Drilling Engineering Assignment

Professor: Dr. E.Dimou

Subject: MOE5012

Registration number: 8168

Submition date: 30/12/2013

Surname: Papapanagiotou

Name: Theodoros

Question 1

You are assigned to describe the process for selecting a drilling unit for drilling a well in a water depth of 100m.

a. What type of drilling rig you are going to select and why?

We are going to select Jack-up drilling rig because:

It is used in relatively low depths.

It is used in water depth that ranges from 15 to 100 m with a maximum depth of 150 m.

It is a very stable platform and has low mobilization costs.

It is designed to move from place to place and afterwards anchor itself by deploying the jack-like legs.

In case of storm we assume that the maximum height of the waves are going to be 10-15m because this is the limit which the hull of the jack up rig can be raised.

b. What is the process for drilling (brief description) for a dry hole completion?

i) The first step is the insertion of the drilling bit along with the collar and drill pipe in the hole.

Due to the weight of the thick walled pipes that are above the drill bit (drill collars), it is cutting inside the rock. There are various types of drilling bits but the main two working principles are the following:

1) The one disintegrates the rock by compression failure,

2) And the other shear slices off the rock as the bit turns.

ii) The second step is to attach the Kelly and the Turntable and begin drilling.

iii) After that the mud circulates through the pipe and out of the bit in order to float the rock cuttings out of the hole.

The so called mud or drilling fluid is pumped inside the drill pipe and reaches the drill bit where it helps it to make the hole.

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Reference: http://www.seadrill.com/tcp/modules/module_123/proxy.asp?D=2&C=72&I=2075&mid=147

It is a relatively complex mixture of fluids solids and chemicals. The mixture is different for every case in order to achieve the right physical and chemical characteristics required to safely drill the well. It can also be water based or oil based. The first one is cheaper and doesn’t need any treatment on the other hand the oil based needs a ship to curry it onshore to treat it and bring it back for reinjection. The main functions of the mud are the cooling of the drilling bit (similar with the case of the lathe), the lifting of the rock cuttings to the surface, the prevention damage of the rock in the wellbore walls, to create a pressure bigger than the pressure of the fluids inside the rock in order for them not to enter the wellbore. When the mud is returned on the surface it goes through shakers that separates the cuttings from the fluid that is going to be reinjected.

Another important parameter during this procedure is to keep monitoring the returned cuttings, pit volume or returning rate because they are essential for the early catch of the “kick”. As mentioned before, when the pressure at the depth of the bit is more than the hydrostatic head of the mud above we have to close the blowout preventers to control it temporarily and then ultimately by increasing the density of the drilling fluid that will allow the formation fluids and mud to come up through the annulus controllably

iv) While the drill pipe goes inside the ground new pipe sections must be added in order to move deeper.

The pipe sections are usually 9m long. This process of adding pipes under the Kelly or topdrive at the surface is called tripping.

v) ‘’Pre-set’’ depth is set from a few hundred to a couple of thousand feet. When it is reached we need to perform the ‘’trip out’’ which is the opposite of the trip in, thus removing the drill pipe, collar and bit.

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Afterwards the casing must be cemented by placing the casing pipe sections inside the hole in order to keep it safe from collapsing. The spacers located outside the casing pipe keep it centered in the hole. The cement is pumped inside the casing pipe using a bottom plug, a cement slurry, a top plug and drill mud. The mud pressure causes the cement to fill the space between the outer of the spacing and the hole. After the hardening of the cement it is tested for hardness, alignment and good seal.

Reference: http://www.bbc.co.uk/news/10370479

The process continues with drilling and new casings are cemented. At the point where the rock cuttings start to appear oil sand from the reservoir rock the wells final depth may be reached.

Reference: http://www.bbc.co.uk/news/10370479

The drilling apparatus is removed and the following testing’s are made after this procedure.

o Well logging – The insertion of electrical and gas sensors into the hole to take measurements from the rock formations.

o Drill-stem testing - The insertion of a device into the hole to measure the pressures, which will reveal whether reservoir rock has been reached

o Core samples - Taking samples of rock to look for characteristics of reservoir rock.

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c. What are the main well controls for preventing accidents from abnormal well conditions and how they work?

An abnormal well condition is the ‘’kick’’ which is explained in the previous question

(1b). Because accidents are the results of these abnormal conditions and have

effects on the equipment and on human life there are some well controls that

prevent this kind of accidents.

To prevent ‘’kick’’ we must:

Use proper equipment.

Practice and employ trained rig crews in the drilling process to control and

maintain density of fluid, pressure, circulation, etc.

An even more dangerous situation is the case of the abandoned well because the

well integrity must be maintained for a longer period of time.

Pressure can be controlled by remote controlled valves with locking mechanisms. A

remote controlled shear-blind ram must be adopted in well intervention equipment.

It must be noted that a number of well controls are used because if one of them fails

the other must be in position to take action.

Kind of well controls:

a) Primary (Inside the wellbore a greater hydrostatic pressure is used than the

fluid pressure which is achieved with the use of mud weight)

b) Secondary (A second option if the primary control fails. It stops with the use

of a Blow out preventer to stop the fluids flow from the well to the wellbore.

The BOP basically: 1) Confines the well fluid to the wellbore 2) Provide means

to add fluid to the wellbore 3) Allow controlled volumes of fluid to be

withdrawn from the wellbore. Apart from these primary functions BOPs are

used to i) Regulate and monitor wellbore pressure ii)Center and hang off the

drill string in the wellbore iii)Shut in the well (seal the void, annulus, between

the drill pipe and the casing) iv) “Kill” the well (prevent the flow of formation

fluid, influx, from the reservoir into the wellbore) v)Seal the wellhead (close

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off the wellbore) vi)Sever the casing or drill pipe ( only in case of

emergencies).

Reference: http://en.wikipedia.org/wiki/File:Blow_Out_Preventer.jpg

BOP

c) Tertiary (A third option if the first two fail. It takes action in underground

situations. Some examples are: i) to drill a relief well to hit an adjacent well

that is flowing and with heavy mud to kill the well ii) rapid pumping of high

density mud iii) pump barite or heavy weighting agents to plug the wellbore

to stop flowing and iv) pump cement to plug the wellbore.

d. You need to submit a drilling plan. What factors you need to consider in order providing a safe drilling activity to the right place?

The factors need to be considered depend mainly on the location, the climate, the noise and generally all the factors that can affect the drilling procedure or equipment. It is essential that the whole procedure is curried out safely.

Equipment is selected with the following way of thinking:

a. Drilling mud

As written in Question 1b it is a relatively complex mixture of fluids solids and chemicals. The mixture is different for every case in order to achieve the right physical and chemical characteristics required to safely drill the well. It can also be water based or oil based. The first one is cheaper and doesn’t need any treatment on the other hand the oil based needs a ship to curry it onshore to treat it and bring it back for reinjection. So depending on the pressure or even better estimated well depth the correct drilling mud is selected and if oil based mud is selected a ship to curry the mud onshore-if the drilling is offshore-for treatment also must be rented. If we cannot find an available ship then we need to order a ship to be constructed especially for our case.

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b. Casing

We need to know the water depth in order to select the correct size and number of casings.

c. Cement

We need to calculate the correct amount of cement needed for cementing the casings selected. It also needs to have the correct specifications.

d. Well control

Primary, secondary and tertiary controls must be selected.

e. Platform type

What kind of platform should I choose depending on water depth etc. All the selected equipment above like (cement, casings, etc.) must have a room on the platform to be placed.

f. Drilling rig

We need to select the drilling rig based on Question 1a answer. We also need to have in mind the selection of power system (diesel engines, electrical generators, mechanical system, hoisting system etc.)

Apart from this important selection of equipment we need to have in mind that geologists and reservoir engineers must provide the correct subsea depth, longitude and latitude of the location etc. Cost must be estimated and an environment study with mitigation options to be available for the protection of both natural environment and human life.

Question2

A company wants to become an operator in UKCS and they want to apply for three licences. The conditions for the licences are that the operator – as a company- has to have a market cap that it is sufficient to cover the expenses of the proposed drilling programme. The government sets up the following criteria for the financial eligibility:

Market cap should be sufficient to cover:

a. 100% of the most expensive well

b. 50% of the cost of the other wells.

Due to the fact that the company wants to apply for three licences, which means to drill 3 wells, they contacted a drilling contractor to estimate the cost for a dry

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hole drilling. The drilling contractors submitted the figures shown to Appendix to the company.

The company had to review the figures and make comments on them since the Market Cap is 25 MM pounds.

Please determine if:

a. The costs submitted from the drilling contractor are sufficient to enable the company to apply for the three licences.

Market Cap = 25MM

For the Fumlar Target, ±13,725 ft.: 100%×22000000=22000000

For the Forties Target, ±6,000 ft.:50%×6350000=3175000

For the Forties Target, ±8,200 ft.: 50%×7500000=3750000

Total=22000000+3175000+3750000=38000000=£ 28.925MM

The costs submitted from the drilling contractor are not sufficient to enable the company to apply for the three licenses because they exceed the market cap by 3.925MM

2b. In the case that the cost is near to the market cap value or exceed it, please propose ways that the cost could be reduced considering data given in Kells environmental statement report.

Some ways to decrease the cost in order to be equal with the market cap are the following:

1. The pricing for every kind of cost stated on the tables are budgetary cost estimates and not a proposal to conduct turnkey operations. A way to reduce the cost, is by estimating the actual cost on a specific type of contract, or even better a different contract for every step of drilling.

2. As stated on the tables ‘’All drilling tangibles supplied by ADTI’’. Since tangible costs are borne by the general partner (manager) while intangible costs are borne by the limited partners (investors), usually to be taken as tax deductions. In the event of a dry hole, however, all costs become intangibles. So with these tax deductions we decrease the market cap even more.

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3. Try to look for alternatives. For example the rig rate has a value of 310000 dollar/day. Is there another cheaper rig available for the specific well that can also meet the requirements? A jack up rig maybe that costs less? Another point with the same way of thinking is the casings. Are there any alternatives? What else is it possible to be changed? Maybe the drilling mud? Is it possible to use the cheaper water based mud? If oil based mud must be used because the well is deep, is it possible to use water based mud and switch afterwards to oil based?

4. Study the market. Try to find other similar estimations made in the past for other wells or use already completed contracts in order to debate the estimations that were given to you. Maybe this is not their final price but the asking starting price.

5. The final point is to try to reduce the working days. Because everything is based on working days it is essential to try to find ways to reduce it.

2c. Define the main parameters that made the cost reduced and explain your approach

The main parameters that made the cost reduced as they did in other cases are the following:

1) Contracts (how to choose the cheaper ones)2) Costs that must not be included in the offer must be subtracted.3) Equipment alternatives (cheaper choice of equipment)4) Better negotiation (in order to get a better price)5) Working days (vital part of the total cost)

2d. Based on your proposed approach, is it possible for the company to apply for the licensing round based on the new data?

Since the ways to drop the final cost are multiple I believe that it is very important to get the new reviewed tables and calculate the new decreased working plan. Also since the deeper well will be 100% included in the market cap special attention must be put there. Also if the final price is not acceptable then it isn’t in the best interest of the drilling contractors not to take the proposed job.

Question 3:

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Whilst drilling a 12 1/4" hole section of the new well the following drilling data is being recorded and provided to the company man. At what point in time would you have suggested that the bit be pulled and why? Assume an average trip time of 8 hrs. , a rig rate of £400/hr. and the bit type selected above had been run in hole. Decision based on the cost of the bit.

TIME ON BOTTOM (hrs)

FOOTAGE DRILLED (ft)

1 342 623 864 1105 1266 1547 1808 2109 216

10 22611 23412 240

C=Cb+(Rt+T t)×C r

F

Where:

C = overall cost per foot (£/foot)

Cb = cost of bit (£), assume 500, 1000, 2000

Rt = rotating time with bit on bottom (hrs)

T t = round trip time (hrs)

C r = cost of operating rig (£/hrs)

F = footage drilled (ft)

Calculations example for bit cost = £500

C=Cb+(Rt+Tt )×Cr

F=500+(1+8)×400

34=120.59 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (2+8 )×400

62=72.58 £

foot

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C=Cb+ (Rt+Tt )×Cr

F=500+ (3+8 )×400

86=56.98 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (4+8 )×400

110=48.18 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (5+8 )×400

126=45.24 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (6+8 )×400

154=39.61 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (7+8 )×400

180=36.11 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (8+8 )×400

210=32.86 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (9+8 )×400

216=33.8 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (10+8 )×400

226=34.07 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (11+8 )×400

234=34.61 £

foot

C=Cb+ (Rt+Tt )×Cr

F=500+ (12+8 )×400

240=35.42 £

foot

Results for all the bit costs:

Cost per footFor bit cost 500

(£/ft)

Cost per footFor bit cost 1000

(£/ft)

Cost per footFor bit cost 2000

(£/ft)120.59 135.2972,5856,9848,1845,2439,6136,1132,8633,8034,0734,6235,42

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Based on the above calculations, we conclude that the point in time that the bit must

be pulled off is at 8 hours, because at that point we have the smallest amount of

overall cost per foot.

Question 4:

4a.Plot the following pore pressure/depth information on a Pressure - depth diagram (pressure x axis, depth at y axis).

DEPTH BELOW DRILLFLOOR(ft)

PRESSURE(psi)

0 01000 4655000 23258000 37208500 68009000 68509500 6900

0 1000 2000 3000 4000 5000 6000 7000 80000

100020003000400050006000700080009000

10000

Pressure vs Depth

4b.Calculate the pore pressure gradients in the formations from surface to: 8000ft, 8500ft, 9500ft.

Gradient pressure=psift

1) Gradient pressure=372008000

=0.465 psift

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2) Gradient pressure=68008500

=0.80 psift

3) Gradient pressure=69009500

=0.726 psift

4c.Determine the mud weight (in ppg) required to drill the hole section down to the following depth assuming that a maximum of 200 psi overbalance on the formation pore pressure is required.

P=0.052×MW×TVD⇒

MW= P0.052×TVD

Where:

P = Hydrostatic Pressure

0.052 = gravitational force

MW = average fluid density

TVD = true vertical depth

1) MW= 39200.052×8000

⟹MW=9.423 ppg

2) MW= 70000.052×8500

⟹MW=15.837 ppg

3) MW= 71000.052×9500

⟹MW=15.171 ppg

4d. If the mud weight used to drill down to 8000ft were used to drill into the formation pressures at 8500ft, what would be the over/underbalance on the formation pore pressure at this depth?

o At 8000 ft

Mud pressure:

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MW=9,423ppg (from part 4c1)

P=0.052×MW×TVD⇒

P=0.052×9.423×8000=3918.72

The formation pressure is 3720 psiAt this case the well is overbalancedMud pressure > formation pressureOverbalance by 198,72 psi We are very close to the maximum of 200 psi overbalance on the formation pore pressure but since we are less we assume that we don’t have a problem.

o At 8500 ft

Mud pressure:

MW=9,423ppg (from part 4c1)

P=0.052×MW×TVD⇒

P=0.052×9.423×8500=4164.966

The formation pressure is 6800 psiAt this case the well is underbalancedMud pressure < formation pressureUnderbalanced by 2635.034 psi

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