Multiple Pulley Power Transfer

[ggcarmichael]

I would like to drive two identical devices off of 1 power source. I would like to use two independent but equal sized V-belts, a double stacked pulley on the drive shaft and a single pulley on each of the device shafts. If the power source has a rating of 36HP @ 3600 RPM, what is the maximum HP rating of the identical devices @ 1200 RPM? The drive pulley/device pulley would have a ratio of 1 to 3. I've looked for an equation or explanation of how to solve for this but have not been successful. Any assistance would be appreciated.

Thanks,
Gilbert

 

[ray b]

HP = power X time = work done
so if you slow the speeds 3600/1200 you don't gain more work
as it takes longer

but you do gain power 3 times the power but at a cost of speed
minus the friction losses

 

[brewnog]

HP = power X time = work done
so if you slow the speeds 3600/1200 you don't gain more work
as it takes longer

but you do gain power 3 times the power but at a cost of speed
minus the friction losses

That's all incorrect, sorry ray.


Power = torque x speed

So (neglecting friction), if your pulley ratio is 1:3, you will increase the torque by 3 times, decrease the speed by 3 times, but power transmitted will remain the same.

With 36hp being developed at your drive shaft, you'll never get more than this at your 'device shafts'.

 

[ggcarmichael]

Thanks for the responses. So the power stays constant, torque increases and speed decreases. So the two devices would have a max HP of 18 each (not accounting for drive train losses)? Is there a way to take advantage of the increase in torque at the device shafts?

 

[brewnog]

Thanks for the responses. So the power stays constant, torque increases and speed decreases. So the two devices would have a max HP of 18 each (not accounting for drive train losses)? Is there a way to take advantage of the increase in torque at the device shafts?

Exactly. Be careful not to underestimate your drivetrain losses with belt/chain type systems.

Taking advantage of the torque increase is exactly what you do when using a lower gear on your bike, or in a car. How this can be applied to give advantage to your 'device' depends entirely on what your application is, and what you're trying to do.

 

[ggcarmichael]

I had anticipated 80% efficiency in the belt drive system. I plan to drive two backward curved centrifugal fans. Will driving these types of fans be more more advantageous than driving another type of device?

 

[brewnog]

I had anticipated 80% efficiency in the belt drive system. I plan to drive two backward curved centrifugal fans. Will driving these types of fans be more more advantageous than driving another type of device?

It depends entirely on what you want to do. Driving centrifugal fans will be a pretty poor solution if what you're trying to do is compress air to inflate car tyres, yet they could work quite well for shifting contaminated air out of a spray paint booth.

What's your application?

 

[ggcarmichael]

I don't need pressure - I need to continuously move a large volume of air. I'm essentially creating a wind tunnel for experimenting with aerodynamics.