1200 kW (12 poles) and 6200 kW (4 poles) (Both 11 kV Motors) windings coil comparison?

[Shark96]

I read a comparison from a rewinding company that the copper used for rewinding of the above mentioned stators is almost of same weight. How come the copper in both stators is almost same (by weight) when their HP ratings are so different? Thanks

 

[davidbeach]

I'm going to guess that the 12 pole machine has a physically larger stator to accommodate the number of pole, or at least not a significantly smaller stator.

 

[jpts]

Probably the 12 pole -machine is a low-speed one (maybe few hundred RPM nominal), which means high torque and thus large active part volume. 4-pole machine is then probably running at higher nominal speed (maybe around 1500RPM), meaning that the torque is still in the same range even though hp rating is much higher. Active material (electric-steel, copper) consumption is into a large degree defined by torque rating instead of power rating.

 

[Shark96]

@jpts but how can it be proven? Your last statement is a bit vague. Thanks

 

[jpts]

I think it is quite well-known fact (for machine designer) that physical size of any electrical machine (and therefore active material consumption) is more or less depending on the torque rather than power. You can for example compare weight power ratio of a low-speed machine (e.g. hydro generator) to a high-speed machine (e.g. gas turbine generator), and you can see that the weight/power of a low-speed machine is significantly higher.

Typically (large) electric machines are designed to have certain (more or less fixed) air gap shear-stress (rotor tangential force per rotor surface area), which depends e.g. on cooling (the higher the stress, the more effective cooling needs to be). When you increase the torque (and hence tangential rotor force), you need at the same time increase rotor surface area, or otherwise you end up with too high shear-stress -->problems with cooling etc. Larger rotor surface area naturally results in physically bigger machine, and needs more active materials

 

[waross]

Vague?? There are many factors which determine the size of a motor of a given HP. Speed, efficiency, type of motor, eg: ODP, TEFC, Wash Down service, Frame rating (Nema frame, U frame, T frame, there is about a 1:2 HP rating for a given frame size between Nema frames and T frames), frequency.
For vagueness, your original question is only one step away from;
"Why is a big motor bigger than a small motor?"

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

I agree with Bill there are a lot of variables to determine motor size of course.

I also agree with jpts, if you want to predict motor size (and copper) from a single parameter (all other things equal) it varies more closely with torque than it does with horsepower.

You asked "why". I'll take a stab at talking through why it's somewhat expected as a first approximation.

If you look at the equation for torque in a textbook you will find something like:
Torque ~ R* L * N * Bgap* Irms * p.f. · Kdp

where
Torque
R is radius of airgap
L is length of airgap,
N is number of turns,
Irms is current per turn
~ represents unit conversions and constants.
Bgap is airgap flux density,
pf represents cos of spatial angle between fundamental B and I distribution (spatial meaning as a function of angle in the airgap).
Kdp is product of pitch and distribution factors.
(By the way you can predict the above textbook equation based on Lorentz force acting directly on conductor IF you make the odd/incorrect assumption that the conductor is in the radial airgap flux density. It is an incorrect assumption and furthermore the torque producing force acts on iron rather than conductor when the conductor is embedded in slots in iron core. Nevertheless these errors "cancel" and interestingly give the right answer. More details here

http://electricpete1.tripod.com/torque_web/attach/TheShortVersion091207.pdf

)


The last three items listed above (Bgap, pf, Kdp) vary within a fairly narrow range. Make a simplifying assumption they are constant. Also assume L~R under assumption that L/R ratio varies only in a small range. These assumptions give:
Torque ~ R^2 * N * Irms

Now define a "linear current density" quantity Lambda as current per distance along the airgap circumference
Lambda = N*I/(2*pi*R).
Lambda ~ N*I/ R

Therefore substitute into previous equation for torque using N*I ~ Lambda * R and we get:
Torque ~ R^3 * Lambda

Consider only rated conditions (rated Lambda, which gives rated torque). Then Lambda (linear current density at rated conditions) does tend to increase with size of the motor as slots get deeper but it increases fairly slowly. If we make a further simplifying assumption that Lambda is constant (this is a grosser assumption than the previous assumptions) then we get simply
Torque ~ R^3

So the Torque (at rated conditions) is roughly proportional to R^3 which is roughly proportional to volume of the motor. This supports the idea that the size of the motor is more directly related to torque than it is to power. At first glance we might think copper weight goes proportional to motor weight. But to be more specific we'd also have to think about the pitch of the windings and how much goes into the endturns. For two motors of same physical size (all other things equal except poles and speed if I can make that fuzzy assumption), the higher-speed motor has wider coil span and has to devote more copper to endturns. This particular factor (more copper in endturns for high speed motors) pushes the balance back in the direction of copper weight proportional to horsepower (instead of copper weight proportional to torque) but probably not all the way back. We can see it's a complicated relationship given assumptions so take it for what it's worth (detailed analysis of specific motors by a designer is more relevant).


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(2B)+(2B)' ?

 

[electricpete]

Consider only rated conditions (rated Lambda, which gives rated torque). Then Lambda (linear current density at rated conditions) does tend to increase with size of the motor as slots get deeper but it increases fairly slowly

The textbook "Design of Rotating Electrical Machines" addresses this behavior. They give an estimate equivalent to:
Lambda ~ sqrt( R)
(they use different symbols: linear current density A instead of my Lambda and characteristic dimension lambda instead of my R).

That book has a much more thorough discussion of estimating size of motors (and their components) than I could ever hope to give in Chapter 6 that you can view on line here

https://books.google.com/books?id=_...ain Dimensions of a Rotating Machine"&f=false

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(2B)+(2B)' ?

 

[edison123]

For any machine, mechanical or electrical, output = k x volume x speed. Think through.

Muthu

http://www.edison.co.in