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Motor Rpm Based on Frequency, Poles and What? 4
- Thread starter daveshei
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2
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I believe the equation, Ns=120f/P, where:
Ns = rpm
f = frequency in Hz(cycles per sec)
P = number of poles
is applicable for single, two-phase, or three-phase(or any-phase) motors, since it is based on the number of poles. I believe that a 4-pole, 3-phase motor has 4 poles per phase.
The 120, I believe, is derived from the change from Hz(cycles per sec) to rpm (revolutions per min). Look at another version of the same formula: Ns = 60f/p, where:
Ns = rpm
f = frequency in Hz(cycles per sec)
p = number of pole pairs
In this case, a 4-pole machine would have 2 pole pairs, and the multiplier is reduced to 60, which is the number used to change seconds(Hz) to minutes (rpm). To complete one revolution, a 4-pole motor has to pass through 2 pole pairs.
I'll admit I can't find any reference to this right away, but I do believe this was the explanation given to me at school.
Ns = rpm
f = frequency in Hz(cycles per sec)
P = number of poles
is applicable for single, two-phase, or three-phase(or any-phase) motors, since it is based on the number of poles. I believe that a 4-pole, 3-phase motor has 4 poles per phase.
The 120, I believe, is derived from the change from Hz(cycles per sec) to rpm (revolutions per min). Look at another version of the same formula: Ns = 60f/p, where:
Ns = rpm
f = frequency in Hz(cycles per sec)
p = number of pole pairs
In this case, a 4-pole machine would have 2 pole pairs, and the multiplier is reduced to 60, which is the number used to change seconds(Hz) to minutes (rpm). To complete one revolution, a 4-pole motor has to pass through 2 pole pairs.
I'll admit I can't find any reference to this right away, but I do believe this was the explanation given to me at school.
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1
- #3
To add to DanDel's reply, the equations give the synchronous speed i.e. the theoretical maximum speed of the motor. On load the speed falls due to slip (maybe only 1 or 2%).
The single phase induction motor speed is found in exactly the same way as the three phase motor, there is no difference. They can also have 2, 4, 6... poles.
The single phase induction motor speed is found in exactly the same way as the three phase motor, there is no difference. They can also have 2, 4, 6... poles.
All:
One point:
The number 120 in the formula refers to the electrical time degrees that seperates the phases . That is why it is a consatant for 50 hertz, 60 hretz etc...
DanDel:
I would guess your young. You have an excellent memory! Enjoy it while it last
One point:
The number 120 in the formula refers to the electrical time degrees that seperates the phases . That is why it is a consatant for 50 hertz, 60 hretz etc...
DanDel:
I would guess your young. You have an excellent memory! Enjoy it while it last
- Thread starter
- #5
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jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: Essentially, 120 = 2 x 60
where
2 is linked to 2/p or 1/(p/2) to consider a pole-pair, where p is number of poles
60 is coming from the conversion of seconds to minutes between rads/sec and revolutions/minutes. Therefore, it holds true for motors of various number of phases.
See for example reference:
1. The Electrical Machine Problem Solver, Staff of Research and Education Association, New York, NY, 1983, (Dr. M. Fogiel, Director),
Problem 10-2, Page 557
where
2 is linked to 2/p or 1/(p/2) to consider a pole-pair, where p is number of poles
60 is coming from the conversion of seconds to minutes between rads/sec and revolutions/minutes. Therefore, it holds true for motors of various number of phases.
See for example reference:
1. The Electrical Machine Problem Solver, Staff of Research and Education Association, New York, NY, 1983, (Dr. M. Fogiel, Director),
Problem 10-2, Page 557
jbartos
I'm not sure I understand your post. The formula for sync speed of an induction motor is:
120 *(F/Poles)
Will your formula return the same result if you have 2, 4 6... pole motors?
If I use your formula would I need to consider the operating frequency? I think your constant number of 60 assumes 60 hertz. Can it be changed to whatever as needed?
The 120 "assumes" (bad word) that you have a 3 phase power supply and the 3 phases are 120 degrees apart. Is your reference assuming a system other than 3 phase supply?
I'm not doubting you or your reference. I just don't understand the way you are presenting in your post.
I'm not sure I understand your post. The formula for sync speed of an induction motor is:
120 *(F/Poles)
Will your formula return the same result if you have 2, 4 6... pole motors?
If I use your formula would I need to consider the operating frequency? I think your constant number of 60 assumes 60 hertz. Can it be changed to whatever as needed?
The 120 "assumes" (bad word) that you have a 3 phase power supply and the 3 phases are 120 degrees apart. Is your reference assuming a system other than 3 phase supply?
I'm not doubting you or your reference. I just don't understand the way you are presenting in your post.
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: I am sorry, this Forum is essentially providing eng-tips for those who have some education and experience in electrical areas. I provided a direct answer to the original questions.
Starting with the simple relationship for motor speed:
wm=(1-s) x (2/p) x ws = (1-s) x (2/p) x 2 x f x pi, in rad/sec
1 rad/sec = 60 /(2xpi) revolution/minute
wm=(1-s) x (2/p) x ws = (1-s) x (2/p) x 2 x f x pi x 60 / (2 x pi) , in revolutions/minute
or
wm=(1-s) x (2/p) x ws = (1-s) x (2/p) x f x 60 , in revolutions/minute
or
wm=(1-s) x (2/p) x ws = (1-s) x 120 f /p , in revolutions/minute
This is where you see 120, this thread is referring to.
Motor phases, are not involved.
The formula is valid for a number of poles p, where p is a positive integer different from zero.
f is sometimes indicated as fs, synchronous frequency.
Starting with the simple relationship for motor speed:
wm=(1-s) x (2/p) x ws = (1-s) x (2/p) x 2 x f x pi, in rad/sec
1 rad/sec = 60 /(2xpi) revolution/minute
wm=(1-s) x (2/p) x ws = (1-s) x (2/p) x 2 x f x pi x 60 / (2 x pi) , in revolutions/minute
or
wm=(1-s) x (2/p) x ws = (1-s) x (2/p) x f x 60 , in revolutions/minute
or
wm=(1-s) x (2/p) x ws = (1-s) x 120 f /p , in revolutions/minute
This is where you see 120, this thread is referring to.
Motor phases, are not involved.
The formula is valid for a number of poles p, where p is a positive integer different from zero.
f is sometimes indicated as fs, synchronous frequency.
d23 - as jbartos says, it is simply the conversion of units of time from seconds to minutes. If we quoted motor speed in revs/sec, there would be no x120 (i.e. 2x60) factor, just x2 (to account for the fact that by p is number of poles and not pole pairs). jb has stated the full formula including the small effect of slip in an induction motor.
I have one very slight addition to jbartos' last post, p must be an even number of course.
I have one very slight addition to jbartos' last post, p must be an even number of course.
d23 - thank you, but most people would not consider my age (44) young, although I do have a good memory.
I don't know why you assume me to be young with a good memory if you don't agree with my analysis of the original question; perhaps you can explain further.
I don't know why you assume me to be young with a good memory if you don't agree with my analysis of the original question; perhaps you can explain further.
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