Re: What happens to armature plates when you double the size of a motor?

Dear Maddy,

I pulled out my old EE concepts book (Electrical Engineering Concepts and 
Applications: Second Edition, by A. Bruce Carlson and David G. Gisser) and read 
up on motors (I am a mechanical engineer).  Since it is tough to exactly 
understand your application from your question, I will try to be broad in my 
answer and hopefully somewhere in there lies the answer you are looking for.

In most motors, the armature is a coil of conducting wires set in the 
stator, which is the motors stationary frame.  The wires of the armature 
run parallel to the axis of rotation of the inner, moving rotor that is 
usually a polar magnet, either permanent or electromagnetic.  The current 
flows through the coil wires and parallel to the rotation axis.  The 
direction of the current dictates the magnetic reaction to the poles and 
gives us our rotational force, or torque.  Now you likely know all of this 
simplified (or butchered, as an EE would say) explanation, but I wanted to 
set up the basics.

There are two equations that I pulled from the book that will give you a 
better understanding of the relationship between the motor's size (and by 
this, I assume you mean power), and the size of the armature coil.  You 
will see that it isn't necessarily size that counts!  Anyway, the first 
equation is for a simple motor with an I shaped permanent magnet rotor and 
armature coils at 180° apart.  This will give a sinusoidal output, but the 
average torque output can be described in the following equation:

Tav = ((NBrA|I|)/sqrt(2))cos(theta)

Tav = average torque
N = number of armature conductors
Br = radial magnetic flux density
A = Area of armature coil
I = current
Theta = impedance angle

Now, I don't expect you to understand all of that, but what I want you to 
see is that if A, the area of the armature, increases then the average 
torque will increase (assuming all else stays the same).  The area of the 
armature A=2lr where l is the length of each slot in the stator that the 
individual wire is in, and r-the radius of the inner stator housing.  But, 
if the area A stays the same and the number of windings N or the current I 
increase, then the torque will also be larger.

Now, here is another example and one that is more real world.  The 
following equation is for the power of a motor with an electromagnetic 
rotor that also has coils.  This is for air-gap power, which is the maximum 
power a motor can produce before you factor in losses from resistance and 
friction.

Pag = k·Na·Ia·Bf·A·n

Pag = Air-gap Power (usually horsepower or watts)
k = constant (frequently coil winding factor, .8-.9)
Na = number of armature conductors
Ia = armature current
Bf = maximum air-gap magnetic flux density per pole
A = 2lr = axial area of rotor
n= rpm speed

Again, I am not expecting you to understand all of this, but to see what 
does affect the power of an electric motor.  If you wanted to increase the 
power of a motor, one or all of the above factors can be increased and the 
power will increase.  We could increase the rotor area A by making the 
rotor longer and wider, which would increase the size of the motor all 
together.  Bf can be increased but there are magnetic saturation limits.  
And for the armature, the number of conductors Na can increase as well as 
the current Ia.  Of course, there are issues with that as well.  The number 
of conductors can't be infinite since they have width.  (See how many 
straws you can place around a soda can that still touch the can.  My guess 
is about 30.  This is similar to the conductor windings).  And, if you 
increase the current, you will need to better insulate the wires and they 
need to be larger in diameter so they don't overheat.  (The greater the 
diameter of a wire, the greater current it can carry without overheating). 
 See how there isn't an easy answer to your question!  I will give a quick 
example to hopefully shed some light on the above confusion.

Say you have an average 1 horsepower motor (1Hp).  The same company has a 
2Hp motor that is bigger, but not exactly twice as big.  The company likely 
took advantage of increasing a few of the above variables.  By increasing 
the size they increase A.  They also have more room so they can increase 
the armature conductors, but not too many as they likely increased the 
current in those conductors as well, so they will need better insulation 
and larger size wire.  (See how current and conductor numbers are 
related!).  Now, if they just increased A, then the motor would have to be 
twice as large.

But what if you need a powerful motor in a small space?  That is where the 
engineers come in and try to maximize all of the variables to give the most 
power in a smallest package without overheating the motor!  I really hope 
this answers your question and that I didn't confuse you too greatly.  If I 
did, you are more than welcome to write me at bradk@jymis.com and I can try 
to confuse you even more:)  Take care.
BK