Exercise 1:
19
This note contains exams inTPG4205 DRILLING TECHNIQUES PRESSURE CONTROL:Exam 2004
Exam 2005
Exam 2006
Exam 2007
Exam 2008
Exam 2009
Exam 2010
Exam 2011
Exam 2012
Exam 2013
and
Solution Exam 2004
Solution Exam 2005
Solution Exam 2006
Solution Exam 2007
Solution Exam 2008
Solution Exam 2009
Solution Exam 2010
Solution Exam 2011
Solution Exam 2012
Solution Exam 2013
NORWEGIAN UNIVERSITY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF PETROLEUM ENGINEERING
AND APPLIED GEOPHYSICS
Contact person during exam:Name:Pl Skalle
Tel.: 94929 (office), or 91897303 (mobile) Date for censorship: Latest three weeks after exam dateEXAM IN COURSE TPG4205 DRILLING TECHNIQUES PRESSURE CONTROLTuesday, June 1st, 2004
Time: 0900 - 1300
Examination supportD:No written or handwritten examination support materials are permitted. Certain, specified calculator are permitted. Additional information found on last page./
NB! This headline repeats itself every year and is not included for later exams. The same is true for Additional Information. Such details are presented only for this first exam, 2004.1. Overlay curvea) Make an overlay curve to the dc data in figure 1. Do it only for two equivalent pore pressures; 1.1 kg/l and 1.2 kg/l. In what depth do you find the equivalent of 1.2 kg/l pore pressure? The example is from onshore drilling where overburden gradient is assumed constant 2.5 kg/l.
Figure 1: dc vs depth
b) Explain why the pore pressure may be different in two different sedimentary, on shore formations but at identical depth.
2.Drillers modified ( moved to exercises3.Water flow( moved to exercisesAdditional information
ANN = annulus
DS = drill string
SCP = slow circulating pressure
SCR= slow circulating rate
SIDPP= shut in drill pipe pressure
SICP = shut in casing pressure
SPM = stroke per minute
SWF = shallow water flow
T= temperature
W & W = wait and weight
Z= compressibility factor of gas
Exam TPG4205, 2005Question 1 ( moved to exercisesQuestion 2 ( moved to exercisesQuestion 3 ( moved to exercisesExam TPG4205, 2006.1. Formation pressure1.1Explain what normal formation pressure is.
There are several reasons for abnormal pressure; explain briefly the following characteristics:
1.2Artesian
1.3Under compaction
1.4Clay diagenesis
1.5Tectonic2. Pressure detection
2.1Explain how the following parameters indicate the well may be drilling into over pressured zones:
Drilling rate
Mud gas
Torque / drag
d-exponent2.2Explain 3 indications that the well actually is in under balance.3. LOT ( moved to exercises4. Primary Well Control ( moved to exercises5.SCR ( moved to exercises6.Gas ( moved to exercises7.Kick Calculation ( moved to exercisesExam PTG4205, 2007.Task 1
Static (not Dynamic) Hold Down
a)Explain the term.
b)What effect does it have on the drilling operation?
c)Can drilling engineers take advantage of this knowledge in any way?
Task 2 ( moved to exercisesTask 3 ( moved to exercisesTask 4 ( moved to exercisesExam TPG 4205 pressure control 2008Task 1: Well Control and SCR ( moved to exercises Task2: Drilling fluid transport modelDuring underbalanced drilling (UBD) the reservoir section is drilled with water and produced gas. Find approximate gas production rate by assuming constant well pressure gradient every 100 m. Other data as follows;
:
1000
m
(constant) :0.1
kg/l
:
1.03
kg/l
0.11
bar/m
20
bar
1000
l/min.
:
22
l/m
=
1.2 vm + 0.2m/s
Cg=
vgs / vg
vg=
qg / A
vg =
m fiksesHints: In order to find the answer we need to guess a surface pressure gradient and find the bottom hole pressure through iteration. qg is related to standard conditions, i.e. 1 bar. Our first guess of flow rate is 0.5 sm3/m. If time does not allow you to find the exact answer; just stop when you feel you have shown how to do it.Task 3: Presented previously ( moved to exercisesTask 4: Miscellaneous
a. Write down the objective of the cementing jobs. Describe in a steps wise fashion the procedure of a squeeze cementing job.b. Shallow water flow and Gas Migration through cement are problematic processes. Define the problem for each of the processes and suggest solutions respectively (og foresl lsning for de respektive problemene).
c. If you need more casing points what will be the technical problem for drilling in deep well besides costs?
d. Gas transport: Make a short note on:
Axial dispersion of two phase flow
Stability criteria of two phase fluid flow through annulus
Exam TPG4205, 2009.Question 1. ( moved to exercisesQuestion 2. Modified Drillers ( moved to exercisesQuestion 3. Pressure detection ( moved to exercisesExam TPG4205, 20101. Pore pressure
a. ( moved to exercisesb. How is it possible to establish a normal trend for parameters in general with respect to estimating pore pressure? The parameters in question are established in sedimentary formations. No equations are necessary.2. Fracture pressure
a + b ( moved to exercises c. Define all parameters in the two equations below and explain especially why the Poisson ratio is varying between approximately 0.25 and 0.5 in sedimentary rocks.
3. Killing Equipmenta. How is it possible to record the SIDPP when a flapper valve is installed in the drill string above the bit?
b. Explain how a 4 opening/ 3 position / type valve inside the Pod (fordelingboks) is operated? Two Pods are installed on the subsea BOP.
4a. Engineers method Friction in annulus ignored ( moved to exercises4b. Modified Engineers ( moved to exercises5. Killing a gas wellA gas producing well is planned to be worked over (bli reparert), and need first to be killed. Explain stepwise how to kill such type of wells.6. Real gas velocity ( moved to exercisesExam TPG 4205, 20111. Formation pressure
a. What geological and material parameters are determining the magnitude of the equivalent pore pressure density (how far above 1.03)?
b. ( moved to exercises2. Conventional killing ( moved to exercises3. Unconventional killing ( moved to exercises4. Deep water issues
a. ( moved to exercisesb. Explain how the casing cement sheet can start leaking at any time from after its initial set point and during the production phase.
c. Can anything be done about the leaking prospect in question b?
Exam TPG4205, 2012Task 1. Pore pressure
The equivalent pore pressure density is found to be typically 1.73 kg/l. Discuss which parameters and which geological processes that have resulted in this number.
Task 2. Poissons ratioa. Prove that the Poissons ratio is equal to 0.5 for elastic materials for small (1 %) deformations (compression). Use a cylinder with diameter d and height h for this purpose.
b. Why is Poissons ratio < 0.5 for sedimentary rocks?
Task 3. Safety margin (SF)a. What is the useful information you get out of MAASP and Kick Tolerance at shut-in during drilling? How are the two SM different?
b. What is Kick Margin and why is it applied?
Task 4. The Drillers MethodThe kick and well data experienced during drilling from a floating platform into a high pressure zones are given here:
TVD / MD=2500 m / 3500 m
TVDcsg / MDcsgWater depth==1500 m / 1 510 m
500 m
SCP = psirc @ 30 spm=52 bar when circulated through the riser
SCP = psirc @ 30 spm=62 bar when circulated through the choke line
mud=1.46 kg/l
pSIDP=20 bar
pSIC=27 bar
Vkick=4.3 m3
Cap pump=20 l/stroke
Cap Ann=20.0 l/m (assumed constant throughout the annulus)
Cap DS=10.0 l /m (0.01 m3 / m)
p LO at shoe=45 bar
ax2 + bx + c=0
x=
a. Sketch the stand pipe pressure (SPP) vs. # of strokes for the complete operation. With respect to annular pressure, give an approximate estimate of its pressure when gas reaches the surface (assume gas behave ideally and follows the mud and no friction during this estimation).
b. What are the SPP and the choke pressure after just having completed filling the drill pipe with kill mud, stopped the pump and closed BOP / choke?
c. Estimate the real gas densityTask 5. Volumetric MethodThe 2 000 m vertical 12 well has to be shut in during tripping out and a pressure of 5 bars is recorded. Experience allows us to assume that the gas is positioned at the depth of 1 500 m. Mud density is 1.2 kg/l, balancing the pore pressure. Please estimate the process up to the second choke bleed-off operations. Assume a capacity of 30 l/m throughout the wellbore, a gas percolating velocity of 0.25 m/s and weightless gas. Bleed-off volume is 500 l. Use a safety margin of 5 bars. Estimate before and after release: Vertical position, volume and local pressure and surface pressure vs. time. Each bleed-off takes 10 s.Task 6. Cementing technologyWhat could be the reason behind all the leaking production wells now a day, on- and offshore?Exam TPG4205, 20131. Pressures in the sediments
Give a physical explanation of why the pressure gradients typically vary between the two extremes as indicated below. Gradients are listed in terms of equivalent density (kg/l). Support your discussion of the two extremes by mathematical relationships if appropriate:a. Pore pressure gradient:
1.025 - frac
Fracture gradient (Eaton).1.33 - (ovb - pore)
Overburden gradient:
2.0 - b. The effect of sea water depth can be commented separately.
During drilling it is essential to detect high pressure zones. We know that the zones porosity has high importance for this task.c. Why is porosity so high in high pressure zones?d. Select 5 methods or tools for pore pressure detection and explain principally how the formations porosity influences the result.
2. Pre-killing considerations
Refer to Appendix for details of an exploration well. Estimate the following factors:
a. MAASP, Riser Margin and Kick Tolerance
b. Discuss the possibility of elevating the fracture pressure in the next wellbore section, in order to improve the margins in question a.
3. Killing operations
On basis of the information stated in Appendix, please compare three different killing methods. Make a drawing of the process of filling the DS with kill mud; SPP vs. strokes (assume ideal gas which follows the mud):a. Standard Drillers method (friction in annulus is ignored).
b. Modified Drillers. Sketch the complete killing process; pump pressure vs. strokes, from the start of pump, at minus 10 strokes, until kill mud comes out of the annulus
c. Modify the Standard Drillers with respect to geometry of the wellbore. Assume that 50 % of the friction occurs through the bit nozzles. For this case a 500 m long HWDP starts 1000 m down in the Drill String (DS), with the same OD as the DS, but the ID is smaller. Assume further, for simplicity, that the pipe friction in the Drill String is linearly distributed over its entire length, and that the friction is the same as in task a. and b. above.
d. Compare the two methods in a. and b. with the volumetric method. No calculations beyond the bottomhole pressure at the start of the method are needed. Include a safety factor of 5 bars for all the three methods. Use here the Depth Pressure graph. Create the three different pressure paths which the kick fluid has to follow.4. Drilling through a gas reservoir
While drilling slightly overbalanced, horizontally through a gas reservoir, the ROP was in average equal to 30 m/h. More details of the operation is found in the Appendix. Assuming these conditions have been stable and constant for several hours. Your task is to present a method to determine the surface choke pressure during drilling. Use the Newton-Raphson forward iteration method.
Solutions:Solution Exam TPG4205, 20041. Overlay curve
a)Assume ppore, normal = 1.03 kg/l.
Eaton: ppore = povb (povb pp,n)(dc/dc,n)1.2Select only two depths since povb is constant; 500 m and 1250 m these two depths
dc,n = 1.0 and 1.75 respectively.
and find dc to become
Gp500 m1250 m
1.10.961.68
1.20.9031.58
Pore pressure of 1.2 is found in 850 m.
b)The sealing mechanisms have been/are different.
Solution Exam TPG4205, 2005.
Solution Exam TPG4205, 2006
1.1Typically water grad
1.2Water level higher
1.3Quick sedimentation of clay
1.4Water expelled, diagenesis
1.5Plate tectonic
2.1ROP Expelling cuttings, hold down pressures
Swabbed at cnx
Gauge hole etc
Deviates from trend
2.2Change in flow rate, change in pit volume and sidpp
Solution Exam TPG4205, 2007
1.a.pmud - ppore
b.Lower ROP
c.Estimate pporeSolution Exam TPG4205, 2008
Task 2
In order to find the answer we need to guess a pressure gradient and improve it though iteration. Qg is related to standard conditions, i.e. 1 bar. Our first guess of flow rate is
0.5 sm3/m.
First guess leads to these parameters:
One step further gives a new gradient (i.e. density) 2When by gradually changing qg, we have found the correct answer.
Task 4: Miscellaneous
a)Objective Protect and support the casing
Prevent the movement of fluid through the annular space outside the casing
Stop the movement of fluid into vugular or fractured formations and
Close an abandoned portion of the well
Squeeze cement
Set the bottom retainer plug just under the zone of interest for squeeze cement
Perforate casing at this zone
Set the top retainer plug above zone. DP set in TOP retainer plug
Pump high hydraulic pressure cement through DP
Pump HCl + HF if needed
Fracture zone of interest
Wait minimum 24 hours
Conduct cement evaluation
Take decision wither its ok or not
b)SWF = shallow water flow
Problem: Water starts flowing from pressurized sands and erode sands, cement and fractures. This causes casing and fixed platforms to collapse, due to loss of support.
Solution: Drill pilot hole without riser, kill dynamically if kick, use kill mud if sand is penetrated, drill to planned depth. Drill extremely carefully with sonar and regular checks for flow. Avoid such areas when detected through radar surveys.
GM
Problem: Cement shrinks and sucks water and gas into cement which then may break through the hydrating cement, erode weak cement, makes worm holes and causes blowouts.
Solution: Avoid suction of gas by a) replace shrinking material (cement) by other no shrinking materials, b) use dispersed nitrogen which will expand on pressure reduction, c) displace gas from near well bore formation before cementing, i.e. use best practice cementing technique.
c)
Top hole can be drilled with only ONE barrier. Potential shallow gas should NOT be penetrated
Possibility for moving installation in case of blowout
Pilot hole shall be drilled through all shallow gas zones
Predicted shallow gas should be drilled with weighted mud
Possibility to kill the pilot hole dynamically
Putting float valve in the BHA
Potential shallow gas zones should be logged with LWD
Returns to sea bed should be observed by ROV
Back-up kill fluid should be available
Solution Exam TPG4205, 2009
Solution Exam TPG4205, 2010
3. Killing Equipment
a.Pump until it opens, seen by the sudden increase in SICP. Or a small hole in the flap. Or use PWD.b.Close upper pipe valve sends a pilot signal to a valve which opens for high pressure oil to the closing piston.
5. Killing a gas well
1. Replace gas with water through lubricate & blood techniques
2. Kill well with standard methods
Solution Exam TPG4205, 2011.
1. Formaton Pressure
a.Figure 1 explains the dynamics. Ovb is compressing the sediments. When a seal exist, compression stops and pressure increases, depending on Ovb. and tightness. Latter is defined by Darcys law. max p = frac.
b. ( moved to exercises4. Deep water issue
a.( moved to exercisesb.Shrinks and have wormholes, exposed to mechanical and temperature induced stresses.
c.Best practice cementing, additives to avoid shrinkage (expanding add.), and to withstand stresses (elastic add.).
Solution Exam TPG4205, 2012Task 1
Geological process: To establish a high pore pressure there are two prerequisites; a reservoir or a sand where the pore pressure can be measured, and a seal that can hold pressure in excess of normal water pressure. The seal is the important factor. It must seal both vertically and in the horizontal direction. In addition to discussing seal mechanisms, it is necessary to state that its ability to hold excess pressure over long periods, the Darcy law is describing this ability. Solve Darcy eqn. with respect to p.
Task 2
a. Poissons Ratio = = 0.5 =
Vo =
After compression the height is reduced to 0.99 h, and the diameter has expanded to d x, where x is the expansion fraction. Should be 1.005 because then:
0.5The compressed volume is therefore:Vc = Letting Vo = Vc we obtain:
= 1.005
b. At a low burial depth the sediments are not behaving elastic, like metamorphic rocks at high depths. At low depths part of the deformation is absorbed by reorientation of the material/reduction of porosity.
Task 3
a. MAASP is indicating the max surface pressure left before fracturing at the shoe. Kick tolerance (KT) is the max kick volume before fracture occurs when the gas kick reaches the shoe. KT looks ahead, MAASP does not.
b. Kick Margin indicates the safety against kicks. The purpose of it is identical with KT, but applies a constant, guessed value of the kick volume, not an exact value like for KT.Task 4
ICP = SIDPP + SCP = 20 + 52 = 72 bars (subtract 10 bar of choke line friction)
FCP = 52 ( 1.54 / 1.46 = 54.8 bars
Strokes to fill DP (and ANN) = strDP = = 1 750 str (3 500 str)
Pressure when gas reaches the surface. (Assume no friction for this part of the evaluation).
ppore = 1460 ( 9.81 ( 2500 + 20 ( 105 = 378 ( 105psurf mud( g (TVD hgas,s) = ppore
hgas ( Capann ( psurf = h1 ( Capann ( pporeh1 ( Capann = 4.3 m3h1 = = 215 mhgas ( psurf = h1 ( ppore ( hgas = psurf - ( TVD h1 ( ppore / psurf ) = pporeps2 ps ps2 ps ( 358 + 378) ( 105 + 8.127 ( 109 = 0
b. Shut in and stop pump: SIDPP = 0, SICP = 20 bar
c. Real gas density
Task 5
ppore = 1200 ( 9.81 ( 2000 = 235.4 ( 105 PaInitial gas volume = Vo
Vo = hgas ( Cap = 42.5 ( 30 = 1274 l
Pressure increases corresponding to 500 l bled off
Initially we let the pressure increase 5 + 2 bars
This corresponds to a height of
1. Which takes 59.5 m / 0.25 m/s = 238 s (5 + 5 + 2)
2. Bled off 1
10s
3. Next 2 bars increase 238 ( = 64 s
4. Bled off 2
10s
After 1. step the volume of the bubble is
V1 ( p1 = Vo ( p1500 V1 = Vo = 1.27 m
After step 2 the volume of the bubble is
V2 p1 = V1 ( p1
V2 ( 1200 ( 9.81 ( 1440 = 1.27 ( 1200 ( 9.81 ( 1500
234
V2 = 1.27 ( Pressure after end of step 2 can now be estimated.
Task 6
The explanation and the solution to this is found in the Text book Chapter 8.3.3Solution P-control exam 20131. a. Pore pressure is normal as long as permeability up to surface allows. Due to closures it may go up till fracture. Magnitude between the 2 boundaries is determined by the overburden (up) and Darcy flow (down).
Fracture pressure is determined by Eaton:
At shallow depths Where
At large depth Overburden matrix loses porosity in the depths, matrix = 2.7 kg/l
b. Large water depth brings frac QUOTE and pore relatively closer together
c. Water is trapped and incompressible. Porosity stays at the level it was when trapped
d.
dc
lower compressive strength (also)
R
high salt water content
Sonic
lower density
Neutron Porosity log
ROP
same as dc
Temperaturemore water higher insulation
2. a. MAASP = LO = QUOTE (45 bar)Riser margin
= 1560 kg/l
Find the balancing fluid:
balance = 1730 kg / m3Riser margin = balance - kill = 170 kg / m3Kick Tolerance:
hgas = 174 m
Kick size when at casing shoe = hgas Capann = 174 0.02 = 2.5 m3
Kick Tolerance = kick size when at bottom = 2.5 * pcsg / ppore = 2.5 * 188 / 306.4 = 1.5 m3c. Increase cohesion with chemicals
Wellbore stress augmentation through generation of small fractures
Increase fracture propagation resistance with filter cake (WBM)
Need to adjust surface mud cleaning process3.a. VDS = 2000 m 0.01 = 20 m3StrokesDS = V/Cappump = 20/0.02 = 1000 strokes or 1000 / 30 = 33.3 minutesICP = SIDPP + SCP = 20 + 62 = 82 bar
FCP = SCP * killmud = 62 * 1.56 / 1.46 = 66.2 bar
(2013: Sensor did not reduce your score ir you had 72 as ICP here because the text was unclear in 2013) b. ICP = SIDPP + SCPup riser = 20 + 52 = 72 bar
FCP = FCPstandard (62 52) = 66.2 10 = 56.2 bar
VANN = 1500 0.02 = 30 m2Vch.line = 500 0.01 = 5 m3Strokes = 30/0.02 + 5/0.02 = 1500 + 250 = 1750 c. Friction is
Strokes to fill Drill String = strokes
TVDMDStrokesSPP
001ICP = 82
1000100050082 10 + 1.05 = 73
1500150062582 15 + 1.58 = 68.6
2000200087482 20 + 2.1 = 64.1
After bit 66.2
d. Compare methods a and b with volumetric. Use here the Depth Pressure graph. Create the three different pressure paths which the kick fluid has to follow.
(2013: All those of you who explained the volumetric method and compared it somehow to a and b got full score from the sensor, since the text was unclear. It got low weight)
4. Initial gas production qgas,o = ROP A
qmud = 30 SPM * 20 l/stroke /(1000 * 60) = 0.01 m3 /s
Initial mixture velocity at bottom: vmix,0 = (qgas,0 + q mud)/ A = 0.00106 ? 0.01) / 0.076 = 0.146 m/s
You must assume a psurf from a guessed average reduced mixture density, e.g. 1.2 kg /l. This gives a psurf of 40 bar. On basis of this assumption Cgas is estimated, and it is possible to estimate the initial mixture density at the surface and accordingly the initial pressure at surface; p0 m,0p0 m,0 s
Estimate p10 mEstimate p20 m
Find average gradient (p0 m + p20 m ) / 2
Iterate until p10 m is acceptable
Add h
Continue till bottom pressure is found; pbottom
If pbottom is different from ppore then add p and iterate from top to bottom until pbottom = ppore 500 m
750
1000
1250
dc
0.5 1.0 1.5 2.0
_1403432117.unknown
_1403432121.unknown
_1403432260.unknown
_1442834062.vsdStandard Drillers Modified Geometrically Modified Drillers
SPP
80
75
70
65
60
55
0 200 400 600 800 1000
Strokes
_1442834063.vsdStandard Drillers Volumetric Modified Drillers
25 -102525 +10
0 50 100 150 200 250 300
0
500
1000
2000
306.4 + 5
306.4 + 5 + 10
_1442834061.vsd0
500
1000
2000
_1403432122.unknown
_1403432123.unknown
_1403432119.unknown
_1403432120.unknown
_1403432118.unknown
_1335783789.unknown
_1403432115.unknown
_1403432116.unknown
_1335783824.unknown
_958814514.unknown
_1146049891.unknown
_1146049966.unknown
_958814517.unknown
_958814511.unknown
_958814510.unknown
