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Kw rating of a motor

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Jk1996

Electrical
Mar 14, 2021
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Hi, is the Kw rating on a motor mechanical output? I’m trying to get my head round if I can use a 7.5kw isolator for a 7.5kw motor?
 
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Jk1996: Strictly speaking, the mechanical output of a rotating machine is torque - not power. For what I think you're trying to accomplish, you need to know the nameplate current as well as the kilowatt rating. This is because the amount of current required to produce 7.5 kW at 120 V is different than how much you need at 460 V (or any other voltage). In general, the higher the voltage the lower the current. Usually the isolator rating includes BOTH power (kilowatt) and current (ampere).

Converting energy to motion for more than half a century
 
Thank you for your reply. I think I was getting confused as from what I believe a motor nameplate displays Kw ratings that are the mechanical output of the motor. The confusion then comes as the isolator is rated for the electrical power. Am I making sense?
 
Isolator? Disconnect switch or something like that?

Yes, they are typically rated in HP or kW at each line voltage even though the motor rating is shaft power. It's not that difficult to figure out input power for a certain shaft power.
 
Dear Mr Jk1996 (Electrical)(OP)26 Jan 23 13:51
" #1. is the Kw rating on a motor mechanical output? ..."

Yes. The kW rating on a motor is the mechanical output in kW. Note: the SI unit is kW , not Kw .

"#2. .... I can use a 7.5kw isolator for a 7.5kw motor? ..."
Take note that the Power (kW) is dependent on [(V x A x pf)/ (efficiency) ]
Same kW will have the current (A) reduced by half, with voltage (V) increased by two.....
In Europe the switching device is rated in Voltage and Current. See IEC documents.
In the US the switching device is rated in hp, voltage and current. See NEC for detail.
Note: the basic rating values are the voltage, current and the application , etc....
Attention: A say 400 V 20 A switching device may NOT be suitable for 100 V 80 A application. A switching device usually can be used at a lower voltage, but usually NOT suitable to take higher current, although the ( V x A) is the same !
Che Kuan Yau (Singapore)
 
Compositepro: Umm ... no. A squirrel cage induction motor will have the speed vary with mechanical load. More load means slower speed, until you get to a point where it stalls out. This means the torque profile is not a true constant (although it varies very little for small changes in load). A wound rotor induction motor behaves similarly, but since it has another power source for the rotor, it can limit the torque variation by varying the power applied to the rotor. Synchronous and direct current machines have even better torque regulation capability since they tend to have even better rotor power control.

Converting energy to motion for more than half a century
 
So when sizing the isolator we can’t just use the Kw rating on the isolator and compare it to the motor nameplate?
 
Dear Mr Jk1996 (Electrical)(OP)1 Feb 23 19:03
"....So when sizing the isolator we can’t just use the Kw rating on the isolator and compare it to the motor nameplate?... "
1. Yes. We can’t just use ......
Take note that kW (NOT Kw) is dependent on the voltage and the current. A switching device has to take the voltage and the current ratings into consideration. For any application, the voltage and the current rating shall NOT be exceeded; even the kW is within the range.
2. In Europe, refer the IEC standards. In the US refer the NEC.

Che Kuan Yau (Singapore)
 
So when sizing the isolator we can’t just use the Kw rating on the isolator and compare it to the motor nameplate?

Usually, yes. But, there are some motors that have a much higher FLA than normal so you can't do that for those.

A 100hp@480V motor is expected to have a FLA = 124A. So, the protection and control devices rated for 100hp@480V are designed expecting 124A. Slow speed 100hp motors might have the FLA in the 150-200A range so you couldn't just match the isolator to the motor HP rating for these motors.
 
kW is the motor output in SI units, equivalent to HP in Imperial units.
The HP or kW rating of an isolator switch is the maximum size of motor that may be disconnected with the isolator.
That is code.
There may be exceptions for very slow speed motors.
That is engineering experience.
And I know of an expensive failure on a 400 HP system.
Control gear completely destroyed.
400 HP motor shaft bent.
Coupling exploded.
Fan shaft bent.
That was ENGINEERING WITHOUT EXPERIENCE.
And the responsible engineer was warned, and chose to stand on his degrees and dismiss and rudely disrespect the man with less degrees but more experience.
Lionel Hutz said:
Usually, yes. But, there are some motors that have a much higher FLA than normal so you can't do that for those.
Well said, Lionel.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
I too have an "expensive" story relating to ratings confusion.

Got called into a client to investigate the repeated failure of some 250HP VFDs. Making a long story short, the client ordered six "250HP" drives, because they had "250HP" dewatering pumps. MOST of the right questions were asked and answered, except the crucial one. The motors were 12 pole slow speed pumps. A 4 pole 250HP pump shows as 302A in the NEC chart, the drives were rated 302A, and USUALLY a standard 4 pole motor comes in at slightly less than the FLC charts in the NEC. But these being 12 pole, were 348A FLA. Ooops... X 6!


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
 
Mechanical power output of the motor could be calculated by using the following formula:

Pmech = Pout = τ * ω

where

Pout: output power, measured in watts (W);
τ: torque, measured in Newton-meters (N•m) or foot-pounds (ft/lbs);
ω: angular speed, measured in radians per second (rad/s).
 
@ Dear Mr bidrohini (Electrical)5 Feb 23 11:44
" ...... Mechanical power output of the motor could be calculated by using the following formula: Pmech = Pout = τ * ω......where Pout: output power, measured in watts (W); τ: torque, measured in Newton-meters (N•m) or foot-pounds (ft/lbs);..."[/color
1. Please clarify " .....τ: torque, measured in Newton-meters (N•m) or foot-pounds (ft/lbs);...".
2. The statement uses the word " or "
Do you mean that SI unit 1 Newton-meter (N•m) = 1 foot-pound ?
Che Kuan Yau (Singapore)
 
Sorry,

P=τ * ω where, τ: torque, measured in Newton-meters (N•m)
1N•m= 0.738 ft-lbs

I want to edit my previous answer but can't find the edit option. Thank you for pointing it out.
 
Thanks guys this has been interesting. So can I just clarify the kW rating on an isolator is equivalent to the motor name plate kW?
 
OP said:
So can I just clarify the kW rating on an isolator is equivalent to the motor name plate kW?
Yes.
To avoid further confusion, the Ampere rating on a switch is intended for resistive loads.
The HP or kW raring on a switch considers the motor starting surge.
The HP or kW raring on a switch will be less that the product of the rated Amps and rated Volts.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
Dear Mr Jk1996 (Electrical)(OP)7 Feb 23 18:40
"..... So can I just clarify the kW rating on an isolator is equivalent to the motor name plate kW? ...."
1. The over simplified answer is NO.
2. In the US , the NEC is the law.
(a) There are numerous conditions and limitations when using a motor-circuit switch as an "isolator"
(b) the motor-circuit switch is rated in hp, NOT in kW.
Note: Motor 7.5kW = 10hp approx. Per NEC typical 3-phase 10 hp motor current [30.8 A at 208 V], [14 A at 460 V] etc....
Attention: A low speed <1200rpm or a high torque motor may requires higher running current. In these cases the nameplate current rating shall be used.
3. Summary: The isolator rated Voltage shall > the system operational voltage and the rated current shall > say 1.2 times the motor nameplate rating.
Caution: use the hp rating with care, as it is dependent on the voltage and the type of motor, see above 2 Note.
Che Kuan Yau (Singapore)

 
To expand on Mr Che Kuan Yau's post:
30.8 Amps at 208 Volts = 11.088 electrical kW.
A 10 HP motor is 7.5 mechanical kW (motor nameplate kW)

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
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