Increased current from speeding up fan

[philly]

If I have a fan being driven by a motor and the motor is connected to the fan by a belt then I would think that by increaseing the speed of the fan by inreasing the belt ratio then the current drawn from the motor would be increased. Do others agree? Here is my reasoning. (We are assuming that head pressure stays the same.)

First I would think that the current would have to increase by the simple fact that because we change the belt or gear ratio to increase the fan speed, we are reducing the avalaible torque at the fan. Therefore in order to keep the same torque requirement at the fan at the increased speed the motor will have to provide an increased torque, and it is this increased torque that will be reduce by the new gear ratio to provide the origonal torque requirement. This increased torque means increased current to the motor.

Secondly since the fan torque increases with the square of the speed the fan torque requirement at the fan will increase. My above paragraph stating that the motor torque would increase with speed increased was based on the fact that the torque would increase linerally for a constant load torque at the fan. But now since the load torque is increasing exponentially I would assume that the motor torque would not only have to increase exponentially for the load torque increase, but would also have to double for the speed increase.

So esentially if you took the exponential fan load torque curve and shifted it upwards by double the amount would this represent the required torque from the motor with the new speed produced by increasing the gear ratio?

Am I on the right track with this or am I way off?

 

[__dan]

current varies with load

current varies with load

Motor current on a fan or pump, yes, according to the parameters you stated, but adjusting the gear ratio is sometimes an intermediate step. Input product, motor current, varies with the work output product, quantity of air moved. There are different ways to modulate the output work product. For a fan or pump, gear ratio changes, but also restricting supply at the suction side, reduces output work product and reduces motor load.

For example, a ducted fan when it's running without ductwork moves a lot more air because there's no restriction at the suction side. Motor running current goes up into the overload trip range. Even though the air is more easily available, moving more air is more work product and more motor current.

Let's say the fan is properly sized at design but at install is not delivering rated CFM. One reason would be restrictions in the ductwork reduce the quantity of air moved by the fan and reduces motor running current. You could try to speed the fan up by gear ratio changes on the adjustable motor sheaves, to bring the system to design air movement and design motor running current. The same effect could be had by increasing suction side air available to the fan.

For the motor that is overloaded into the trip range, restricting the pump or fan (throttling) at the suction side greatly reduces load and motor current by reducing the quantity of air or water moved by the pump, less work output.

Trying the same thing at the discharge side is the hard way to do it, could be asking for trouble.

 

[philly]

I guess the point I was trying to make is that if the load were a constant torque load then speeding it up with sheaves would increase the speed and decrease the torque at the load. So even know the load torque is staying the same the motor torque would have to increase to produce the same load torque due to the origonal torque ratio being reduced by the new gear ration.

I then tried to look at the standpoint of what happens with a variable torque load and thus was my resoning in my previous post.

 

[__dan]

Hp goes up the rpm

Hp goes up the rpm

Let's say torque is flat at the motor or at the load. Increasing speed, rpm, increases load Hp and motor load kW.

Theoretical or actual?

 

[Besoeker]

I then tried to look at the standpoint of what happens with a variable torque load and thus was my resoning in my previous post.

The torque for a centrifugal fan is generally proportional to the square of the speed. Thus, the power, being speed times torque goes up as the cube of the speed.

 

[gar]

100217-1549 EST

Philly:

Do not get all wound up in the changes in torque as you change things.

Dan has said what you need to consider and it is simple.

I will rephrase it.

HP = Constant * RPM * Torque

If you know that your fan has constant torque vs RPM, or at least in the range of interest, then for the fan speed to be increased you need to supply more HP.

Assuming 100 % efficiency of your speed changer, then whatever you do between the motor and the fan, if you increase the fan speed, then you increase the power the motor must supply, and the required power input to the motor.

.

 

[philly]

The torque for a centrifugal fan is generally proportional to the square of the speed. Thus, the power, being speed times torque goes up as the cube of the speed.

Yes I understand this, and if the speed was being increased with a VFD or something similar I would say that the torque would increase exactly with square of speed.

But since we are changing the speed by way of a gear ratio dont we have to account for motor torque increase due to changing gearing? Wont this add to the overall torque and thus change the profile of the squared torque due to simply the speed increase?

Let me ask this question. If we had a constant torque load and we sped the load up by changing a gear ratio, wouldn't the motor then have to supply a larger torque to account for torque loss by increasing gear ratio?

 

[One-eyed Jack]

The torque for a centrifugal fan is generally proportional to the square of the speed. Thus, the power, being speed times torque goes up as the cube of the speed.

This is very true. A simpler way to look at fans is "IF" it moves more air it takes more hp. Contrary to gin equipment installers idea that air does not weigh anthing.

 

[Besoeker]

Yes I understand this, and if the speed was being increased with a VFD or something similar I would say that the torque would increase exactly with square of speed.

For a centrifugal load, the torque requirement increases as the square of the speed regardless of how you generate that torque. It could be by different ratio belt drive, a mechanically variable gearbox or a VSD. The torque requirement doesn't change.

Let me ask this question. If we had a constant torque load and we sped the load up by changing a gear ratio, wouldn't the motor then have to supply a larger torque to account for torque loss by increasing gear ratio?

Yes it would.

 

[winnie]

Let me ask this question. If we had a constant torque load and we sped the load up by changing a gear ratio, wouldn't the motor then have to supply a larger torque to account for torque loss by increasing gear ratio?

Yes, you are correct. If you are supplying a fan load by means of a variable ratio gear, and your motor output is fixed speed, then the _motor_ torque will vary approximately as the _cube_ of the fan speed.

The equation that gar gave is a good way to look at this:

HP = Constant * RPM * Torque

Whatever your load is, its power changes with speed. Different loads will have different characteristics, but you can figure it out for each load.

With your fixed speed motor, the RPM is constant, so the only thing that can change to provide the correct power is _torque_.

-Jon

 

[gar]

100217-1645 EST

philly:

Read my post that I made just before your last one.

By making torque constant I was trying to get you to look at the problem from a power (energy) perspective. Afterwards you can go back and see what the motor torque does without getting tangled up with what the gear box is doing.

Once you see what happens with constant torque, then you can change the way load torque varies with speed.

Ultimately you can write equations that would bring the power loss in the speed changer into the whole system.

.