# Nominal full-load speed



## jeanbj2000 (Jun 15, 2011)

a) Calculate the approximate values of full load current, starting current and no load current of a 75KW, 4000, 3-phase, 900 r/min, 60Hz induction motor.

B) Calculate the nominal full-load sped knowing that slip is 2 %?

My question is what do they mean by *Nominal Full-Load-Speed*?

Thanks!!


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## wilheldp_PE (Jun 15, 2011)

Dude, seriously? You posted this 12 TIMES to the same forum.


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## Supe (Jun 15, 2011)

He REALLY wants to know the answer!


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## Guest (Jun 15, 2011)

Fixed it for ya ... slow down on that &lt;enter&gt; key!!

JR


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## palvarez83 (Jun 15, 2011)

jeanbj2000 said:


> a) Calculate the approximate values of full load current, starting current and no load current of a 75KW, 4000, 3-phase, 900 r/min, 60Hz induction motor.B) Calculate the nominal full-load sped knowing that slip is 2 %?
> 
> My question is what do they mean by *Nominal Full-Load-Speed*?
> 
> Thanks!!


Jeanbj2000,

The nominal full load speed is the speed that you would see on the motor's name plate. That is the speed simple calculated from the synchronous speed and the the slip alone. In this case it would be: 900 -).02x900 = 882 rpm. Therefore 882 rpm is the "nominal full load speed". Nominal means that that actual full load speed for that specific motor will vary around there. Each motor even same make an model can be slightly different.

I suggest you read this for more info: Nominal Full Load Speed


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## Dolphin P.E. (Jun 15, 2011)

jeanbj2000 said:


> a) Calculate the approximate values of full load current, starting current and no load current of a 75KW, 4000, 3-phase, 900 r/min, 60Hz induction motor.B) Calculate the nominal full-load sped knowing that slip is 2 %?
> 
> My question is what do they mean by *Nominal Full-Load-Speed*?
> 
> Thanks!!


This is my shot for this problem, approximate Values:

FLA = 75000W/4000V/1.732 = 10.8A

NLA is about 30% for big motors = 3A

Starting current is about 5.7 times of the FLA = 62A

Nominal speed = 900- (900*.02) = 880 rpm


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## palvarez83 (Jun 15, 2011)

Dolphin P.E. said:


> jeanbj2000 said:
> 
> 
> > a) Calculate the approximate values of full load current, starting current and no load current of a 75KW, 4000, 3-phase, 900 r/min, 60Hz induction motor.B) Calculate the nominal full-load sped knowing that slip is 2 %?
> ...


Yes, I agree 880 rpm would be would you would see on the name plate (rounding).


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## BamaBino (Jun 15, 2011)

palvarez83 said:


> I suggest you read this for more info: Nominal Full Load Speed


From that book "For example, four-pole 60Hz fractional horsepower motors have a synchronous speed of 1800 rpm, a nominal full-load speed (as shown on the nameplate) of 1725 rpm and an actual full-load speed ranging from 1715 to 1745 rpm."This motor would have a slip of 4.2% and that would be it's only slip value (regardless that it has a full-load range), correct?


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## Dolphin P.E. (Jun 15, 2011)

BamaBino said:


> palvarez83 said:
> 
> 
> > I suggest you read this for more info: Nominal Full Load Speed
> ...


Yes, slip value should be computed from the Sync speed 1800rpm and the Full-load speed 1725rpm which is 4.2%


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## BamaBino (Jun 15, 2011)

Dolphin P.E. said:


> Yes, slip value should be computed from the Sync speed 1800rpm and the Full-load speed 1725rpm which is 4.2%


Thanks.

I was flipping thru the NCEES 2009 sample exam and noticed there isn't a problem dealing with slip.

I was surprised by that.


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## Dolphin P.E. (Jun 15, 2011)

BamaBino said:


> Dolphin P.E. said:
> 
> 
> > Yes, slip value should be computed from the Sync speed 1800rpm and the Full-load speed 1725rpm which is 4.2%
> ...


I thought there is one dealing with speed regulation. Speed regulation in this case will be (1800-1725)/1725 = 4.34%


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## BamaBino (Jun 15, 2011)

jeanbj2000 said:


> a) Calculate the approximate values of full load current, starting current and no load current of a 75KW, 4000, 3-phase, 900 r/min, 60Hz induction motor.B) Calculate the nominal full-load sped knowing that slip is 2 %?
> 
> My question is what do they mean by *Nominal Full-Load-Speed*?


I sorta have a problem with the way that problem was stated. It didn't say what the "900 r/min" is. The other info seems to be from the motors nameplate whereby the 900rpm should be the full-load speed, correct?

Also, is the sync speed always the same as the no-load speed?

Thanks.


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## palvarez83 (Jun 15, 2011)

BamaBino said:


> jeanbj2000 said:
> 
> 
> > a) Calculate the approximate values of full load current, starting current and no load current of a 75KW, 4000, 3-phase, 900 r/min, 60Hz induction motor.B) Calculate the nominal full-load sped knowing that slip is 2 %?
> ...



No, 900 rev/min would be the sychronous speed. That is the speed at which the magnetic field is spinning. For an induction motor the actual mechanical speed is always less than the synchornous speed due to slip. In other words the motor is always trying to "catch up" with the magnetic field.

Synchornous speed is found by the following formulat N =(120Xf)/(#P poles), where N is rpm , f is frequency in Hz, and poles is the quanity of magnetic poles that the motor has. Poles on a motor come in pairs (one North and one south), so that number can only be 2, 4, 6, 8, ect. In USA frequency is 60 hz. Therefore if you plug in 900 rpm for N and 60 for f, you will find that this is an 8 pole motor.

Look at this table: http://www.engineeringtoolbox.com/synchron...ors-d_1448.html

The 8 pole motor indicates a 900 rpm synchornous speed and a full load speed less than that. In this case 870 rpm which corresponds to roughly a 3% slip.


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## jeanbj2000 (Jun 17, 2011)

palvarez83 said:


> BamaBino said:
> 
> 
> > jeanbj2000 said:
> ...


Thanks everyone for the reply.


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## jeanbj2000 (Jun 17, 2011)

palvarez83 said:


> jeanbj2000 said:
> 
> 
> > a) Calculate the approximate values of full load current, starting current and no load current of a 75KW, 4000, 3-phase, 900 r/min, 60Hz induction motor.B) Calculate the nominal full-load sped knowing that slip is 2 %?
> ...


Thanks for the link. Very useful!!


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## palvarez83 (Jun 17, 2011)

For the exam be very careful not to confuse "slip" with "speed regulation" they are very closely related cousins of one another. Basically the difference is which value is used on the denominator.


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## yuyii (Jul 5, 2011)

palvarez83 said:


> For the exam be very careful not to confuse "slip" with "speed regulation" they are very closely related cousins of one another. Basically the difference is which value is used on the denominator.


palvarez83

I'm currently working on a problem shown below. Bear with me, I need to go through my calculation first before I can ask my question...

"A generator provides 400 VAC at 500 Hz turning at 14,000 rpm under full load. Calculate (a) How many poles does it have? (B) What is the speed regulation, SR(%)?"

Here's how I solved the problem:

(a) How many poles does it have?

sync rpm = 120 * f / p

Since sync rpm is not provided, I used the full-load rpm and solved for 'p' . Once solved, I rounded it down to the nearest multiple of 2.

14,000 = 120 * 500 / p

therefore, p = 4.28 = 4 (rounded down, since the number of pole is always a multiple of 2)

(B) What is the speed regulation, SR(%)?

First, I calculate the sync rpm using the 'p' calculated in (a).

sync rpm = 120 * f / p

sync rpm = 120 * 500 / 4

sync rpm = 15000 rpm

Now, to solve the SR, I used the equation:

SR = (no-load rpm - full-load rpm) / full-load rpm

... now here's my question... since we were not given the no-load rpm in this problem, is it safe to assume that the sync rpm can be used instead? In other words, here's how I solved the SR

SR = (sync rpm - full-load rpm) / full-load rpm

SR = (15000 - 14000) / 14000

SR = 0.0714 = 7.1%

Is my SR answer correct? Or is it wrong because I used the sync rpm, not the no-load rpm, to calculate for SR.

Sincerely,

Ruly

I understand that sync rpm is different from no-load rpm.


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