Runaway armature? 2

[THooper]

Can anyone help me with figuring out a formula? Let me give a brief background of my problem. We have been rewinding some dc traction motors for many years. We have a customer that lost eight motors off the same side of their car hauler. Each of these failures on the motors are classic runaways, the centrifugal force has caused the amatures to flare excessively taking out the field coils and armatures.

The nameplate on these motors are: 25/13hp, 2100 rpm, 110 vdc, series winding. We use cc1118LV Dolph's epoxy resin in our vpi process which is rated at 10,100 psi tensile strength, along with banding the coil head that is rated at 240,000 psi. The motors are load tested at our facility. The customer said they can not find anything wrong with the car and is questioning are rewinding process. We have not had this problem with other customers using the same motors and car haulers. Can anyone help me come up with a formula or ball park figure, using the epoxy and banding ratings, how fast would this armature have to be spinning above the 2100 rated rpm to flare out and mushroom??

thanks for your help in advance......

 

[aolalde]

Certainly the information you are providing is incomplete to make any calculation.
The diameter, width and thickness of the cured banding are required, as well as the weight and distribution of the copper supported and the nominal operating speed.

The 225, 000 psi tensile strength is for brand new banding at room temperature. After 1000 hrs operation under high temperature ( 150°C ave) the figure of tensile strength drops as low as 70,000 psi if properly applied and cured.

For a banding 3/4” W x 0.015” thickness this means 790 LB of tension to reach the tensile strength limit. Note that at tensile strength limit the banding is broken.

1 Lb of copper rotating at 1000 rpm with 10 inches diameter will produce a centrifugal force of:
F= 1(LB)/32.2(ft/sec^2)*(2*3.1416*1000/60)^2*5/12 = 142 LB

1 Lb of copper rotating at 3000 rpm with 10 inches diameter will produce a centrifugal force of:
F= 1(LB)/32.2(ft/sec^2)*(2*3.1416*3000/60)^2*5/12 = 1278 LB

 

[THooper]

This is good information to know and look at, I will get more details as you have describe, I also posted this in Mechanical Engineering and other topics, which also had some good input.

thanks again aolalde!

 

[UKpete]

motorhead1

The following formula as used on dc series wound rail traction motors (generally larger than the machine you have):

total csa of tape = (0.00034*w*r*RPM²)/(24*pi*30000) in²
(glass area only)

where w = weight of overhang copper (lbs)
r = radius of gyration at overhang (ins)
30000 = retained stress (psi) obtained from production tests and corresponds to initial applied stess of 75000psi. Note this is even lower than aolalde's figure.

We then added 5% to the number of turns as an additional safety margin. The formula was used for polyester resin/glass banding tape 0.3mm x 20mm and is successful on these larger machines (eg all the dc motors still in service on London Underground, typically over the range 60-100hp, 3000-5000rpm), subject to proper preparation of the wound armature i.e. all the gaps between the coil noses filled with epoxy loaded felt and gap filling putty to give a smooth even surface for the banding. Final VPI was with epoxylite 347, though I don't think the choice of resin has much effect on the strength.

The motors were specified with typically a 25% overspeed capability (where the vehicles were fitted with an anti-slip system; I believe there is an even bigger margin allowed where there was no such system fitted).

What is your applied tension?

 

[cbarn24050]

Hi Motorhead, the speed of a traction motor is controlled by the load on it. If your client is losing motors due to overspeed it's because they are losing load, so it's not your fault.

 

[edison123]

Aolalde,

Can you give your formula in metric units (kgs, metres etc.)?

 

[aolalde]

Edison. For MKS system.

In a circular movement w=2*Pi*rpm/60 (rad/sec)
w (omega) is the angular speed
rpm = revolutions per minute.
Pi = 3.1416..

the mass m = W/g
W = weight in Kgf
g = 9.81 m/sec^2 gravity acceleration.

Centrifugal force (Kgf) of an object spinning at a radius R (meters)

F=m*R*w^2

 

[edison123]

Thx aolalde.

 

[edison123]

aolalde,

I want to cross check this calculation with you

Coil weight - 650 kgs

Rotor Radius - 1475 mm

Speed - 600 RPM

I got centrifugal force as 386 tons. Am I right ?

 

[aolalde]

Your calculation looks right (385,830 Kgf).
This application has very high tangential speed, 18,243 Foot per Minute. Regular machines do not exceed 10,000 FPM.
This machine will need special steel for the spider and lamination, the banding is very critical.
You should distribute the holding force between two coil heads and the slot embedded portion.

 

[edison123]

Thx aolalde.