Horse power and such…….

[Rik ShawParticipant @rikshaw]

[Rik ShawParticipant @rikshaw]

Is it possible to calculate the horse power of a single phase electric motor given the following spec?

450w

930 rpm

240v

2.8A

50Hz

The motor will need fitting with a capacitor/s. The plate shows 12uF. I assume that is the value of the capacitor/s required – right?

Is there much involved in wiring in a suitable capacitor/s or should I be letting a "competent" do it for me?

Rik – shockingly ignorant in the "sparks" department.

[Anonymous]

For electric motors the following is defined:

1hp = 746W

So you only need the first number and a simple division.

Andrew

[Ian ParkinParticipant @ianparkin39383]

Generally work on 750w to 1 hp so your motor is just over 1/2hp

electrical input does not equate to mechanical output

Cap should be wired in as diagram which sometimes is in terminal cover

[Jeff DaymanParticipant @jeffdayman43397]

746 W = 1 HP

if a motor uses 450W of electric power at 50 % efficiency it will produce 225 W of mechanical power, or 225/746=.302 HP, nominally a quarter HP motor.

Your mileage may vary. Efficiency can be difficult to predict and is rarely stated. Wattage on data plates of motors is usually the electrical power consumption rating. JD

[Swarf, Mostly!Participant @swarfmostly]

Hi there, Rik,

If I remember rightly, 746 watts equal 1 horsepower so your motor is apparently 0.603 HP.

BUT, that's input power and the motor is not 100% efficient. I'd say it's a nominal ½ HP mechanical output power. It might just be ⅓ HP but that would require the efficiency to be so low it'd cook!!!!

The speed is interesting, 930 RPM indicates it's a six-pole machine, fairly rare in low power single phase motors in my (limited) experience.

I shall duck the question of the capacitor and see what others suggest.

Best regards,

Swarf, Mostly!

PS: Wow, Ian & Jeff beat me to it, not to mention Andrew!!!!

Edited By Swarf, Mostly! on 29/04/2015 17:02:13

Edited By Swarf, Mostly! on 29/04/2015 17:03:12

Edited By Swarf, Mostly! on 29/04/2015 17:03:43

[Les Jones 1Participant @lesjones1]

Hi Rik,
The wattage rating of some motors refers to the output power on others it refers to the input power. Normally the start capacitor (And run capacitor if used.) is fixed to the outside of the motor. 12 uF sounds small for a start capacitor for that motor but it would be a good starting point in the absence of more information. 12 uf would be what I would expect for a run capacitor. Pictures of the motor, its label, the connection box and the cover if it contains a diagram or any other information would be helpful.

Les.

[Rik ShawParticipant @rikshaw]

Thanks all for your replies but it seems as though this motor will be a little under powered at 1/2 hp or thereabouts.

Someone on here told me I should be looking for a 3/4 horse. It needs to be running at approx. 900 / 950 rpm but it seems as though I might be hard pressed to find one at a reasonable price. I have seen one for sale described as "very rare" at £135 but that seems a bit steep to me. Maybe though, I'll have to bite the bullet and fork out.

BTW it's destined to replace the 1444 rpm (approx.) motor on my Alba 1A shaper.

Rik

PS I must say I am impressed by all the replies to my OP – once again, thank you all.

[RobboParticipant @robbo]

Rik,

Just a "beware".

I thought I'd found a single-phase motor for my Elliott shaper, ran at the right speed, right horsepower, and I was told it had been bought at an ME show and similar had been used for shapers.

However, I discovered when run that it had no integral cooling fan, and investigation revealed that it was an American design used to power large fans, and it lived in the draught from the fan so didn't need its own. Consequently it got too hot to touch in minutes when run, and was quickly switched off before smoke and flames emerged.

Consequently I left the original 3-phase motor on and bought an inverter.

Phil

[Rik ShawParticipant @rikshaw]

Robbo – excellent input – I will definitely bear in mind your comments. Thank you.

Rik

[Clive FosterParticipant @clivefoster55965]

Yup, 900 /950 rpm single phase motors are rare and, if you do find one, frequently well used. Only saving grace is that they are invariably a decent brand and well made. Some of the bargain basement imports have been of terrifyingly bad quality and poor efficiency. Northern Tools once sold me a 3 HP cap start – cap run motor that proved massively underpowered and very hot running on installation. Fortunately the chinglish instruction and specification leaflet was in the box proudly proclaiming 3 HP input power, 1 1/2 HP output power. I suspect the output was ahem, optimistic. Pretty effective shop heater tho'. After a deal of remonstration with Northern Tools I got my money back and was intrigued to discover that the next but one catalogue had the specifications corrected to 3 HP input, 1 1/2 HP output. I don't think may were sold as the item had disappeared from the next issue.

I second Phils suggestion of retaining the standard 3 phase motor and controlling it via an inverter. My Elliot 10M runs very happily off a Mitsubishi box bought a fair few years back from Gavin Oseman, a very nice man, who occasionally advertises on Homeworkshop.org. There are times when its really nice to bring the speed up from a crawl. Especially in the higher speed ranges.

Clive

[bodgeParticipant @bodge]

Hi Rik

have a look at the post shaper speed reduction it does what your looking to do ! IT does work Ive been running 10 inch shaper like this for the last 5-6 years IT dont matter that poly v belt runs on top of the old fan belt all you need to do is pull old fan belt ends together tight as you think the wire / clip can stand and fasten , as long as you bury wire or clip so the poly v dos not pick up on any sharp ends

motor pulleys straight forward enough 1 inch & 11/5 inch grove to take poly v ribs bore to suit motor shaft.

I think i should have posted to here rather than make new thread .I think your six pole motors fine for use on your shaper and will be power enough ! Cant help with wiring though. I have 1hp cap start cap run 1 phase 1425 rpm motor driving the shaper and its way over powered now the stroke is down 26 spm on lowest 52-ish on fastest . Have got a half horse cap start to go on some when! One more thing might help the big pulley wheel on my machine the dia is 8inch-ish at bottom of v pulley 8 1/2 with old fan belt filler in place and the 4 rib poly v runs very sweet on top of the old fan belt , Depends which way you want to go about it really !!

I like cheap & quick , makes me very cheer full !! and i do like a good bodge ..should say this first post & email ? to any one any where and im still trying to figure it out [ am thinking machining EN hard as witches tits or rebar is easier know for sure it is] will try a post or on what did you do today thread ,see how it goes

steve

[David LittlewoodParticipant @davidlittlewood51847]

Going back to Rik's original question, no one has mentioned the fact that motors are a mostly reactive load. The current and voltage are not in phase. For a perfectly reactive load, the two are 180 degrees out of phase and no power is consumed, but the current rating is still important as it determines the size of conductor wire required for its connection to the power source. When the motor is at no load, it is nearly purely reactive, but as it is loaded the current and voltage move closer in phase and real power consumption takes place.

Thus, in Rik's example, the specification is 240 V and 2.8 A, which would (for a resistive load) mean a power consumption of 672 W. At no load, the motor would (theoretically) consume no power, but under load the specification suggests it actually consumes 450 W. The efficiency of synchronous AC motors is reasonably high, but I can't say what the mechanical output power is; I believe the power rating quoted for motors is usually the actual power consumption at full load, but I stand to be corrected if someone has more expertise.

David

[John HaineParticipant @johnhaine32865]

David, er, no. For a reactive load the voltage and current are 90 degrees out of phase. The current will be rather low at low load and nearly in quadrature, and build up to the rated current largely in phase as the motor is loaded. And this almost certainly isn't a synchronous motor but an induction motor.

[Rik ShawParticipant @rikshaw]

………and fairly ancient as well John is my guess with its oil cups at either end of the shaft. Something I've only seen once before on a very large motor driving a line shaft in a factory many years ago.

Rik

[MuzzerParticipant @muzzer]

The AC power drawn (VA) would be composed of an in-phase ("real&quot component (Watts) that is responsible for generating the useful shaft power and a reactive ("imaginary&quot component (VAr) that is due to the inductive nature of the windings. The latter will be largely fixed and 90 degrees out of phase to the voltage – you pay for this but it doesn't generate any useful mechanical work. The former will be largely proportional to the mechanical load and some losses will be incurred along the way.

If you use the numbers at the top of the thread, you would expect to see a load of 240V * 2.8A = 678VA. The power figure is most likely to be the rated shaft power – not a lot of use to you otherwise. If you assume a typical efficiency of 85%, the real component drawn will be about 450W / 0.85 = 530W. That suggests that the inductive (no-load) draw would be around 423VAr, which would be seen as a reactive current of around 1.75A. So the load would vary from around 1.75Arms (no load) to around 2.8Arms at rated load.

Sanity check: 678^2 = 530^2 + 423^2

Murray – bit jet lagged, so apologies if some of the sums are plain wrong.

[Anonymous]

Posted by Muzzer on 01/05/2015 16:01:56:

Murray – bit jet lagged, so apologies if some of the sums are plain wrong.

Murray: The sums look ok to me. I appreciate that commercial outfits pay for poor PF, but I thought that households didn't? I also thought that this was because the older meters, with the disc that rotated, didn't register VArs. However, looking at Wikipedia that doesn't seem to be case, even with the older meters; another illusion shattered.

Still, at least the lady who came to read my meter this week, at the unearthly hour of 9am, getting me out of bed, said that the workshop (aka garage) was the most extreme mancave she'd seen.

The most important question though is; did you sneak any workshop goodies back from the trip to China?

Andrew

[MuzzerParticipant @muzzer]

Disappointingly, no I didn't have time this visit to get anything. I was run ragged on a whistle stop tour of suppliers and customers of my new employer. However, I will be returning several times each year, next in June, so should have time for field prep.

Murray

[David LittlewoodParticipant @davidlittlewood51847]

Posted by John Haine on 01/05/2015 07:52:55:

David, er, no. For a reactive load the voltage and current are 90 degrees out of phase. The current will be rather low at low load and nearly in quadrature, and build up to the rated current largely in phase as the motor is loaded. And this almost certainly isn't a synchronous motor but an induction motor.

John,

Mea culpa, you are quite right on both points; serves me right for posting so late after sharing a bottle of wine! I was thinking of AC induction motors being (almost) synchronous with the AC frequency, but the term synchronous motor is something else. I do think the underlying point is valid though.

David

[Anonymous]

Posted by Muzzer on 02/05/2015 12:53:55:

However, I will be returning several times each year, next in June, so should have time for field prep.

Murray

Better start looking for a triple/quadruple garage…………

Andrew

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