Why there are two KW ratings of Three phase induction motor?

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Muhammad Umer ansar

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I came accross a three phase induction motor with two KW ratings written on its nameplate against two frequencies 50hz and 60hz. My question is why KW rating increases with frequency? shouldnt the KW rating depend upon design of motor? and why does current rating too?
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To a rough approximation, the output limit of an induction motor is _torque_. Motor heating is roughly proportional to the square of the current flow though the machine, and torque similarly scales with current.

_Power_ is the product of torque and speed, and drive frequency changes _speed_. So when you operate the motor at higher frequency, you get higher speed, and thus higher power for the same torque.

If you look closely at the name plate, the 400V 50 Hz delta rating and the 460V 60Hz delta rating show similar current ratings, very different speed ratings, and very different kW ratings.

-Jon
 
Even if you have same torque at each speed, think about it, that higher speed is moving the driven load farther in a same amount of time period - therefore it is doing more work in the same amount of time.
 
winnie said:
... If you look closely at the name plate, the 400V 50 Hz delta rating and the 460V 60Hz delta rating show similar current ratings, very different speed ratings, and very different kW ratings.

-Jon
Click to expand...
Agree.
And the current and kW in the second rating shown for a 460V 60Hz delta is lower than the first one because it's under a smaller mechanical load with a 1.1% slip at 1780 RPM, vs. 1.4% at 1775 RPM in the first rating (a 1800 RPM synchronous speed at 60 Hz).
The 400V 50 Hz delta is specified at 1475 RPM which is a 1.67% slip.
 
First you have to understand that the kW rating on the nameplate of a motor is not the electrical kW, it is the maximum steady state MECHANICAL output power. The definition is;
Mechanical kW = Torque (in N-m) x RPM / 9550. So if the RPM changes, so does the mechanical kW.

This is a time when having a unit like HP for the mechanical output power is a little less confusing.
HP = Torque (in lb-ft) x RPM / 5250

To find the maximum ELECTRICAL kW you would need to know the rated Efficiency (Eff) of the motor at full load.
 
In terms of the data on the nameplate, electrical kW is V*A*sqrt(3)*cos. The cos term, whose full notation is cos(phi), is the power factor, and the other terms are the voltage and current respectively. For the 460 delta 60 Hz, you get 13.5kW. Compare to the 12.6 kW mechanical, and you see that the efficiency is 93%, which is consistent with what you expect from the 92.4% that the nampelate indicates.
 
winnie said:
To a rough approximation, the output limit of an induction motor is _torque_. Motor heating is roughly proportional to the square of the current flow though the machine, and torque similarly scales with current.

_Power_ is the product of torque and speed, and drive frequency changes _speed_. So when you operate the motor at higher frequency, you get higher speed, and thus higher power for the same torque.

If you look closely at the name plate, the 400V 50 Hz delta rating and the 460V 60Hz delta rating show similar current ratings, very different speed ratings, and very different kW ratings.

-Jon
Click to expand...
Thanks Jon. You wrote current rating will stay same as it is proved by nameplate. But does normal current drawn increase with frequency?
and what about efficiency? because i see a slight upstream in Efficiency value at higher frequency ie 60hz
 
Jraef said:
First you have to understand that the kW rating on the nameplate of a motor is not the electrical kW, it is the maximum steady state MECHANICAL output power. The definition is;
Mechanical kW = Torque (in N-m) x RPM / 9550. So if the RPM changes, so does the mechanical kW.

This is a time when having a unit like HP for the mechanical output power is a little less confusing.
HP = Torque (in lb-ft) x RPM / 5250

To find the maximum ELECTRICAL kW you would need to know the rated Efficiency (Eff) of the motor at full load.
Click to expand...
Got it thanks buddy
 
Muhammad Umer ansar said:
Thanks Jon. You wrote current rating will stay same as it is proved by nameplate. But does normal current drawn increase with frequency?
and what about efficiency? because i see a slight upstream in Efficiency value at higher frequency ie 60hz
Click to expand...

The current drawn by the motor depends on the connected mechanical load. If the load at the two frequencies presents the same torque, and the supply voltage is adjusted to maintain the same magnetization, then the current should be about the same at the two frequencies.

The dominant losses in the motor are associated with torque production. At higher speed you convert more power for the same torque, so at higher speed you generally see better efficiency.

Jon
 
Muhammad Umer ansar said:
Thanks Jon. You wrote current rating will stay same as it is proved by nameplate. But does normal current drawn increase with frequency?
and what about efficiency? because i see a slight upstream in Efficiency value at higher frequency ie 60hz
Click to expand...
Many things can come into play depending on what the load type is and how much it actually is loaded.

Constant torque loads, say a conveyor belt, if you connect it to 60 Hz lets say that belt moves at 1 foot per second. Doesn't matter a lot what the total load is on the motor it draws whatever it takes to move the load at that speed . Now connect same machine with same load on it to 50 Hz, first thing that is noticeably different is it is moving at about 83% the speed it was at 60Hz, which means it will be moving product on that conveyor at 83% the rate it was before. Keep in mind the rated volts @ 50Hz is also only 400 instead of 460 so total VA input drops in similar proportion.

Variable torque loads however draw more load in proportion to speed as speed increases. So a blower that draws right at motor nameplate rating if run at 400V 50 Hz will very likely be overloaded if you switch it to 460 V 60 Hz.

If you run a 60 Hz motor on a VFD at 30 Hz you basically are operating it at half the speed, half the voltage, but still up to full load rated current - which basically gives you half the nameplate HP rating as max allowable in this situation. short term overload is acceptable just like it is at full voltage and full frequency.
 
winnie said:
The current drawn by the motor depends on the connected mechanical load. If the load at the two frequencies presents the same torque, and the supply voltage is adjusted to maintain the same magnetization, then the current should be about the same at the two frequencies.

The dominant losses in the motor are associated with torque production. At higher speed you convert more power for the same torque, so at higher speed you generally see better efficiency.

Jon
Click to expand...

The 1/min ratings on this nameplate, are very close to half the grid frequency (i.e. 1500 RPM at 50 Hz, and 1800 RPM at 60 Hz). This implies that the topology of the motor magnets and solenoids is set up so it rotates half a turn for every grid voltage cycle.

What is the reason why this nameplate RPM ratings aren't exactly 1500 RPM and 1800 RPM respectively?
Also, it seems to be a loss of 25 rpm at both 50 Hz and 60 Hz, rather than a common percentage loss in RPM. What is the reason the difference in rotation rate from theoretical rotation rate is subtraction, rather than division?
 
Carultch said:
The 1/min ratings on this nameplate, are very close to half the grid frequency (i.e. 1500 RPM at 50 Hz, and 1800 RPM at 60 Hz). This implies that the topology of the motor magnets and solenoids is set up so it rotates half a turn for every grid voltage cycle.

What is the reason why this nameplate RPM ratings aren't exactly 1500 RPM and 1800 RPM respectively?
Also, it seems to be a loss of 25 rpm at both 50 Hz and 60 Hz, rather than a common percentage loss in RPM. What is the reason the difference in rotation rate from theoretical rotation rate is subtraction, rather than division?
Click to expand...

Slip.

An induction motor will always have slip, unless the load is in a condition where inertia of the driven load is greater than output of the motor, usually temporary, in that case the slip is negative or inverted slip and the motor is temporarily operating like a generator and putting energy back into the source instead of drawing from it.

Basic induction motors have no torque if there is no slip and therefore can not operate at synchronous speed.
 
Carultch said:
What is the reason why this nameplate RPM ratings aren't exactly 1500 RPM and 1800 RPM respectively?
Also, it seems to be a loss of 25 rpm at both 50 Hz and 60 Hz, rather than a common percentage loss in RPM. What is the reason the difference in rotation rate from theoretical rotation rate is subtraction, rather than division?
Click to expand...
Because without slip, there would be no torque developed. The drag of the load causes the motor torque to increase to compensate, which causes the motor's current to increase.
 
Carultch said:
The 1/min ratings on this nameplate, are very close to half the grid frequency (i.e. 1500 RPM at 50 Hz, and 1800 RPM at 60 Hz). This implies that the topology of the motor magnets and solenoids is set up so it rotates half a turn for every grid voltage cycle.

What is the reason why this nameplate RPM ratings aren't exactly 1500 RPM and 1800 RPM respectively?
Click to expand...

It's obviously a 4-pole motor and so it takes two cycles of the applied AC for each revolution of the rotor at a synchronous speed.

As others have said there needs to be slip for torque to be produced. If the rotor is spinning in synchronism with the rotating magnetic field created by the stator, then from the vantage point of the rotor the magnetic field it experiences is constant (as if it was created by a DC current). But a changing magnetic flux is necessary to induce any current in the shorted turns of the rotor's squirrel cage. And so without any current in these turns, they will produce no magnetic field that can interact with the magnetic field of the stator and thereby create torque.

It might be helpful to think of the rotor being "strobed" by the rotating magnetic field from the stator. So when there's a positive or negative slip it's like the wagon wheels in an old movie appearing to spin slowly one way or the other, depending on the difference of the wheel rotation rate and the movie frame rate. Strictly speaking for the stobe effect to occur the rotation rate can be close to an integer multiple of the strobe frequency (aka "aliasing"), and not necessarily at the same frequency.
 
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