Formula for determining the output power in a electric motor shaf 1

[petronila]

Dear all,

Does anybody have the formula to correlate the shaft dimensions (diameter and length) with the rated output (kW/hp) in one electric motor?

Thanks in advance

Petronila

 

[Gr8blu]

petronila There is a lot of missing information in that question. To determine shaft diameter requires knowledge of the average and transient peak torque requirements - which means not only do we need to know power, but also speed. Different applications have different values: for example, a general industrial duty machine has a different shaft diameter than one in metal rolling, which is different from a reversing duty application. The material of the shaft is required to be known as well. Shaft length is another story altogether. Are you looking for shaft EXTENSION length (the length on which the coupling fits? Or the length of the bearing journal(s)? Or the end-to-end length?

Converting energy to motion for more than half a century

 

[petronila]

Gr8blu

Shaft EXTENSION length is what I mean. I am referring to the SHAFT dimensions. A [highlight #FCE94F]standard[/highlight] 300 hp 1755 RPM 449T frame 3-phase NEMA motor should have a 3 3/8" diameter and 8 1/4"extension length shaft.

My specific question is whether there is a formula for determining both SHAFT dimensions (diameter and extension length) based on the 300 hp and the 1755 RPM.

Thanks

Petronila

 

[macmckim]

Shaft diameter & length is based upon the NEMA Frame size in question, 143T ~ 449T. There's no formula for shaft length.
The only formula I am aware of is determining shaft height. Divide by 4 the first two digits of frame size.

Mac

 

[Gr8blu]

petronila The shaft diameter (for any application) is determined by how much mechanical stress is applied to the material and how much stress (tensile and ultimate yield) the material can take. Ideally, the applied stress is no more than 75-80 percent of the mechanical limitations based on material used. Keeping things relatively simple, the designers tend to use rated torque and a multiplier, then compare the result to the material yield limit. More difficult applications (metal rolling and/or reversing duty) have higher multipliers than general industrial (which is the NEMA "standard" sizing approach). This is why you see shaft sizes increase as torque throughput increases.

The shaft extension diameter is also a function of torque, in that the ability of the fit between shaft and coupling determines the amount of torque that can be successfully transferred. The maximum torque transmission capability is also a function of the contact pressure (usually a function of coupling wall thickness) and fit length (i.e., shaft extension length). Since the majority of coupling geometries for "standard" general industrial ratings are well defined, the resulting fit length is as well.

Also, since you're looking at "standard" NEMA ratings, chances are good the shaft material is a medium carbon steel (such as AISI 1040, 1045, or even A36) formed into a bar stock and then machined as required. Some applications (food industry for example) might use a different shaft material such as stainless steel, but how the calculation is made is the same - just apply the properties of the stainless steel instead of the n=more normal carbon steel.

A quick glance through Shigley would give you the basics for the calculation: all you'd then have to do is apply the appropriate material properties and "application" factors.

Converting energy to motion for more than half a century

 

[electricpete]

It seems to me macmckim answered op's question. I'm sure petronilla is familiar with NEMA frame dimension charts, example here

Edit - it includes both shaft diameter and shaft extension length.

=====================================
(2B)+(2B)' ?