Hemisperical 150.docx

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Hemisperical 150.000 bbl, berdasarkan appendiks E item 3 pada buku Brownell 150.000 bbl kita menetukan kapasitas tangki sebesar 153.500 bbl

2. Material yang dipakai pada pembuatan tangki ini adalah SA-283 Grade C karena….

3. Volume tangki dan jumlah course dilihat dari appendiks E 13.1 pada buku brownel

4. Diperoleh untuk D=140 ft dengan H=56 ft, dan volume tangki, V=153.500 bbl. Dilihat dari kapasitas tangki dari 150.000 bbl. Jumlah course (bagian shell pada circum ference) = 7 course Maka tinggi setiap course = 5.

Htangki Jumlah course

=

56 ft 7

= 8 ft

Shell plate thickness

Pada table item 4 dapat dilihat bahwa untuk tangki dengan D=140 ft & H=56 ft, yaitu minimum = 0,31 in dan maximum=1.13 in dengan ΒΌ in

Pada persamaan 3.18

t=

Pin D 2f E

Densitas fluida harus lebih kecil dari water joint efficiency= 0,85 (didapat pada table 13.3) karena pada soal itu double-welded maka water jount efficiency nya 0,85.

Shell=12650x0,80=10120 psi = 10000 psi, minimal bagian bidang (shear) Shell=12650x1.60=20240 psi = 21000 psi, maximal bagian bidang (bearning) Allowable stress nya adalah 12650 ο‚·

𝑑=

Course 1 (Buttom shell course)

𝑃𝑖𝑛 𝐷 πœŒπ‘€ Γ— 𝑔⁄𝑔𝑐 Γ— (𝐻 βˆ’ 1)𝐷 = 2𝑓 𝐸 2 𝑓 𝐸 Γ— 144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—98,1⁄9,81Γ—(56 π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑝𝑠𝑖 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

+𝐢

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—98,1⁄9,81Γ—(56 π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑙𝑏⁄𝑖𝑛2 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

+𝐢

= 0,0969 𝑓𝑑 + 𝐢 = 1.1628 𝑖𝑛 Untuk allowable corrosion, C= 0 Setiap plate ada 2x course = 2x7 = 14 (brownell)

6.

plate length in course. L=

πœ‹π‘‘βˆ’π‘› 𝑑𝑀𝑒𝑙𝑑𝑖𝑛𝑔 12𝑛

= =

3,14(π·π‘š+𝑑)βˆ’14(3⁄16 𝑖𝑛) 12(14) 3,14[(12𝑖𝑛⁄𝑓𝑑×140 𝑓𝑑)+1,1628 𝑖𝑛]βˆ’14(3⁄16 𝑖𝑛)

= 31,406 𝑖𝑛

168

ο‚·

Course 2 . H=48 ft

𝑑=

𝑃𝑖𝑛 𝐷 πœŒπ‘€ Γ— 𝑔⁄𝑔𝑐 Γ— (𝐻 βˆ’ 1)𝐷 = 2𝑓 𝐸 2 𝑓 𝐸 Γ— 144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—(98,1⁄9,81)Γ—(48 π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑝𝑠𝑖 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—98,1⁄9,81Γ—(48π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑙𝑏⁄𝑖𝑛2 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

+𝐢

+𝐢

= 0,0828 𝑓𝑑 + 𝐢 = 0,9936 𝑖𝑛 L=

πœ‹π‘‘βˆ’π‘› 𝑑𝑀𝑒𝑙𝑑𝑖𝑛𝑔 12𝑛

= =

3,14(π·π‘š+𝑑)βˆ’14(5⁄32 𝑖𝑛) 12(14) 3,14[(12𝑖𝑛⁄𝑓𝑑×140 𝑓𝑑)+0,9936 𝑖𝑛]βˆ’14(5⁄32 𝑖𝑛) 168

= 31,4055 𝑖𝑛

ο‚·

Course 3. H=40 ft

𝑑=

𝑃𝑖𝑛 𝐷 πœŒπ‘€ Γ— 𝑔⁄𝑔𝑐 Γ— (𝐻 βˆ’ 1)𝐷 = 2𝑓 𝐸 2 𝑓 𝐸 Γ— 144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—(98,1⁄9,81)Γ—(40 π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑝𝑠𝑖 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—98,1⁄9,81Γ—(40π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑙𝑏⁄𝑖𝑛2 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

= 0,0687 𝑓𝑑 + 𝐢 = 0,8244 𝑖𝑛

+𝐢

+𝐢

L=

πœ‹π‘‘βˆ’π‘› 𝑑𝑀𝑒𝑙𝑑𝑖𝑛𝑔 12𝑛

= =

3,14(π·π‘š+𝑑)βˆ’14(5⁄32 𝑖𝑛) 12(14) 3,14[(12𝑖𝑛⁄𝑓𝑑×140 𝑓𝑑)+0,8244 𝑖𝑛]βˆ’14(5⁄32 𝑖𝑛) 168

= 31,4028 𝑖𝑛

ο‚·

Course 4. H=32 ft

𝑑=

𝑃𝑖𝑛 𝐷 πœŒπ‘€ Γ— 𝑔⁄𝑔𝑐 Γ— (𝐻 βˆ’ 1)𝐷 = 2𝑓 𝐸 2 𝑓 𝐸 Γ— 144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—(98,1⁄9,81)Γ—(32 π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑝𝑠𝑖 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—98,1⁄9,81Γ—(32π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑙𝑏⁄𝑖𝑛2 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

+𝐢

+𝐢

= 0,0546 𝑓𝑑 + 𝐢 = 0,6552 𝑖𝑛 L=

πœ‹π‘‘βˆ’π‘› 𝑑𝑀𝑒𝑙𝑑𝑖𝑛𝑔 12𝑛

= =

3,14(π·π‘š+𝑑)βˆ’14(5⁄32 𝑖𝑛) 12(14) 3,14[(12𝑖𝑛⁄𝑓𝑑×140 𝑓𝑑)+0,6552 𝑖𝑛]βˆ’14(5⁄32 𝑖𝑛)

= 31,3992 𝑖𝑛

168

ο‚·

𝑑=

Course 5. H=24 ft

𝑃𝑖𝑛 𝐷 πœŒπ‘€ Γ— 𝑔⁄𝑔𝑐 Γ— (𝐻 βˆ’ 1)𝐷 = 2𝑓 𝐸 2 𝑓 𝐸 Γ— 144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—(98,1⁄9,81)Γ—(24 π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑝𝑠𝑖 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—98,1⁄9,81Γ—(24π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑙𝑏⁄𝑖𝑛2 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

+𝐢

+𝐢

= 0,0405 𝑓𝑑 + 𝐢 = 0,486 𝑖𝑛 L=

πœ‹π‘‘βˆ’π‘› 𝑑𝑀𝑒𝑙𝑑𝑖𝑛𝑔 12𝑛

= =

3,14(π·π‘š+𝑑)βˆ’14(1⁄8 𝑖𝑛) 12(14) 3,14[(12𝑖𝑛⁄𝑓𝑑×140 𝑓𝑑)+0,486 𝑖𝑛]βˆ’14(5⁄32 𝑖𝑛) 168

= 31,3986 𝑖𝑛

ο‚·

Course 6. H=16 ft

𝑑=

𝑃𝑖𝑛 𝐷 πœŒπ‘€ Γ— 𝑔⁄𝑔𝑐 Γ— (𝐻 βˆ’ 1)𝐷 = 2𝑓 𝐸 2 𝑓 𝐸 Γ— 144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—(98,1⁄9,81)Γ—(16 π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑝𝑠𝑖 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—98,1⁄9,81Γ—(16π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑙𝑏⁄𝑖𝑛2 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

= 0,0264 𝑓𝑑 + 𝐢

+𝐢

+𝐢

= 0,3168 𝑖𝑛 L=

πœ‹π‘‘βˆ’π‘› 𝑑𝑀𝑒𝑙𝑑𝑖𝑛𝑔 12𝑛

= =

3,14(π·π‘š+𝑑)βˆ’14(1⁄8 𝑖𝑛) 12(14) 3,14[(12𝑖𝑛⁄𝑓𝑑×140 𝑓𝑑)+0,3168 𝑖𝑛]βˆ’14(5⁄32 𝑖𝑛) 168

= 31,3955 𝑖𝑛

ο‚·

Course 7. H=8 ft

𝑑=

𝑃𝑖𝑛 𝐷 πœŒπ‘€ Γ— 𝑔⁄𝑔𝑐 Γ— (𝐻 βˆ’ 1)𝐷 = 2𝑓 𝐸 2 𝑓 𝐸 Γ— 144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—(98,1⁄9,81)Γ—(8 π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑝𝑠𝑖 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

=

62,37𝑙𝑏⁄𝑓𝑑 3 Γ—98,1⁄9,81Γ—(8π‘“π‘‘βˆ’1)140 𝑓𝑑 2 Γ—20240 𝑙𝑏⁄𝑖𝑛2 Γ—0,85 Γ—144 𝑖𝑛2 ⁄𝑓𝑑 2

+𝐢

+𝐢

= 0,0123 𝑓𝑑 + 𝐢 = 0,1476 𝑖𝑛 L=

πœ‹π‘‘βˆ’π‘› 𝑑𝑀𝑒𝑙𝑑𝑖𝑛𝑔 12𝑛

= =

3,14(π·π‘š+𝑑)βˆ’14(1⁄16 𝑖𝑛) 12(14) 3,14[(12𝑖𝑛⁄𝑓𝑑×140 𝑓𝑑)+0,1476 𝑖𝑛]βˆ’14(5⁄32 𝑖𝑛)

= 31,3975 𝑖𝑛

168

COURSE 1 2 3 4 5 6 7

t 0,0969 ft 0,0828 ft 0,0687 ft 0,0546 ft 0,0405 ft 0,0264 ft 0,0123 ft

1,1628in 0,9936 in 0,8244 in 0,6552 in 0,486 in 0,3168 in 0,1476 in

L 31,406 in 31,4055 in 31,4028 in 31,3992 in 31,3986 in 31,3955 in 31,3975 in

3/16 5/32 5/32 5/32 1/8 1/8 1/16

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