completions calcs

CAPACITY

The Diameter of a circle = 2 x RadiusThe Radius= Diameter

2

Area of a circle= 3.14 x (Radius)^2

Area of a circle= 3.14 x Diameter ^22

Area of a circle = 3.14 x Diameter^24

Volume of a pipe = Area of circle x Height

Volume required to pressure up tubing = Volume of tubing x Pressure x 0.0000045

Eg. Vol of tubing 130 BBLSPressure testing to 3500 psi

Volume required to pressure up tubing = 130 x 3500 x 0.0000045 = 2.0475 BBLS

Volume required to pressure up tubing = 3500psi / 100 x 0.0003 x 130bbls = 1.365 BBLS

Capacity

1 BBL = 42 Gallons1 BBL = 5.6146 ft^31 ft^3 = 7.4805 Gallons1 ft^3 = 0.1781 BBLS1 Gallon = 0.02381 BBLS1 Gallon = 0.1337 ft^3

The capacity of a size of tubing can be worked out using the Cementing tables book.

Capacity = Depth x Capacity factor for tubing

Eg. If you have a well with 8500 ft of 5 1/2" 17 lb/ft tubing.

Capacity = (Factor obtained from Red Book)

This is a guide for refreshing your memory on some of the calculations in completions. The formula's are in red and bold text for ease.

8500 x 0.0232 = 197.2 BBLS

Diameter

Radius

CAPACITY

5 1/2" 17#

8500 ftTotal capacity = 197.2 + 47.68 = 244.88 BBLSRound up to 245 BBLS 4 1/2" 13.5#

11700 ftThere is also a rule of thumb for calculating the barrels per foot if red book isn't at handBBLS per ft = 0.0009714 x Diameter^2

To calculate an annular capacity : 9 5/8" 47#

7" 29#

9800 ftEg.

(Factor obtained from Red Book)251 BBLS

The Red book also has sections for calculating casing capacity and hole capacity, these are worked out using the same methods.

If there is more than one string in the hole, (step- down completion), calculate each string individually and then total the results.

Capacity of 5 1/2" = 8500 x 0.0232 =197.2 BBLS

Capacity of 4 1/2" = (11700 - 8500) x 0.0149 = 47.68 BBLS

Note: Diameter is in inches (OD) and is not as accurate since the weight of the tubing is not known

Capacity = Depth x Capacity Factor of Annulus

Capacity = 9800 x 0.0256 = 250.88 BBLS

DISPLACEMENT

Open ended Plugged

5 1/2" 17#

8750 ft

For open ended pipe :-

Displacement of tubing = displacement factor of tubing x depth

For plugged pipe :-

Total displacement = displacement of tubing + capacity of tubing

Total displacement = 54 + 203 = 257 BBLS

Displacement of plugged tubing = Capacity of hole = capacity factor of hole x depth


Displacement occurs when running tubing into a well, the metal displaces some of the completion fluid. It's handy to know so that the gains across the trip tank are expected and not the beginning of a kick.

When calcualting displacement you MUST know if the tubing is open ended or plugged as it affects your calculation.

Displacement = 0.00619 x 8750 = 54.1625 BBLS

The displacement of the actual tubing is the same as for open eneded in this case 54 BBLS, but to find the total displacement you have to work out the capacity of the tubing and add the two together.

Capacity = 0.0232 x 8750 = 203 BBLS

Note : Another way to work out displacement for plugged tubing is to use the formula for capacity of hole. This is an easier way as you have only one formula to work out.

Capacity of 5 1/2" hole = 0.0294 x 8750 = 257.25 BBLS

HYDROSTATICS

Hydrostatic pressure

Fresh water = 8.33 PPGFresh water = 0.433 psi/ftFresh water = 62.4 lbs/cu.ftFresh water = API gravity 10Fresh water = Specific Gravity 1

Hydrostatic pressure = gradient x true vertical depth

Eg. A well is full of 10.4 PPG mud, what is the hydrostatic pressure at 6700 ft ?Since we know the weight of the fluid we can work out it's gradient.Gradient of mud = (Properties of fresh water)

8.33Gradient of mud = 0.5406 psi/ft (round up to 0.541 psi/ft)Hydrostatic = 0.541 x 6700 = 3624.7 psi

Eg. A well is filled with 50 lbs/cu.ft oil, what is the hydrostatic pressure at 7400 ft ?Gradient = (Properties of fresh water)

62.4Gradient = 0.3469 psi/ft (round up to 0.347 psi/ft)Hydrostatic = 0.347 x 7400 = 2567.8 psi


0.433 x 10.4

0.433 x 50

TENSILE STRENGTH

Tensile strength of pipeThis is in your red book but if you don't have one this is how you would work it out

Yield = grade x cross sectional areaEg. 5 1/2" 17# tubing L80

OD area =3.14 x (5.5)^2 = 23.74 sq.ins4

ID area = 3.14 x (4.892)^2 = 18.78 sq.ins4

C.S.A. = 23.74 - 18.78 = 4.96 sq.ins

Yield = 80,000 x 4.96 = 396800 lbs


TUBING STRETCH

Pipe Stretch

From the red book there is a table for calculating pipe stretch using the tables provided :-

Stretch = L x P x C L = Length of pipe in feet1000 x 1000 Stretch is in inches

P = pull on the pipeC = constant from red book

Eg. 9640 ft of 5 1/2" 17# tubing with a 25000 lbs pull on it will stretch ?

Stretch =1000 x 1000

Stretch = 19.37 inches

Using Hooke's Law

Stretch = F x L Stretch is in inchesA x E F = force pulled in pounds

A = cross-sectional area

Stretch = 25000 x (12 x 9640) = 2892000000 = 19.43 inches4.96 x 30000000 1.49E+08

Stretch due to changes in temperature

To calcualte this change you must first work out the average temperature :-

Ta = Ts + BHT2

Ta = average temperatureTs = surface temperatureBHT = bottom hole temperature

Eg. If the surface temperature is 65F and BHT is 225F, what's the average temperature ?

Ta = 65 + 225 = 145F2


9640 x 25000 x 0.0804

L = length of pipe in inches

E = elasticity of steel ( 30,000,000)

Temperature affects the tubing as an increase in temperature will cause it to increase in length or stretch, similarly a decrease will cause it to contract.

TUBING STRETCH

Now that we know the average temperature we can work out the change in length :-

e = L x C x Tc

e = change in pipe length due to temperature (inches)L = length of pipe (feet)C = coeeficient of expansion of steel = 0.0000828 in/ft/FTc = change in temperature = Ta - Ts If we have 12,000 feet of tubing, how much will it expand ?

Tc = 145 - 65 = 80Fe = 12000 x 0.0000828 x 80 = 79.488 inches

Stretch due to it's own weight

e = 72(L x L) [g-2w(l-m)]E

e = change in length due to weight ( inches )L = measured length of pipe ( feet )E = Modulus of elasticity of steel ( 30,000,000 )g = density of steel (lbs/ cu.in ) = 0.2832061w = density of fluid (lbs/cu.in )m = constant ratio ( 0.28 )

If we have 12000 ft of tubing in 10.3 PPG mud, how much will it stretch ?

w = 10.3 = 0.0445887 cu. Ins231

L x L = 12000 x 12000 = 144000000

e = 72(144000000) [0.2832061-(2 x 0.0445887)(1 - 0.28)]30000000

e = 345.6 [ 0.2832061-( 0.0891774)(0.72)]e = 345.6 [ 0.2832061- 0.064207728]e = 345.6[0.218998372]e = 75.686 inches

SLACK-OFF

Slack -off

Slack = Weight down x packer depth x Slack-off factor1000 1000

Slack = 15000 x 50001000 1000

Slack =Slack = 12.75 inches

If the size of tubing is not in the red book it can be worked out by using this formula :-

Factor (unknown pipe) cross-sectional area (known pipe) x factor (known pipe)cross-sectional area (unknown pipe)

Eg. What is the slack off factor of 5 1/2" 17# tubing?

Cross sectional area of 3 1/2" 9.3# tubing = 2.59 sq. ins.Cross sectional area of 5 1/2" 17# tubing = 4.96 sq. ins.

Factor = 2.59 x 0.17 = 0.0894.96


If you are setting the completion in compresion you have to calculate how much slack off is required to get the desired weight down.

Eg. How much slack off is required to obtain 15000 lbs weight down on a packer at 5000 ft with 3 1/2" EUE tubing?

x 0.17

15 x 5 x 0.17

BUOYANCY

Buoyancy

Density of steel = 65.4206

The air weight of a tubing string is calculated by the length of the tubing multiplied by it's lbs/ft.Eg. 10,000 ft of 5 1/2" 17# tubing, if suspended in air would weigh :-Air weight = 10,000 x 17 = 170,000 lbs

When the tubing is suspended in a well, the weight is reduced due to the tubing partially 'floating' To calculate the buoyancy factor we need to know the weight of the fluid in PP

Buoyancy Factor = 65.4206 - weight of fluid (PPG)65.4206

B.F. = 65.4206 - 10.365.4206

B.F. = 0.843

Buoyant weight = buoyancy factor x air weight

Buoyant weight of 5 1/2" = 170,000 x 0.843 = 143310 lbs

5 1/2" 17#9 5/8" casing

10.3 PPG 5 1/2" 17# tubing

10,000 ft

Eg. Using the same example from above :-

Cross-sectional area of 5 1/2" 17# = 4.96 sq.ins (worked out previously)Hydrostatic pressure of 10.3 PPG at 10,000ftGradient = 10.3 x 0.433 = 0.535 psi/ft

8.33Hydrostatic = 0.535 x 10000 = 5350 psiForce pushing upwards = 5350 x 4.96 = 26536 lbsThe air weight is the force acting down = 170000 lbs


Eg. If the well is full of 10.3 PPG mud, what is the buoyancy factor and what is the buoyant weight of the 5 1/2" string?

Another way to calculate the buoyancy is by what is known as the pressure x area method. This is calculated from the hydrostatic pressure at the bottom of the string acting across the cross-sectional area of the tubing.

BUOYANCY

Buoyant weight = Air weight of tubing - Force acting on CSA at depth

Bouyant weight = 170000 - 26536 = 143464 lbs

5 1/2" 17#

10.3 PPG

10,000 ft

If the tubing is plugged then the second method must be used to work out it's buoyant weight.

Eg. Same conditions as previously except the tubing is plugged.

Hydrostatic pressure = 5350 psi.Air weight = 170000 lbsArea = (5.5 x 5.5) x 3.14 = 23.75 sq.ins

4Force pushing upwards = 5350 x 23.75 =127062 lbs

Buoyant weight = 170000 - 127062 = 42938 lbs

The difference in the two answers is due to the first method taking into account the upsets at every joint, for this calcualtion the difference can be considered negligible.

SPACEOUT

HOP 0.86 ft

Pup Joint 18.96 ft

Pup Joint

6.5 ftDrillfloor

37.83 ft

97.6 ft38.97 ft

39.96 ft Wellhead (HOP)

36.44 ft

Pipe In Pipe OutHanger 25.74 ft 31.33 Joint length below drillfloorElevation 97.6 ft 38.97 (70.3 ft)

39.96 (101.26 ft)36.44 (137.7 ft)

123.34 ft 137.7 ft


The HOP (Hang-off Point) is where the hanger lands out in the wellhead. It is used as a reference point for working out the spaceout as it is a known distance from the drillfloor.

5.92 ft ( Make up loss has been taken off length)

The length that we require is from the HOP to the end of the assembly *Note. Remember to take the make up loss off of the pin end. In this example the total length is 25.74 ft.

When running the completion the customer may require to tag a restriction in the well for correlation so that the completion is spaced out, for example 5 feet from that restriction.

In this instance the tubing would be run until the restriction is tagged and then picked back up the 5 feet required, the pipe would then be marked and the spaceout calculated from there.

So the spaceout required is the difference between the pipe in and pipe out, in this case 137.7 ft - 123.34 ft = 14.36 ft. A pup joint of around this length or a combination of pups to make up this difference is required.

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