How to calculate the energy required to cut a specific material?

[DarkDare]

I was wondering how I would go about working out the energy required to cut a material (e.g. stone or steel) if given a volume, and all the necessary values pertaining to that specific material?

For example, how would I work out the amount of energy required to cut a steel block of dimensions 1 x 1 x 1m. At a depth of 0.2m about the center of the block? Hypothetically, let's say it was cut with a knife of regular kitchen knife dimensions. What steps and formulas would you use?

 

[Simon Bridge]

Welcome to PF;
Basically I'd take a shortcut by considering material properties and rules of thumb.
I can look these up in tables.

i.e. the energy needed to put a decent scratch into the steel cube of your example would be more than that required to destroy the knife. I'd probably want to use a purpose-built cutting tool - in which case I'd consult the manufacturer's notes on the tool.
(Looked it up - probably an automated hacksaw. Easier to melt the entire block and pour it into shape.)

The basic physics is to do with binding energies - the binding energy depends on the configuration. The difference in the binding energies before and after the cutting would be the minimum amount of work needed to cut it. After that it is a matter of considering the efficiency of the specific process.

Doing it from scratch like that is hugely complicated - hence the shortcuts.

i.e. if the knife is hot and the substance butter, then the energy is that required to heat the knife, plus that required to lift the knife above the butter. The weight of the knife does the rest.

 

[DarkDare]

Hello,

I was looking for help more along the lines of calculations, and calculation methods to work out the energy needed for the task. Formulas etc..

 

[Simon Bridge]

I just gave you the methods.
The basic question is too general for me to give you equations for... do you not know how to calculate binding energy? It is the amount of work needed to assemble the configuration. W=Fd.

The specific example you had could be solved very easily just by comparing the hardness of a steel kitchen knife with the hardness and volume of the steel cube. Since the one was overwhelmingly more than the other, no detailed calculations were needed.

The specific example I gave you: butter - the equation is "E=mgh+mcΔT" where h is the height of the block of butter, m is the mass of the knife, c is the specific heat of the knife, and ΔT is the temperature range it was heated through. The temperature part of the knife has to be enough energy to liquify the butter the knife has to pass through ... so [mcΔT]_knife = [mcΔT]_butter+m_butterL ... where m_butter - is the mass of butter in the width of the knife, and L is the latent heat of melting for the butter.

See how the details matter?
So I'm afraid you'll have to be specific.
But you cannot cut through a cubic meter block of steel with a kitchen knife.

 

[DarkDare]

I see. Suppose there was a diamond cutting tool then? Because all I really want to know is supposing there's a hard enough cutting tool (even marginally harder than the steel block), how would the amount of energy to make a cut on the block of x dimensions be calculated?

 

[tygerdawg]

This type of information is readily available in textbooks on "Manufacturing Processes" that explain material removal processes. The mathematics & formulas are primarily concerned with metal removal, but I suspect the formulas can be adapted. These calculations are used to determine what motor power is needed to remove a specified amount of a specified material in a specified amount of time in a specified material removal process.

The formulas developed in these texts are specific to the TYPE of machine or process: sawing, lathe work, milling, drilling, etc.

What you must decide before going any further is what type of material removal process you will be using. THAT will determine the ultimate amount of energy required.

 

[Simon Bridge]

tygerdawg is right ... which is why I'm asking for specifics.

However, one can do a back-of-envelope.
Imagine the process completely melts the material to be removed.
Then the minimum energy is $U=\rho V (c\Delta T + L)$ ... this does not include the energy needed to ensure removal of the molten steel.

A 1m^3 block of steel, cut to leave 20cm square in the middle, using (say) a CO2 laser (0.15mm width cut)

V=0.000144m^3 ... volume of material removed to make the cut.
\Delta T = 1350C ... approx to heat from room temp to melting point

These are very approximate values because there is more than one kind of steel.
\rho =8050kg/m^3 ... density
c = 490J/kg/C ... specific heat
L= 272000J/kg ... latent heat of fusion (unsure of this figure)

So... about 1.1MJ by that approach - this would be a low value.

For a diamond tool - say a saw - you'd need to find the cutting rate and multiply by the power of the saw and the dimensions of the cut. I had a quick look, and it takes a diamond saw quite a long times to cut quite small samples of steel ... like 8-12hours for a 15mm deep cut depending on the saw. The exact figures will depend on the saw (and how hard you press etc) though and I don't have those to hand.

Get the idea?