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Combination Motor/Gearbox

carrabta

Industrial
Mar 17, 2025
3
Hi,
On inspection of a motor/gearbox assembly that drives a large axial fan, we've found that the 110kW rated motor was attached to the Siemens H2 NV110 gearbox, which sounded just right (assuming the '110' in the name was the rating of the gearbox).
However, looking at the Acceptance Certificate in the documentation of the gearbox, it shows a power rating of 92,5 kW.
I'm getting confused, here, as I would have expected (for safety reason and mechanical "soundness") that the gearbox would be designed such as to take the max power that the motor could deliver, isn't that right or am I mis-reading something ?
Thank you !
Tanguy
 
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OP
a gear box must be designed with the torque and the rpm rating. problem is from previous
discussions comercial manufactures are not properly rating the gear boxes.
plus the above gear box should include a safety factor. and the precision based on SN data for the required number of cycles.
 
Are you 100% sure this fan and drive were fitted this way from the manufacturer? (If I had a dollar for every time someone thinks that having a drive service factor of 1.5 or 2.0 is adequate diligence to upsize the motor later...). 92.5kW / 125HP is the next lower standard motor power and from your documentation that seems to be the case.

I don't know Siemens/Flender drive codes with any special insight but I see their cooling tower drives come in sizes 105 through 112. Increments of gear drive size / torque capacity are typically at least 20% between case sizes and increasing geometrically. So it doesn't make sense to have 8 different case sizes with ratings 1% apart.
 
What about a load curve for the fan?
The gearbox should have a torque rating.
It the max load that the fan can generate is 80kW then might have said good enough.
 
Thank you all for your answers and comment.
So, effectively, I was wrong and EdStainless and dgeesaman point in the right direction: determine the power on load side, then multiply by x (1.3, 1.4, 1.5...) to compensate losses in the gearbox then you have the power required for the motor.

All good for me, I've got my answer !
Thanks all !
Tanguy
 
Thank you all for your answers and comment.
So, effectively, I was wrong and EdStainless and dgeesaman point in the right direction: determine the power on load side, then multiply by x (1.3, 1.4, 1.5...) to compensate losses in the gearbox then you have the power required for the motor.

All good for me, I've got my answer !
Thanks all !
Tanguy
I don't think you do.

Determine the power on the load side. Make sure that the drivetrain losses and expected power are less than the motor load. Drivetrain losses for a gear drive like this are typically assumed to not exceed 10%. That's one check to make sure you don't overload the motor.

Nominal drive ratings are based on continuous, perfectly smooth, non-stop torque transmission without any additional external loads. Service factor (extra margin over the nominal rating) is required for basically all real-world applications. The amount of service factor required depends on the application. 1.5x is a minimum reasonable service factor in my industry and it can go up to 2.0 or 2.5 pretty quickly depending on the amount unsmooth input power, fluctuating output load, when the unit is started and stopped more than daily, etc. Having an appropriate service factor ensures some degree of reliability and acceptable life.

There is also a thermal rating. In my industry that's rarely a limiting factor but with a cooling tower fan this could be important also.

I don't think you know your service factor for this application yet, but I can tell you it decreased when the larger motor was installed and it's very important. I think it's time to make a call to the manufacturer (not Siemens, probably the company that provided the fan system) and have your tag data handy. You should be able to tell them there is a larger motor and ask if they were involved in that change. Then confirm the actual service factor of the current configuration and recommended service factor for your application.
 
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In my world a system is calculated with a built in safety factor for reliability. And number of cycles.
But within that calculation there is also power loss, and efficiency. Which is different than a service factor. Which I defer because as I generally don't get involved. But there is a loosely used term which for the consumer heavy duty cycle, or light duty. On aerospace is all way super heavy duty. But we don't use that term . So that may be the diffrence.
 

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