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NM957 Offshore Engineering Practice - Coursework - Angelos Papavasileiou NM957 Offshore Engineering Practice Coursework 1 of 3 - Well design and construction: Design exersise question 1 a) From the notes we know that for 17 1/2inch (13 3/8 inch casing) section in the central North Sea we must use drilling fluid with seawater and high viscosity sweeps and displace after drilling with 10.5 ppg bentonite WBM. In case of reactive shales , the formation pressure may be higher than our initial estimation.As a result, we have to increase the mud density in order to prevent blowout. b) We should use TCI drill bit because it has adequate penetration rate and it can also drill chert. question 2 P1e 0 pmud 11.4 lb gal ltotal 80 ft a 0.052 gal psi lb ft P1i = a ( ( pmud ltotal) ) 47.424 psi we use the values for Pc from the table from page 24 NEXEN notes Pc 2671 psi t = 0.514 in 0.043 ft do = 13.375 in 1.115 ft SF = Pc P1e P1i 1 2 t do −61.011 pwater = 1025 kg m 3 8.554 lb gal h2e 325 ft a 0.052 gal psi lb ft P2e = a pwater h2e 144.563 psi pmud 11.4 lb gal ltotal = + 80 ft 325 ft 405 ft P2i = a ( ( pmud ltotal) ) 240.084 psi Pc 2671 psi t 0.514 in do 13.375 in SF = Pc P2e P2i 1 2 t do −34.658 Page 1 of 5

NM957-2014-Angelos Papavasileiou-CW1

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Page 1: NM957-2014-Angelos Papavasileiou-CW1

NM957 Offshore Engineering Practice - Coursework - Angelos Papavasileiou

NM957 Offshore Engineering PracticeCoursework 1 of 3 - Well design and construction: Design exersise

question 1

a) From the notes we know that for 17 1/2inch (13 3/8 inch casing) section in the central North Sea we must use drilling fluid with seawater and high viscosity sweeps and displace after drilling with 10.5 ppg bentonite WBM.In case of reactive shales , the formation pressure may be higher than our initial estimation.As a result, we have to increase the mud density in order to prevent blowout.

b) We should use TCI drill bit because it has adequate penetration rate and it can also drill chert.

question 2

≔P1e 0 ≔pmud 11.4 ――lb

gal≔ltotal 80 ft ≔a 0.052 ―――

⋅gal psi

⋅lb ft

≔P1i =⋅a (( ⋅pmud ltotal)) 47.424 psi

we use the values for Pc from the table from page 24 NEXEN notes

≔Pc 2671 psi ≔t =0.514 in 0.043 ft ≔do =13.375 in 1.115 ft

≔SF =―――――――Pc

−P1e ⋅P1i⎛⎜⎝

−1 2 ―t

do

⎞⎟⎠

−61.011

≔pwater =1025 ――kg

m3

8.554 ――lb

gal≔h2e 325 ft ≔a 0.052 ―――

⋅gal psi

⋅lb ft

≔P2e =⋅⋅a pwater h2e 144.563 psi

≔pmud 11.4 ――lb

gal≔ltotal =+80 ft 325 ft 405 ft

≔P2i =⋅a (( ⋅pmud ltotal)) 240.084 psi

≔Pc 2671 psi ≔t 0.514 in ≔do 13.375 in

≔SF =―――――――Pc

−P2e ⋅P2i⎛⎜⎝

−1 2 ―t

do

⎞⎟⎠

−34.658

Page 1 of 5

Page 2: NM957-2014-Angelos Papavasileiou-CW1

NM957 Offshore Engineering Practice - Coursework - Angelos Papavasileiou

≔plead 13.2 ――lb

gal≔h3e 2828 ft ≔a 0.052 ―――

⋅gal psi

⋅lb ft

≔P3e =+⋅⋅a plead h3e P2e 2085.702 psi

≔pmud 11.4 ――lb

gal≔ltotal =++80 ft 325 ft 2828 ft ⎛⎝ ⋅3.233 10

3 ⎞⎠ ft

≔P3i =⋅a (( ⋅pmud ltotal)) 1916.522 psi

≔Pc 2671 psi ≔t 0.514 in ≔do 13.375 in

≔SF =―――――――Pc

−P3e ⋅P3i⎛⎜⎝

−1 2 ―t

do

⎞⎟⎠

8.44

≔ptail 16 ――lb

gal≔h4e 500 ft ≔a 0.052 ―――

⋅gal psi

⋅lb ft

≔P4e =+⋅⋅a ptail h4e P3e 2501.702 psi

≔pmud 11.4 ――lb

gal≔ltotal =+++80 ft 325 ft 2828 ft 500 ft ⎛⎝ ⋅3.733 10

3 ⎞⎠ ft

≔P4i =⋅a (( ⋅pmud ltotal)) 2212.922 psi

≔Pc 2671 psi ≔t 0.514 in ≔do 13.375 in

≔SF =―――――――Pc

−P4e ⋅P4i⎛⎜⎝

−1 2 ―t

do

⎞⎟⎠

5.821

the minimum collapse factor during the cementing procedure is the SF=5.82

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Page 3: NM957-2014-Angelos Papavasileiou-CW1

NM957 Offshore Engineering Practice - Coursework - Angelos Papavasileiou

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Page 4: NM957-2014-Angelos Papavasileiou-CW1

NM957 Offshore Engineering Practice - Coursework - Angelos Papavasileiou

question 3

≔pwater =1025 ――kg

m3

8.554 ――lb

gal≔h1 325 ft ≔a 0.052 ―――

⋅gal psi

⋅lb ft

≔P1 =⋅⋅a pwater h1 144.563 psi

≔ppore 8.7 ――lb

gal≔h2 3328 ft ≔a 0.052 ―――

⋅gal psi

⋅lb ft

≔P2 =+⋅⋅a ppore h2 P1 1650.15 psi

≔ppore 8.7 ――lb

gal≔h3 7179 ft ≔a 0.052 ―――

⋅gal psi

⋅lb ft

≔P3 =+⋅⋅a ppore h3 P2 4897.93 psi

≔FP =P4e 2501.702 psi ≔Pinflux =P3 4897.93 psi

≔pgas 0.1 ――psi

ft≔h3 7179 ft

≔Pint2 =−Pinflux (( ⋅pgas h3)) 4180.03 psi

=if >Pint2 FP‖‖ −FP (( ⋅pgas h2))

2168.902 psi

≔Pb 5380 psi ≔Pint 2168.902 psi ≔Pext P1

≔SFbrust =――――Pb

−Pint Pext2.658

the minimum burst factor for the casing full of gas density is the SF=2.658

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Page 5: NM957-2014-Angelos Papavasileiou-CW1

NM957 Offshore Engineering Practice - Coursework - Angelos Papavasileiou

=if >Pint2 FP‖‖ −FP (( ⋅pgas h1))

2469.202 psi

≔Pb 5380 psi ≔Pint 2469.202 psi ≔Pext P2

≔SFbrust =――――Pb

−Pint Pext6.569

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