energy drilling prospects - myoilandgascareer.com...fraser offshore ltd is a drilling project...

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Nurturing skills for a newgeneration in Oil & Gas

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Energy Drilling ProspectsFraser Offshore Ltd is a drilling project management company. It designs, plans and drills oil wells for clients who are typically oil & gas companies or large utilities companies similar to British Gas and EDF Energy.

Your task is to use your mathematic skills to assist Fraser Offshore with the design of a new drilling project. In completing this project you will develop your skills in the following areas of mathematics:

n Number: adding: subtraction, multiplication, division, rounding and percentages, converting imperial to metric

n Algebra: linear equations, formulae, straight line graphs y = mx + c

n Shape & space: prism volumes, circle theorems, bearings, scaledrawings, angles, Pythagoras/trigonometry, coordinates, loci.

n Statistics: pie charts, scatter diagrams/line of best fit.

n Probability and averages are very easy to apply to oil andgas e.g. likelihood of discovery based on tabular data etc.

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TASk 1

We have been asked by an oil company to drill a well (“Prospect 1”) for them. In order to do so we need to move the drilling rig from its current position (“Prospect 2”) to the new well site (“Prospect 1”) as shown below

The client pays a day rate for each day that the rig is in their possession (this can be anything between £130k at present to £245k at peak times). When the rig is not drilling, a period known as downtime, the client is losing money. The client therefore wishes to expedite the move in order that drilling can commence as soon as possible and they have asked for an estimate of cost for the move to be produced. In order to help us provide them with a cost we require answers to the following items:

Energy Drilling Prospects

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Energy Drilling Prospects continued

a. What are the approximate coordinates of both drill sites; Prospect 1 and 2?

b. Now assume that the rig can move in a straight line, use measurements together with the scale at the bottom of the map (8.5cm:100m) to find what distance does it need to be moved.

c. Check and refine your measurement in part (b) by using Pythagoras and your coordinates to find a more accurate figure.

d. The rig moves at a rate of 100m per hour. Using your answer from (b) calculate the length of time that it will take to manoeuvre the rig to its new position. Remember s = d/t.

e. Given that the cost of moving the rig is given as £12,500 per 50m, calculate the cost of the rig move.

f. Two wrecks can be seen lying on the seabed near the well site at Prospect 2. The rig requires 50m of clear seabed around it in order to operate safely. Will the wrecks inhibit drilling operations?

g. Supply boats will be run twice a day to the rig, weather dependant. In order that they may navigate their way to the rig calculate the bearing of each wreck relative to Prospect 1.

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h. The heli-pad on the rig is circular with a radius of 7.5 metres. The largest helicopter used requires an area of 175m2 to land. Please confirm, or otherwise, if the heli-pad on the rig is large enough for this helicopter to land.

i. The rig team are considering reducing the operational spend by using even larger helicopters so that they may deploy greater numbers of staff more cheaply. This means that they must increase the diameter of the existing heli-pad by 50%. What will be the area of the new heli-pad?

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j. It is later found that there is a wreck exactly half way between Prospect 1 and Prospect 2 and that upon moving, the rig must remain more than 50m clear of the wreck at all times. Use loci, measurement, Pythagoras and circle formula to find the shortest path of movement and associated time and cost to move.

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k. It is later found that the radius of the heli-pad in part (h) was measured to the nearest 2 significant figures and the landing area required to the nearest 5m2. Considering this new information can we be certain that the heli-pad is large enough for the helicopter?

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well deSigN moNTAge

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TASk 2

Once the rig is safely in situ drilling can commence. Drilling through the seabed requires a number of different approaches dependent upon the geology that is encountered as some materials are much harder than others.

The attached ‘well design montage’ PAge Above details a lithology column which shows the expected depths of the different rock formations and how deep the sea is above the seabed (which is included in the total depth measurement).

Looking at the lithology column of the ‘well design montage’;

a. What percentage of the total depth (3000ft) is made up of the Mercia Mudstone Group? What are the percentages of the remaining other rock types?

b. Construct a pie chart demonstrating the quantity of the total depth that is occupied by sea water, Tertiary/Dowbridge Mudstone, Mercia Mudstone Group and Ormskirk Sandstone.

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The well is drilled in sections as shown in diagram b. As each hole is being drilled, casing (a cylindrical hollow pipe) is run down the hole and cement is injected around it to hold it in place and maintain the cavity. In this case the first drilled hole is 12 1/4 inches in diameter and the casing is 9 5/8 inches in diameter.

c. In order that we can know how much cement is required, convert the imperial measurements into metric diameters and calculate the circumference and area of each hole.

d. Calculate the volume of cement that would be required to fill the gap between the open hole and the 9 5/8” casing from a depth of 2436 feet to the sea bed 100ft from the top (see diagram A).


diAgRAm A

diAgRAm b

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The nature of cement that is used depends upon a variety of factors such as the depth of the well, the pressure that is encountered and the changes in temperature that occur with depth. Often it is necessary to add chemicals or other materials to the cement in order to make it stronger, when under pressure, or to prevent it from hardening too soon, as it is pumped into deep wells.

e. Use your answer to (d) to calculate how much of the following additives taken from the cement recipe would be required to mix the cement properly:

Additive (a) at 2.3 litres per cubic metre (answer in litres)

Additive (b) at 4.6 gallons per cubic metre (answer in gallons and convert to litres, remember 1 gallon = 4.5 litres).

Sometime during drilling a void is discovered which needs to be filled.

f. How much cement would be required if the drilling engineer requests that a 30% excess on the original volume (d) is required? What is the volume of this void? If a void of this volume were to be spherical, what would be the radius of the sphere? (Remember volume of sphere = 4/3 π r3)

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TASk 3

It is absolutely crucial in well design to know at what pressure the oil/gas is stored in the formation that is being drilled into. This in order that an escape of pressure (known as a ‘blow out’) through the well bore may be avoided. Once the cement has set, the well is drilled deeper by passing a smaller drilling assembly through the casing and into the new formation.

A new formation is drilled and tests are performed continuously to measure the pressure at each new depth (see attached information). Unfortunately, the Measure Whilst Drilling tool has provided erratic readings so the constant pressure cannot be determined. It is therefore necessary to plot a scattergraph of pressure vs depth to determine the line of best fit.

a. Use the information shown below to plot a scattergraph of depth (feet, x axis) vs pressure (psi, y axis) and include a line of best fit.

Pressure (psi)

450

1000

1250

1750

2500

3000

depth (ft)

3000

3500

4000

4500

5000

5500

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b. For your line of best fit, find its gradient and the point where it crosses the y axis. Use this to write the equation of your line of best fit in the form y=mx+c (where m is the gradient and c is the point where it crosses the y axis).

c. Use your line of best fit to estimate the pressure when the pressure is 4750ft


20000

1000

2000

3000

3000 4000 5000 6000

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The normal formation pressure gradient is given as 0.452 psi/ft. This means that, for every foot of depth dug by the well, the pressure increases by 0.452 psi (pounds per square inch).

d. What would the pressure be at a depth of 2000ft?

As the well is being drilled, mud is pumped down through the bore. Keeping the hole full of drilling mud prevents oil and gas from the formation from flowing into the drilled well. The mud must be kept at a higher pressure, and therefore pressure gradient, than the formation to prevent blow outs. It is usual for 10ppg (pounds per gallon) mud to be used.

To arrive at the pressure gradient of the mud, we multiply the mud weight (10ppg) by a constant conversion factor of 0.052.

e. Find the mud pressure gradient and then use this to check whether the well is safe by comparing with the formation pressure gradient of 0.452 psi/ft. Remember, for the well to be safe the mud pressure gradient must be higher than the formation pressure gradient otherwise a blowout may occur!

diAgRAm C

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Once drilling has commenced a survey shows that the well has been drilled at an angle of 3° to the vertical, as shown in diagram C. The well was supposed to have been drilled vertically (0°) in order to hit a target 30m deep and of horizontal radius 10m.

f. If the well is 3000ft deep, has it missed the target? Refer to attached diagram and note that the planned well was designed to hit the middle of the target.

g. What is maximum angle of error that the well can be drilled at in order to hit the target 3000ft deep?

h. Given that the well is drilled at 3° to the vertical, what is the greatest depth that the target area could be at for the well to be successful? And what is this as a percentage of the planned well depth?

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TASk 1 ANSweRS

a. Prospect 1 (13.9, 12.1), Prospect 2 (21.2, 17.1)

b. 15.2 ÷ 8.5 = 1.788 178.8m

c. √[(21.2-13.9)2 + (17.1-12.1)2] = 8.8 coord units = 14.96cm : 176m

d. 1hr 46 minutes

e. (176 ÷ 50) x 12,500 = 3.52 x 12,500 = £44,000

f. No, they don’t inhibit the move

g. (g – depends on position of rigs)

h. Area = π x 7.52 = 176.7m2 heli-pad is large enough for helicopter to land.

i. Scale factor = 1.5 area factor = 1.52 = 2.25.Therefore new area = 176.7 x 2.25 = 397.6m2

Energy Drilling Prospects answers

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j. Difficult! See ‘image 2’. Loci required:

Circle radius 4.25acm, centred on midpoint of Prospect1/Prospect2. Arcs on circle radius 6.18bcm from each Prospect.Straight lines from each Prospect to intersections of each arc with circleMeasure anglec of arc of circumference around which movement required

workings:a 50m clear of wreck, 8.5 cm : 100m 50m = 8.5cm ÷ 2 = 4.25cmb Right angled triangle required, hypotenuse end points on Prospect 1(or 2) and midpoint, shortest side = radius of circle. Pythagoras to find remaining side: √(7.62 – 4.252) = 6.18cm c Measure angle as 68°. Arc of circumference = (2 x 4.25 x π) x (68/360) = 5.04cm

Shortest movement distance =

6.18 + 5.04 +6.18 = 17.4cm

Using scale of 8.5cm : 100m 204.8m

k. Limits are radius min = 7.45m, Area required max = 177.5m2

Min poss area = π x 7.452 = 174.37m2

174.37 < 177.5, therefore not certain safe for landing.


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TASk 2 ANSweRS

a. Mercia mudstone: 3.3cm ÷ 14.7cm = 22.4%Tertiary & Dowbridge mudstone: 7 ÷ 14.7 = 47.6%Ormskirk sandstone: 3.4 ÷ 14.7 = 23.1%

b. Angles for pie chart are:Mercia mudstone: 22.4% x 360 = 81°Tertiary & Dowbridge mudstone: 47.6% x 360 = 171°Ormskirk sandstone: 23.1% x 360 = 83°Remaining angle for sea = 360-81-171-83 = 25° (6.9%)

c. 12 1/4 inches x 2.54 = 31.115cm, area = 760cm2, circumference = 97.57cm9 5/8 inches x 2.54 = 24.4475 cm, area = 469cm2, circumference = 76.8cm

d. 2436 – 100 = 2336ft = 28032 inches = 71,201cmVolume = (760-469) x 71201 = 20,719,572cm3 = 20.72m3 (Better method is to convert area into metres first)

e. Additive (a): 2.3 x 20.72 = 47.656ltAdditive (b): 4.6 x 20.72 = 95.312gallons = 428.904lt

f. Additional 30% excess: = 1.3 x 20.72 = 26.936m3

Volume of void: 0.3 x 20.72 = 6.216m3

Radius of sphere = 3√[(6.216 x 3) ÷ (4π)] = 1.14m


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TASk 3 ANSweRS

a) See image below:

b. Gradient = (3000-450) ÷ (5500-3000) =2550 ÷ 2500 = 1.02

Crosses y axis: approx -2650

Equation: y=1.02x – 2650

(Equation of y on x regression line: y=1.014x – 2652)

c. Pressure = 4750 x 1.014 – 2652 = 2164.5psi

d. 0.452 x 2000 = 904psi

e. 10 x 0.052= 0.52.

0.52 > 0.452 well is safe.


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f. 3000ft = 914.4m

Using trigonometry, opp = hyp x sinθopp = 914.4 x sin3 = 47.86m

Well target was circle of radius 5m, 47.86 > 5 well missed target (by a long way!)

g. Using trigonometry, tanθ = opp ÷ adj

tan-1(5 ÷ 914.4) = 0.31°

h. Using trigonometry, hyp = opp ÷ sinθhyp = 5 ÷ sin3 = 95.5m = 313.4ft

313.4 / 3000 = 10.4%


energy drilling prospects - myoilandgascareer.com...fraser offshore ltd is a drilling project...

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