How to calculate how fast i can speed up a fan with remaining full load amperage 1

[AirMD]

Hello all,

I am trying to calculate how much faster I can speed up a fan without going over the nameplate amperage. It is a 20hp 3ph 575v motor with eff. and power factor of 0.8. The motor is currently drawing and average of 16.0 amps across 3 phases nameplate is 18.9. If someone could help me out with this it would be much appreciated

 

[MintJulep]

Not very much.

P1 / P2 = (n1 / n2)³

P is power, n is RPM.

 

[mikekilroy]

how does this help him know how much faster he can go to same speed?

N1=N2

He does not know P1 (just knows it is LESS than 20hp) and knows P2=20hp.

And he wants to solve for a term t (time in seconds). I don't see t in the equation either.

 

[mikekilroy]

I would take a different approach... there is a missing data that is required to exactly calculate this: Isd magnetizing current. With that for this motor, it can be calculated. I also assume you are not running above base speed (in constant HP area).

If one has access to the fan and can run it at some speed like 10% of base speed and measure the then basically no load current, they could use that as very close approximation of Isd.

Lacking that, a close approximation can be calculated if it is assume Isd= 1/2 nameplate current.

So 18.9a means 10 amps magnetizing and therefore the remaining vector current is Isq or torque producing value. from Inameplate^2=Isd^2+Isq^2, 18.9^2=10^2+Isq^2 or Isq=16amps

since load is now 16.0amps, then 16^2=10^2+Isq^2 or you are using only Isq=12.48amps worth of the LINEAR TORQUE PRODUCING CURRENT AVAILABLE OF 16 AMPS.

So, since T=Jw/t where T=torque, J=total inertia, w=speed change, & t=time to do it in, if we use the remaining available current we can accel at a faster rate of 16/12.48= 28% faster than now.

 

[mikekilroy]

and this may be so much hogwash too! since your fan load goes up by the cube of the speed, it means if you used the same 16 amps the whole way up, more torque is available at the lower speeds to accel so it could be quicker.

Or your 16 amp steady state speed probably means nothing to accel time and torque since for accel you may be using the drives 125 or 150 or 200% overload capacity.

Seems like best answer should be go set accel ramp faster and faster until it faults then back off a tad - that is the fastest you can accelerate your fan.

 

[dvd]

Engineers sometimes ask questions that are quite clear to themselves, but no one else. I read the question similar to MintJulep and btrueblood. The OP has a fan that is not moving enough air at its current rpm and wants to increase the rpm up to the point of using full motor power at a higher rpm. Furthermore, since the OP is asking this question, the fan in question is not on a vfd, but has a sheave ratio that produces the current fan rpm.

But, if the question is one of getting up to a particular angular velocity sooner, another aspect to consider would be dampering the airflow so that the fan is not moving any air, but just overcoming inertia.

A good paper on motors is The Cowern Papers which contains a section on fans.

 

[dvd]

... but another question that needs to be asked is if the density of the airstream is constant. If the fan gets re-sheaved to max out the HP at some particular speed and then the temperature of the airstream dramatically decreases, the fan motor is going to trip out.

A nice source of information on fan affinity laws is New York Blower's Engineering Letters 2 and New York Blower's Engineering Letters 4

 

[mikekilroy]

wow, I feel really dumb this morning after re-reading op and I can see the question as you guys did! guess it is my servo background seeing it as an acceleration question instead of a cubed constant load issue!

 

[mikekilroy]

it actually makes the answer easier tho using my method too... he is still using 28% less than max torque from the motor, so since the load should go up by the cube of the speed as mintjulep pointed out, it would mean speed could go up approx (28%)^.33 or only 3% faster. But it sure seems that would not be a good move when your points are considered dvd.

 

[MintJulep]

Efficacy. It's important.

 

[zekeman]

"t actually makes the answer easier tho using my method too... he is still using 28% less than max torque from the motor, so since the load should go up by the cube of the speed as mintjulep pointed out, it would mean speed could go up approx (28%)^.33 or only 3% faster. But it sure seems that would not be a good move when your points are considered dvd.
Your 28% is incorrect, since we are talking SS.

I get
N2/N1=1.056
assuming the PF remains 0.8
(18.9/16)^.33

 

[mikekilroy]

zekeman, the 18.9 & 16 amp numbers are the vector sum of torque producing current Isd & magnetizing current Isq, which are at 90 ele degrees to each other. Only the Isd contributes to the torque while the Isd remains constant; this is why you must separate the two and only work with the Isq portion. that is why you cannot simply do 18.9/16.

 

[TheTick]

18.9 amps/16 amps = 1.18

1.18 ^ 1/3 = 1.057

You can increase speed 5.7% from whatever value it was that you failed to mention. Be sure to measure accurately when you are this close to 100% capacity on the motor.

Sincerely,
Former air balancing technician.

 

[zekeman]

Aren't we missing the point of how he controls the voltage and the frequency to get to the nameplate current and satisfy the load speed line of the fan?

The fan torque is proportional to N^2 and the power proportional to N^3. so you have to vary both V and f to get it and at the same time get to the rated current.

 

[TheTick]

Depends on how speed is controlled. If it's a variable pulley, it's just a matter of dialing speed up and down, and amperage follows by a factor of n^3

 

[electricpete]

zekeman, the 18.9 & 16 amp numbers are the vector sum of torque producing current Isd & magnetizing current Isq, which are at 90 ele degrees to each other. Only the Isd contributes to the torque while the Isd remains constant; this is why you must separate the two and only work with the Isq portion. that is why you cannot simply do 18.9/16.

I'd say if you work out the numbers assuming exciting current magnitude is on the order of 20 - 30% of FLA, the answer will be pretty darned close. Within 1% or so.

=====================================
(2B)+(2B)' ?

 

[electricpete]

To clarify my point, it should be fine to neglect the complications of magnetizing and load components of current in this case.

=====================================
(2B)+(2B)' ?

 

[Artisi]

I would say why bother, the margin of 2.9 amps should be left just about cover any voltage fluctuations / upset conditions and how accurate is the amp meter anyway.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[mikekilroy]

I agree from the data it seems too close to try to squeeze any more speed out of this fan load with the potential load variables.

but that said, we still can calculate rather than guess at the exact speed increase possible assuming no outside forces. the magnetizing current is significant, particularly in this case due to the high pf, so using nameplate amps ratio alone does not get us close. I had a few false starts so redid the calcs properly using pf to determine the exact magnetizing exciting current for this motor - rather than 20-30% it is well over 50% of nameplate. Here is the calcs that I believe are accurate for this motor:

.8 pf means Isq, Itorque producing portion, of 18.9a vector current =18.9*.8=15.12a full load.

so this means Isd, magnetizing current, which is a constant from 0 to base speed, is

(18.9^2-15.12^2)^.5=11.34a (*note 60% of nameplate)

present measured vector current is 16a, so Isq=(16^2-11.34a^2)^.5=11.29a used now

so fan is using 11.29/15.12= 75% of the motor torque.

btw, this means the pf at this present load is 11.29/18.9=.60

so since fan torque goes up by square of speed (power goes up by cube of speed) this fan theoretically can increase present speed by the ratio (15.12/11.29)=1.339 .... this means since we can increase torque producing current 33.9% to get to nameplate rating. so I think this means we can increase speed by 33.9^.5= 5.8%

 

[Artisi]

Looks to me to be the cube root of 18.9 / 16 = 1.057 -- seems that's a 5.7% increase.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)