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Project done during Completions Engineering Course. Production casing string was designed based on casing material properties and various potential failure scenarios
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Daevin Dev 2015 PEGN 361 Project #1
Due: Friday, February 27, 2015 at 23:59 via Blackboard
This page will be the cover page for your project.
Name: DAEVIN DEV
Determine and document the following for a production casing string;
β’ Design equation(s) and graph(s) for collapse conditions
β’ Design equation(s) and graph(s) for burst conditions
β’ Design equation(s) and graph(s) for tensile loading
Design and select the least expensive 5-1/2β production casing string that does not fail.
Do not use more than four sections. You can use fewer sections if it is the least expensive string Report
your design in the following table for the production casing string (top to bottom as in the
wellbore):
Weight (lbf/ft) Grade Connection Bottom depth (ft) Top depth (ft) Cost ($)
23 N80 LTC 11500 0 288,535
20 Q125 LTC 11500 12000 17,045
- - - - - -
- - - - - -
Total Cost $305,580
Do not use any casing other than what is listed in the inventory.
Deliverables:
β’ Write a one page memo addressed to Dr. Ermila and myself describing the situations and scenarios
and the selection of casing string. It should be in a narrative format.
β’ Four graphs
o One for all collapse scenarios
o One for all burst scenarios
o One for all tensile scenarios (pipe body)
o One for all tensile scenarios (joint strength)
β’ Show your casing selection collapse, burst, and tensile capabilities on your respective graphs.
β’ Be certain to show your work.
β’ Show and describe all design criteria equations used
o Add sample calculations.
MEMO
To : Dr. Ermila & Dr. Eustes
From : Daevin Dev
Subject : Production Casing String Design Criteria
Date : 2/27/2015
An inexpensive and safe 5.5in production casing string was to be designed based on the casing material properties and several
potential failure scenarios. A table of available pipe inventory was used as the main basis of comparison and reference for t he
surface material properties. Three cases were considered as potential failure events. These were collapse, burst and tensile. Each
case was then analyzed based on two of the worst possible scenarios that could cause the casing to fail by the specified case.
Based on the analysis performed, the final casing design is estimated to cost $305,580 in material costs.
The two worst possible cases for which the casing could collapse is when the casing is evacuated and during a cementing job. In
an evacuated casing, there isnβt any fluid inside the casing but the last run mud weight, 16.5ppg in this case, is in the annulus of
the production casing. This means that there is an outer radial pressure acting inwards on the casing string and no backup lo ad to
mitigate the outer pressure. As for the cementing job, different types of fluids will be present in the annulus and one type of fluid
will be present on the inside. In our scenario, cement, spacer fluid and original mud were present in the annulus and
displacement fluid (brine) was present on the inside. Because of the varying fluid types, the casing strength requirements vary at
the different fluid depth intervals. Figure 1 graphically i l lustrates this scenario. From the graph, it is apparent that the evacuated
casing scenario has a higher casing strength requirement than the cementing scenario. Next corresponding real tension
equations for both the aforementioned scenarios were determined and its corresponding adjusted collapse pressure rating was
computed.
The two worst possible cases for which burst could occur is when there is a leak in the production tubing (especially near the
hanger) and when a pressure test is done on the casing. Both these cases account for the situation in which the fluid density on
the inside of the casing is significantly higher than the fluid density in the annulus, thus creating an outward pressure differential
that could burst the casing. In the case of leaky tubing, the completions fluid and leaked gas will exert an outward pressure on
the casing. The leaked gas pressure is exacerbated by the reservoir pressure which pushes the gas out of the tubing and into the
annulus of the tubing. The backup load here is assumed to be connate water (0.465psi/ft). In the case of pressure testing, a well
head pressure is applied onto a predetermined mud weight with the same backup load as the previous scenario. Figure 2
i l lustrates these situations.
As for tensile scenarios, the two worst cases were assumed to occur when the original mud was on the inside & outside of the
casing and when a pressure test was being done. Real tension equations were computed for both scenarios and a corresponding
pipe body strength and joint strength value were computed by multiplying the real tension by its required design factor and then
adding that product to the required margin of overpull. A few assumptions were made. First, the margin of overpull was added to
the minimum casing strength requirements. (Product solved before addition) Second, the outside mud weight for the pressure
test was assumed to be 16.5ppg.
The initial casing design selection was done by plotting all the minimum casing strength requirements , Sc (Figures 1 through 4)
and then referencing the inventory table for a casing that satisfied all the minimum requirements for collapse, burst, pipe b ody
and joint strength. The initial casing design was a 5.5in N80 23ppf LTC case which extended from surfac e to 11,500ft followed by
a 5.5in Q125 20ppf LTC which extended from 11,500ft to 12,000ft. After analyzing this initial design with the biaxial stress effects,
it was deduced that the initial design could be used as the final design. See sample calculations for computations performed to
arrive at the aforementioned conclusion of casing design.
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
11000
12000
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 11000 12000 13000 14000
De
pth
, D (
ft)
Minimum Casing Strength, Sc (psi)
Figure 1: Minimum Casing Strength Requirements For Collapse.
Evacuated Casing Minimum Strength Requirement
Cement Job Minimum Strength Requirement
Casing Collapse Strength:Cement Job
Casing Collapse Strength:Evacuated Casing
Uniaxial Stress
Different Scenarios
Worst Case
Ideal Case
Neutral Point Of Tension & Compression
Neutral Point Of Tension &
Compression
0
1,000
2,000
3,000
4,000
5,000
6,000
7,000
8,000
9,000
10,000
11,000
12,000
0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000 10,000 11,000 12,000 13,000 14,000
De
pth
, D (
ft)
Minimum Casing Strength, Sc (psi)
Figure 2: Minimum Casing Strength For Burst Prevention.
Leaky Tubing
Pressure Testing
Casing Burst Strength
Different Scenarios
0 100,000 200,000 300,000 400,000 500,000 600,000 700,000
De
pth
, D (
ft)
Minimum Casing Strength, Sc (psi)
Figure 3: Minimum Casing Strength Requirements Based On Pipe Body Strength.
OMW Inside & Outside
Pressure Testing
Casing Pipe Body Strength
Different Scenarios
0 100,000 200,000 300,000 400,000 500,000 600,000
De
pth
, D (
ft)
MInimum Casing Strength, Sc (psi)
Figure 4: MInimum Casing Strength Requirements Based On Joint Strength.
OMW Inside & Outside
Pressure Testing
Casing Joint Strength
Different Scenarios
Project 1 PEGN 361 2/27/2015
Sample Calculations
Collapse Scenario: Evacuated Casing
πΏπππ = 16.5
19.25π·
Backup = 0
ππ β₯ 1.1 Γ (0.857π·)
ππ β₯ 0.9429π·
πππππ1 = 20 Γ (12000 β π·) β (16.5
19.25Γ 12000) Γ
π
4Γ (5.52)
πππππ1 = β4371 β 20π· 0 β€ π· β€ 11500
πππππ2 = [23 Γ (11500 β π·) + (β4371 β 20(11500)) ]
πππππ2 = 30129 β 23π·
ππ =30129 β 23(0)
π4
Γ (5.5.2β 4.672)
ππ = 4544.6 ππ π
πππ = 80000 Γ β[1 β 0.75 Γ (4544.6
80000)
2] β (
1
2Γ [
4544.6
80000])
πππ = 77630.8ππ π
πππ€ πΆππππππ π ππππ π π’ππ = 10832ππ π
Note: VBA Function used to find adjusted collapse
pressure.
Project 1 PEGN 361 2/27/2015
Collapse Scenario: Cement Job
πΏπππ1 = 16.5
19.25π·
πΏπππ2 = 16.5
19.25(5711) +
15
19.25(π· β 5711)
πΏπππ3 = 16.5
19.25(5711) +
15
19.25(10000 β 5711) +
16.4
19.25(π· β 10000)
π΅ππππ’π = 8.7
19.25π·
ππ β₯ 1.1 Γ (16.5
19.25π· β
8.7
19.25π·)
ππ β₯ 1.1 Γ (16.5
19.25(5711) +
15
19.25(π· β 5711) β
8.7
19.25π·)
ππ β₯ 1.1 Γ (16.5
19.25(5711) +
15
19.25(10000 β 5711) +
16.4
19.25(π· β 10000) β
8.7
19.25π·)
ππ β₯ 0.4457π· 0 β€ π· β€ 5711
ππ β₯ 489.51 + 0.36π· 5711 β€ π· β€ 10000
ππ β₯ 0.44π· β 310.486 10000 β€ π· β€ 12000
SIMILAR PROCEDURES TO COMPUTE REAL TENSION IN THE CEMENT JOB WERE REPEATED HERE.
Project 1 PEGN 361 2/27/2015
Burst Scenario: Leaky Tubing
πΏπππ = 9.5
19.25π· + ππΌππ
π΅ππππ’π =0.465ππ π
ππ‘
BHP = 16
19.25Γ 12000 = 9974ππ π
BHT = [70 +1.8β
100ππ‘Γ 12000ππ‘] + 459.67 = 745.67Β°R
πππ = 9974 + 14.7
673= 14.842
πππ =745.67
343= 2.174
Z = 1.36
ππππ =9974
1.36 Γ154416.04
Γ 745.67= 0.10217ππ π/ππ‘
ππΌππ = 9974 β (0.10217 Γ 12000) = 8747.933ππ π
πΏπππ =9.5
19.25π· + 8747.933
ππ β₯ 1.1 Γ (9.5
19.25π· + 8747.933 β 0.465π·)
ππ β₯ 0.03136π· + 9622.73
Burst Scenario: Pressure Testing
πΏπππ = 3500ππ π +9.5
19.25π·
π΅ππππ’π = 0.465ππ π/ππ‘
ππ β₯ 1.1 Γ (3500 +9.5
19.25π· β 0.465π·)
ππ β₯ 0.03136π· + 3850
Project 1 PEGN 361 2/27/2015
Initial Casing Design:
5.5" 23πππ π80 πΏππΆ 0 β€ π· β€ 11500
5.5" 20πππ π125 πΏππΆ 11500 β€ π· β€ 12000
Tensile Scenario: Same OMW Inside & Outside
πππππ1 = 20 Γ (12000 β π·) β (16.5
19.25Γ 12000) Γ
π
4Γ (5.5.2β 4.7782)
πππππ1 = 180052.63 β 20π· 11500 β€ π· β€ 12000
πππππ2 = [23 Γ (11500 β π·) β (16.5
19.25Γ 11500) Γ
π
4
Γ (4.7782 β 4.672)] + [180052.63 β 20(11500)]
πππππ2 = 206653.035 β 23π· 0 β€ π· β€ 11500
Axial Stress
At D = 0 (surface)
ππ =206653.035 β 23(0)
π4
Γ (5.5.2β 4.672)
ππ = 31171.2psi
Project 1 PEGN 361 2/27/2015
Adjusted Yield Point
At D=0 (surface)
YP = 80,000psi
πππ = 80000 Γ β[1 β 0.75 Γ (31171.2
80000)
2] β (
1
2Γ [
31171.2
80000])
πππ = 59722.25ππ π
New Collapse Pressure Rating (Biaxial Collapse Stress)
A = 2.8762 + (0.0000010679 Γ 59722.25 ) + (0.000000000021301 Γ 59722.252)
β (5.3132E β 17 Γ 59722.253)
A = 3.00463
B = 0.026233 + 0.00000050609 Γ 59722.25
B = 5.6458E-02
πΆ = β465.93 + 0.030867 Γ 59722.25 β 0.000000010483 Γ 597222.252 + 3.6989πΈ β 14
Γ 59722.253
πΆ = 1348
π =5.6458E β 02
3.00463
π = 1.879πΈ β 02
πΉ = (46950000 β ((3 β π) / (2 + π)) ^ 3) / (59722.25 β ((3 β π / (2 + π)) β π) β (1
β ((3 β π) / (2 + π))) ^ 2)
πΉ = 1.983
πΊ =1.983
1.879πΈ β 02
πΊ = 3.7266πΈ β 02
πΆππΏ1 = (((3.00463 β 2) ^ 2 + 8 β (5.6458E β 02 + 1348 / 59722.25)) ^ 0.5 + (3.00463
β 2)) / (2 β (5.6458E β 02 + 1348 / 59722.25))
πΆππΏ1 = 14.462
Project 1 PEGN 361 2/27/2015
πΆππΏ2 = (59722.25 β (3.00463 β 1.983)) / (1348 + 59722.25 β (5.6458E β 02 β 3.7266πΈ
β 02))
πΆππΏ2 = 24.456
πΆππΏ3 = (2 + 1.879πΈ β 02) / (3 β 1.879πΈ β 02)
πΆππΏ3 = 35.813
OD = 5.5β
ID = 4.67β
ππ·
π‘=
5.5
5.5 β 4.672
ππ·
π‘= 13.253
ππ·
π‘β€ πΆππΏ1
ππΆππππππ π_πππ€ = 2 β 59722.25 β (13.253 β 1) / (13.253^ 2)
ππΆππππππ π_πππ€ = 8332.59ππ π
Pipe Body Strength
ππ β₯ 1.5 Γ (206653.035 β 23π·) + 100000
ππ β₯ 409979.55 β 34.5π· 0 β€ π· β€ 11500
ππ β₯ 1.5 Γ (180052.63 β 20π·) + 100000
ππ β₯ 370078.945 β 30π· 11500 β€ π· β€ 12000
Joint Strength
ππ β₯ 1.8 Γ (206653.035 β 23π·) + 100000
ππ β₯ 471975.5 β 41.4π· 0 β€ π· β€ 11500
ππ β₯ 1.8 Γ (180052.63 β 20π·) + 100000
ππ β₯ 424094.73 β 36π· 11500 β€ π· β€ 12000
Project 1 PEGN 361 2/27/2015
Tensile Scenario: Pressure Testing
πππππ1 = 20 Γ (12000 β π·) β (16.5
19.25Γ 12000) Γ
π
4Γ (5.52)
+ [8.7
19.25Γ 12000 + 3500] Γ
π
4(4.7782)
πππππ1 = 155625.8 β 20π· 11500 β€ π· β€ 12000
πππππ2 = [23 Γ (11500 β π·) β (8.7
19.25Γ 11500 + 3500) Γ
π
4
Γ (4.7782 β 4.672)] + [155625.8 β 20(11500)]
πππππ2 = 197096 β 23π· 0 β€ π· β€ 11500
Note: Similar procedures to determine the biaxial collapse stress in the first tensile scenario were done
for this tensile scenario as well.
Pipe Body Strength
ππ β₯ 1.5 Γ (183155.6 β 23π·) + 100000
ππ β₯ 374733.4 β 34.5π· 0 β€ π· β€ 11500
ππ β₯ 1.5 Γ (155625.8 β 20π·) + 100000
ππ β₯ 333438.7 β 30π· 11500 β€ π· β€ 12000
Project 1 PEGN 361 2/27/2015
Depth, D (ft) Real Tension, Treal (lbf) Axial Stress, Οa (psi) Adjusted Yield Point, YpaC (psi) New Collapse Pressure Rating, Pc (psi)
0.0 131392.3 19819.0 68227.6 9519.3
10096.2 -100820.3 -15207.5 86512.2 11987.1
12713.6 -161020.4 -24288.0 89329.3 12303.9
13049.6 -168748.4 -25453.7 89629.9 12337.5
13085.3 -169568.9 -25577.4 89661.0 12341.0
13089.0 -169653.6 -25590.2 89664.2 12341.4
13089.3 -169662.4 -25591.6 89664.5 12341.4
13089.4 -169663.3 -25591.7 89664.6 12341.4
13089.4 -169663.4 -25591.7 89664.6 12341.4
13089.4 -169663.4 -25591.7 89664.6 12341.4
13089.4 -169663.4 -25591.7 89664.6 12341.4
13089.4 -169663.4 -25591.7 89664.6 12341.4
13089.4 -169663.4 -25591.7 89664.6 12341.4
13089.4 -169663.4 -25591.7 89664.6 12341.4
13089.4 -169663.4 -25591.7 89664.6 12341.4
Table 1: Iteration To Find The Theoretical Depth Of Failure For The Top Casing In Cement Job Scenario
Joint Strength
ππ β₯ 1.8 Γ (183155.6 β 23π·) + 100000
ππ β₯ 429680 β 41.4π· 0 β€ π· β€ 11500
ππ β₯ 1.8 Γ (155625.8 β 20π·) + 100000
ππ β₯ 380126.44 β 36π· 11500 β€ π· β€ 12000
ITERATION TO FIND THEORETICAL FAILURE DEPTH
Iteration using the real tensile equations and adjusted collapse pressure ratings for cement job, since
that is the worst case for collapse.
Find Neutral Point of Tension & Compression
πππππ1 = 132711.2 β 23π·
D = 5770.05ft
Theoretical Failure Point = 13089.4ft
Project 1 PEGN 361 2/27/2015
Given the new Collapse Pressure = 9519.3psi
&
ππ β₯ 0.9429π· for the evacuated casing scenario which has a higher casing strength requirement than
the cement job scenario
Next Iteration Depth Would Be:
π· =9519.3
0.9429= 10096.2ππ‘
Keep repeating these steps until the depths & new collapse pressure ratings converge.
Final Casing Design
5.5" 23πππ π80 πΏππΆ 0 β€ π· β€ 11500
5.5" 20πππ π125 πΏππΆ 11500 β€ π· β€ 12000
COST ANALYSIS
N80 Casing Price per foot: $25.09
Total Length: 11500ft
Q125 Casing Price per foot: $34.09
Total Cost = (11500 x $25.09)+(500 x $34.09)
Total Cost = $305,580
