From: Bobby Joe on
On Jul 24, 12:29 pm, Tim Wescott <t...(a)seemywebsite.com> wrote:
> On 07/24/2010 10:03 AM, Jamie wrote:
>
>
>
> > Bobby Joe wrote:
>
> >> 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-4....
>
> >> Possible replacements,
> >>http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-4M1....
>
> >>http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-3LU....
>
> >> BTW, what is the difference between a Y and YZ frame?
>
> >> Thanks for the help,
> >> Bob
>
> > Did you replace or test the Capacitor ? Those motors like cooking the
> > 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..
>
> 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?

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.

> I don't think this is a problem with the motor, assuming that it is the
> correct motor for the blower.

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.

> --
>
> Tim Wescott
> Wescott Design Serviceshttp://www.wescottdesign.com
>
> Do you need to implement control loops in software?
> "Applied Control Theory for Embedded Systems" was written for you.
> See details athttp://www.wescottdesign.com/actfes/actfes.html

From: Bobby Joe on
On Jul 24, 12:28 pm, Tim Wescott <t...(a)seemywebsite.com> wrote:
> On 07/24/2010 09:18 AM, Bobby Joe wrote:
>
> > If I have a motor that is being overloaded ~2x it's rated will
> > doubling the HP improve this?
>
> 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.
>
> > 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]
>
> (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?
>
> > 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,
>
> 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.
>
> > 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]
>
> But it isn't all that simple...
>
nothing is every "that simple" but generally most things follow simple
principles. See Pauls response.

> _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.

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.

>
> 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.
>

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

> > 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-4...
>
> > Possible replacements,
> >http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-4M1...
> >http://www.grainger.com/Grainger/DAYTON-Direct-Drive-Blower-Motor-3LU...
>
> > BTW, what is the difference between a Y and YZ frame?
>
> Fix your blower and ductwork.  You'll be happier.
>

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.
From: Paul Hovnanian P.E. on
Bobby Joe wrote:

> 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]

If I were you, I'd start by diagnosing why the motor is running slow and
drawing too much current. Its highly unlikely that a motor as a component
of some appliance would have been designed to run in such a manner. The
only other possibility is that something has gone wrong.

A run cap has failed, centrifugal starting switch is stuck, excess voltage
drop in the contactor or supply ckt., frozen bearing, etc.

Find the problem and fix it. The motor may still be OK. If you just drive a
stuck fan with a bigger motor, eventually that bearing will really freeze
and burn the new motor up as well.


--
Paul Hovnanian paul(a)hovnanian.com
----------------------------------------------------------------------
Have gnu, will travel.
From: Jamie on
Tim Wescott wrote:

> On 07/24/2010 10:03 AM, Jamie wrote:
>
>> Bobby Joe wrote:
>>
>>> 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
>>>
>> Did you replace or test the Capacitor ? Those motors like cooking the
>> 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..
>
>
> 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.
>
Yes, I know of those motors, mostly series motors/universal
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.



From: David on


"Bobby Joe" <bobbyjoe23928(a)gmail.com> wrote in message
news:4f644a69-7562-4334-9c5b-ffc4075fe5a7(a)w30g2000yqw.googlegroups.com...
<snip>
> 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.
>
Do you know the power factor on the motor? Some of that
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


First  |  Prev  |  Next  |  Last
Pages: 1 2 3 4 5 6 7
Prev: bluffing fools
Next: stampede