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

 

[Tmoose]

I assume more airflow/static pressure is desired.
If the buss voltage and voltage at the motor are not low, and the motor SF is over 1.0, and there are no instances of alarm in operation I'd be inclined to go right to FLA if needed.
it doesn't hurt to look at the fan system to see if there are "issues" there.

It may be hard to find sheaves/pulleys with a 3 or even 6% different ratio.
An "adjustable sheave" might do it, but they often have standard incremental adjustments greater than 3%, so might need modification to allow teeny speed adjustments.

http://www.fremontindustrialsupply.com/catalog/62611 pics 024.JPG

Adjustable sheaves are justifiably hated by those responsible for vibration measurements, because of their inherent runout and resulting 1X tugging. When combined with wrapped belts and typical airhandler frame "eNgIneerInG" they can create impressive amounts of vibration.

 

[zekeman]

"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"


So why don't you stop here.

According to this ,the speed can increase 33.9%

Why are you now getting into power?

 

[zekeman]

mikekilroy ,

The OP wants to change the gear ratio to speed up the fan and as it increases, the torque on the
the motor will go up proportional to the speed squared.

So , using your input,
the speed ratio becomes

1.339^.5=1.157
or an increase of 15.7%

 

[zekeman]

I suspect the the motor will not handle the increased power which would be
1.157^1.5=1.244
or an increase of 24%

 

[zekeman]

Coorrection
Power ratio would be
1.157^3=1.54
or 54% increase in power
Could an induction motor handle that increase?

 

[MiketheEngineer]

I am one of those try it and see if it flies guys. Crank it up until you start blowing breakers - then back down about 10%??

 

[electricpete]

the magnetizing current is significant...so I think this means we can increase speed by 33.9^.5= 5.8%

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

Yes, that proves my point. Including magnetizing current in this calculation is not necessary and has no significant effect in this particular calculation.

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[electricpete]

...or more precisely, considering the magnetizing and load load components separately has no significant effect on this particular calculation (total power is very nearly proportional to total current in the range near full load amps).

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[rb1957]

admitting my ignorance ... how would you achieve this (speed increase) ?
i mean this is about getting the motor turning faster (and not a gearing issue), yes?
so what's limiting the speed now ? i'm assuming the OP is just flipping the switch, and the motor's doing what it can (under the load given).
it seems to me that to get the motor turning faster you have to reduce the load, yes?

 

[TheTick]

Fan laws are basic and quite reliable.

Air flow is 1:1 proportional to RPM (N): double RPM and double air flow
Pressure goes up by square: 2x RPM = 4x pressure
Power goes up cubes: 2x RPM = 8x power (and 8x amperage if voltage is same)

This WORKS. I've done it dozens of times on pulley-driven air handlers with single- and 3-phase motors. Need 5% more air flow? Increase speed by 5%, pressure goes up 10.25%, amp draw goes up 15.8%. Like magic (except that it's science and actually works).

If you are having trouble getting straight answers from engineers, go straight to the bottom and ask an air balancing technician.

 

[zekeman]

mikekilroy (Electrical)

"8.9^2-15.12^2)^.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."

Why not analytically?

Once more,

The PF=0.6 at present I=16
At full load
PF=.8 I=18.9

So
P2/P1=sqrt(3)*V*18.9*.8/[sqrt(3)*V*16*.6]=1.575
From the cubic proportionality, Power proportional to the cube of speed
(n2/n1)=1.575^.33=1.16
or 16 % rise is my final answer

 

[electricpete]

To Zeke and MikeE

Change in power factor from 0.8 at 100% FLA (18.9A) to 0.6 at 85% FLA (16A) is not anywhere close to realistic.

How did we come up with such a low number 0.6?
Two items

1 - There was an error in the calculation:

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

should be

btw, this means the pf at this present load is 11.29/16=.70

2 - The model is not perfect. There is the simplifying assumption of constant magnetizing component... a pretty good start and I’m not criticizing it for rough calcs but not 100% right. But there are vars consumed in the leakage reactance that go up with load. So we shouldn't vectorialy remove the same amount of reactive current at 16A that we had at 18.9... it would be lower reactive current at 16A (and higher prediction of load current component at 16A)

Both effects act in the same direction when we reverse them or consider them more precisely they will bering the predicted 16A power factor up toward 0.8 and the predicted speed change down toward 1.057 (in other words I'm suggesting the assumption of constant power factor and efficiency pretty good in this range).

We could do a detailed calculation, but why not look at a typical motor data sheet 575 volts, 20hp

http://www.reliance.com/pdf/pdf/aced/E09317ACDJ023.pdf

Curve 2 is motor power factor.
It looks like 84% at 20hp and 82.5% at 16hp.
If we had started at 80%, we might have dropped to 78.5%.
It’s a drop, but not much to worry about.
P2/P1 = [0.8*18.9] / [0.785*16] = 1.204
[note we efficiency considered constant, a good assumption by the curve)
N2/N1 = (P2/P1)^0.333 = 1.064
Not to far 1.057 that Artisi came up with simply using (18.9/16)^0.333


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(2B)+(2B)' ?

 

[electricpete]

A terminology correction (wish we had an edit button)

There is the simplifying assumption of constant magnetizing component

should've been

There is the simplifying assumption of constant reactive component

The reason for the clarification is that reactive includes both magnetizing and leakage contributions.

And how good/bad is the simplifying assumption of constant reactive component with load (item 2 above)?
....let's look at a reference:

http://www.energy.siemens.com/us/po...topics-application-notes/techtopics20rev0.pdf

. For a range of medium voltage machines sampled, the ratio between full-load reactive current and no-load reactive current varied from 140-260% (depending on machine design, speed, and voltage)

So according to Siemens, the vars approximately double going from no-load to full load.
So the reactive component of the current will approximately double going from no-load to full load.
This particular reference is addressing medium voltage motors, but the same principle holds for low voltage machines like this as well.

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[electricpete]

Lest I be accused of complicating something simple, let me follow up my longwinded detour with summary of relevant principles most others in this thread already agree with:
1 - P1 / P2 = (n1 / n2)^3 (first response of this thread)
2 - P~I (usually a reasonable approximation when over 50% FLA).

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[Artisi]

Correct, power increases as the cube of the speed change - no other argument on that point.

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]

funny how coincidence can make wrong answer come out same as right answer....

I really have enjoyed learning on this thread. I do not have much experiences in fan/pumps, and so have to research and take on faith stuff posted to a large extent.... I have known power in fan is cube of speed from my windmill design/build experience, but never messed with torque vs speed on fan..... at least one previous post has said torque increases with square of speed vs power with cube of speed... seems reasonable... i found docs via googling that agreed... so to cube amps (torque) vs square them is in question in my mind - obviously both are not right - I have seen curves showing torque goes up by square of speed on same chart as power going up by cube of speed so seems right. yet if voltage is constant, then amps is all that can change with speed increase, and so is it sq or cube? but then again, using full vector result current that includes magnetizing current is obviously wrong. So I am still confused on what fan torque/amps will actually do with speed increase... some have shown cube, some square...

I can though respond on magnetizing current vs reactive current comment... Again I have no experience in mv motors so cannot comment on the siemens article saying VAR current changes by almost 2:1 factor no load to full load. but I can say from 20+ years experience in vector control of magnetizing and torque producing current that this reactive current change based on load does NOT happen in 230v or 460v 3 ph motors. just does not. i have seen for 20 years that there is NO significant change in reactive current portion of vector sum current in a 230/46v motor in the range of 1-200hp. none. I can produce years of scope picture of reactive current vs real current that shows this to be true from no load to full load. so not sure how to react to this siemens article.... before the quoted section shown, it did say "To understand how to compensate for the poor power factor of a motor, we need to look at the components of the motor current. The real-power producing work is done by the resistive component of the current, which varies with the load on the motor. The reactive current of the motor consists of two components. The first is the magnetizing current that establishes the magnetic flux in the core which allows the motor to function. The magnetizing current is essentially constant regardless of load. The second component of reactive current is the leakage reactance current, and this component varies according to the load on the motor. The leakage reactance current is relatively small, so that the total reactive current is relatively constant (compared to the kW variation) over the range of motor no-load to motor full-load."

I do not know how to turn this relatively constant comment in 140-260% they go on to mention - I wonder if perhaps that is a typo and should have been 14-26% or something? or since it is "relatively small" this 140-260% turns out in reality to be as they say, so small as to be ignorable?

anyway, I can produce many scope pictures of reactive current real time vs load but perhaps this one on a 200kw system we did shows it all:

(I'd be happy to put a jpg scope pix here showing this clearly but have no idea how to include it, sorry)
xxxxxxxxxxxxxxxxxxxxxxxxx

it is very obvious that the total reactive current does not budge one bit from 0 to almost 90% load. so again, I do not know about mv motors and perhaps for some reason reactive current does change in these. but not in this 20hp motor per my experience.

i do not mean to offend anyone but i would like to understand the real relationship between torque and speed and power in these fan applications.

 

[electricpete]

I can say from 20+ years experience in vector control of magnetizing and torque producing current that this reactive current change based on load does NOT happen in 230v or 460v 3 ph motors. just does not.

i have seen for 20 years that there is NO significant change in reactive current portion of vector sum current in a 230/46v motor in the range of 1-200hp. none. I can produce years of scope picture of reactive current vs real current that shows this to be true from no load to full load. [/quote]
I assume that you (like me) are talking about a 3-phase motor fed from constant frequency, constant voltage. (not vfd).
In that case, you are mistaken in thinking the reactive current is constant with load. It is not the case.
As Siemens stated, the reactive component of current will increase on the order of 200% from no-load to full-load, give or take 60% or so.

so not sure how to react to this siemens article.... before the quoted section shown, it did say "To understand how to compensate for the poor power factor of a motor, we need to look at the components of the motor current. The real-power producing work is done by the resistive component of the current, which varies with the load on the motor. The reactive current of the motor consists of two components. The first is the magnetizing current that establishes the magnetic flux in the core which allows the motor to function. The magnetizing current is essentially constant regardless of load.

The second component of reactive current is the leakage reactance current, and this component varies according to the load on the motor. The leakage reactance current is relatively small, so that the total reactive current is relatively constant (compared to the kW variation) over the range of motor no-load to motor full-load."

I do not know how to turn this relatively constant comment in 140-260% they go on to mention - I wonder if perhaps that is a typo and should have been 14-26% or something? or since it is "relatively small" this 140-260% turns out in reality to be as they say, so small as to be ignorable?

You don’t need to invent a factor of 10 typo in two different places to understand what the authors meant.
It makes sense as is. Let’s review the statement in question. I’m going to add some emphasis, and I’m going to clip the entire paragraph (including the two sentences immediately following the ones you focused on).

The leakage reactance current is relatively small, so that the total reactive current is relatively constant
(compared to the kW variation) over the range of motor no-load to motor full-load. For a range of medium
voltage machines sampled, the ratio between full-load reactive current and no-load reactive current varied
from 140-260% (depending on machine design, speed, and voltage). For perspective, the ratio between
full-load kW and no-load kW is of the order of 4000%!

Notice that the “relatively constant” is compared to the kw variation (otherwise there is no reason to include the parenthetical “compared to the kW variation). This is further confirmed by the fact that the remaining two sentences of the same paragraph go on to discuss both reactive current variation and kW variation. So yes, I think you’ll agree the purpose of the statement is to contrast the vars variation to the watts variation (although they certainly could have said it better) and yes I think you’lll also agree that the watts variation is much more than the vars variation.

But you don’t have to take Siemens word for it...
Let’s look at some cold hard data.
Look at the same 20hp 575vac motor data sheet I linked above, repeated here for convenience:

http://www.reliance.com/pdf/pdf/aced/E09317ACDJ023.pdf

I have attached an excel spreadsheet analysing the data provided in the datasheet. You are welcome to double check my calculations, but I included cross check columns that prove to me that my numbers are correct.

From my spreadsheet, here is the punchline for this 20hp 575 volt motor:
HP VARS
0 5963
5.01 6321
10 6988
15 8316
20 10256
25 12644

Notice, the vars are not constant as you said, but increasing with load as Siemens said.
And the ratio full-load to no-load vars is 10256/5863 = 172%, well within the 140-260% range we were told by Siemens to expect.

I really hope you are convinced by now, the numbers don’t lie.

You are not the first one to assume reactive current component is approximately constant with load. It is much more the case for transformers than for motors, so sometimes the transformer thinking creeps into the motor world where it doesn’t belong. Several members made the same incorrect assumption of constant vars with load, including myself to a certain extent at beginning of the following linked thread. By the end, I had shown it was not the case.
thread237-262325
The thread is longwinded, but I would direct your attention to my post dated 11 Jan 10 10:21 where the vars expression Q = I^2*X is used to demonstrate that the vars consumed in the leakage reactances at full load are approximately equal to the vars consumed in the magnetizing branch (which as you know is ~ load independent). That is the same thing as saying that the total vars doubles from no-load to full load. i.e:
No-load vars ~ magnetizing vars (no significant leakage vars present at no-load)
Fulll load vars = magnetizing vars + leakage vars
Full load vars = 2*magnetizing vars (since we showed magnetizing=leakage).
Full load vars = twice no-load vars
The circuit parameters in that post are certainly typical among what I have seen and if you study them it is not in the least surprising that vars can double from no-load to full load. If you have another set of equivalent circuit parameters in mind, it can easily be analysed.


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[mikekilroy]

Electricpete, guess it is never too late to teach an old dog a new trick after all. more research and I see how this leakage current varies with load and thus increases total VARS. where I have been coming from is years of working with magnetizing current and torque producing current in sophisticated vector vfds. years of watching magnetizing current remain constant at differing loads such as attached 200hp example showing 0 and nearly 100% load on/off in cycles, while magnetizing current is constant. Since these drives controlled mag current Isq and torq current Isd in independent separate PID loops, and since final vector sum motor current is (Isq^2+Isd^2)^.5, it left no room for mystery additional VARS. So guess they are compensated for elsewhere and I never bothered to learn where or how; I do not see how to attach 2 links so will follow up this post with another with link showing the basic vfd control algorithms I have been used to for both terms. I will say I always wondered why as load went up so did motor voltage, higher than it should have, on these current controlled systems; it obviously was pumping in the additional mystery volts to provide the required additional reactive leakage current... Anyway, thanks for the wake up.

 

[mikekilroy]

follow up to link to base equations for Isd & Isq

 

[electricpete]

Hi Mike – I can see how you came to the conclusions you did from this data (vector-controlled vfd motor). It looks like constant speed, varying load, varying stator Iq, constant stator Id. Ordinarily I would associate Iq with real current component and Id with reactive current component, but that is based on an ASSUMPTION that Vd is 0 (and Vq is of course non-zero). I’m not particularly familiar with drives, especially vector controlled drives, but I have a hunch that it controls both Vd and Vq in a manner that may results in Vd being non-zero, which would invalidate the assumption and therefore invalidate the association of Isq and Isd with real and reactive components respectively.

You might try posting in the motors forum if you’re still interested. I would be interested to hear explanation for these patterns as well.


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