The following questions and answers have been compiled as supplemental exercises to be completed after each relevant section, to help consolidate your learning experience
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 1
1. Convert 486 inches to yards ft and inches.
2. Convert 486 square inches into square ft.
3. How many US gallons would fill a tank with a capacity of 450 cubic ft?
5. What would be the equivalent in P.S.I. of 15 tons resting on a square 2ft by 2ft 6. How many pounds force would be exerted on a hatch 2ft by 1.5ft if the
pressure behind it was 3 P.S.I.
7. Convert 240 US gallons per minute flow into litres per minute flow.
10. Convert 90 ees rankin.
4. Convert 10.3 ppg into P.C.F.
8. Convert 1000kg into pounds
9. How many pounds difference between 1000kg and one long ton?
oF to degr
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 1 : ANSWERS
1. 13 yds, 1 ft 6 inches
2. 3 square ft 54 square inches 3. 3366.4 US gallons
4. 77.05 P.C.F.
5. 15 x 2240 = 58.33 PSI 24 x 24
6. 24 x 18 x 3 = 1,512 lbs force 7. 908.4 ltr/min
8. 2,204 lbs
9. 2240 – 2204 = 36 lbs 10. 90 + 460 = 550°R
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 2
1. Express the following fractions as decimals.
2. Round off the following to 2 places of decimal.
a) .6356 b) .7945 c).7987 d) .8429 e) .6464
3. Calculate the circumference of a circle with a diameter of 6 inches.
4. Calculate the area of a circle with a diameter of 6 inches.
5. Calculate the annular area between a 13 inch inside diameter pipe and a 5 inch outside diameter pipe.
6. Calculate the square roots of.
a) 69 b) 138 c) 276 d) 552
7a. Calculate the capacity of a tank in US BBLS with the following dimensions.
a) 5/8 b) 11/16 c) 7/9 d) 23/32 e) 5/24
15’ long by 6’ wide by 8’ deep
7b. What volume would be in the tank if the liquid height was 1 foot?
7c. How much volume has been added to the tank if during drilling operation the lever rose by 1 foot 5 inches?
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 2 (Cont)
8. What would be the capacity in BBLS of a sand trap?
SAN Sand
Trap D TRAP
8’ deep 10’ long by 10’ breath and is triangular in shape.
9a. What percentage of 186 does 42 make up?
9b. What percentage of 93 does 56 make up?
9c. What percentage of 56 does 60 make?
10.What would be the volume of a 1,500 foot annulus between 5 inch pipe and 17 ½ inch hole?
DRILLING CALCULATIONS COURSE
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3
Casing shoe depth 10,200
e) What would be the volume in the drill pipe?
2. Calculate the pump output per stroke of a triplex pump with a 12” stroke and liner size of 6” at 98% volumetric efficiency.
1. From the following well information calculate the volumes:
Hole size 12 ¼” hole Hole depth 12,650’
Casing size 13 3/8” ID 12.46”
5” pipe ID 4.2”
a) What is the hole volume with no pipe in the hole?
b) What would be the liquid volume with 5 inch pipe in the hole from top to bottom?
c) What volume would be in the drill pipe open hole annulus?
d) What volume would be in the drill pipe casing annulus?
3. Calculate the pump output in BBL/STK of a triplex cement pump with 5 inch liners and an 8 inch stroke.
Use a volumetric efficiency of 95%.
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 (Cont)
4b. What is the annular capacity of 8” drill collars in 12 ¼” hole?
5. How long would it take to circulate an annular volume of 950 BBLS using the pump in 4a?
7. Determine the new pressure required by increasing the pump rate from 60 to 65 strokes/min.
8. Determine the new pump pressure required to pump a lighter fluid at the same rate. 10.5 PPG mud is being pumped at 80 STKS/MIN at 3,000 PSI.
The weight is being reduced to 9.8 PPG.
4a. What would be the fluid output per minute of a triplex pump running at 80 strokes per minute, with 6.25” liners and 98% volumetric efficiency?
4c. What would be the annular velocity of the fluid passing round the drill collars?
6. What would be the maximum pressure that could be reached pumping at 400 gallons/minute with a pump of 750 hydraulic horse power?
Pump pressure at 60 STK/MIN was 2,650 PSI.
9. Calculate the total cross sectional area of three jet nozzles 16/32, 16/32 and 14/32.
DRILLING CALCULATIONS COURSE
1c. Drill pipe open hole annular capacity length (12.252 - 52) x 2,450 = 297.8 BBLS 1029
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 ANSWERS (Cont)
1d. Drill pipe casing and capacity x length (12.422 – 52) x 10,200 = 1,281.3 BBLS 1029
1e. Drill pipe capacity x length
(4.22) x 12,650 = 216.9 BBLS 1029
2. Volume in BBLS for 1 ft of 6 inch diameter
= 62 = .034985 BBLS/Cylinder 1029
For 3 cylinders = .10495 BBLS/STK At 98% volumetric efficiency
.10495 x 9.8 = .103 BBLS/STK 100
DRILLING CALCULATIONS COURSE
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 ANSWERS (Cont)
5. Annular volume (BBLS)
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 ANSWERS (Cont)
8. New Pressure = Old Pressure x New Mud Old Mud = 3,000 x 9.8
10.5 = 2,800 PSI
9. Area = .7854 x (Diameter2) 16/32 Jet = .7854 x (16/322)
= .19635 sq”
Two = . 3927 sq”
14/32 = .7854 x (14/322) = .1503 sq “
Total area = .3927 + .1503 = .543 sq “
DRILLING CALCULATIONS COURSE
2. Calculate for a balanced cement plug.
a. Slurry volume.
c. Pump stroke to displace.
Drill pipe size 5” cap .01738 BBL/FT Water ahead of cement 10 BBLS Hole depth 10,350’
Calculate:
Slurry volume with 10% excess
of Sacks
c. Pump stroke to pump the plug for the following single stage cement job.
Hole size 17 ½”
Casing size 13 3/8” set at 4,600 FT Yield 1.05 cubic FT/SACK
Float set 80’ above shoe
b. Volume of water behind the cement.
Hole size 8 ½”
Cement plug height – 500 FT Pump output .103 BBLS/STK
DRILLING CALCULATIONS COURSE Volume between float and shoe
= 80 x (12.422) 1029
= 12 BBLS
Total CMT requirements with no excess = 581 BBLS With excess = 639 BBLS
DRILLING CALCULATIONS COURSE
2b. Height of the water placed ahead when in annulus.
= 10 BBLS ÷ Annular Cap
= 10 ÷ (8.52 – 52) 1029
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 4 ANSWERS (Cont)
2b (cont)
= 10 ÷ .0459
= 218’
= 218 x .01738 Volume of water behind
= 218’ x DP CAP
= 3.78 BBLS
2c. Height of plug with pipe in the hole.
= 35.1 (0.459 + .01738)
= 555’
Add the height of the water in the pipe.
555 + 218
= 773
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 4 ANSWERS (Cont)
2c.(Cont)
= 166.45
Mud will be pumped to a depth of 10,350 – 773
= 9,577
Capacity of 9,577 of DP
= 9,577 x .01738 BBLS
Nos of STKS = 166.45 ÷ .103
= 1616 STKS
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 5
b. Depth 8,500’ (TVD) weight 15.2 PPG
From surface to 2,000’ – 9.2 PPG mud
1. Calculate the hydrostatic pressure exerted by the following columns of fluid.
a. Depth 12,000’ (TVD) weight 10 PPG c. Depth 17,200 (TVD) weight 17.8 PPG
2a. Calculate the mud weight in PPG that would give the following pressures at:
i) 5,000 PSI at 10,000 ft (TVD) ii) 2,325 PSI at 5,000 ft (TVD) iii) 10,950 PSI at 16,450 ft (TVD)
2b. What would be the increase in mud weight required to exert an additional 350 PSI hydrostatic pressure for the examples in (2a)?
3a. Calculate the hydrostatic pressure exerted by column of fluid made up of different densities
From 2,000’ to 3,000’ – 8.9 PPG From 3,000’ to 5,000’ – 16.4 PPG 3b. What mud weight would exert this pressure?
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 5 (Cont)
4. Calculate the relevant values and construct a step down chart for pumping kill mud down the drill pipe.
Well information:
Depth 8,200
b. Initial circulating pressure e. Chart
5. Calculate in cubic feet the volume a 10 BBL gas kick would occupy on surface. If the
6a. Calculate the formation strength (fracture pressure) from the following data.
Leak off pressure 1,200 PSI Mud weight 10 PPG
Shut in drill pipe pressure = 250 PSI
Slow circulating pressure = 850 PSI at 30 STK/MIN
Strokes required to pump from surface to the bit = 860 STKS a. Kill mud weight
c. Final circulating pressure d. Pressure drop per 100 STKS
original formation pressure was 5,300 PSI and atmosphere pressure is 14.75 PSI.
Shoe depth 6,200’ (TVD) Mud weight – 9.6 PPG
Give answers as a pressure and a pressure gradient.
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 5 (Cont)
6b. Calculate the maximum allowable annular surface pressure for 10 and 12 PPG mud respectively.
DRILLING CALCULATIONS COURSE
DRILLING CALCULATIONS COURSE
DRILLING CALCULATIONS COURSE
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 5 ANSWERS (Cont)
6. Fracture Pressure = Leak off plus hyst to shoe
= 1200 + 6200 x 9.6 x 0.52
= 4,295 PSI
Fracture pressure gradient = Fracture Pressure TVD Shoe Depth
= 4295 6200
= .693 PSI/FT
MAASP = Shoe depth (formation fracture gradient – mud gradient) For 10 PPG = 6,200 (.693 - .52)
= 1073 PSI
For 12PPG = 6,200 (.693 – 624)
= 428 PSI
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 6
1. Calculate the maximum hook load that can be applied with 12 lines
2. With the rig up in question if the string weight was reading 350,000lbs.
With depth 12,500’
300’ DC weight in mud = 40,000 lbs Block weight = 28,000 lbs
13/8” wire with a breaking strain of 167,000 and a design factor of 3.
What would be the loading on the fast line?
3. Calculate the ton miles for a round trip.
12,200’ of DP weight in mud = 200,000 lbs
DRILLING CALCULATIONS COURSE
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 6 ANSWERS (Cont)
3. For a round trip
Move 12,200 before they reach surface.
= 2.31 Miles
Ton Miles = 40,000 x 2.31
2,000
= 46.2 Ton Miles
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 6 ANSWERS (Cont)
R/Trip 300’ Weighing 40,000 40,000 x 300
2,000 5,280
= 1.14 Ton Miles
Round trip ton miles for the blocks.
Weight = 28,000 = 14 Tons
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 7
1. Calculate the buoyancy factor for:
a. 12 PPG
b. 14 PPG
c. 16 PPG
2a. How many 30’ drill collars would be required if 60% of the available collar weight is 20,000 lbs?
8” drill collars in 11.8 PPG mud.
Weight = 146 lbs/ft
2b. Where in the drill calculations would be the neutral point if 18,000 lbs was being applied to the bit?
DRILLING CALCULATIONS COURSE
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 7 ANSWERS (Cont)
Length = 40,600
146
278’ of DC
Drill Calcs = 278 = 9.26 30
Run 9
=
2b. Length required to make up 18,000 lbs 18,000
Buoyed Wt of one Ft of DC = 18,000
.821 x 146
= 150’
Neutral point is 150’ above the bit.
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 8
26,500 ft lbs
1a. What volume of pill is required to fill the drill collar annulus and leave 20% of that volume in the pipe?
Hole size 8 ½” Depth 12,200’
Drill Collars 6 1/4” OD 2 ¾ ID Drill Collar Length 360’
1b. How many strokes would be required to spot the pill then displace it out of the string?
Drill Pipe 5” Cap .01738 BBLS/FT Pump Output .102 BBL/STK
2. Calculate the line pull to apply the following torques using an effective tong length of 3.5 ft.
64,000 ft lbs 92,000 ft lbs
3. What would be the resulting density if 200 BBLS of 15.2 PPG mud was mixed with 150 BBLS of 12.6 PPG?
4. How much barite would be required to increase the mud weight in a system of 950 BBLS from 11 PPG to 11.6 PPG?
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 8 ANSWERS
1a. DC/OH Annular Cap = 8.52 – 6.252
1029
= .0323 BBL/FT
Total DC/OH Annular Cap = 36 x .0323
= 11.6 BBLS
Plus 20% = 13.9 BBLS
1b. String Capacity = 360 x (2.752) 1029
Plus = 11,840 x .01738
= 2.65 BBLS + 205.78 = 208.4 BBLS
Volume to leave 20% in string
= 208.4 – 2.3
= 206.1 BBLS
DRILLING CALCULATIONS COURSE
Plus 23 STKS to displace 20%
2. 7,571 lbs Pull