Grainger Problem #3.1

[kduff70]

Hi I tried working problem 3.1 in Grainger but not sure how they came up with the correct number of pole I have attached the problem if someone can figure out how they obtain the pole # ? I understand the formula but not how they came up with the # of poles.

thank you

 

[MyBeardAndMe]

To obtain the highest speed, you would need the lowest number of poles. For the 25 Hz gen, 10 poles is the lowest even number that would satisfy the equation while also producing an even number of poles for the 60 Hz gen.

 

[kduff70]

MyBeardAndMe

thank you but I stuck for some reason on how to find the pole # at 25hz ? not given the speed I assume the rpm at 25hz is 25*60= 1500 rpm and then you divide that by 120f = 120*60HZ=7200 is the the path ? not sure but when I divided by i get the pole # close to 10 and then I can find the poles at 60 hz?

 

[kduff70]

sorry MyBeardAndMe I mess this one up can you please clarify this one better for me

 

[JB66money]

Kduff,

The solution skipped a couple of steps and made the solution unnecessarily difficult. The equations in the solutions were really derived from the basic equation for determinining the speed of synchronous machines.

1. The two equations were set equal because the two generators on the same shaft which means that they have the same speed, even if they operate at different frequencies.
2. You use the equations for speed (ns=120fe / P = 60*2*fe / P)that are illustrated better in the attachment.
3. ns = 60*2*fe1 / P1 = 60*2*fe2 / P2
   
4. fe1 = 60Hz, P1 = # poles for generator 1, fe2 = 50Hz, P2 = # poles for generator two.

 

[kduff70]

Thank you JB66money

I understand the setup but not how you get 10 pole I'm stuck at that part

 

[dayrongarcia]

Thank you JB66money

I understand the setup but not how you get 10 pole I'm stuck at that part

Remember that we need a minimum of 2 pole per motor or multiples of 2.

2 poles x 2.4 = 4.80 (need multiple of 2)

4 poles x 2.4 = 9.60 (need multiple of 2)

6 poles x 2.4 = 14.4 (need multiple of 2)

8 poles x 2.4 = 19.2 (need multiple of 2)

10 poles x 2,4 = 24.0 (BINGO)

 

[kduff70]

Thank You !!! So the 10 poles is the Max you can go with out going over the the 25Hz right!

 

[dayrongarcia]

Thank You !!! So the 10 poles is the Max you can go with out going over the the 25Hz right!

Nope, the 10 poles is the lowest amount of poles you will have for the first motor in order to be synchronized with the second, both will run at 300 rpm. The frequency is fixed for these motors at 60 and 25. Plug the numbers in the equation Ns = 120*60/P so you see for yourself.

 

[kduff70]

[SIZE=10.5pt]Dayrongarcia,[/SIZE]

[SIZE=10.5pt]I think it finally click !!!![/SIZE]

[SIZE=10.5pt]You start out with your 2 poles at 25hz which is 4.80 pole at 60hz but since its only goes by even pairs of poles you use 4 poles and so on while getting your speed RPM for each which then can find your frequency until you get to the lowest speed (RPM) and the lowest amount of poles in order to be synchronized with the second while at the same time not going lower than 25hz or greater the 60hz .Is that about right? I think see what's going on now[/SIZE]

[SIZE=10.5pt]thank you[/SIZE]

 

[dayrongarcia]

kduff70,

The speed is given by the problem, it will not change unless there is a change in the prime mover of the generators. 10 poles and 24 poles respectively will give you the max speed both motors can run connected to the same shaft. If you use 20 poles and 48 poles,which is the 2nd best possible scenario, the speed will be halved. The problem is asking you for the highest speed. Plug the numbers of poles in the formula it will become clear to you.

 

[kduff70]

dayrongarcia

i plug in the numbers and i see you have to half and I see how your second choose could be 20 poles and 48 poles just as your third chose could be 30 and 72 ?

 

[dayrongarcia]

dayrongarcia

i plug in the numbers and i see you have to half and I see how your second choose could be 20 poles and 48 poles just as your third chose could be 30 and 72 ?

I dont understand your question

 

[MyBeardAndMe]

dayrongarcia

i plug in the numbers and i see you have to half and I see how your second choose could be 20 poles and 48 poles just as your third chose could be 30 and 72 ?

Yes, those are both options for the number of poles in the two generators to be on the same shaft. However, the only answer to the question is 10 poles and 24 poles because you want the highest rpm possible

 

[kduff70]

I if you start with 2 poles and plug them into the equation when you get to 10 poles then you will get 24 poles at the max speed. and is there was a 3rd bet possible set of poles would 30 and 72 ?

 

[dayrongarcia]

You got it.

 

[kduff70]

:Locolaugh:

Thank you for your information and your patient with me on this problem. I was trying to gather a few problem from Grainger that are a little different from spin,ci, graffeo and Ncees to help me be prepare as best as I can for the exam once again thank you

 

[JB66money]

One main principle to learn from this excercise is that the machine's speed varies inversely with the number of poles that it has. That is, the more poles a machine has the slower its speed will be, and the less poles a machine has the faster its speed will be.

 

[JB66money]

One main principle to learn from this excercise is that the machine's speed varies inversely with the number of poles that it has. That is, the more poles a machine has the slower its speed will be, and the less poles a machine has the faster its speed will be.

 

[MyBeardAndMe]

:Locolaugh:

Thank you for your information and your patient with me on this problem. I was trying to gather a few problem from Grainger that are a little different from spin,ci, graffeo and Ncees to help me be prepare as best as I can for the exam once again thank you

Feel free to post more examples from Grainger. I don't have Grainger and I ran out of new problems to do with NCEES, Spin, CI, and Graffeo.