correlation between torque and magnetic field

A

athlon

100 mW
Joined
Aug 15, 2011
Messages
44
Iron Losses

hello , I'm trying to understand how iron losses work in a motor for getting more power out of the motor with better efficiency

I was looking to the different type of electrical steel and found out that the name ( M330-35A , M400-35A , M400-50A , M700-100A) is composed for the first part
of the Watt per Kg of heat generated at 1,5 tesla at 50Hz , for example
M330 = 3,30 Watt / Kg
M400 = 4,00 Watt/ kg
M700 = 7,00 Watt /kg

the second part of the name is the thikness
35A = 0,35 mm
50A = 0,5 mm
100A = 1 mm


in this document
http://www.sura.se/Sura/hp_products.nsf/29a268683cdcefd4c12569ed004229a2/03a8b2433fae16c4c1256aa8002280e6/$FILE/datasheets.pdf
at page 18 there is a table with different losses of an average M400-50A on different magnetic field and different working frequency.


I don't know if motor torque is directly correlated to magnetic field strenght , in that case looks like the best solution to have power is to use small magnets , low current and spin the motor as fastest as possible.

Loss at 1,2 T and 50 Hz = 2,09 W/kg
Loss at 0,6T and 100Hz = 1,73 W/kg
Loss at 0,3T and 200Hz = 1,35 W/kg

I DON'T KNOW if my hypotesis are true ( waiting for some ES expert) but looks like if you halved the field but double the speed you get a cooler motor with the same power


.
 
of course other type of loss goes up linearly with RPM ( bearing loss for example) , and also faster spinning motor need greater reduction ratios ( more loss) so the whole thing is not so straight forward
 
Well if you go down to half the magnetic field, you need twice the winding current for the same torque, which makes 4 times more heat in windings.
 
bearing said:
Well if you go down to half the magnetic field, you need twice the winding current for the same torque, which makes 4 times more heat in windings.
Click to expand...

Or twice the magnet area (bigger motor).
 
bearing said:
Well if you go down to half the magnetic field, you need twice the winding current for the same torque, which makes 4 times more heat in windings.
Click to expand...
Yes. Not a great trade for a modest reduction in iron losses.... :)
 
bearing said:
Well if you go down to half the magnetic field, you need twice the winding current for the same torque, which makes 4 times more heat in windings.
Click to expand...

I don't need to keep the torque , Power is torque x rpm .

If I cut in half the torque but double the speed I still have the same power output.

on a 4 pole motor switching from 50Hz to 100 hz mean going from 1500 rpm to 3000 , I think about two motor with the same power , one spinning at 1500 rpm with full torque , the second spinning at 3000 rpm with half torque
 
athlon said:
bearing said:
Well if you go down to half the magnetic field, you need twice the winding current for the same torque, which makes 4 times more heat in windings.
Click to expand...

I don't need to keep the torque , Power is torque x rpm .

If I cut in half the torque but double the speed I still have the same power output.

on a 4 pole motor switching from 50Hz to 100 hz mean going from 1500 rpm to 3000 , I think about two motor with the same power , one spinning at 1500 rpm with full torque , the second spinning at 3000 rpm with half torque
Click to expand...

Yes, your logic is correct. But what you are doing is making the motor larger to make it more efficient. If you can tolerate 3000 RPM, then a motor with that speed and the original flux density would be half size. But less efficient. Larger motors operating at lower flux density and/or lower current density can be more efficient. Reducing flux density and increasing frequency will have off setting effects on efficiency. Consider the controller and mechanical transmission sides to see if it is a system gain.
 
I want to understand how to reduce the iron losses ,

iron losses depends on frequency and magnetic field , from the datasheeh of the M400-50A iron looks like there is a sweet spot between 200 and 400 hz in trading speed vs field .

but if every industrial motor is tuned for 50 or 60 hz maybe i'm wrong
 
athlon said:
I don't need to keep the torque , Power is torque x rpm .

If I cut in half the torque but double the speed I still have the same power output.
Click to expand...

But look at the numbers. You are gaining fractions of watts per kg of iron. Maybe 1-2 watts on a 10kg motor, which probably have nominal losses in the order of 1kW. There are many things you can do to gain those 2 watts without cutting the motors potential in half.
 
motor potential still the same , is just a matter of working point , or in other words to choose higher or lower Kv.

when you get closet to the maximum every watt count.

On sealed motor is very hard to cool the active parts because you don't have almost any air moving inside the motor , so going from the worst case 2,09W/kg to the best case 1,35 W/kg it saves 0,74 W/kg , given an average efficiency of 94% on electric motor it means you can have 12W more for every Kg, for sure not much , but why not ;)

I know this is only theory but I like to know what is going on inside a motor , not just watch it spinning
 
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