Motor comparisons [CA120 et al.]

[bearing]

In transformers you get the most power if you make iron losses the same as copper losses, and I assumed it would be the case with motors too. But that may be wrong.

In a PMSM motor peak efficiency occurs when copper losses are the same as parasitic losses. Right?

I don't know.

My thinking was very simple. If the 20-pole motor had it's nominal power at 2800 rpm and 100A, the 28-pole motor would have it's nominal power at 2000 rpm 100A. So the power would be the same, and the ironlosses and copper losses too, only a different RPM.

 

[Miles]

My thinking was very simple. If the 20-pole motor had it's nominal power at 2800 rpm and 100A, the 28-pole motor would have it's nominal power at 2000 rpm 100A. So the power would be the same, and the ironlosses and copper losses too, only a different RPM.

But Km has increased.......

 

[bearing]

Yes, but Km doesn't take into account the ironlosses. Obviously, without iron losses, the 28-pole is better, because then there is nothing limiting RPM. And with the ironlosses, the power at the same RPM is better. But the nominal power may be the same. And the peak power may be the same. And I think it makes sense, since we are not changing the stator.

Km has got to do with specific torque, not specific power. Pole count will change specific torque in the motor, but I think not specific torque at the wheel, if the gearing is changed proportionally.

 

[Miles]

I didn't say it was a free lunch

Specific power, may or may not go up.

As long as copper losses are dominant for the rpm at which max. cont. power occurs, then the max. cont. power of the motor will have gone up.

A higher pole count at the same rpm might need thinner laminations. So, cost of materials will increase.

 

[bearing]

I still don't see the benefit of going to 28 poles. Why not just rev the stock motor 40% higher?
Sounds like exactly the same as using stock RPM and putting more poles in the rotor.

 

[Miles]

I still don't see the benefit of going to 28 poles. Why not just rev the stock motor 40% higher?
Sounds like exactly the same as using stock RPM and putting more poles in the rotor.

If you're happy with gearing it down the extra 40%, sure...

For the same rpm, taking that as a given, you'll get a more powerful motor by increasing the number of poles and reducing lamination thickness.

 

[bearing]

If you're happy with gearing it down the extra 40%, sure...

OK, it's about making it more useful in practice. Sure, I can accept that.

For the same rpm, taking that as a given, you'll get a more powerful motor by increasing the number of poles and reducing lamination thickness.

If you reduce lamination thickness, you could rev the low pole count further, and make the same power, so I still want to argue that pole count doesn't make a more powerful motor.

 

[Miles]

If you reduce lamination thickness, you could rev the low pole count further, and make the same power, so I still want to argue that pole count doesn't make a more powerful motor.

So, it will depend on whether the max. feasible rpm is limited by the physical integrity of the motor, or the dominance of eddy current losses.

 

[Miles]

In transformers you get the most power if you make iron losses the same as copper losses, and I assumed it would be the case with motors too. But that may be wrong.

In a PMSM motor peak efficiency occurs when copper losses are the same as parasitic losses.....

Maybe it's true then? That the maximum possible continuous power occurs at the rpm where parasitic losses are equal to copper losses? In other words, when peak efficiency coincides with maximum continuous power.

 

[bearing]

Could be. Not that sure anymore. I remember now that I never read a mathematical evidence on the transformer thing, only accepted it because it seemed good. So I feel we need math here, to prove the theory. I'm not up for the challenge a.t.m.

 

[Martini]

Hello friends ,

I am looking to sell my Kelly KHB72701C and CA 120-100 ( 20Kw 80Volt 250A continue ) motor because I will change my project from
75Volt BLDC motor to 120Volt motor so i have to buy a HV Kelly

please take a look at :

http://www.ebay.it/itm/141041217144?ssPageName=STRK:MESELX:IT&_trksid=p3984.m1555.l2649

and

http://www.ebay.it/itm/141041220609?ssPageName=STRK:MESELX:IT&_trksid=p3984.m1555.l2649

Saluti

GIanni

 

[crossbreak]

Say I want high power and high efficiency from 100-400rpm at the wheel.

A 20pole with 0.2mm lams can take 6000rpm which is one Khz or 60.000e-rpm. It has a very wide high-power-high-efficiency-rpm-band (10.000-60.000 e-rpm or 1000-6000rpm, 600% spread). I want 400 wheel rpm, so 1:15 reduction. This needs at least 2 reductions. Such a 60pole would have 2000rpm, only single reduction.

A 20pole with 0.33mm lams is cheaper, still has a good efficiency but a much smaller high-power-high-efficiency band (1000-3000rpm). It has almost half the power to weight ratio due lower speed, same torque. Still, reduction would be half Such a 30 pole would have 2000rpm, only single reduction.

Or a 42pole DD-Hub that has 0.5mm lams in a DD-Hub-middrive like that does 250Hz/15000e-rpm or 714rpm max and almost does not need any reduction like 1.78:1 just as in a DD-Hubdrive, which has very limited high-power-high-efficiency-rpm-band (10.000-15.000e-rpm or 476-714rpm). For high power and high efficiency I would need 267% gear range.

Does this sound close to the truth?

Conclusion: If a shifter gearbox is involved, 0.5mm laminations are okay. Like using the derailleur with limited power.
If I want a very low weight-high-power-high-efficiency drive, I will go for a motor with thin lams, high pole count and no shifter gearbox at all. Now lets discuss what rpm is best for a middrive without shifting gears!

 

[speedmd]

How big of a hill, at what speed you want power/ acceleration and what load? These factors are most likely what determines the ratio/ minimum reduction ratio.

 

[boostjuice]

Now lets discuss what rpm is best for a middrive without shifting gears!

Accepting that more than about 5:1 is very difficult per reduction stage when using chain or belt (from a reliability/power transfer/chordal action minimization perspective)

The way I see it is that if you can't avoid moving from a single reduction stage to a 2-stage reduction system (for a single drive ratio drive-train), then the faster you can spin your motor as close to 10KRPM - the better (efficiency losses and mechanical limitations aside), so as to take advantage of the massively compounded extra reduction accessible by going to 2-stage (5:1*5:1 = 25:1 = easily accomplished)

So if by choosing 20 magnets (10 pole pairs) over 28 magnets (14 pole pairs), torque stays much the same but the added angular velocity yields an acceptable net gain after iron losses are subtracted from spinning it [EDIT: 40%] faster, then it's the way to go. Toolman2 has explained max power is circa 4000rpm for his 14 pole pair/24 slot CA120-70, so assuming changing this motor to 10 pole pairs up-shifts this max power point to 5600rpm, then why wouldn't you want to do that unless 0,35mm laminations are just too lossy.

 

[Miles]

Such a 60pole would have 2000rpm, only single reduction.

Bear in mind that there is a limit to the number of poles you can cram in (for a given rotor diameter) before the reduction in effective magnet area becomes a significant factor.

 

[Miles]

Toolman2 has explained max power is circa 4000rpm for his 14 pole pair/24 slot CA120-70, so assuming changing this motor to 10 pole pairs up-shifts this max power point to 5140rpm, then why wouldn't you want to do that unless 0,35mm laminations are just too lossy.

It doesn't really matter whether you increase the fundamental frequency by angular velocity or by pole count.

An increased pole count requires a smaller reduction ratio.

 

[Martini]

Hello ,

Someones can help ( or do for me ) to know if is possible rewind my CA 120-100 Kv70 >>> Kv95 ?

Actual power is 80Volt 250A but I would rewind motor to 60Volt 320A ( max power for 60 second )

This application is for ultralight motorglider

Props: 39''12D 2 balde or 34''10D 3 balde

FROM : 21s3p A123 prismatic 20AH to 17s3p

FROM : Kelly KHB72701C to HBC 25063-3

Thanks in advance

 

[crossbreak]

Guess you motor has 6turns per coil. Just rewind it with 5turns, the KV will rise.

How big of a hill, at what speed you want power/ acceleration and what load? These factors are most likely what determines the ratio/ minimum reduction ratio.

Does not matter. An ebike is around 150kg (with rider) and has to climb a 30% hill with good efficiency. Still goes 50kph. That is what the 99% want. This needs a 400% spread to get up the steepest hill @12.5kph and still run high eff @ 50kph.
Peak Power around 4kW for a single speed drive or 1.5kW for a 9-speed middrive match these needs just perfect, that is my point.

Such a 60pole would have 2000rpm, only single reduction.

Bear in mind that there is a limit to the number of poles you can cram in (for a given rotor diameter) before the reduction in effective magnet area becomes a significant factor.

Indeed, that was just an experiment of thought. A larger dia leads to more unused space in the center, so power/weight will suffer. A Goldenmotor hub has 56poles and 230mm dia, still fits between the cranks, this is the largest package that still fits in a bicycle.

What I want to say is, a cheap motor with 0.5mm lams can be ok for a middrive, since it has gears and does not need a wide spread (150% motor x 267% derailleur = 400%)

A single speed has to use 0.2mm lams to gain the 400% rpm spread.

Pictures tell a thousand words, here is an example graph for a 0.5mm lams motor and one for a 0.2mm lams motor:

 

[Martini]

Thanks ,

my primary question is originally CA120-100Kw70 run 80Volt 250A

how theoretically I can rewind motor and runnung to 60Volt 300A Kv95 ?

Battery pack is 17s3p A123 60Ah

Gianni

 

[Miles]

how theoretically I can rewind motor and runnung to 60Volt 300A Kv95 ?

70 / 95 = 0.74

You need to reduce the number of turns in the winding by a factor of 0.74.

 

[Martini]

Hello ,

Sorry for thie " spudid " questions but is not my matter

there are some software application in order to evaluate or simulated performance of the BLDC motor ?

any ideas or suggestions for rewind in order to improve motor power and performance ?

I need all power for 90 / 120 second after this power will be cut to 60%

Gianni

 

[Miles]

http://www.drivecalc.de/

Use the thickest wire that you can (singular or parallel strands) for the required number of turns.

 

[Martini]

Hi & thanks ,

But I am talking of professsional / semi professional application

Mine is NOT a model but is a true glider 13.5meter span 160Kg empty

Gianni

 

[Miles]

Hi & thanks ,

But I am talking of professsional / semi professional application

Mine is NOT a model but is a true glider 13.5meter span 160Kg empty

Gianni

Hi Gianni,

In that case, you need professional advice... :wink:

Seriously, it doesn't really matter for the purpose of analysing the motor that DriveCalc is aimed at modellers.

 

[Martini]

Hello ,

I didn't offend anyone and I sincerely hope that no one in the forum may have offended,

I'm just looking for accurate data calculation process approach

thanksa gain to all

( So I will try to add on the Database mine CA120-100 motor and HBC 25063-3 MGM compro driver )