Posted 13 October 2006 - 02:52 AM
We have a vacuum blower working at a corrugated machine. It has an eddy current motor 400vac, 50HZ, 15KW and 1390RPM. Vcc blower is driven through V belts. The size of both pulleys is same. Now the life of vcc blower is over and we want to replace it. I want to increase the rpm of blower up to 20% by decreasing the size of its pulley. What is the formula for calculating the desired KW of motor? Someone told to me that this new application requires more than 50% more power of motor (22.5KW). Is it true? I want to know the proper formulas.
Any help will be highly appreciated.
Posted 13 October 2006 - 02:34 PM
As such, POWER is proportional to the cube of the speed.
In your case, you want to increase the speed by 20% : from 1390 rpm to 1688 rpm.
Here's the math:
1688 ÷ 1390 = 1.2
1.2 ^3 = 1.728
15 kW x 1.728 = 26 kW
A 30 kW motor should handle the additional power requirements without any concern.
Verify that the Eddy Current Coupling is rated for the additional speed and power transmission.
Posted 14 October 2006 - 02:20 AM
Thanks for your useful information. I would apply and inform members in this forum for this kind of experience.
0 user(s) are reading this topic
0 members, 0 guests, 0 anonymous users