Calculating speed and volume of a drilled hole

Question

I'm trying to represent a hole being drilled using a web based application and I'm having difficulties.

It's essentially a well that is being drilled. I'm in control of the inputs and various variables but unsure of how best to approach the issue.

I have the size of the drillbit, the speed at which it rotates, the time since it started running and the permeability / strength of the material it's drilling into.

I thought I'd be able to calculate the volume of the drillbit and then apply a percentage of sorts that would represent the strength of the material so as to slow the progress of the drill.

Any help would be fantastic. In it's simplest form I'm trying to figure out how best to represent a hole being drilled and then calculate the area of the hole as it's being drilled.

Thanks!

Edit: Further Info.

The well I'm trying to model is that of a subsea oil well. It's in its infancy at the moment so even though I'd love for it to be perfect - at the moment I'll settle for functional and displaying an output.

The simulation currently has these values, they're for test purposes only.

Radius of drill bit= 15 inches Length of drill = 1000ft RPM of drill = 100

The stratigraphic layers have their own properties nd in this instance have:

Name = seabed Depth = 100ft (used to model start and finish of each layer - in this case from 0 to 100ft down.) Permeability = 10 (currently unsure how best to model)

I think I may also need to take into account the upward pressure of the well as the drill gets deeper.

I don't know at the moment how to model the pressure being applied to the drill but a constant can be used if need be.

Answer 1

This isn't so much a math problem as a modeling problem: you're trying to make a model of the real world. To do that, you need information about the real world. For instance, if you're drilling stainless, then as speed increases (for a given (modest) pressure on the drill face), there's a good chance that the amount of material removed will go DOWN, because the stainless steel will tend to work-harden. If you're drilling mahogany, the amount removed will instead go up. You say you've got "permeability/strength" of the material, but those two things don't tell the whole story (and they're not that well defined, either) -- there might be a mild steel that's as strong as some other stainless, and equally "permeable" (when you use the right speeds/feeds), but there's still going to be a difference in how the two materials respond at other speeds/feeds.

The complete physical properties of the material being drilled are really important if you want an honest answer to your question.

Short answer: there's not enough data given to provide an answer. If you can say a bit more clearly what you mean by permeability, and how approximate you're willing to be in your solution, we might be able to help a little.

After additional information about the situation was filled in:

Well, there's still not a lot to go on here. Here are some assumptions that I'm going to make:

1. As you drill, the amount removed is proportional to (i) the area of the drill bit, (ii) the speed in RPMs, and (iii) the pressure on the drill minus the "back pressure". As you drill deeper, the back-pressure is the pressure associated to pushing material back up past the drill.

2. The material removal rate is inversely proportional to the thing you've called "permeability" -- permeability 5 gets twice as much material removed for a given area, RPM, and force, as does permeability 10.

3. The layers you've already drilled through don't matter.

Now I'm going to make a totally wild-guess model about material removal from the well:

1. The friction involved in pushing some material (the cut-away seabed, etc) back up the hole you've made (what I've called the back pressure), through channels in the drill, is linear in the depth.

2. The friction involved in the lateral surface of the drill being moved against the wall of the hole is linear in depth as well.

And I'm also going to make a modeling assumption about the way this whole thing is operated:

1. There's a fixed amount of power available, so that when the drill is deep, you can no longer provide enough power to overcome the lateral-wall friction, the getting-stuff-out friction, the grinding-surface friction, etc.

Now part 5 isn't consistent with your inputs: you say you're given drill-speed, but if there's not enough power, your user can't specify an arbitrary drill speed. Anyhow, let me work with what I've got. There are going to be several unknown coefficients, because they depend on things I just don't know about.

First, some variable names: \begin{align} r &\text{ = radius of the drill bit} \\ A &\text{ = area of the drill bit} \\ d &\text{ = depth of the drill bit below ocean bed} \\ s &\text{ = speed of bit in RPM } \\ P &\text{ = pressure on the drill} \\ q &\text{ = permeability of material at depth $d$ } \\ L_1 &\text{= frictional load at drill face } \\ L_2 &\text{= frictional load along lateral surface of the drill} \\ L_3 &\text{= load associated to pushing out cut-away material, i.e., backpressure} c_1, c_2, \ldots &\text{ = coefficients of the model that depend on material properties I don't know.} \\ M &\text{ = material removed in one unit of time} \\ u &\text{ = how far drill advances in one unit of time} \end{align}

With those in mind, here are some equations: \begin{align} A &= c_0 \cdot \pi \cdot r^2 \\ L_1 &= \frac{1}{q} c_1 \cdot s \cdot \max(0, (A \cdot P - L_3)) \\ L_2 &= c_2 \cdot s \cdot 2 \cdot \pi \cdot r \cdot d \\ L_3 &= c_3 \pi \cdot r^2 \cdot d \\ M &= c_4 L_1 \\ u &= c_5 \dfrac{M}{A} \end{align}

The first says that the area of the drill face is proportional to $\pi r^2$ (because the drill tip might be conical, or even more complex).

The $L_2$ formula says that the lateral area of the drill is $2 \pi rd$, and the load on the motor due to friction along this surface is proportional to area ... and to the speed of the drill, $s$.

The $L_3$ formula says that the amount of material being pushed back up the drill bit is proportional to the area of the drill bit and the length you're pushing along, so the total load is proportional to that number.

The $L_1$ formula uses the notion that the effective pressure on the drill face is the difference between $P$ (the pressure applied at the top) and the backpressure. The rate of advance in response to this effective pressure is inversely proportional to the permeability (hence the $1/q$ factor) but proportional to the speed, $s$, of the drill.

The amount of material removed $M$ is proportional to $L_1$, and the advance of the bit is proportional to the material removed...but inversely proportional to the bit's surface area. (A big bit that removed a cubic foot of material does so in much less distance than a small bit.)

All this gives you a method to simulate: You start at depth $d = 0$, look up the permeability $q$ for the current depth $d$, and compute $L_2, L_3, L_1, M,$ and $u$. You then increase $d$ by amount $u$, and repeat the process.

If I were doing this, I'd also display the value $L_1 + L_2 + L_3$, which represents the total load on the system. If that gets larger than your motor's capacity, then you're not doing a realistic simulation.

As I've indicated, this whole model is based on a bunch of wild conjecture. But at least the wild conjecture is all listed in one place, so you can have an expert evaluate it. I hope this is of some help to you.