Current Draw

D

David

Guest
I just wired up my cyclone with the 5hp Lesson motor for CW rotation. When I metered the current draw I got 15 amps at 244 volts. This is without a dust bin, filters, or ducting installed. Does anyone have an idea on what may be wrong, I thought I should be drawing around 20 amps. The 15 amps is what I metered on one side of the 240 volt service, the other side is at 14.8 amps.

Any help would be great!

David
 
David,
We just hooked up a new blower housing and 15" impeller on my son Matt's system. I took a current reading and got 18.3 amps @ 245 volts. I don't know what the difference is other than the meters. Maybe mine or yours is off.

Ed
 
When I got my unit the motor was set up to be wired for CCW rotation so thats the way that we wired it. Is there a difference between our two units. I haven't tested any current draws yet, ,plan to do that this weekend.
 
Jon,
The motor needs to be wired using the CW wiring diagram. That will give you a Counter Clockwise rotation when looking from the shaft end of the motor.

Ed
 
Sounds like I am on the right track then. I hope to have some amperage readings by this weekend. Hope evrything continues to work out.
 
Motors are rated for ampacity at full load. With a 5Hp load the motor should draw the rated current 20A. With a smaller load the current will be less, as you add filters and more restriction the current will increase because the motor is working harder. There is nothing wrong with running it below the rated current/Hp but it does make the motor less efficient and the power factor drops. If you want to know and play with some math 746Watts = 1 HP. I can dig up the formulas for figuring all the specs but it is not necessary. What you are seeing is perfectly fine and expected.
 
Originally posted by bbenkaz@Mar 9 2006, 05:21 PM
Motors are rated for ampacity at full load. With a 5Hp load the motor should draw the rated current 20A. With a smaller load the current will be less, as you add filters and more restriction the current will increase because the motor is working harder. There is nothing wrong with running it below the rated current/Hp but it does make the motor less efficient and the power factor drops. If you want to know and play with some math 746Watts = 1 HP. I can dig up the formulas for figuring all the specs but it is not necessary. What you are seeing is perfectly fine and expected.
Click to expand...
I'm probably wrong, but what I read on Bill Pentz site doesn't agree with that statement. I think I read that less restriction means more work{ airflow} and that makes for more amperage.
 
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
 
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
Click to expand...
Actually I did test the amperage with a clamp meter with the piping and filter disconnected and with the filter and piping connected and also with the piping blocked. In contrast to your statement, the highest amperage was with the piping and filter off and the lowest was with the piping blocked. I bought the meter just to test my Clear Vue to protect the motor. I'm sure you know what you are writing about, conversely I know what the Bill Pentz site says and the results of my own testing. Perhaps your version would be more valid in a closed system, such as a hydraulic system. I think there is a lot of slippage in the Clear Vue "pump" so it can be "dead headed" and the power consumption goes down. But--I'm not an authority and these things, I just read about them. Good Luck to you and I hope you are enjoying your Clear Vue as much as I am.
Bob S
 
well I guess that is possible, I am not sure why but it is something to do with the air and the cyclone and how it works. Electrically it dosen't matter regardless if you load the motor it will draw more current. Whether this happens with things connected or not, it must be a mechanical air flow reason which is fine. Either way the orignal question is answered. As long as you are not overloading the motor and drawing too much current the motor will not overheat and the world will be happy. :)
 
Just an educated guess, but possibly the impellor blades are stalled when the gates are closed (so there's little flow). That would mean the motor is running almost unloaded, and hence not draw much current.

-Rob-
 
HuH?

HuH?

""but possibly the impellor blades are stalled when the gates are closed ""

That just made me wonder If you have all of the gates on and the dc running, how does it not keep from burning up the motor? OH, I guess it is because of the design of the impeller with extra space around it so it does not strain it if nothing is open to let air in. I was thinking it was like the exhaust fan in my warehouse. If it is on, we have to open a door on the other end or it noticablly pulls down the motor because of the increased load as it strains to pull in air from the cracks around our doors.

V
 
The reason the motor doesn't burn up is because it has 1 6" inlet so opening 1 6 inch gate lets in the maximum amout of air minus piping so you will get a little extra by opening more but still the restriction is the 6 inch intake into the cyclone. Also the more air(less restriction) the harder the motor works and as you restrict the airflow the less air the motor has move hence a lower amp draw. As an example turn on your shop vacuum or house vacuum and listen to rpm of motor then close it off and RPM will go higher(no airflow). This higher rpm with no airflow to cool the motor is usually what burns the vacuum up although newer models may have seperate fans for this.

Hope this helps to clear up the confusion,

Matt
 
So the net results is that I will not damage the motor if the unit is on but no gates are open, correct?
 
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