Which is the best option to go with?

A) Using a turbine with higher RPM and then using reduction gears to bring it down the RPM equal to the alternator. (from 22,000 to 1500 RPM)
B) Using a turbine having equal RPM as an alternator. (1500-1600 RPM to compensate any losses or anything)

09-03-2014 03:21 PM

Top #2

Ing. Ole Knudsen

09-03-2014 03:21 PM

That mostly depend on the power range you are looking at.
It would not be reasonable to run a 600MW turbine at 22,000RPM, and then use a gear to bring it down to 1500RPM, while it for a smaller turbine could be quite feasible, or you could even run a specially designed alternator at 6000RPM, and bring out 50HZ AC power from both rotor and stator.

Your comment under B) about 1600-1500 RPM to compensate for losses can only be relevant if you use some standard induction machine as your generator, and you can only use that to generate into some existing network, if you do not want to invest in some elaborate voltage control scheme.

09-03-2014 05:23 PM

Top #3

Alan Maltz

09-03-2014 05:23 PM

Kamran,

If you can afford to add the additional complexity, cost, maintenance, etc. required for a high powered gear box, then go for it. If there is some compelling reason such as tapping power off of a turbine driving a high/off synchronous speed process, then it may make sense rather than let the energy go to waste; reduction gear driven generators coupled to high speed stages of gas turbine engines comes to mind.

Alan

09-03-2014 07:43 PM

Top #4

ANAND KUMAR GUPTA

09-03-2014 07:43 PM

Mr. Kamran Yusuf, Basically all machines, be they electrical or mechanical follow the principle of "Mechanical Advantage".
Output a machines per revolution can be multiplied by increasing its RPM. For example in case of a winch several rotations of the driving handles result in lifting the load (which is heavy) just by a bit.
Similarly, let us take the case of a small size Gas Turbine can be run at very high RPM to drive a Generator through a reducing gear box for the required output. If a Gas Turbine runs at 22,000 RPM to drive a Generator of say 300 kW at 1500 RPM through a reducing gear box, If gear box is removed and the Gas Turbine is directly coupled to this Generator, theoretically the Generator output capacity would get limited to 300x(1500/22000) = 20.45 kW because the Gas Turbine's mechanical output would get reduced to what can produce 20.45 kW only. However, the matter is not so simple because all Gas Turbines have a shaft driven air compressor and the air compressor designed to run at 22,000 RPM would not produce the require quantity of compressed air at the required pressure.
Choice of RPM of Gas Turbine for driving Generators is that of the manufacturer rather than that of the purchaser. For very large capacity units no manufacturer can produce the required gear box. therefore in those cases the Gas Turbines are run at same RPM as the Generator, 3000 RPM for 50 Hz systems and 3600 RPM for 60 Hz systems.
Overall cost goes high if speed of gas turbine is low because Mechanical Advantage cannot be availed of without Gear Box.
For more scholarly matter on Mechanical Advantage you may please refer to.

www.princeton.edu/~achaney/tmve/wiki100k/docs/Mechanical_advantage.html

https://www.khanacademy.org/science/physics/work-and-energy/mechanical-advantage/v/introduction-to-mechanical-advantage

Best Wishes.

09-03-2014 10:06 PM

Top #5

ANAND KUMAR GUPTA

09-03-2014 10:06 PM

I would like to further add that each and every machine is designed to endure Out per Revolution, stresses at finally designed RPM, capability of handling Input and Output at design RPM, being mechanically balanced at design RPM ... etc.

09-04-2014 12:14 AM

Top #6

Kamran Yusuf

09-04-2014 12:14 AM

Thank you all!
Sir Anand good to see you commenting. Your response pretty much answers my question. Thank You!

09-04-2014 02:59 AM

Top #7

Ranveer singh

09-04-2014 02:59 AM

Dear Mr Ing

How can be design a alternator running at 6000 rpm and giving output ac at 50 hz?
As we know that N = 120 f/P where N = 6000 rpm, f = 50 hz. so the pole,P we are getting from this is 1.how through this one no. of pole ,we can generate?

If you please give a brief over it, It would be very helpful for me.

09-04-2014 05:43 AM

Top #8

Ing. Ole Knudsen

09-04-2014 05:43 AM

This one is easy!
In a standard wound rotor motor, the frequency Fr in the rotor circuit is proportional to the difference between synchronous speed Ns and actual speed Na, multiplied by the system frequency Fs: Fr = Fs * (Ns - Na) / Ns. By driving this machine at twice its synchronous speed, ( Na = 2 * Ns ), the frequency of the rotor voltage will then be: Fr = Fs * (Ns - 2Ns)/Ns, or in other words, the rotor frequency will be the same as the system frequency, but rotating "backwards". We can easily reverse this rotation by just swapping two leads.

I know it works also in practice, for I have tested it!