1. A well can be induced to flow by swabbing which happens due to the reduction of bottom hole pressure when pulling pipe. List 3 conditions that can cause swabbing.

a) Pulling speed - If a drill string is being pulled out of hole too fast, and the drill string is almost the same hole size, it will create a temporary piston like condition, which cause swabbing.

b) Welbore Geometry - The size of Drilling Equipment being tripped is almost the same size as hole

c) Viscosity, gel strength and density - The Viscosity, gel strength and density of mud is high, causing the high friction.

2. List at least 2 causes of the increase in rate of penetration during drilling.a) Increase in bit weightb) Increase in rotary speedc) Increase in fluid circulation rate, the rate of mud flowing from nozzle

3. Mention at least 5 components of drill stem.a) Drill Stringb) Swivelc) Subd) Kellye) Drill Collarf) Packerg) Drill bith) Tool joint

8. Given the following data:

Depth 10000ft TVDBit size 8 ½”Shoe depth 8500ft TVDMud weight 12.6 ppg

Collars – 600ft. Capacity = 0.0077 bbl / ftMetal displacement = 0.03 bbl / ftDrill-pipe 5” capacity = 0.0178 bbl / ftMetal displacement = 0.0476 bbl / ftCasing / pipe annular capacity = 0.0476 bbl / ftCasing capacity = 0.0729 bbl / ftOne stand of drill-pipe = 94 ft

Assuming the 12.6 ppg mud givens an over-balances of 200 psi

Problems with question. Metal displacement is same as Casing/pipe annular capacity. Supposely, Drill Pipe capacity + Metal displacement + Annular = Casing capacity

a. If 10 stands of pipe are removed “dry” without filling the hole, what would be the resultant reduction in bottom-hole pressure?

Total depth = number * one stand drill-pipe length = 10 * 94

= 940 ft

Pulling dry pipe (psi/ft) = MudGradient∗Volumeremoved

Volumeremaining

Since it is removed dry, no mud is being carried out.

Pulling dry pipe (psi/ft) = MudGradient∗metal displacement

casing capacity−metal displacement

Mud Gradient = 0.052 * Mudweight

= 0.052 * 12.6

= 0.6552 psi/ft

Metal Displacement = 0.0476 bbl/ft

Casing Capacity = 0.0729 bbl/ft

Pulling dry pipe (psi/ft) = 0.6552∗0.04760.0729−0.0476

= 1.2327 psi/ft

Reduction in Pressure = Pulling dry pipe * Depth

= 1.2327 * 940

= 1158.74 psi

b. If 5 stands of pipe had been pulled “wet” without filling the hole, the resultant reduction in bottom-hole pressure would be.

Total depth = number * one stand drill-pipe length = 5 * 94

= 470 ft

Since it is removed wet, mud is being removed too.

Pulling wet pipe (psi/ft) = MudGradient∗Volumeremoved

Volumeremaining

Pulling wet pipe (psi/ft)

= MudGradient∗metal displacement+drill pipe capacity

annulus capacity

= MudGradient∗metal displacement+drill pipe capacity

casing capacity−metal displacement−drill pipe capacity

Metal Displacement = 0.0476 bbl/ft

Casing Capacity = 0.0729 bbl/ft

Drill pipe capacity = 0.0178 bbl/ft

Pulling wet pipe (psi/ft) = 0.6552∗0.0476+0.01780.0729−0.0476−0.0178

= 5.713 psi/ft

Reduction in Pressure = Pulling dry pipe * Depth

= 5.713 * 470

= 2685.11 psi

c. If prior to tripping a 20 barrel slug of 14.6 ppg mud was displaced to prevent a wet trip, what would be the expected volume return due to the U-tubing of the heavy mud?

Dry pipe volume = SlugVolume∗( SlugweightMudweight−1)

= 20∗( 14.612.6

−1) = 3.175 barrel