1000 kilo Newton meter (K/Nm) torque

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To achieve 1000 kNm of torque at 30 RPM, a motor's power output must be calculated, typically in kVA, which is essential for determining the motor's specifications. A 150 kVA, 3-phase motor running at 960 RPM can be paired with a double-stage gear system to reduce RPM effectively, with the first stage reducing by a ratio of 10:1 and the second by 3:1. The output torque can be calculated based on the motor's power and the gearing system, ensuring that the motor can handle the required load. The need for a flywheel depends on the load's variability, and engineers may consider alternatives like hydraulic motors for weight efficiency. Proper gear design, such as using an internal gear or epicyclic system, can also optimize performance while minimizing gearbox mass.
parsci
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i want to produce 1000 K/nm of torque using motors and gear , output rpm needed is 30 and 1000 k/ nm of torque

1) motor kva? and rpm ?

2) gear ratio and size ?

3) does it also require flywheel ?
 
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You mean you want to produce 1000 kNm and a frequency of 30 RPM. The way you wrote it makes no sense.
 
I also have no idea what a motor's kva is. No doubt another botched unit usage?
 
dauto said:
I also have no idea what a motor's kva is. No doubt another botched unit usage?

No, the units are correct (although they should be written kVA). Motors (and e.g. transformers) are always rated in VA, not W. The former is the unit for apparent power whereas the latter is the unit for real power. They are related via the power factor of the load.
 
parsci said:
i want to produce 1000 K/nm of torque using motors and gear , output rpm needed is 30 and 1000 k/ nm of torque
1) motor kva? and rpm ?
2) gear ratio and size ?
3) does it also require flywheel ?

You will get better answers if you solve the problem yourself. Here's a hint to get you started: You already have enough information to calculate the power output, so you know how much power you need the motor to deliver. That will tell you what the motor kva must be; and when you're considering a motor it will be specced to deliver that power at a particular RPM. If that RPM is not 30, then you will need gears in whatever ratio will get you from the motor RPM to your desired 30RPM output.

Whether you need a flywheel or not will depend on how variable the load is.
 
Nugatory said:
You will get better answers if you solve the problem yourself. Here's a hint to get you started: You already have enough information to calculate the power output, so you know how much power you need the motor to deliver. That will tell you what the motor kva must be; and when you're considering a motor it will be specced to deliver that power at a particular RPM. If that RPM is not 30, then you will need gears in whatever ratio will get you from the motor RPM to your desired 30RPM output.

Whether you need a flywheel or not will depend on how variable the load is.

Thnks for reply , help me in getting output torque
150 kva motor 3 phase 960 rpm using double stage gear 1st stage 78 inch dia Reducing rpm 10:1 2nd stage gear 36 inch dia reducing 3:1 ,how much the output torque
 
parsci said:
Thnks for reply , help me in getting output torque
150 kva motor 3 phase 960 rpm using double stage gear 1st stage 78 inch dia Reducing rpm 10:1 2nd stage gear 36 inch dia reducing 3:1 ,how much the output torque

You want 1000 kNm of torque at 30 RPM. What power (in watts) does that represent?
If your motor delivers that power while turning whatever revs are needed to get 30 RPM after going through the gearing, it's generating the necessary torque.

(Although wise engineers will choose a motor that provides some excess capacity).
 
The practical implementation of such a system will be limited by the mass of the gear wheels and the box that holds them together. The gear box will weigh less if you use an extra stage with a final bull gear ratio closer to 3:1 than 10:1.
You might consider using an internal gear which has more tooth contact. An epicyclic with multiple planets could also help reduce total weight.

An obvious alternative to such a heavy gearbox would be to use a hydraulic motor. A hydraulic “gear motor” for example exerts the hydraulic force on many teeth. The gear ratio is determined by the relative fluid volume per revolution of the motor and the pump. It is self lubricating and the hydraulic fluid can be cooled if necessary.
 
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