Starting Equipment and Gd^2 Value of the Equipment's Rotor

Please excuse my ignorance but what is Gd squared? The National Institute of Science and Technology does not list this as a SI unit. If you mean a Giga dyne squared this makes no sense to me for a motor unit. It's a good practice to define an acronym, even a common acronym at least once. To not define an acronym is CRIMINAL

Gd^2 is usually used in stead of inertia moment it has same units J has as units kg*m^2.

In Gd^2 the G is not the weight but the mass. Before the weight was used but after the introduction of ISO new system it remained as presentation form. What I do NOT understand is the way the "machine" inertia is written. May be who wrote the message can give an explanation. I never saw a number as he put it down.

So is Gd^2 just an arcane term for the rotational moment of inertia (J) or is there some additional dimensionless conditioners that translate J → Gd^2? Also do you know what dimension system is this from?

Oh and I gave you a GA for the other post, but my questions seemed more pertinent here.

The inertia moment of a cylinder is J= ∫2*π*r*dr*r²*l*ρ where

r= current radius l= length and ρ= specific mass J=2*π*ρ*l*∫r^3*dr from Ri to Re

J= 2*π*ρ*l* (Re^4-Ri^4)/4 = π*ρ*l/2*(Re^2+Ri^2)*(Re^2-Ri^2)

But π*ρ*l*(Re^2-Ri^2)=M or in old way G If you consider a full cylinder then Ri =0

the result will thus be J= M*Re^2/2 = M*D^2/8 here you see the direct relationship between Gd^2 (or Md^2) and J. The fact that you are not accustomed to this notation surprises me since as I saw you are at top in electrical and mechanical domain.

Dear Mr. Redfred,

Gd^2 is related to the INERTIAL EFFECT OF A ROTATING MACHINERY. IT DEPENDS UPON SEVERAL FACTORS such as mass, radius of rotation, speed to be attained in a given time. Higher the value, more torque is required, to rotate from zero speed to given speed.

dhayanandhan.s,

The rotor moment of inertia (Gd^2) can be much smaller than the maximum rated load inertia. Do you have a maximum load inertia specification from the motor manufacturer?

Will assume your load inertia is 100,000 [kg-m^2] .

The calculated run-up time seems long, but without more data it is difficult to be certain if your motor can reliably start this large inertial load or not.

ref...

http://www.eng-tips.com/faqs.cfm?fid=1131

The motor mechanical side equation is

T_Motor = T_load+ Jdw/dt+ ωdJ/dt

J is the system instantaneous moment of inertia, and w is the instantaneous speed.

J may or may not be dependant on speed. If the application is such that it is independent

T_Motor = T_load+ Jdω/dt

In your case

Motor RPM = 750 = 78.5 Rad/Sec

Full Load Torque = 19.1 KNm

Assuming 50% as pull up torque

PUT = 9.55 KNm

Angular acceleration at starting – assuming no load torque (T_load = 0)

a = dw/dt = 10.5 rad/sec2 = 1.7 rev/sec2

This comes out to be too slow (even without any type of load torque, it will take about 8 sec to bring the equipment to the speed, and definitely there will be a significant amount of load torque.

Best is if you take the load torque characteristics and add the inertial load (Jdw/dt) over this then you may get the system OC.

Then with the motor OC you may get the system operation and check whether it will work or not.

As far as the prime mover Moment of inertia being less than the load, it is quite common than not isn't it?

Dear Guest,

thanks for your comment.

dhayanandhan.s

Where is my mistake ?

Motor 750 RPM, 1500 kW

1) Total inertia = motor inertia + load inertia (assuming motorspeed equals load speed)

• J total = J motor + J load
• = 687 + 100 000
• = 100 687 kg m2

2) Motor angular speed = RPM x 2 x pi / 60

• = 750 x 2 x 3.1415 / 60
• = 78.54 rad / sec

3) Mechanical power = torque x angular speed

• P mech = T x w

or

Torque = mechanical power / angular speed

• T = P mech / w
• = 1 500 000 / 78.54
• = 19 099 Nm

4) T motor = T load + J dw/dt + w dJ/dt

• assuming that application is starting without any load T load = 0
• assuming that inertia does not change when it accelerates w dJ/dt = 0
• assuming that the angular acceleration is constant during dw/dt = a
• assuming that starting torque equals 50% of nominal torque T start = 9 550 Nm
• this results in Motor torque equals total inertia times angular acceleration
• T start = J x a
• or
• a = T start / J
• = 9550 / 100687
• = 0.0949 1/sec2

5) angular speed equals angular acceleration times acceleration time

• w = a x t

or

acceleration time equals angular speed / angular acceleration

• t = w / a
• = 78.54 / 0.0949
• = 827.60 seconds

In replay number 4, Guest says acceleration time equals about 8 sec

Where is the mistake ?

Even if the motor starting torque equals 3 times the nominal torque

the accelaration time is +/- 138 sec, which is to long ...

Your calculations are basically correct which is why I posted response #3. Guest response of ~8 second run-up time is for rotor only with no additional load torque or load inertia. You can see from your own numbers why this motor seems inadequate for the OP's application based on the limited data supplied. Additional motor specs and information regarding installed starters and/or VFDs are needed for further analysis.

The link supplied in #3 has some good information on how to handle large inertial loads.

We still don't know what sort of a number 1,00,000 (or 1,00,687) is....

Is it supposed to be 1,000,000; or 100,000; or something else?

Also, is the load direct-driven, or does it turn at a lesser rpm than the motor?

The kV of the motor is irrelevent, but the kW is useful to know.

That is the nomenclature used in this part from where OK is.

After the thousands (,000) it is in hundreds so it is 1000; 10,000; 1,00,000; 1,00,00,000 etc)

(Here the sequence is hundreds, thousands (10³), lacs (10⁵) , Crores (10⁷) ...

Hope this clarifies.

Warning! That is completely nonstandard elsewhere, and it is likely to be highly confusing to others. I would urge your entire country to abandon this goofy practice within the next ten minutes. Please start spreading the word.

In any prime mover there is accelerating torque and the retarding torque. Difference of there torques (if positive) is available for accelerating the mass coupled to it. Now inertia of the mass and the torque available would decide the rate at which the motor would accelerate and attain its rated speed. One has to consider any change in retarding torque with change of speed (increased windage loss with speed, change in friction and off course the torque speed characteristic of the motor)

A couple of general remarks:

- Gd^2 has nothing to do with speed but speed increase over time is function of Gd^2 as it is mentioned in the answer to redfred, it is a mechanical property of the system depends on its mass and how the mass is distributed with respect to rotational axis

- at start the motor and the sytem has no rotation so that for the motor the stall moment has to be considered since n=0 This moment is between 2 and 3 x nominal moment not 50%. Now this is valid for a AC motor which I presume being the type of motor used in this case. When the rpm increase the moment decreases to become =0 at the synchronous rpm. In normal function the motor works on the falling moment curve and its value is roughly proportional to the relative difference between rpm and current frequency.

- it seems in this particular case that due to the high inertia (if the motor is coupled to the load without any gear reduction) the time to reach a "normal" working domain is too long even if there is no resistant torque. During this period a lot of power is generated which leads to high local temperatures since the rpm are low (thus convection low, low air flow). Usually when the drives arrive at such big powers the motor manufacturer is able to supply specific informations (not only catalogue values) about the load which will be accepted by the motor without harm. the most "dangerous" effects are the high temperature in the coils' core which drastically reduce the life expectancy of wire insulation. Of course the over al life reduction is function of the relative number of starts versus "normal" function time.

- I further think that such a subject should be treated by the interested person in a much more professional way not by asking for opinions on a site as ours where he can only get indications and more or less superficial comments.