pulley rpm math
pulley rpm math
(OP)
Please check my math. 3400rpm mtr with 5" pulley connected to a 10" pulley on compressor. We will replace 3400rpm mtr with a 1725rpm mtr. So, either the mtr pulley needs to be enlarged or the compressor pulley needs to be smaller:
(5 x 3.14 x 3400)/(10 x 3.14) = 1700.9 comp rpm
(9.85 x 3.14 x 1725)/(10 x 3.14) = 1700.0 comp rpm
(5 x 3.14 x 1725)/(5.07 x 3.14) =1701.2 comp rpm
tks, paul
(5 x 3.14 x 3400)/(10 x 3.14) = 1700.9 comp rpm
(9.85 x 3.14 x 1725)/(10 x 3.14) = 1700.0 comp rpm
(5 x 3.14 x 1725)/(5.07 x 3.14) =1701.2 comp rpm
tks, paul





RE: pulley rpm math
Check your torque/power requirements before doing this.
RE: pulley rpm math
You can leave it out. You have the numbers right.
Don't get too hung up on them. A few RPM will not make a big difference. Usually you just go for the closest off-the-shelf size to get you back to where you want to be.
I would caution you to not use a small pulley as below 5 inches starts to be very hard on belts.
Keith Cress
kcress - http://www.flaminsystems.com
RE: pulley rpm math
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
RE: pulley rpm math
RE: pulley rpm math
Dm/Dc = Nc/Nm
D = Diameters (motor, compressor)
N = Speed (compressor, motor)
I agree with pete. Go with a 10" pulley for the motor.
RE: pulley rpm math
Mike Halloran
Pembroke Pines, FL, USA
RE: pulley rpm math
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: pulley rpm math
On third thought, you may need to make the larger motor pulley extra light, so the extra inertia won't prevent starting.
On fourth thought, given the extra inertia of the larger motor pulley, even a lightweight one, you may be forced to go to somewhat smaller pulleys and a high torque belt, i.e., you may have to re-engineer the drive.
In which case, a new two-pole motor may be cheaper.
Why are you replacing the motor?
Mike Halloran
Pembroke Pines, FL, USA