Gear Box and Energy Saving

At the first look, answering on can energy be saved, I say yes. You will eliminate the gearbox and add net power to the fan. The SCARY part now: I think you have to check if the fan it is designed for a higher speed. The fan must be very big and heavy, and I am wondering why the gearbox is there in the first place. 67 rpm more could be critical. You might have to slow down your motor or have it max.RPM limited and protected. If that fan breaks loose, I don't want to around. Al depends on how smooth the operation goes, if the fan vibrates too much, you might damage: bearings, structural elements. And the centrifugal force may not be higher than the design RPM. Wish you the best

Maybe also this: if the fan is coupled on the shaft of the gearbox, it might be possible the gearbox has been designed to deal with the axial forces.

Data you have provided is bit confusing to me at least, I guess it is a 50Hz motor and this means 995 is free running RPM of motor (6pole 50Hz motor), Also you have suggested that gear box output RPM is 928 Which again I am assuming is underloaded conditions, based on these assumptions I can conclude that gear box you are using is of the ratio of 1/1 or very near to 1/1, And if that is the case, I am not clear whats the point in using gear box.

Motor Nameplate:
SIEMENS

Type: 1RR1713-6

3-Phase Motor

V A KW Duty Cycle P.F 1/min

600 Y (+- 10%) 284 2500 S1 0.84 993

Rotor 1265 V 1215A

Frequency is 50 Hz

It is special designed gear box gives output of 928 RPM on the input speed of 993 RPM

Process is controlled by the Speed of This Fan (Raw Mill Fan) and Seperator Fan of Raw Mill. Second is Frequency Drive. We want to save some energy by removing this Gear box and for process this speed is compensated by reducing seperator Fan speed

What is the exact measured voltage on the motor? And the current drawn? In operation.

Dear Sir,

I will be happy to know is your voltage is 600 V star or 6000 v (6 Kv) since

at 284 A ( 600 v ) your kw will be 247.9 Kw and not 2500 Kw.This will clear a lot

of assumptions.

Thanls,

Actually it is 6300V not 600 V. It is MV motor

According to the fan affinity laws, power varies as cube of speed. 2250*(995/928)³ ≈ 2773 kW. This will overload the motor unless it has a service factor > 1.092.

I am not getting this clear. IMO the fan will just be supplied with the available power of the motor. If the setup is that critical, it should definitely not have been used with the gearbox. I think it will not loose synchronism. That gearbox is a lot more expensive than a slightly more powerful motor (in the basic design)

We have two plants 4000 TPD(tons Per Day)( Since 1998) and 6700 TPD ( Since 2008). In the newer plant (6700 TPD) Raw mill fan which is 5300 KW (By VEM)induction motor is directly connected o the Raw Mill fan without using gear box.

I am presently talking about 2500 KW motor driving Raw Mill fan of 4000 TPD plant. It has gear unit contrary to the 6700 TPD of 5300 KW Motor.

We want to remove this gear unit in 4000 TPD and I want to calculate that how much torque this gear box deliver to the load. This will be our Load torque.

"I does't know how to calculate this"?

When I calculate this Load torque, then I will calculate how much Power would take the 2500 KW motor. Also I check the Seperator Fan power. From There I calculate the energy saving

Thanks a lot for Ur interest

Waiting for Ur Help and Direction

Tornado gave you the correct load difference. Maybe the motor will be just a little bit light, maybe not. The fan will produce more air. If you want to keep this motor when it shows too light. you can lower the frequency or modify the blades. (shorter) and balance it out afterwards. What are the currents on the nameplate of the motor?

Also what we do not know is where is the fan supplying to and what pressure it builds up. Looks like a big fan, no turbine with a lot of pressure, to me.

In Cement Industry we have Raw Mill where Raw material are bring into power form. through rollers they are fine. Raw mill creates a negatives pressure lift the light fine particles of size up 12 Micron(Decided by Seperator fan Speed). Through this raw mill negative pressure they are lifted rom raw mill and put into the raw material silos from where they are passed through kiln exposing them to 1400 C temperature resulting in the formation of clinker

In the present system, N₁T₁ = N₂T₂; 995*21615 = 928*T₂; T₂ = 21615*995/928 ≈ 23176 N-m.

However, if you speed up the fan, the torque it requires will increase as the square of the speed (affinity laws again); thus 23176*(995/928)² ≈ 26643 N-m.

I don't know if the fan could be trimmed in the manner that pump impellers can be trimmed for tuning purposes.

"In the present system, N₁T₁ = N₂T₂; 995*21615 = 928*T₂; T₂ = 21615*995/928 ≈ 23176 N-m.

However, if you speed up the fan, the torque it requires will increase as the square of the speed (affinity laws again); thus 23176*(995/928)² ≈ 26643 N-m."

This is what I want

The Load Torque in case of Gear is 23176

When the gear is removed, Presently the fan is rotating at 928 RPM. When it is coupled with the Motor directly, motor tries to move it at 995 RPM

To Move the fan at 995 RPM Torque that is required is 26643 N.m

For this we require about 2780 KW (Minimum motor)

Thank U very much for Ur Time and attention

Aghvel Niazi

Maple Leaf Cement Factory

You're welcome. I think you should investigate more on the possibility of trimming down the fan. The original design seems quite strange to me, but there may be a reason--I just don't know what.

#include<iostream>
using namespace std;
void load_torque();
main()
{


load_torque();





return 0;
}

void load_torque()

{

/*********Initialization ********/

float N1,N2,N3,P1,P3;
/*
Output Power
Rotational speed of motor
Rotational speed of Gear (output)
Required Power without Gear
*/

float T1,T2,T3;
/*
Load Torque

*************************************************************************/

cout<<endl<<endl;


cout<<"\t *** Motor Side *** ";

cout<<endl<<endl;

cout<<"\tOutput Power (P1) in KW :"; cin>>P1; cout<<endl<<endl;

cout<<"\tShaft Speed (N1) in RPM :";cin>>N1; cout<<endl<<endl;

/*
T1 = P1 / W* 2* (pi/60)

T1=Torque in N.m

P1=Power in Kilowatt

N1 in Revolution Per Minute

*/


T1=(1000*P1*60)/(N1*2*3.14159); cout<<"\tOut Put Torque (T1) in N.m :"<<T1;


/*************************************************************************/


cout<<endl<<endl;
cout<<"\t *** Gear Unit Side *** ";

cout<<endl<<endl;

cout<<"\tInput RPM to Gear (N1):";cout<<N1; cout<<endl<<endl;

cout<<"\tOutput RPM of Gear (N2):";cin>>N2; cout<<endl;

/*
N1*T1 = N2*T2
T2=T1*(N1/N2)

T1 is torque at Speed N1
T2 is the torque at Speed T2

*/



T2=(N1*T1)/N2;cout<<"\tOut Torque (T2) of the Gear unit in N.m :"<<T2<<endl<<endl;


/*************************************************************************/


cout<<endl<<endl<<endl;
cout<<"\t *** Change speed Case *** ";

cout<<endl<<endl;


cout<<"\tSpeed of Fan (N2) in RPM is :"<<N2<<endl<<endl;

cout<<"\tTorquue at This Speed T2 (N.m) :"<<T2<<endl<<endl<<endl;

/*
T3/T2 = Square(N3/N2)
T3 = T2 * (Square(N3/N2))

T3 = Torque at the speed N3

T2 = Torque at the speed N2
*/

cout<<"\tEnter New Desired Speed (N3) of Fan in RPM :"; cin>>N3;
T3 = T2 * ((N3/N2)*(N3/N2));
cout<<endl<<endl;

cout<<"\tTorquue (T3) in N.m at Speed N3 :"<<T3<<endl<<endl;

/*

T3 = P3 / N3* 2* (pi/60)

P3= (T3*N3*2*pi)/(60*1000)

T3=Torque in N.m

P3=Power in Kilowatt

N3 in Revolution Per Minute


*/

P3= (T3*N3*2*3.14159)/(60*1000);

cout<<"\tPower (P3) in Kw for Load of Torque ( T3) :"<<P3<<endl<<endl;
cout<<endl<<endl;



}

I am assuming it is a straight (spur or helical, likely single stage) almost 1:1 gear box (precisely 1:1.07)

I am really not able to comprehend the usage of such a small ratio gearbox.

Also remember this gearbox will be around 98% or even more efficient.

Before coupling the fan to the motor please consider the forces that are coming on to the motor bearings. Hope the bearings may be over loaded causing the failure of the motor.

Once look back on the motor design also.

Contact the fan manufacturer. They went to a lot of trouble to put a gearbox there. There are also special coupling devices that can do what you need.

http://www.magnadrive.com/

mount the fan on double row spheircal roller bearing one axial locking and one non-locking. The device above can regulate power to the fan requardless of the motor.