Moment of Inertia? 1

[veritas]

Hi

I have an 11kV, 50Hz, 4.818MVA, 16-pole hydro generator. Shaft height = 1m, width = 1.95m. Any idea what the moment of inertia in J in kg.m2 is please? or what is a typical value for this type of machine. Have tried the manufacturer but it's like trying to pull chicken teeth!

Thanks in advance.

 

[FredRosse]

in the vicinity of 5000-7000 kg-M^2, but the OEM should have this information.

 

[veritas]

Thanks Fred.

 

[7anoter4]

In my opinion, if the rotor weight is known and rotor diameter also then
J=GD^2=1/8*W*D^2 W=weight in kg ; D=rotor diameter [m]; J [kg.m^2]
If it is not known ,we may use the H constant formula:
J=2*H*Srated/(2*π*Frated/N)^2[kg.m^2]
Where H constant [2-4 Ws/VA]; Srated [4818000 VA]; Frated =50 Hz N=number of pole pairs[8]
In your case it could be 12500 to 25000 kg.m^2.
The above formula it is the same as per IEEE Std 399/1997 Brown Book 4.9.3.3.2 The H constant.
See[for instance]:

http://www.mathworks.com/help/physmod/sps/ref/machineinertia.html?requestedDomain=

http://www.mathworks.com

file:///D:/D_My%20Documents/Engineering/Generators/Synchronous%20generator%20dynamics%20Lund%20Un.pdf

 

[veritas]

7anoter4

Still working through your equations. First one no problem. Busy with 2nd one.

I did finally manage to get the rotor data from the OEM. The stated J = 2723 kg·m² is substantially lower than the predictions in this post and I'm wandering why?

However, if I use the equation for J = m·r²/2 (identical to your first equation) = 3080kg·m2 as follows:

Add up the weights of the fan + shaft + magnetic wheel + rectifiers + exciter = 14582kg

Assume radius, r = 0.65m (approximately) to get

J = 14582*0.65²/2 = 3080kg·m² which is in the ballpark of 2723kg.m²

 

[itsmoked]

I'm waondering why?

Missing decimal place?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[veritas]

What I also find intriguing is the following:

H = K/S_r = (5.48e-6·J·N²)/S_r =(5.48e-6·2723·375²)/4818000 = 0.436 J/VA.

where K = kinetic energy at rated frequency in Joules
S_r = rated power in VA
N = speed in rpm

Also a bit outside the range of 2-4 Ws/VA.

 

[veritas]

7anoter4,

Your 2nd formula, J = 2*H*S_r/(2*Π*f/N)² is the same as mine. I used rpm instead of frequency and pole pairs. If H = 0.436, I get J = 2723 again using your formula.

Still a bit perturbed by the substantial discrepancy between the OEM values for J and the values suggested in this post.

 

[davidbeach]

Have you left out the hydraulic end of the shaft? All of that rotating mass counts too.

 

[Hoxton]

It does not help here, I was about to say what you need is Fred, but we have already a post from a Fred, so you need a joe. He's the little old guy in maintenance who jealously guards all the original plant data. This is because he knows how difficult it is going to be to replace it - from manufacturers who merged or went out of business.

Joe retires and the new guy, usually with an MBA or what ever throws "all that old junk" out.

I have seen this time and time again!

I have seen the inertia checked from a free run down. It was done graphically, but was 40 years ago, so sorry cannot remember the details. It was done at low speed, to minimize windage. I don't know the practicalities of doing it in your case.

 

[7anoter4]

I think the error stays in the moment of inertia units. If 2-4 refers to gravimetric units then H has to be divided by 9.8 then 2 to 4 correspond to 2/9.8=0.204 to 4/9.8=0.408
J=GD^2/4/g g=9.81 m/s^2
See[for instance]:

http://www.nissei-gtr.co.jp/english/gtr/pdf/mini_e07.pdf

 

[veritas]

Ok, will check it out, thank you. The rather low J for the machine is in itself good and bad. Bad because the machine will "react" relatively more to system disturbances than a higher J machine which of course is not desired. Good in that it is easier detectable with something like ROCOF which I am setting up for Loss of Mains. Wonder where's the ugly...

 

[veritas]

7anoter4,

Sorry, could not open the attachment. Needs some Adobe Acrobat Extended Asian Language font pack which I unsuccessfully tried to install. Maybe an alternative way to resend?

 

[dArsonval]

An older Adobe Reader X (10.1.4) version opens the file correctly here on line.
When saving the file and attempting to read it, it does not open.

Capturing two screen shots and inserting them into a word document, I came up with this:

 

[veritas]

Many thanks dArsonval, got it.

7anoter4 - makes sense now.

Hoxton - I know what you mean. Thankfully, this is a new generator and the OEM is still around. Just not very good at communicating.

Regards.

 

[electricpete]

However, if I use the equation for J = m·r²/2 (identical to your first equation)

I don't know if that was a typo, but there is no /2 involved here.

=====================================
(2B)+(2B)' ?

 

[cswilson]

A decade ago, I wrote an application note on calculating moments of inertia starting from the most basic concepts. Partly I was working through things myself to make sure I understood every step, but also I was seeing people make mistakes by plugging into formulas they didn't fully understand. Using weight values instead of mass and then forgetting to use "g" is a common mistake.

Anyway, I offer the app note here. Comments (even corrections!) welcome.

Curt Wilson
Omron Delta Tau

 

[veritas]

electricpete,

J = m·r²/2 is the formula for the inertia of a solid cylinder of mass m and radius r rotating on its axis. Units are kg·m² (SI). I have seen this in more than one reputable text on the subject. However, I will also be first to admit that I have taken it on face value as I have never delved into deriving it from first principles.

Reading through the excellent text of Curt, also verifies this formula.

So I am still interested to know why you say there is no /2?

 

[electricpete]

My formula would apply for thin-walled hollow cylinder of radius R.
Yours would apply for solid cylinder of radius R.
I don't know which context you were working with.


=====================================
(2B)+(2B)' ?

 

[veritas]

Yeah, the rotor of a generator is approximated by a solid cylinder, hence the formula I used.