Tugas 2 Drilling [PDF]

Zitiervorschau

Tugas 2 Drilling Nama : Arnas Hardianto NPM : 1815051028 2.26 The density of 800 bbl of 14-lbm/gal mud must be increased to 14.5 lbm/gal using API barite. The total mud volume is limited to 800 bbl. Compute the volume of old mud that should be discarded and the weight of API barite required. Answer: Discard 19.05 bbl, add 28,000 lbm of barite. Jawab: Diketahui: V 2=800bbl ρ1=14 lbm/ gal ρ2=14 .5l bm/gal Ditanya:

V 2−V 1 =… ? mB=…?

Penyelesaian: ( ρB −ρ2 ) ( 35−14.5 ) V 1=780.95 bbl V 1=800 V 1=V 2 ( 35 – 14 ) ( ρB – ρ1 ) Volume old mud that should be discarded: V 2−V 1 =800−780.95¿ 19.05 bbl Weight of API barite: mB=(V ¿ ¿ 2−V 1 ) ρB ¿¿ 19.05∗42∗35¿ 28 , 000 lbm Jadi, volume old mud yang harus dibuang adalah 19.05 bbl dan berat API barite yang ditambahkan adalah 28,000 lbm.

2.27 The density of 900 bbl of a 16-lbm/gal mud must be increased to 17 lbm/gal. The volume fraction of low-specific- gravity solids also must be reduced from 0.055 to 0.030 by dilution with water. A final mud volume of 900 bbl is desired. Computer the volume of original mud that must be discarded and the amount of water and API barite that should be added. Answer: Discard 409 bbl, add 257.6 bbl of water and 222,500 lbm of barite Jawab: Diketahui: V 2=9 00bbl ρ1=1 6 l bm/gal ρ2=1 7 l bm/ gal f c1 =0.030 lbm/ gal f c2 =0.055 lbm/ gal Ditanya:

V 2−V 1 =… ? V w =… ? mB=…?

Penyelesaian: f c2 0.03 V =491 bbl V 1=V 2 V 1=900 0.055 1 f c1 Volume of original mud that should be discarded: V 2−V 1 =9 00−491¿ 409 bbl and the amount of water: ( ρ B− ρ1) V 2− ( ρB −ρ1 ) V 1 ( 35−1 7 ) 900−( 35−16 ) 491 ¿ V w= ( 35−8.33 ) ( ρ B−ρ w ) ¿ 257.63 lbs Weight of API barite: mB=(V ¿ ¿ 2−V 1−V w ) ρB ¿¿ 151.37∗42∗35¿ 2 22, 51 3lbm Jadi, volume original mud yang harus dibuang adalah 409 bbl, jumlah air adalah 257.63 lbs dan berat API barite yang ditambahkan adalah 222,513 lbm.

2.28 Asumsing a clay and chemical cost of $10.00/bbl of mud discarded and a barium sulfate cost of $0.10/lbm, compute the value of the mud discarded in Problem 2.27. If an error of +0.01% is made in determining the original volume fraction of low-specific-gravity solids in the mud, how much mud was unnecessarily discarded? Answer: $16,697; 191 bbl 2.29 Derive expressions for determining the amounts of barite and water that should be added to increase the density of 100 bbl of mud from rho 1 to rho2. Also derive an expression for the increase in mud volume expected upon adding the barite and the water. Assume a water requirement of 1 gal per snack of barite. Answer: MB = 109,000 (𝜌2 - 𝜌1)/(28.08- 𝜌2); Vw = MB/4,200; V = 0.00091/MB 2.30 A 16.5-lbm/gal mud is entering a centrifuge at a rate of 20 gal/min along with 8.34 lbm/gal of dilution water, which enters the centrifuge at a rate 10 gal/min. the density of the centrifuge underflow is 23.9 lbm/gal while the density of overflow is 9.5 lbm/gal. The mud contains 25 lbm/bbl bentonite and 10 lbm/bbl deflocculant. Compute the rate at which bentonite, deflocculant, water, and API barite should be added downstream of the centrifuge to maintain the mud properties constant. Answer: 6.8 lbm/min of clay, 2.7 lbm/min of deflocculant, 7.4 gal/min of water, and 3.1 lbm/min of barite.