Originally posted by bbenkaz@Mar 11 2006, 02:21 PM
No
What most people think is a 5hp motor always produces 5hp. This is not true, it means that the motor is rated to handle a 5hp load. The current for the motor is rated at full load so if there is a smaller load it will draw less current because there is a smaller load like maybe it works out to be 2.5hp load. The big thing is not to operate the motor over the rated load. The leeson motor has a 1.0 service factor which means it cannot be used over its rated load. A lot of motors have a 1.15 service factor rating which means that they can be run at full load +15% and not break down. But trust me more restriction in the sytem will cause the current to go up because the motor will be working harder. If you want to test it open all your blast gates and run the motor then close all but one and you should see the current rise. Since electrically current and power are related roughly by
P=IE, or
power=voltage*current or
current=power/voltage.
I say roughly because since it is an AC motor there are actully some other things going on because the current and voltage are out of phase but the concept is the same. Since the voltage is staying the same a larger amount of power is going to always going to equal a larger current by the laws of electricty i.e
I=100watts/120V = .833A /
I=1000W/120V = 8.33A .
Since 746Watts = 1Hp you can get close these motors actully are 3Hp but say 5Hp, something to do with the way they rate compressors. anyway (746Watts*3Hp)/120Volts =
2238Watts/120Volts = 18.65Amps.
Why is this not the 20....amps the motor is rated? Because it does not take into account the power factor and efficiency which will cause the current to be higher. This is a very simplifed example and the math will not exactly work for your motor because you have to know the efficiency and the power factor at the load you are running. There are ways to calculate it but it is a pain in the butt and not neccesary. You can not use the efficiency and power factor that is on the motor nameplate because that is at rated load and will be different with a smaller load.
As for the less restriction equalling greater air flow yes that is true. Because the RPM of the motor is indirectly proportional to the current and output power. As the load increases the RPM will go down. So in your pipe to your machine you want max airflow = max RPM. This works only to a point because the motor cannot go faster than its sychronous speed which is determined by the number of poles inside the motor and the frequency of the electrical supply.
Hope that helps