I've run across a term in some grinding manufacturing research papers called the specific material removal rate. The units given for this term are [mm$^3$/mm$\cdot$s] (I don't really like this style of formatting because it makes it unclear as to whether the [s] is in the numerator or denominator, but that is the style that has been used in all of the papers that I have seen with this term). I am familiar with the general machining term material removal rate (or $MRR$), which is the volume of material removed per second from a workpiece in [mm$^3$/s], but what is the "specific" part of the "specific material removal rate"?

Usually in machining when we say "specific" we're talking about per unit volume removed. So the specific energy of machining is normally expressed as unit energy per unit volume removed (e.g. [J/mm$^3$]). But for the specific $MRR$ it is a volume per second per unit length (after all, it wouldn't make any sense to have volume removed per unit volume removed). So I guess the real question is, what is this length that we are dividing the $MRR$ by to get the specific $MRR$? Why is this term useful?

This seems like a pretty basic question, but I have been unable to find a definition for the specific material removal rate anywhere. I've looked at several papers and from context in the papers it is rather unclear. I've checked my textbook, Fundamentals of Modern Manufacturing 5th Ed. by Mikell P. Groover, no answer there. It may be a term that is specific to grinding operations.

I would appreciate it if any answers could be backed up with a source.

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    $\begingroup$ From the units you show : [mm^3 /mm.s], it is the volume removed per mm of travel seconds. $\endgroup$ – Solar Mike May 12 '17 at 15:07
  • $\begingroup$ @Solar Mike Do you know the answer or are you guessing based on the unit analysis? There are many lengths by which we could divide the MRR. It could be the length of the workpiece, the width of the grinding wheel, the radius of the wheel, the depth of cut, etc. That is the point of my question: what length are we dividing by? $\endgroup$ – ConjuringFrictionForces May 12 '17 at 15:29
  • $\begingroup$ Do you often cut with the radius of the wheel? Most common is the depth of cut... As I said in my comment "from the units you show" ... Perhaps you should go back to the research papers and check how they defined the units. $\endgroup$ – Solar Mike May 12 '17 at 16:49
  • $\begingroup$ Let me try to explain what I mean. The units of kinetic energy are [J] but the equation for kinetic energy is $0.5mv^2$. The units of potential energy are also [J] but its equation is $mgh$. Similarly I know the units for the specific MRR are [mm^3/mm.s] but what is the equation? The units do not give us an answer to this question. The length in question could be the distance from Earth to the moon. Sure that's ridiculous but there is no way to tell from the units alone. Your guess that the answer is length of travel is not helpful unless you can back it up with a source or machining theory. $\endgroup$ – ConjuringFrictionForces May 12 '17 at 18:42
  • $\begingroup$ My best guess is that it is the MRR divided by the wheel width, but it is only an educated guess. I have been unable to find any sources or justification for that. That is why I posted the question. $\endgroup$ – ConjuringFrictionForces May 12 '17 at 18:46

The specific material removal rate appears to be the product of the depth of cut and the feed rate. "Specific" refers to normalization by the width of the cut. From a quick search at Google Books, this definition appears, for example, in:

  • Li's "Efficiency of Manufacturing Processes"
  • Klingelnberg's "Bevel Gear: Fundamentals and Applications"
  • Bach et al.'s "Properties of Optical Glass"
  • Kimber's "Quality Management Handbook"
  • Shen's "Minimum Quantity Lubrication Grinding Using Nanofluids"

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