Motor Load vs HP vs Current

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If I have a motor that is being overloaded ~2x it's rated will doubling the HP improve this?
I have a 1/2HP Direct drive 1P, 120V, 1075RPM blower motor that is drawing 12A while it is rated at 6.5A Full load, 9A max and only spinning at about 700RPM. [No load it draws 3A]
The problem I was having is any time the AC kicked in I would lose connection to the net and the power would sag. I went to the attic and measured the AC and found it was drawing 2x the rated current at full load. I thought the motor was bad so I got an identical replacement but almost exactly the same measured specs and same problem.
I figured I could replace it with a 3/4 or 1HP motor and get better results but I'm not sure how much. I want to save power and possibly increase the rpm's,
The main thing I would like to know is how HP, current, and load are related. If I double the HP I should effectively be doubling the max load and probably the current at max load? Basically, if I have an x HP motor using a certain load and I move to y HP motor then what can I expect the RPM's and current to be? [simple estimates are ok. I understand that it depends on a lot of factors but there should be general principles involved]
As I said, I would like to be able to determine if a 1HP is effectively going to allow me to increase the RPM's and reduce the current[since the motor shouldn't be overloaded].
Motor used,
http://www.grainger.com/Grainger/AO-SMITH-Direct-Drive-Blower-Motor-4KA36?Pid=search
Possible replacements, http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-4M183?Pid=search http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-3LU91?Pid=search
BTW, what is the difference between a Y and YZ frame?
Thanks for the help, Bob
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Bobby Joe wrote:

http://www.grainger.com/Grainger/AO-SMITH-Direct-Drive-Blower-Motor-4KA36?Pid=search
http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-4M183?Pid=search
http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-3LU91?Pid=search
caps that come with them.. Made in Mexico! also, the bearings are crappy! Make sure you are not voltage sagging at the leads.
Also, have you pull the motor out in the open with blades on and tested the load? You could have a restricted vent some where. That does not sound normal..
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On 07/24/2010 10:03 AM, Jamie wrote:

Some blowers load the motor more heavily when the duct work isn't restricted -- have you ever noticed how your vacuum cleaner speeds up when the intake is clogged?
I don't think this is a problem with the motor, assuming that it is the correct motor for the blower.
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The AC unit is not a vacuum cleaner. It is much much bigger. The ducts are 2 feet wide and node 2 inches. I don't know what you think could be clogging up the ducts but it seems you've never done much air duct worked for central heat and air as then it should be pretty obvious.

That's a big assumption. The motor is drawing 12A simple as that. The only way it can draw this much current is if it is suppose to or if it is overloaded.

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On Sat, 24 Jul 2010 11:20:38 -0700 (PDT), Bobby Joe

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Tim Wescott wrote:

types(brushes), this isn't what he's using. He's using a standard run of the mill motor found on many $200+ floor fans and AC units, used in mill's etc... They slow down when restricted and use more energy.. The bearings are cheap, the motor runs hot and heats the cap because many of them have the cap mounted to the side of the motor. Over time cap then bearings goes. When the cap goes, it slows or exerts more current to get up to speed or just sits there and hums when it gets bad enough. Normally the motor will burn it self out after that happens when lefton.
He may even have a unit made in China which are worse... we get an average of 1 year's use from the chinese made units.
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Now, let's think about that. If a motor speeds up, is it more loaded or less loaded?
That's right. For an induction motor, the heavier the load, the more it slips, and the slower it turns.
When you put your hand over the vacuum hose, you *unload* the motor (no airflow = no work) and, thus, it speeds up. Still don't believe me? Put your ammeter on a vacuum and try it!
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On Jul 24, 12:03pm, Jamie

Of course. The capacitor was good. Everything is fine except the motor. I took the blower out of the duct so none of that is an issue. It is simply an overloaded motor. This was probably done intentionally to cut costs. From what I have read most motors are underrated anyways. This motor has been drawing 12A for about 10 years and hasn't burned up yet so it's max current is a bit conservative.
The real question I had was not about what I have done but about the relationships between HP, rpm, and current. It seem Paul was the only one that got that.
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<snip>

current is reactive current so measured amperage times voltage is not the way to find the real watts consumed. I would suspect that the PSC capacitor value may not be correct.
David
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Bobby Joe wrote:

most likely wasn't made in China or Mexico..
Its funny because the one's we have at work have their Caps made in Mexico and the motors in China..
And personally, motors running over loaded like that would make me look at the engineers that designed the system. Makes for a nice camp fire when it decides to sit there and smoke for a while.
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Bobby Joe wrote:

Brand name of the motor? Sounds like El Cheapo. Decent blower motors are actually under rated.
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Bobby Joe wrote:

How do you know the cap is good? How do you know everything is fine?

Almost 2x rated current for 10 years. I don't really think so.

And you could find that the HP is proportional to the cube of the RPM. Not really likely that the electrical side is OK. Exact replacement motor? Is there a run cap? Did you replace it? That is my bet.
If the duct was totally blocked the fan would spin a hole in the air. As the duct gets blocked, the HP does not go up drastically - this is not a piston pump.
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On Sat, 24 Jul 2010 13:03:41 -0400, Jamie

cheap motor). Start with checking all filters, dampers, heat exchangers, evaporators, condensing unit, also all of the rest of the electrical work.
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On Sat, 24 Jul 2010 09:18:14 -0700 (PDT), Bobby Joe

    look up 'fan laws'.
http://www.engineeringtoolbox.com/fan-affinity-laws-d_196.html
http://www.americraftmfg.com/PDFs/Engineering.pdf

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On Jul 24, 12:15pm, .p.jm.@see_my_sig_for_address.com wrote:

Thanks, at least someone here seems to be able to provide some information.
If that pdf is right then
rpm2 = (hp2/hp1)^(1/3)*rpm1
which will increase my case about 200 extra rpm to about 950. This is closer to no load. If I new what the no load current was[or current to rpm graph] then I could determine if this would be a significant savings. It would definitely push out more air which would be nice but I just have to make sure it will take less current.
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On Sat, 24 Jul 2010 11:16:21 -0700 (PDT), Bobby Joe

    You expect to move MORE air with LESS current by changing motors ? Assuming the motors are nominally similar in efficiency ( and no other changes to the system ), that ain't gonna happen.
    1st Law of Physics applies : 'There ain't no free lunch".
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On Sat, 24 Jul 2010 11:16:21 -0700 (PDT), Bobby Joe

So apparently this is 370 W 6 pole asynchronous motor (synchronous speed 1200 RPM at 60 Hz). Assuming 120 V at motor and 12 A consumption, the apparent power is 1440 VA. The slip is also huge, since the motor is running at 60 % of synchronous speed. The losses will most likely be large, since it is quite far from nominal specifications.

Replacing the motor with a 8 pole design (no load speed about 800 RPM) would at least allow the motor to run closer to the nominal power and hence reducing the losses.

In wind turbines, the power of the wind is proportional to the third power of wind speed, so assuming the air speed in the duct is proportional to the RPM (no blade stall etc.), that formula makes sense.

If the fan rotates at 700 RPM, it will require the same mechanical axial power regardless what the motor nominal power is.
What the motor real power depends on the efficiency vs. load curve and in this case, it is not even possible to calculate the real power taken by the motor, since there was a voltage sag and no voltage measurement was taken _at_ the motor, neither was the angle between voltage and current measured.

Anyway, if you want to increase the blower speed from 700 RPM to say 1075 RPM, the axial power requirement would be (1075/700)^3 or 3.6 times the current motor output mechanical power. Be prepared that the increase in input power requirement would be similar.
If the current motor is driving a simple fan, changing the fan with smaller blades or using blades with lower angle of attack to match the current motor better.
Alternatively, use a slower 8 pole motor with the current blower, so that it will reach nearly the nominal speed, thus operating with a reasonable slip and not heating up as badly.
If you want to increase the RPM with the current blower, be prepared to use a significantly larger motor and be prepared to deliver the current it wants to consume.
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On 07/24/2010 09:18 AM, Bobby Joe wrote:

Not necessarily, particularly since you don't know what the torque is at the shaft of the motor, just that the motor is running slowly.

(a), there's no guarantee that doubling the horsepower of the motor will spin the blower at the right speed.
(b), if the blower and motor are in good shape, is properly installed, and all the duct work is designed correctly, then you shouldn't have a problem. Fix the blower and motor, correct the installation, and fix the duct work.
(b1), if your car had frozen rear brakes and always went slower than before, would you want to just put a bigger motor in it?

See my "car with frozen brakes" analogy. More power out means more power in -- if you actually manage to stick a motor on there that can turn the blower without making the motor unhappy, it'll still need enough current to do the job.

But it isn't all that simple...
_If_ the motor isn't being bogged down, and if turning the shaft at rated speed took 1/2 HP, then your 1/2 HP motor could turn the shaft at it's rated speed. Going to a 2 HP motor with the same rated speed would only consume slightly more current, because motors are pretty efficient, and the underlying physics makes sure that they don't pull more current than they need to generate the torque they're delivering.
If it takes 2HP to turn that shaft at rated speed, then the 1/2HP motor simply won't cut it. The 2HP motor will, and should take roughly four times as much current (at the same voltage) to do the job as the 1/2HP motor would to deliver it's 1/2HP.
Within limits, a motor's power in will be equal to the motor's power out, plus enough power to make up the motor's losses. If the motor shaft is loaded enough that it starts turning at much less than rated speed then the power requirements go up, both for the shaft (obviously) and the motor losses. Even if efficiency doesn't go down, the motor gets hot -- and at some point, the efficiency starts to go down.

http://www.grainger.com/Grainger/AO-SMITH-Direct-Drive-Blower-Motor-4KA36?Pid=search
http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-4M183?Pid=search
http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-3LU91?Pid=search
Fix your blower and ductwork. You'll be happier.
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Wescott Design Services
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principles. See Pauls response.

Duh. HP = 746W(IIRC) which means if you increase the HP at the same voltage you must increase the current. That is not rocket science.
But I'm interested in RPM's verses HP. Paul was intelligent enough to at least show me where to get the formulas(assuming they are correct).
It's not a simple relationship as it depends on the cube root.
The way I see it is by using a larger motor but the same load I am making the larger motor less loaded so it get's closer to no load conditions. Therefor by using a larger motor with the same blower fan I should approach the larger motor's no load current. How much was the question. I think I have my answer thanks to Paul.

Yes, I think this is based on the conservation of energy law?

Jesus. You could have just said you had no clue what the relationships are between HP, current, and rpm's. Simple as that. Instead you subvert attention away from your ignorance on the subject with demands, claims, and useless statements.
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On Sat, 24 Jul 2010 11:29:14 -0700 (PDT), Bobby Joe

    Bzzzt. Fail. Go to jail, do not collect $ 200.
    You will come closer to approaching the nominal speed, but that has nothing to do with a 'no load' condition, or the current of 'no load'.
    By your statement above, you could take the wheel off and expect the same current as if there were a wheel on it turning at the same speed. Obviously, not gonna happen.
    And I'm not even gonna mention 'fan curves', since that's over yuor head. Let's just say 'they ain't straight lines, that's why they're called 'curves'.

    You might try learning something about the man before insulting him. He knows a damn sight more than you do.
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