Motor Load vs HP vs Current
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From: Tony Hwang on 25 Jul 2010 16:14 Bobby Joe wrote: > 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. Hmmm, Not just HP vs. RPM. Think winding. Doesn' tha motor has any speed taps? If it is running at hi speed tap and bogging down, The motor is too small. Byt other factors are duct size, blower blade pitch/size, mothr efficiency. Not all is created equal. Also why is the voltage sagging? If you turn off the motor, it does not happen? Thin simple.
From: Sparks Fergusson on 26 Jul 2010 00:03 Tim Wescott <tim(a)seemywebsite.com> 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? 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!
From: bud-- on 26 Jul 2010 13:45 Bobby Joe wrote: > On Jul 24, 12:03 pm, Jamie > <jamie_ka1lpa_not_valid_after_ka1l...(a)charter.net> 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.. > > Of course. The capacitor was good. Everything is fine except the > motor. How do you know the cap is good? How do you know everything is fine? > 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. Almost 2x rated current for 10 years. I don't really think so. > > 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. 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. -- bud--
From: Grumpy on 26 Jul 2010 16:48 Any chance that wire has loop where you are taking current reading? "Bobby Joe" <bobbyjoe23928(a)gmail.com> wrote in message news:f4495dba-62d3-4a83-93d7-6b91f0b58834(a)f6g2000yqa.googlegroups.com... > 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 > --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: JosephKK on 30 Jul 2010 22:59
On Sat, 24 Jul 2010 13:03:41 -0400, Jamie <jamie_ka1lpa_not_valid_after_ka1lpa_(a)charter.net> 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.. > Definitely. Way more is wrong than the fan motor (and never buy a cheap motor). Start with checking all filters, dampers, heat exchangers, evaporators, condensing unit, also all of the rest of the electrical work. |