid-slip" design (e.g. 3% < slip < 5%) has a fairly horizontal slope: no real difference, although it is very slightly higher at the right compared to the left. A "high slip" design (e.g. slip > 5%) will generally slope from upper left to lower right.
Converting energy to motion for more than half a century
In answer to your original question I offer this graphic which supports the ideas described above.
Now
Assume your motor is driven by a VFD (also called an inverter drive).
Set the drive to a specific speed (ignore the time lag from the speed control loop)
at any set constant speed, slip angle is approximately constant for a particular shaft torque.
Squirrel cage motors are not designed for high slip continuous operation, just for starting. The rules above get relaxed somewhat with wound rotor motors which are designed for high slip operation depending on the amount of resistance inserted into the rotor circuit.
There are some other uncommon motor winding arrangements that can tolerate high slip. It is unlikely that any of us will encounter them, as the VFD drive arrangement can mimic almost all of the unusual winding arrangements.
Now
Assume your motor is driven by a VFD (also called an inverter drive).
Set the drive to a specific speed (ignore the time lag from the speed control loop)
at any set constant speed, slip angle is approximately constant for a particular shaft torque.
Squirrel cage motors are not designed for high slip continuous operation, just for starting. The rules above get relaxed somewhat with wound rotor motors which are designed for high slip operation depending on the amount of resistance inserted into the rotor circuit.
There are some other uncommon motor winding arrangements that can tolerate high slip. It is unlikely that any of us will encounter them, as the VFD drive arrangement can mimic almost all of the unusual winding arrangements.
- Thread starter
- #9
Thank you guys I’m really trying to get my head round these concepts so thanks for helping. One in struggling to understand is why does slip occur is this something to do with the RMF of the windings inducing an emf in the rotor bars and this causes some kind of lagging effect that makes the rotor never reach the synchronous speed?
electricpete
Electrical
without slip there is no current induced into the rotor. Without current in the rotor there is not rotor field and no torque (the rotor and stator fields interact to produce torque).
Imagine you are in a synchronuosly rotating reference frame. Then the stator is at rest so it can be represented as a stationary (in your frame) set of permanent magnets. You have a crank that you can use to apply torque to the rotor. If rotor is stationary in your frame (rotating synchronously) you don't push and there is no torque. As you start to push the rotor it passes the stator field and induces a voltage in the rotor which causes current to flow roughly proportional to the speed (slip speed in normal ref frame) which creates I^2*R losses in the resistance of your rotor proportional to slip speed squared. As you increase speed (propostional to s) the power is increase proportional to speed squared (s^2) so you have to apply roughly linearly increasing torque. All of this is intuitive in the synchronous ref frame and can be transformed to the normal stationary ref frame using standard rules (current and torque don't change between reference frames, speeds, voltages and powers do) and you can derive the same results as the equivalent circuit, except that this thought experiment is valid within the range of slips far below the slip at which breakdown slip occurs... at this point the role of series stator inductance starts to become importance and it is not included in that synchronous ref frame model.
All of the above is simply an attempt to explain the torque vs speed (or torque vs slip) curve in the range of low slips. You could probably look in textbooks and find a variety of explanations.
=====================================
(2B)+(2B)' ?
Imagine you are in a synchronuosly rotating reference frame. Then the stator is at rest so it can be represented as a stationary (in your frame) set of permanent magnets. You have a crank that you can use to apply torque to the rotor. If rotor is stationary in your frame (rotating synchronously) you don't push and there is no torque. As you start to push the rotor it passes the stator field and induces a voltage in the rotor which causes current to flow roughly proportional to the speed (slip speed in normal ref frame) which creates I^2*R losses in the resistance of your rotor proportional to slip speed squared. As you increase speed (propostional to s) the power is increase proportional to speed squared (s^2) so you have to apply roughly linearly increasing torque. All of this is intuitive in the synchronous ref frame and can be transformed to the normal stationary ref frame using standard rules (current and torque don't change between reference frames, speeds, voltages and powers do) and you can derive the same results as the equivalent circuit, except that this thought experiment is valid within the range of slips far below the slip at which breakdown slip occurs... at this point the role of series stator inductance starts to become importance and it is not included in that synchronous ref frame model.
All of the above is simply an attempt to explain the torque vs speed (or torque vs slip) curve in the range of low slips. You could probably look in textbooks and find a variety of explanations.
=====================================
(2B)+(2B)' ?
Induction motors work on alternating current.
AC devices depend on the presence of some frequency of AC to operate.
The generation of magnetic flux in an induction motor's rotor depends upon the difference in speed between the rotating magnetic field generated in the stator windings and the rotor itself, meaning that at the instant of start-up from rest the squirrel cage will, by transformer action, have an AC current induced in it identical to the frequency of the AC supply from the mains.
In the case of starting a motor not connected to any mechanical load, the frequency of the rotor AC will drop as the speed of the motor rises.
Once the motor reaches its idling, barely sub-synchronous no-load speed, the current generated in the rotor's squirrel cage bars will be alternating current with a frequency that will be a fraction of one Hertz, which some sources describe as a quasi-DC current. The rotor speed curve will be asymptotic, since by definition the closer to synchronous speed the rotor speed becomes the less transformer action there will be to induce a magnetic field in the rotor; if the motor in question were coupled to a DC machine that could supply rotative power to it, it would be possible to increase the speed of the AC motor so that the frequency in the rotor would become zero and the motor itself would generate zero torque.
Indeed if at this point more rotative power was supplied to the AC induction motor, driving it into super-synchronous territory, it would begin to function as an induction generator.
Read your textbooks, and if they don't explain all this, find other ones.
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
AC devices depend on the presence of some frequency of AC to operate.
The generation of magnetic flux in an induction motor's rotor depends upon the difference in speed between the rotating magnetic field generated in the stator windings and the rotor itself, meaning that at the instant of start-up from rest the squirrel cage will, by transformer action, have an AC current induced in it identical to the frequency of the AC supply from the mains.
In the case of starting a motor not connected to any mechanical load, the frequency of the rotor AC will drop as the speed of the motor rises.
Once the motor reaches its idling, barely sub-synchronous no-load speed, the current generated in the rotor's squirrel cage bars will be alternating current with a frequency that will be a fraction of one Hertz, which some sources describe as a quasi-DC current. The rotor speed curve will be asymptotic, since by definition the closer to synchronous speed the rotor speed becomes the less transformer action there will be to induce a magnetic field in the rotor; if the motor in question were coupled to a DC machine that could supply rotative power to it, it would be possible to increase the speed of the AC motor so that the frequency in the rotor would become zero and the motor itself would generate zero torque.
Indeed if at this point more rotative power was supplied to the AC induction motor, driving it into super-synchronous territory, it would begin to function as an induction generator.
Read your textbooks, and if they don't explain all this, find other ones.
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
As the shaft torque increases, the slip increases (assuming constant gap flux)
As the magnetic flux reduces (assuming constant shaft torque) the slip increases.
So an increasing slip can indicate an increasing shaft torque and/or a reducing flux (Stator voltage or V/Hz ratio)
Mark Empson
Advanced Motor Control Ltd
As the magnetic flux reduces (assuming constant shaft torque) the slip increases.
So an increasing slip can indicate an increasing shaft torque and/or a reducing flux (Stator voltage or V/Hz ratio)
Mark Empson
Advanced Motor Control Ltd
- Thread starter
- #13
Thank you for all your in-depth replies. I’m trying to get my head round back emf reducing the current I know when the shaft is up to speed and unloaded current is low but as we load the shaft slip increases and more current is drawn from the supply. Is this due to the shaft emf not interact with the stator as much due to less speed so less back emf is produced?
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