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





RE: How to calculate how fast i can speed up a fan with remaining full load amperage
P1 / P2 = (n1 / n2)3
P is power, n is RPM.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.)
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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%
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.)
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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/626...
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.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
So why don't you stop here.
According to this ,the speed can increase 33.9%
Why are you now getting into power?
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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%
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
1.157^1.5=1.244
or an increase of 24%
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
Power ratio would be
1.157^3=1.54
or 54% increase in power
Could an induction motor handle that increase?
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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?
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
"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
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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:
should be
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...
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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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
should've been
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/pool/us/services/...
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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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.)
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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.
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).
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...
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: No load current on a motor
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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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
http://literature.rockwellautomation.com/idc/group...
You probably already knew that piece, but I didn't. So now it makes more sense to me. There are two different rectangular coordinate systems by which we can express the stator current vector:
(Ireal, Ireactive) decomposes the total current vector into orthagonal components using the applied voltage as a reference. Ireal is in phase with applied voltage, Ireactive is the vector remainder resolved in the perpendicular direction. This coordinate system is useful for determining power flow (real and reactive).
(Itorque, Imagnetizing) decomposes the total current vector into orthagonal components using the magnetizing current as a reference. (The magnetizing current is Im = Is - Ir' where Ir' comes from an model within the drive). Imag is in the direction of Imag (same thing). I torque is the vector remainder resolved in the perpendicular direction. This coordinate system is useful for examining/controlling motor behavior.
It may be tempting to equate (Itorque, Imagnetizing) to (Ireal, Ireactive) since they are both rectangular expressions of the stator current vector, and they're very close at low load. But they are different as discussed above. While Imagnetizing is itself reactive, it is not the only contributor to Ireactive (there is also a component due to current flow in leakage reactances). The net result is that the two sets of coordinate axes (Ireal, Ireactive) and (Itorque, Imagnetizing) are rotated by some angle with respect to one another (and that rotation angle is not a constant but would vary with motor conditions, including load).
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
That sentence is wrong, all else looks good to me.
I'm done now.
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
Im(V x Is*) = Im^2xXm + I1^2xX1 + I2^2xX2
The left side is reactive power that flows into the motor, the right side is reactive power dissipated in each reactance.
Notation above: x represents vector multiplication (not cross product) and * represents conjugate
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
Let Zpara be parallel impedance of magnetizing branch and rotor branch. Compute it as a product/sum of impedances:
Zpara= (R2/s+j*w*X2) j*w*Xm/ (R2/s+j*w*X2+j*w*Xm)
By voltage divider, we can find voltage accross magnetizing branch in terms of supply voltage and circuit parameters:
Vm = Vs * [Zpara / (Zpara + j*w*X1+R1)]
The phase angle of the quantity in the square brackets is the angle shift between the two coordinate systems.
I'm really done now.
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
A real vector drive controls current and current only; voltage is basically ignored: ie., we don't control voltage at all - it does whatever the heck it needs to do to make the correct two currents - which are always and forever 90 degrees apart.
So if you leave voltage totally out of your descriptions then they are correct. We almost totally ignore voltage - it simply gets jerked where it needs to in order to make the two PID controlled currents. With that blinder on, what I ignored all these years was the voltage changing with load. I know can look back upon many many systems and see WHY the voltage mysteriously increased 10-20% WITH load at a given speed. Never paid attention to that leakage current before; to the point of ignoring its term in the motor model that IS there but I never had a use for it. Guess I still don't have any use for it since I control my vector motor with Isq & Isd and to hell with the leakage current. But it is there. If my motor model is not close then it does screw things up and I have to go back and tweak my model.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
Merry Xmas.
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RE: How to calculate how fast i can speed up a fan with remaining full load amperage
since real torque producing current is 100% in phase with motor voltage, I have always just ignored the voltage as it is not required for any calcs like PF since PF is given with Isq & Isd's vector sum.
I'll tell you next systems we do I will be sneaking scope pixs as before both unloaded and in heavy metal cut loads, but also include vectored output current too, since it will show total VARS!
Agreed, tho this thread moved from how much faster can I go to this discussion, I think the bottom line for the OP question is still valid to separate both currents and calculate based on them since the motor will be almost fully loaded to fully loaded and in any case the PF given was for full load motor so this separation of currents should still be valid since it would include the full reactive currents anyway & much more accurate than just messing with composite vector current and its inherent error. My confusion remains whether torque producing current in fan app goes up by square as one post and some google searching shows or by cube like power. both seem to make sense and I have not been able to determine yet which is right. obviously only one. since speed also goes up, seems like it needs to be sq function so that the increasing speed can add the cube to it....
Merry Christmas to all.
RE: How to calculate how fast i can speed up a fan with remaining full load amperage
Either way, motor output power goes up as cube of fan speed (under assumptions mentioned: density constant, system does not change resistance)
If motor speed varies together with fan speed N, then
MotorTorque ~ MotorPower/MotorSpeed ~ N^3/N ~ N^2
If motor speed is constant as fan speed N varies, then
MotorTorque ~ MotorPower/MotorSpeed ~ N^3/Constant ~ N^3
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