electical draw from increasing fan speed

[cmaki]

Wasn't exactly sure where to put this thread. If I'm wrong please move it.

We are currently upgrading our dust collection system and part of that is increasing the speed of the fan. We are running a 75hp motor to drive a fan at about 1275rpm. We want to increase that speed to about 2000-2500rpm. (not sure yet on the final speed) The motor is controlled by a driver that I've never seen before. It seems to charge up to 100% then pulls in the starter. I haven't looked at it that closely and I can if someone has questions.

My question is that if we increase the driven speed will that have any great effects on the motor's current draw? Will it have other effects? Just want to cover all my bases.

 

[GoldDigger]

The power required to turn an axial fan (blade type) will go up as the cube of the speed if the airflow is not restricted.
I would worry a lot more about the motor being underpowered than about the current.
Doubling the speed could require 8 times the HP.

 

[mgookin]

If your system is doing more work, it will require more (power) input. At constant voltage, amperage will of course increase.

But can you elaborate on how you are "upsizing" your system? The design really should be done by a qualified engineer unless you're getting a new off the shelf dust collector system which already prescribes collector duct sizes & quantities, lengths, bends, etc.

edit: looks like digger & I were typing at the same time; great minds think alike! :thumbsup:

 

[petersonra]

making the fan go faster sounds like a great idea until you crunch the numbers.

some fans have blade shapes that do deal well with having the fan rpm doubled.

 

[cmaki]

The design really should be done by a qualified engineer

And as soon as we get one we will be good to go. I looked closer at the drive. Its a baldor digital soft starter.

So what you are saying is that if we take our 75hp motor running at 1800 spinning a fan pulley at 1275 and increase it to 2000 we will need a motor that is possibly 8 times as large?

*I just ran a calculator at freecalc.com for fan speed CFM and RPM. a 75 hp motor going from 1275 to 2000RPM will require a 289.5 hp motor. Holy crap. don't think the boss has any clue about this.

 

[cmaki]

ok how about this. Is there a way for me to tell if I am using the close to the full motor HP? I would think taking an amp draw and seeing how close I am to FLA but I'm not that great of an electrician to know. motor is rated at 86 amps. If I do an amp draw and get 15 amps I would think I can put more of a load on the motor. if I'm at 75 amps I can't do anything. Am I correct in my thinking?

 

[cmaki]

But can you elaborate on how you are "upsizing" your system?

Forgot about this question. We have 3 dust collectors. 2 of them are quite old and we have a lot of problems with them. The middle one is the largest and newest. It looks like we can easily tie this one into the ducting of the other 2 and run the entire plant off of 1 dust collector. We had a contractor look at everything and said it was possible. the boss likes this guy but I know he is just a salesman that will sub every aspect of the job and mark everything up a ton.

 

[Besoeker]

And as soon as we get one we will be good to go. I looked closer at the drive. Its a baldor digital soft starter.

So what you are saying is that if we take our 75hp motor running at 1800 spinning a fan pulley at 1275 and increase it to 2000 we will need a motor that is possibly 8 times as large?

*I just ran a calculator at freecalc.com for fan speed CFM and RPM. a 75 hp motor going from 1275 to 2000RPM will require a 289.5 hp motor. Holy crap. don't think the boss has any clue about this.

But is it drawing 75 hp at 1275 rpm?

 

[cmaki]

But is it drawing 75 hp at 1275 rpm?

I don't know. How do I find this out?

 

[Besoeker]

I don't know. How do I find this out?

Ideally by measuring the power.
But a simple measurement of supply current would be a good start in checking whether the motor drawing full load current.

 

[cmaki]

I checked the amps after I posted. we are drawing an average of 44amps (42, 43, 46 each leg) My mechanical side would say that if our FLA is 86 amps we are using just over half of our power but I know that isn't true for electrical items. Is there a way I can figure out how much I can increase the pulley size without buying 8 new pulleys and trial and error-ing it to figure it out?

 

[GoldDigger]

If you are really at half power on the motor (probably less than that when you account for power factor), you could probably get away with a ratio change of sqrt(2), which would be close to 1:1.4 change.
But much better to find an engineer.
The fan calculator you are using should give you close to the same answer if you look for the speed change that will double the motor size from 37 to 75.
But add more air flow from the other two systems, as well as possibly different pressure requirements for the other two duct systems and you are right back to needing an engineer.

 

[gadfly56]

I checked the amps after I posted. we are drawing an average of 44amps (42, 43, 46 each leg) My mechanical side would say that if our FLA is 86 amps we are using just over half of our power but I know that isn't true for electrical items. Is there a way I can figure out how much I can increase the pulley size without buying 8 new pulleys and trial and error-ing it to figure it out?

Use the following: P1 / P2 = (n1 / n2)³. All things being equal, just ratio the amps (44/86 = 0.51) and take the cube root so n1/n2 = .8; you can run at 1,594 rpm (thereabouts).

 

[Besoeker]

I checked the amps after I posted. we are drawing an average of 44amps (42, 43, 46 each leg) My mechanical side would say that if our FLA is 86 amps we are using just over half of our power but I know that isn't true for electrical items. Is there a way I can figure out how much I can increase the pulley size without buying 8 new pulleys and trial and error-ing it to figure it out?

Not easily without motor characteristics.
The FLC of 86A looks about right for a 75 hp (56kW) motor.

But you can't simply pro rata output power based on current.
The motor will typically take 30% of FLC just idling. Doing no work at all.

 

[gadfly56]

If you are really at half power on the motor (probably less than that when you account for power factor), you could probably get away with a ratio change of sqrt(2), which would be close to 1:1.4 change.
But much better to find an engineer.
The fan calculator you are using should give you close to the same answer if you look for the speed change that will double the motor size from 37 to 75.
But add more air flow from the other two systems, as well as possibly different pressure requirements for the other two duct systems and you are right back to needing an engineer.

This can't be overemphasized. You've got three systems, each with it's own characteristic "resistance" and goosing up the fan motor is no guarantee that you'll see the the correct flow in each. Just like electricity, you'll see flow in all three "conductors" but it will be partitioned according to the "resistance".

 

[mgookin]

Is it airflow or airflow + woodflow?

 

[gadfly56]

Is it airflow or airflow + woodflow?

Likely "B". From flour-fine to thumbnail-sized or larger chips.

 

[ptonsparky]

Use the following: P1 / P2 = (n1 / n2)³. All things being equal, just ratio the amps (44/86 = 0.51) and take the cube root so n1/n2 = .8; you can run at 1,594 rpm (thereabouts).

Nice.

Now if everything always works as today and no one goes "why is that damper closed?"

 

[Besoeker]

Use the following: P1 / P2 = (n1 / n2)³. All things being equal, just ratio the amps (44/86 = 0.51) and take the cube root so n1/n2 = .8; you can run at 1,594 rpm (thereabouts).

You simply can't just ratio the amps.
On that basis you would take the power at 30A to be 35% load. It could be 30A at zero load.

 

[cmaki]

The way we are redoing it, it is not 3 systems it will become one large system. Here is what I got from an engineer friend of mine. He is possibly the smartest person I've ever met because he does this type of math in his head for fun.

Power (P) = Volts (V) x Amps (i) x Efficiency (h) x Power factor (f), or i = P/Vhf. The value of f is strongly dependent on load, varying from about 0.4 at 25% power to 0.8 at full power. Your motor is drawing about 50% of full load amps, so it is producing about 0.5 x (0.4/0.8) = 25% of its rated power. You wish to draw 77 amps, or 90% of its full load current, and I'd estimate its power factor at about 0.75 at that load. An affinity law states that power is proportional to the cube of speed, so N2 = N1 x [(77)(0.75)/(44)(0.4)]^(1/3), where N2 is the desired speed and N1 is the current speed. If my assumed power factors are correct, this is about 1857 rpm. Before buying any sheaves, contact the motor manufacturer for a power factor curve, and the fan manufacturer to verify that the fan is rated for this speed.

Any objections to this?