A fresh water mud is used to drill 12-1/4\" hole. The hole is 10,000\' deep, sur
ID: 297617 • Letter: A
Question
A fresh water mud is used to drill 12-1/4" hole. The hole is 10,000' deep, surface casing (13-3/8", 12.5" I.D.) is set at 4200 fit. The drill string has 650' drill collars. Linear steel volume of D.C. & D.P. is 0.32, 0, 032 cuff/ft respectively. Mud has initially 10% by weight bentonite. It is required to raise the mud weight to 14.0 ppg. Determine the initial mud weight and mud volume. Find the amount of barite in lb/bbl to be added to the initial mud. If the rig has 3 tanks, each of cross section area 150 sq.ft., mud height in the tanks are 7, 8, & 9 ft. Find the total amount of barite to be used for this purpose. Determine the total mud volume after mixing of barite (Density of barite is 35 ppg and Sp. Gr. of bentonite is 2.7).Explanation / Answer
A fresh water mud is used to drill 12.25” hole. The hole is 10,000’ deep, surface casing (13.3/8, 12.5” I.D. ) is set at
4,200’. The drill string has 650’ drill collars. Linear steel volume of D.C. and D.P. are 0.32. 0.32 cuff/ft respectively.
Mud has initially 10% by weight betonite. It is required to raise the mud weight to 14.0 ppg
a.Determine the initial mud weight and mud volume
Volume of drill hole (cylinder) = V = r² h
Hole dia 12.25 “
Radius of hole r = 12.25/2 = 6.125” = 0.51 ft.
Depth of the hole = h = 10,000 ft.
So volume of mud = V = 22/7 x 6.125² x 10000 = 3.14 x 37.51 x 10000 = 1,177,814 cu,ft.
Specific gravity of Bentonite = 2.7
The, weight of the mud = volume x sp. Gr. = 1177814 x 2.7 = 3,180,097 lb
b.Find the amount of barite in lb/bbl to be added to the initial mud
Mud has 10% barite
Barite weight = 3180097 x 10/100 = 318009.7 lb
318009 lb of barite added to mud initially
c.If the rig has 3 tanks, each of cross section area 150 sq.ft. mud height in the tanks are 7, 8 & 9 ft.
Find the total amount of barite to be used for this purpose.
Volume of each tank = 150 ft. x 7 = 1050 cu. ft.
= 150 ft. x 8 = 1200 “
= 150 xv9 = 1350 “
Total mud volume of 3 tanks = 3600 cu,ft.
Bentonite sp.gr. = 2.7 lb/ft³ and Barite density = 35 ppg (262 lb/ft³)
To use equal barite against benite we need barite = 2.7/262 = 0.10 lb/ft³
Wight of the barite mud in 3 tanks = 3600 x 0.10 = 360 lb/ft³
Wight of the barite in mud = 10% = 360 x 10/100 = 36 lb