Brook Motor power

[garethp]

Hi,

I've got a spare 3phase brook motor which came off a little used boxford TUD - I've decided to think about using it on my super7 through an inverter, the thing is I dont know what power it produces - I'm guessing from the plate its 1/2 or 3/4hp, does anyone have any ideas? The (i think!) important stuff on the plate is;

Brook Crompton Parkinson
VMA544CM1B

380/440 1.5A
220/250 2.6A

I've noticed the base plate is welded to the case, I assume this is due to 3ph vibrating less so not needing resilient mounts?

Gareth.

[the_viffer]

3/4 hp

[houstonceng]

Gareth wrote "I've noticed the base plate is welded to the case, I assume this is due to 3ph vibrating less so not needing resilient mounts?"

Yes. 3-phase motors are - usually - smoother than 1-phase so they don't - usually - have resilient mountings.

If you are thinking of an inverter drive make sure it's rated at 0.55kW (0.75hp) or more.

[steam4ian]

G'day Gareth.

Your other correspondents are very conservative. By my calculations the motor is 1.0 horse power. I base this at 380 volts(three phase) with an efficiency of 90% and a power factor of 0.85, these figures are typical.

For me I would not use a VFD of less than 1kw capacity.

And yes, three phase motors inherently have much, much less torsional vibration than single phase motors.

If you go VFD remember the motor cooling falls of with speed. Theoretically you can get full torque down to 10% full speed, practically you will melt copper if you do this for a long time.

regards,
Ian

[Tony K]

G'day Ian and Gareth.
To me it looks like this (IMHO).
Taking your figures, the power required by the motor is Volts x Amps x power factor.
Therefore power supplied to the motor = 415 x 1.5 x 0.85 = 529Watts
The useful power produced by the motor = 529 x 0.9 = 476Watts (it is 90% efficient)
When I went to school there were 746 Watts in one horsepower, Therefore the useful power produced by the motor will be 476/746 = 0.64 h.p.
The 3 phase inverters sold in UK are rated in horsepower (for some reason).
The inverter will have to supply 529/746 = 0.71 h.p.
Therefore, buy a ¾ h.p. inverter.

As Ian says, beware running the motor at slow speeds and high load since the cooling will be lost (unless you push in some air from somewhere else which is a bit of a fad!)

Now, although in theory the full torque could be available down to slow speeds, with many inverters this is not the case.
The way I see the UK inverter marketplace, if you pay just over £100 for a ¾ h.p. inverter, the torque will have dropped substantially at 50% of normal speed. At 10% of normal speed the torque will be very low.
If you pay £150 - £175 the torque at low speeds will be much better - will have torque vector control.
If you pay £250 the torque at low speeds will be very good - better torque vector control.

So folks, if you are fitting one of these to your lathe and want to join the utopian society of no belt or gear changing, you could be disappointed. With the cheaper ones maybe you should look at it as giving greater flexibility of your system to plus and minus 50% of normal speed (25 - 75 Hz.). With the more expensive ones - considerably more.

The speed is proportional to the frequency, normal mains being 50Hz. As one of the suppliers says - it is better to achieve a 50 rpm shaft speed from a 100 rpm geared speed at 25Hz than 250 rpm geared speed at 10Hz.

There you are Gareth, I think you need a 0.75 h.p. inverter with torque vector control and I would pay £150 - £175 for it.

There you are guys, IMHO - head above the parapet - await it being shot off!!

[steam4ian]

G'day Gareth

Unfortunately tonytrans is WRONG. He has forgotten the 1.732 factor which multiplies the volts (415) times amps (1.5) to give the volt-amps (VA) input to the motor. The 1.732 factor is the adjustment for three phase connection.
My figures were based on 380volts (the lowest on the name plate) multiplied by 1.5amps multiplied by 1.732 (for 3 phase) for a VA = 987. Adjusting for power factor =0.85 and efficiency = 90% gives a shaft output of 755 watts which is equivalent to 1 horsepower.

I stick by my earlier recommendation regarding the size of VFD. As Tony says you can run the motor above its nominal speed. DO NOT through out you belt changing facility.

Regards,
Ian

[Tony K]

steam4ian said:

G'day Gareth

Unfortunately tonytrans is WRONG. He has forgotten the 1.732 factor which multiplies the volts (415) times amps (1.5) to give the volt-amps (VA) input to the motor.
Regards,
Ian

I have not forgotten the 1.732 (square route of 3) factor - I expected it to have been included in the 1.5A quoted rating. I would expect the 1.5A to be the total current taken from the supply, not the current per phase.

I am still prepared to be wrong if an arbiter knows better!

[steam4ian]

G'day Tony.

I could say "trust me"! I am a professional electrical engineer and for over 35 years have made my living getting this equation right.

Three phase VA is line volts times line current times 1.732 (Root 3); power is VA time power factor = watts.

Motor nameplates show line volts and line amps. BTW line volts are the phase to phase volts.

Regards,
Ian

[Tony K]

G'Day again

OK - final shot from me, then finished arguing, the other guys must be getting bored with it.

The Brooke Crompton Parkinson 3 phase motor on my Super 7 is marked as follows....

370W
380/440V 1.2A, 220/250 2.1A

I reckon 370W is pretty close to 0.5 h.p. (1 h.p. = 746W)
Taking the median supply volts of 410V and a current of 1.2A and power factor of 0.9....
Power = 410 x 1.2 x 0.9 = 442.8W
Power out with efficiency of 85% = 376W (pretty close to 370W)

No factor of 1.732 involved here. Gareth is OK with 0.75h.p.

Maybe rating plates are different in Oz?

[steam4ian]

G'day Tony

Tedious aint it?
Tony's motor definitely seems to be a 1/2 HP. If the motors are nameplated in the same manner the Gareth's motor is 3/4HP. BUT there is no mistaking the VA input to Gareth's motor which is close to 1000VA, the VFD must be able to handle this. I don't think you will find much price difference up to about 3HP.

"Maybe rating plates are different in Oz?" This could be so, but why? also small motors are relatively inefficient compared with say 10kW and over, there is a lot of slip and power factors are lower.

As I said, I stick with 35 years of experience in motors and power distribution, admittedly many times larger.

Over and out!

Regards,
Ian

[garethp]

Hi all!

Thanks for the replies, I'm toying with getting a .75 or a 1hp inverter - the 1hp would nicely fit my milling machine if I ever change the motor on that. The main reasons for getting the inverter are;

1) I already have a decent motor
2) I like the idea of reduced vibration
3) Soft start / stop sounds good
4) I've got a dewhurst reverser which I dont like - the lever looks too easy to catch (not done it, yet....)

I'm quite happy to keep changing the belt positions and use back gear for wide speed adjustment, one idea which I did think about was looking at combining the inverter with a dro to give constant surface speed - dont know if anyone's ever tried it?


Regards,
Gareth.

[houstonceng]

Gareth et al.

Being as I'm a Chatered Engineer, I took the very efficient route of looking in a current Brook-Crompton catalogue.

Although the actual model is so old as not the feature anymore, the basic characteristics of newer moters were used to assess the rating. (Why try to re-invent the wheel or make rash assumptions about the "root three factor" ! - included or not ?)

For 1450 RPM (4-pole) three-phase fractional hp motors, the following data aws listed :-

A 0.75kW (1hp) Premium Efficiency 3-Phase motor is shown as taking a full-load current of 1.97A @ 380V and 1.88A @ 415V.

A 0.55kW (3/4hp) T range motor is listed as taking 1.6A @ 380V and 1.45A at 415V.

Therefore, I concluded that a data-plate showing 1.5A @ 380/440V was indicative of a motor rated at 3/4hp max.

Rather than being "conservative" in my assumption that it's a 3/4hp motor (as Ian has accused me of being) I think that I've actually been slightly the opposite.

I'm sure, from 40+ years in the profession and my qualifications as a CEng, FIET (pka IEE), that the manufacturer is (nearly) always correct and that, as Tonytrans said, the "root-three" factor is already included.

Andy

[steam4ian]

G'day Andy.

Thanks for looking up the BC catalogue which has to be the AUTHORITY. I had an old handbook which suggested a 1HP motor drew 2 amps @ 400V. It appears the root 3 factor is lost in low power factor and lesser efficiency.

I did a quick check on a current job; a 1250kW (1700HP) motor at 3300 volts draws 260 amps, power factor 0.92, efficiency 95%; the 1.732 factor definitely has to be included in the equation. The voltage is phase to phase or line volts, the current is the nameplate current, running current is about 95% of this.

Cheers
Ian

[circlip]

Garethp, as a PRACTICAL engineer of more than 40 years perhaps you could consider the following. Should you really wish to reach a practical solution to your drive requirements conversion kits to change the drive system to treadle power via a Singer sewing machine leather belt system SHOULD be available although it will now be classed as rather retro. The alternative would be to buy a matched drive (VFD/motor) from a recognised supplier. Where are we, oh yes it is getting round to the rutting season drjohn.

[houstonceng]

Nah circlip. Why treadle when he could couple it to a horse jin or water wheel. Better still, to a steam engine.

Seriously. Buying a matched VFD/motor is, obviously, the ideal solution for anyone with nil electrical knowledge - - - but it ain't cheap. I was quoted £600.00 for a 1.5kW (2hp) system for my 1327 lathe and told it couldn't be controlled (on/off & fwd/rev) from the existing saddle switch. I managed to do it - and for circa £150.00 - by buying the necessary bits (motor, VFD & contactor) as individual item from e-bay.

We were trying to help Gareth to save money since he's already got the motor - and a good make at that.