# 1000 kilo newton meter (K/Nm) torque

1. Apr 16, 2014

### parsci

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 ?

2. Apr 16, 2014

### dauto

You mean you want to produce 1000 kNm and a frequency of 30 RPM. The way you wrote it makes no sense.

3. Apr 16, 2014

### dauto

I also have no idea what a motor's kva is. No doubt another botched unit usage?

4. Apr 16, 2014

### f95toli

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.

5. Apr 16, 2014

### Staff: Mentor

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.

6. Apr 16, 2014

### parsci

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

7. Apr 16, 2014

### Staff: Mentor

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).

8. Apr 17, 2014

### Baluncore

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.

9. Apr 17, 2014

### Baluncore

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