Thank you and I would love to share my inputs & be in touch.
Regards,
Hi Venkat and all,
Not sure if the attached document will be helpful. Please let us know the out come of your model/experiment.
you already got your first customer (Me 
).
thanks
VK
ijeemv4n5_09.pdf (622 KB)
Hi Sarvana,
I have heard manufacturing cost of low RPM alternator will cost more , when compared to same alternator with higher RPM.Is that so?
Venu,
This is kind of DIY project for me,will keep you all posted on the progress.
Regards,
Venkat
We know that the frequency of a supply is the number of times a cycle appears in one second and that frequency is measured in Hertz. As one cycle of induced emf is produced each full revolution of the coil through a magnetic field comprising of a north and south pole if the coil rotates at a constant speed a constant number of cycles will be produced per second giving a constant frequency. So by increasing the speed of rotation of the coil the frequency will also be increased. Therefore, frequency is proportional to the speed of rotation, ( ƒ ∝ Ν ) where Ν = r.p.m.
one north and one south pole, giving just one pair of poles. If we add more magnetic poles to the generator so that it now has four poles in total, two north and two south, then for each revolution of the coil two cycles will be produced for the same rotational speed. Therefore, frequency is proportional to the number of pairs of magnetic poles, ( ƒ ∝ P ) of the generator where P = is the number of “pairs of poles”.
Then from these two facts we can say that the frequency output from an AC generator is:
generator frequency f = NP/60
Where: Ν is the speed of rotation in r.p.m. P is the number of “pairs of poles” and 60 converts it into seconds.
@ Venu
Thanks for the doc.It will ofcourse help beginners and give a insight as to what is what.
@ Venkat
True. Lower RPM machines cost a bomb.
@ Krishna Prasad
May be I wasnt clear earlier.
Let me explain with an example;
Lets consider an alternator whose name plate rating is as below
Rated Power - 5Hp
Rated Output - 415V 3phase @ 50Hz
Rated RPM - 500rpm
The above means that the alternator shaft when rotates @ 500rpm will generate a voltage of 415V at 50Hz interval.
This “rated” frequency is determined by the machine construction namely no of poles, coil winding across the poles etc.
For the above case,
P = 60 * 50/500 = 6 meaning we need a 6 pole machine which when run at 500rpm generates 3.7Kw power @ 415V 50Hz.
Now this 6 poles effectively concerns how coil is wound across stator core.
Motor manufacturers use the equation the same way you have put across, but alternator manufacturers see the same equation as,
P = 60f/N
As the poles is inversely proportional to the speed, the moment we decrease the RPM the no of poles required shoots high , but note"f" remains constant the whole case.
If it is rated 60hz,
P= 60*60/500 = no of poles would be 7, note RPM is still 500RPM.
Regards,
Saravana
Mr. Saravana
Yes the same formula can be used in both the ways.
It looks like i am back to my Engg Class. I wonder if i have followed the subject as diligently as i am doing now,i wouldn’t need to start from the scratch like now.
But any ways its good learning.
Regards,
Venkat
Hi,
I’m not technically or Mechanically strong so Please apologies if my question is wrong.
Is it possible to run an alternator whic is capabale of generating power to run 5HP motor using 1 HP or 2 HP motor?
My idea is
Let us consider an alternator which is capable of producing power to run 6 HP motor.
1HP Motor —> Alternator —> 5 HP motor
Since my alternator is capable of running 6 HP, I can use the power genetated by Alternator to run the 1 HP motor and 5 HP motor above.
Thanks,
Dhana.
No. This is not possible.
The driving mechanism should be equal to or more powerful than the machine being driven.
Considering losses, an alternator requires more than 5Hp of mechanical energy to generate an electrical energy of 5Hp.
Regards,
Saravana