# Monitoring While Drilling Problem

1. Apr 21, 2015

### Claudio_Baldo

I am trying to program a MWD (monitoring while drilling) software for a Exploration Driller and I was wondering if the forum could confirm what I have found out.
I need to calculate the WOB (weight on bit) and Head Torque, I do the following direct reading:
WOB:
Pushdown pressure (PSI). = Ppd
HoldBack pressure (PSI). = Phb
String weight is easily calculated as the sum of the number of rods+head weight+hammer weight. Sw
Theoretically the depth of the hole does not matter, air is used to take out the cutting from the bottom hole, so I neglect any buoyancy factor.
The WOB result should be WOB=Sw+Fpd-Fhb.
Where Fpd is the pushdown thrust calculated as:
Fpd=Ppd*(2*pi*Rp1^2) Rp1 is the radious of piston1
Fhb=Phb*(2*pi*Rp2^2) Rp2 is the radious of piston2

For the Torque on the Hammer, I will be equal to the torque generated by the hydraulic motor which rotates it:
Pressure of the hydraulic motor (PSI) = Pm
Flow Rate of motor=Fm
RPM
Torque= (Fm*Pm*36.77)/RPM

I am confident in the calculation of the torque, not 100% sure about the WOB.
Did I semplify the model too much?
Are there any other contributes to be taken into account?
Thanks to everyone who can help me out with this model.

2. Apr 22, 2015

### Claudio_Baldo

Some more data, the type of drilling is RC (reverse cycled).
There is a system of two traverse cylinders, one on each side of the must, the diameter of the piston is 5'' and the diameter of the rod is 2.5''.
As I said I have some empirical data as the weight of string, PSI of Holdback (rod side of the cylinders) and pushdown (butt side) pressures.
The cylinders travel half way the head, the ratio is 2:1.
This machine has installed a monitoring system and I need to evaluate the reliability of the reading.
With: weight of string=755Kg; Holdback=345.2PSI, Pushdown=675.7PSI. (measured)
The system gives me in output WOB=3747.37 while my calculation results in WOB=6504

Weight force due the Holdback:=2 * 345.2 Pound/inch^2 * 3.14*(5^2-2.5^2)=7875 Pound=3571 Kg
Weight force due the Pushdown=2 * 675.2 Pound/inch^2 * 3.14*(5^2)=20553 Pound=9321 Kg

The factor 2 at the beginning is due to the ratio 2:1 of the travel between the head and the cylinders.
I am actually wondering if I should also consider the Pressure inside the Hammer due this is a RC drilling, the drilling is both rotational and percussive.
The traditional way to calculate the WOB is just considering this three components so I dont understand if I did something wrong or the system is not accurate at all.
The percussive hammer shouldnt matter as the mechanical advantage of the cylinders.

I am a bit lost, any idea, suggestion?
Thanks

3. May 3, 2015

### Baluncore

You need a factor of two for travel and a factor of two for number of cylinders. I expect they may cancel to become one.
What is the travel mechanism between the drill string and the cylinders?

Buoyancy of the string in the hole will depend on water, mud or volume of compressed air used to drive the hammer.

You are mixing force and weight. Are you drilling vertical holes only?

4. May 3, 2015

### Claudio_Baldo

Hi Baluncore,

In regards to your answer, yes, at the moment we are only considering the vertical drilling, this is the reason why I considered force weight the same factor as pull down and hold back pressure.
In case of angular drilling I need to use the cos factor of the drilling angle, the pull down and hold back force will always be axial.

The drill string is connected to the head which is connected to the cylinders through a system of ropes.
The travel factor is given as 2:1, the head travel double distance of the cylinders.

The drilling is for RC drillers which means dry hole, no mud or water, about the buoyancy for the hammer, I should calculate it as the PSI by the section of the hammer and sum the result to the Hold Back Pressure?
The manual of the system which is currently installed on the machine says the WOB is only calculated with Hold Back/Push Down and WOS.

So you say to use the factor 4 instead of 2, 2 for the travel of the head and 2 for the double cylinders.
I had considered the double cylinders but the data I see on the current system differ sensibly from what I did calculate.
This system was done by a company which is a involved in general PLC programming and does not have specific experience with drilling machines.
Still, this is a mid size company so I would be surprised to find out the calculation has such a big error.
The point is that if I semplify the calculation not taking into account the buoyancy caused by the volume of the air within the hammer, the final formula should be the one reported above (with the factor 4 instead of 2).
What do you think?

Claudio

5. May 4, 2015

### Baluncore

No, I believe the factor should be 1.000

So the string moves twice as far as the cylinder change in length. The distance is doubled so the force to the string is halved.
Now you have two cylinders, that doubles the force which exactly compensates for the system of ropes.

I therefore say the 2:1 linkage from the hydraulics to the drill string is compensated by the second cylinder. The factor should therefore be 1.000

6. May 5, 2015

### Claudio_Baldo

I am probably missing the point of the ratio two between cylinders and head.
If I apply a force to an object which through a system of ropes makes a second object moving:
if they move the same distance means the force has the same ratio.
If the second object moves half distance of the first object means we have a not advantageous system, we need to apply double force to have the same movement.
if the second object moves double distance then the first means we have an advantageous system, we need to apply half force to have the same movement.

But at the same time if we consider the force coming from two cylinders, it means each cylinder will generate half of the resulting force of the head.
So the factor of 1.000 may be right.
Did I get the problem in the wrong prospective?

So, if I have a cylinder with the diameter 5 (push down side) and the diameter of the rod is 2.5''.
With a push down pressure 834.6 PSI; hold back 323.2 PSI and WOS=755 Kg
The resulting WOB should be 7465.134 Kg while the result from the system is 8964.95 Kg with a delta of roughly 1499 Kg!! (considering the factor 1)
The delta between my data (calculated with the given WOS and pressures) and the result given by the system goes up and down, sometime with a difference of few hundred Kg sometime with 2 tons of difference.

I am not a mechanical engineer, I just want to make sure I understand what is going on before to go back to tell them the system they have programmed is not accurate at all.

7. May 5, 2015

### Claudio_Baldo

Ok, you mean that due the principle of conservation of energy, the work done by the cylinder and the head must me the same.
So if we calculate the work as the force by the distance and the head travels double, the force applied must be half.
Considering the force applied by two cilinders the force due the pressure is doubled.
So the resulting is the 1.000 factor.

8. May 5, 2015

### Claudio_Baldo

I agree with it, the fact is still the result does not match.

9. May 5, 2015

### Claudio_Baldo

As you can see I have adjusted the calculation of the area, from:
3.14*(5^2) to 3.14*(2.5^2) and from 3.14*(2.5^2-1.25^2)
Due to the fact the 5'' and 2.5'' are diameters.

10. May 5, 2015

### Baluncore

push down 834.6 psi * π * 2.5^2 = 16387.3 lbs
hold back 323.2 psi * π * (2.5^2 – 1.25^2) = 4759.5 lbs
difference is 11627.8 lbs = 5274.27 kg (pushing down)
plus the weight of string 755.kg, so WOB = 6029.27

Unless I see a diagram of the cylinder and rope connection to the head I cannot be sure what is happening.

11. May 5, 2015

### Claudio_Baldo

They are traverse cylinders and I have asked for a diagram or datasheet but didnt get anything yet.
I want to double check the effective area for hold back and push down.
About the ropes, please correct me if I am wrong, they should be considered only to calculate the ratio of the distance between Cylinders and Head.
So in this case you want to double check it is effectively 2:1?

12. May 5, 2015

### Baluncore

The cylinders and the path taken by the ropes decides the ratio of displacement of the head to the cylinder movement.
The force ratio is the inverse of the movement ratio.
Draw a diagram that shows cylinder position and the path of the rope between pulleys on the head, chassis and cylinder attachments.

13. May 5, 2015

### Baluncore

The technical term for the cylinder - rope system employed in a drill rig is "hydraulic jigger".
A hydraulic jigger is used to lift something faster and further than the hydraulic cylinder is capable of by itself.
It also reduces the maximum tension in the rope and so permits lighter wire rope to be used.

14. May 11, 2015

### Claudio_Baldo

I am try to get the data you have asked.
Thank you very much for your help so far.

15. Oct 12, 2015

### Claudio_Baldo

Hi Baluncore,

I cant get hold of this information and I need to proceed with the software design, I will leave the movement ratio head/cylinders as 2:1 for now.
These are the actual formulas I will use (in case anyone is trying to do the same or wants to comment on it):
Fpd=Ppd* π*Rp1^2 (lbs). Rp1=2.5''-1.25'' is the radius of cylinder less the radius of the rod.
Ppd=Push down pressure(psi)
Fhb=Phb* π*Rp2^2 (lbs) Rp2=2.5'' is the radius of cylinder (other side)
Phd= Pressure holding back (psi)
Sw(String weigth)= sum of all rods weight, head, hammer, etc...
WOB(Weigth On Bit) = Fpd-Fhb+Sw.

Pressure of the hydraulic motor (PSI) = Pm
Flow Rate of motor=Fm
Torque= (Fm*Pm*36.77)/RPM

ASSUMPTIONS:
I will take away the factor of buoyancy due to the fact the hole should be dry most of the time.
I will also not consider the hold back force caused by the air inside the Hammer, I would need specific data from each Hammers manufacturer and this would take too long, I may fix this part once the first stage is finalized.

How can I adjust the Sw in case of the hole is not vertical?
I would need a functional formula, just to keep it easy for the software..
Thanks

16. Oct 13, 2015

### Claudio_Baldo

Also, any other idea of measures I can do with these sensors?
I should have installed two pressure sensors (hold back/push down pressures), rotation speed, vibration sensor.

I may consider to install additional sensors in case of any interesting suggestion.

17. Oct 13, 2015

### Baluncore

The vertical component of the drill string weight is simply the cosine of the declination = angle from vertical, = Cos(dec).

But there is then friction between the drill string and the rock. It will take the frictional force before the string can move in the hole. A wild guess at the coefficient of friction between steel and wet rock is about Cf = 0.3, so I would expect a factor of Sine(dec) * 0.3 * string_weight, due to friction for non-vertical holes.

Do you have a make and model number for the drill rig so I can look it up on the web ?

Your Rp1 = 2.5'' – 1.25'' worries me if you are working out areas. Should you not be subtracting the squares of the radii to compute the area of the annulus ?

I'll get back to you again when I have time to think.

18. Oct 13, 2015

### Claudio_Baldo

Yes, sorry, my mistake, that will be the subtraction of the outer and the inner areas... :)

The monitoring should be made up independently from the Rig manufacturer and this should be for RC rigs.
RC rigs do not use water down the hole but compressed air instead, this is to say, the hole should be pretty dry.
The coefficient of friction I guess will depend on the material (ground) as well, is there any way to calculate an empirical value?
I don`t think this factor will have a major impact on the WOB while drilling but this is just to get the most accurate measurement as possible.