# Appropriate power

1. Apr 24, 2017

### Bassam Salman

Hi
I have a personal project, In link bellow in figure A I need to calculate appropriate power motor (HP) and torque (N.M) to drive flywheel at maximum speed 2000 rpm and what is value of resulting power on 750 rpm pulley.
In figure B if we use two flywheel the weight for each half weight of flywheel in figure A do motor power and resulting power equal to figure A

2. Apr 24, 2017

### Dr.D

At steady state, in an ideal (lossless) system, the power out at 750 rpm will be equal to the power in at 3000 rpm.

Since no real system can be built without losses, you will need to account for the losses in the system to see how much power can be drawn out. Much depends on the quality of your bearings, the alignment of shafting, and the windage of the several elements.

3. Apr 24, 2017

### anorlunda

With zero friction it takes zero power to maintain a rotation at constant speed.

Are you sure that the question is related to constant speed rather than acceleration?

4. Apr 24, 2017

### anorlunda

Real life flywheels have friction and thus power loss.

5. Apr 24, 2017

### Bassam Salman

Is that mean no power loss on flywheel

6. Apr 24, 2017

### Bassam Salman

Now I need calculate without friction and at constant speed

7. Apr 24, 2017

### Dr.D

An ideal flywheel requires no power to run at constant rotational speed. Real flywheels (and every other machine part as well) always involve friction at bearings and windage. How much these friction losses are depends upon the shapes, the speeds involved, and the surrounding fluid medium.

8. Apr 24, 2017

### anorlunda

@Dr.D is telling you the same thing I am.

Edit: rearranged the above.

9. Apr 24, 2017

### Bassam Salman

Ok thank you Dr. D and anorlunda
I need value of motor size (calculation) and also calculate out power

10. Apr 24, 2017

### Staff: Mentor

If there is no friction, the output power is equal to the input power; the flywheel doesn't change that.

You aren't thinking that the flywheel let's you get more power out than you put in, are you? That would violate the laws of physics and therefore the rules of this forum...