BOP TESTING PROCEDURES

I was asked about BOP testing procedures from my junior staff on the rig. This is very simple but important question; therefore, I would like to share with you as well about the BOP testing procedure.

Note: Reference is based on API RP 53.

General BOP testing procedures

• Use water to test BOP’s. The reason is that water has less compressibility than mud.

• Make up testing assembly and set in into a wellhead profile. Ensure that the casing valve must be left opened and there must be personnel monitoring the outlet of casing valve all time while testing. You must ensure that personnel who monitor the outlet must stay for from the BOP while it is being tested. The reason behind this step is to prevent pressure build up in the casing if the test plug is leaking.

• Circulate through choke/kill lines, choke manifold, standpipe manifold, and valves to ensure that all lines are full with water. This practice is for preventing pressure dropping off while testing.

• Line up cement unit and rig team shut rams and valves as per each rig specific testing sequence

• Pressure test must be low and high, respectively, and the pressure should be stabilized with minimum bleed off at least 5 minutes. Ensure that pressure recording on a chart is recorded correctly.

• Ensure that any equipment doest not pass a pressure test requirement must be reported to supervisors.

• Continue pressure testing until all equipment is tested as per each rig specific.

• Rig down testing assembly.

BOP testing periodFunction test BOP: Every week, personnel must perform function test BOP alternating between remote panels.

Full BOP test:There are 3 categories that you should consider for full BOP test.

1. Prior to supping the well or the first time that BOP is installed on the well.

2. After repairing or disconnecting of any pressure sealing elements of BOP.

3. As per MMS, you can use BOP for 21 days (3 weeks) before you need to test it again. Or you must follow the local regulations. For example, in the North Sea, they allow you only 14 days before next BOP test.

High pressure test for RAMS and auxiliary equipmentHigh pressure test to rated working pressure of RAMS BOP and auxiliary equipment OR rated working pressure of your wellhead. Select the lower number. For example, you BOP is rated for 10,000 psi and your wellhead is rated for 6500 psi. Your high pressure test for the RAMS BOP and auxiliary equipment must be only 6500 psi.

High pressure test for Annular PreventerHigh pressure test should apply to 70% of rated working pressure or RAMS test pressure. Select the lower number. For example, you test your ram at 5,000 psi and the annular working pressure is 6,500 psi. You need to test your annular to 3500 psi (70% of rams testing pressure).

Low pressure test for both annular and RAMs preventer

200 – 300 psi should be applied for a low pressure test and the period is 5 minutes.

Kick Tolerance Concept and Calculation for Well Design

Kick tolerance is the maximum gas volume for a given degree of underbalance which the circulation can be performed without exceeding the weakest formation in the wellbore. This article is the extended version of Kick Tolerance Calculation  which will explain more on this topic. It is very critical that drilling personnel understand its importance to well design and drilling operation.

There are two important factors used for determining the kick tolerance

• Kick Intensity – It is the different between the maximum anticipated formation pressure and planned mud weight. For example, the planned mud weight is 13.0 ppg and the possible kick pressure is 13.5 ppg. Therefore, the kick intensity is 0.5 ppg (13.5 – 13.0).

A zero kick intensity (swabbed kick scenario) should be used for a know area where you have less uncertainty about an overpressure zone.

 • Kick Volume – It is a gas influx entering into the wellbore from the formation. Gas kick is always used for well control calculation because it is the worst case scenario. The kick volume should be realistic figure which personal can detect the influx on the rig. In a larger hole, it allows bigger influx volume than a small hole.

Maximum Allowable Annular Surface Pressure (MAASP) and Kick Tolerance

Weakest formation point in the open hole is assumed to be at the shoe depth of the previous casing. The well bore will be fractured if a summation of hydrostatic and surface pressure exceeds the weakest pressure (Leak Off Test pressure). The maximum surface pressure before breaking the formation is called “Maximum Allowable Shut In Casing Pressure” (MASICP).

Make it simpler for your understanding. MASICP is the total of kick tolerance budget. It consists of pressure from kick intensity and hydrostatic pressure loss due to gas.

Kick Tolerance Example Calculation

Previous casing shoe (9-5/8” casing) at 6,000’ MD/ 6,000’ TVD

Predicted formation pressure at TD (10,000’MD/10,000’TVD) = 14.0 ppg

Pore pressure uncertainty = 1.0 ppg

Planned mud weight = 14.5 ppg (0.754 psi/ft)

Gas gradient = 0.1 psi/ft

LOT = 16.0 ppg

Hole size = 8-1/2”

Drill Pipe = 5”

BHA + Drill Collar = 7”

Length of BHA+Drill Collar = 400 ft

Annular capacity between open hole and BHA = 0.0226 bbl/ft

Annular capacity between open hole and 5” DP = 0.0459 bbl/ft

Calculation Steps

Maximum anticipated pressure = 14.0 + 1 = 15.0 ppg

Maximum Allowable Shut In Casing Pressure (MASICP) = (LOT – MW) x 0.052 x Shoe TVD

Maximum Allowable Shut In Casing Pressure (MASICP) = (16 – 14.5) x 0.052 x 6,000 = 468 psi

Kick Intensity = 15.0 – 14.5 = 0.5 ppg

Underbalance due to kick intensity = 0.5 x 0.052 x 10,000 = 260 psi

As you can see, when the well is in underbalance condition (260 psi), the shoe will not be broken because the MASICP is more than underbalance pressure (468 > 260).

We know that 0.5 ppg kick intensity we will have 208 psi (468 – 260 = 208 psi) before shoe broken.

It means that gas bubble can replace mud in equivalent to 208 psi before fracturing the shoe. With this relationship, we can determine height of gas kick by the following equation.

Height of gas kick = remaining pressure, psi ÷ (mud gradient, psi/ft – gas gradient, psi/ft)

Height of gas kick = 208 ÷ (0.754 – 0.1) = 318 ft.

Determine gas kick volume base on height of gas kick

We need to separate into two cases and compare the smallest volume.

1st case – Gas at the bottom

Volume of gas kick = Annular capacity between open hole and BHA x Height of gas kick

Volume of gas kick (bbl) = 0.0226 bbl/ft x 318 ft = 7.2 bbl

 2nd case – Gas right below casing shoe

For this case, we need to convert gas at the shoe to the bottom condition by applying Boyle’s Laws.

Volume of gas kick = Annular capacity between open hole and 5” DP x Height of gas kick

Volume of gas kick (bbl) = 0.0459 bbl/ft x 318 ft = 14.6 bbl

Convert to the bottom hole condition

Volume at the bottom = (volume of gas kick at shoe x Leak off test) ÷ formation pressure

Leak off test = 0.052 x 16 x 6,000 = 4,992 psi

Formation pressure (gas kick condition) = 0.052 x 15 x 10,000 = 7,800 psi

Volume at the bottom = (14.6 x 4992) ÷ 7800 = 9.3 bbl

We can compare the kick volume from two cases like this.

1 st case : kick volume = 7.2 bbl

2nd case : kick volume = 9.3 bbl

The smallest number must be selected to represent maximum kick volume therefore kick volume is 7.2 bbl.

Drill Pipe Elongation Due to Temperature

Drill pipe elongation will occur because the higher bottom hole temperature.

Rule of Thumb

Pipe will elongate approximately 0.83 inc per 100 ft of length, per 100 degree F increase in Temperature.

We need to know the surface temperature and bottom hole temperature in order to determine pipe elongation.

The following formulas are used to determine pipe elongation.

Average Temperature = (Bottom Hole Temperature + Surface Temperature) ÷ 2

∆ Temperature = Average Temperature – Surface Temperature

Pipe Elongation = (L ÷ 100) × (∆ Temperature ÷ 100) × 0.83

 

Where

Average Temperature, Bottom Hole Temperature and Surface Temperature are in F.

Pipe Elongation is in inch.

L is total length in feet.

Example

Determine pipe elongation from the following information.

Total length of pipe is 10,000 ft.

Surface temperature is 80 F.

Bottom hole temperature is 375 F.

 

Average Temperature = (80 + 375) ÷ 2 = 227.5 F

∆ Temperature = 227.5 – 80 = 147.5 F

Pipe Elongation = (10,000 ÷ 100) × (147.5 ÷ 100) × 0.83 = 122.43 inch

Cement Transition Period in The Oil Well Can Cause Well Control Situation

When cement is in transition period (forming the bond), you will lose some hydrostatic pressure because cement becomes solid phase therefore water in the cement will provide hydrostatic pressure. In many cases happened, there is no issue while performing the cement job however once the cement is set after period of time, there is a casing pressure indicating that there is hydrocarbon in the annulus.

This example will demonstrate you why the well is in underbalance condition while waiting on cement.

Well information: Previous 9-5/8” casing shoe at 4000’ ft.

The vertical well (8.5” hole) is drilled to TD at 10,000ft with 12.1 ppg mud and the pay zone is at 9,800’ TVD with reservoir pressure of 11.6 ppg. The cement is planned to cover 3,500 ft in the annulus above the casing shoe. Water used to mix cement is 8.3 ppg weight.

 

 

Will the well go underbalance during the cement in transition period?

 Formation pressure at 9,800 ft

Formation pressure at 9,800 ft = 0.052 x 11.6 x 9800 = 5,911 psi

Total hydrostatic pressure in the annulus

Hydrostatic pressure of drilling mud at 6,500 ft = 0.052 x 12.1 x 6,500 = 4,090 psi

When cement in transition period, only water in cement will provide hydrostatic pressure. So we can calculate hydrostatic pressure of water in cement

Hydrostatic pressure of water in cement = 0.052 x (9800 – 6,500) x 8.3 = 1,424 psi

While cement is in the transition period, total hydrostatic pressure in the annulus is equal to hydrostatic pressure of mud plus hydrostatic of water in cement

Total hydrostatic pressure in the annulus = 4,090 + 1,424 = 5,514 psi

 

You will see that during the transition period total hydrostatic pressure in the annulus is less than formation pressure therefore the well. For this case, the well will flow during transition period.

 

Equivalent Circulating Density (ECD) Using Yield Point for MW More than 13 ppg

Another equation to determine equivalent circulating density is to use yield point.

This formula below is used to calculate the ECD and it is good for mud weight more than to 13.0 ppg

Where:

ECD is equivalent circulating density in ppg.

MW is mud weight in ppg.

Hold ID is inside diameter of hole in inch.

Pipe OD is outside diameter of pipe in inch.

YP is mud yield point.

YP can be calculated by these following equations.

YP = Reading at 300 rpm – PV

PV = Reading at 600 rpm – Reading at 300 rpm

AV is annular velocity in ft/min

AV can be determined by the following equation.

AV in ft/min = (24.5 x Q) ÷ (Dh2 – Dp2)

where

Q = flow rate in gpm

Dh = inside diameter of casing or hole size in inch

Dp = outside diameter of drill pipe, drill collars, or tubing in inch

Determine the ECD with the following information

MW = 13.5 ppg

Reading at 300 = 25

Reading at 600 = 40

Hole diameter = 6.2

Pipe diameter = 4

Flow rate = 200 gpm

 

AV in ft/min = (24.5 x 200) ÷ (6.22 – 42) = 218.4 ft/min

PV = 40 – 25 = 15

YP = 25- 15 = 10

ECD = 14.2 ppg

Equivalent Circulating Density (ECD) Using Yield Point for MW less than 13 ppg

Another way to determine equivalent circulating density (ECD) is to use yield point.

This formula below is used to calculate the ECD and it is good for mud weight less than or equal to 13.0 ppg.

Where:

ECD is equivalent circulating density in ppg

MW is mud weight in ppg

Hold ID is inside diameter of hole in inch

Pipe OD is outside diameter of pipe in inch

YP is mud yield point

YP can be calculated by these following equations

YP = Reading at 300 rpm – PV

PV = Reading at 600 rpm – Reading at 300 rpm

Determine the ECD with the following information

MW = 9.2 ppg

Reading at 300 = 25

Reading at 600 = 40

Hole diameter = 6.2

Pipe diameter = 4

PV = 40 – 25 = 15

YP = 25- 15 = 10

ECD = 9.7 ppg

Do you have too much slug in the well?

Slug is heavy drilling mud that is used to pump when you want to pull pipe dry. However, excessive volume of slug in the well can create higher mud weight and cause the problem. Today I would like to share my experience regarding slug in the well.

The situation happened on the drilling rig like this.

The well was TD with 13.0 ppg mud and circulation was conducted until the shale shakers. After that we pull 10 stands wet and pumped 40 bbl of 15.0 ppg slug and came out of hole without any problem. Then we run the logging and we got stuck. We picked the fishing gear and grabbed the fish. We pulled out 10 stands wet and hole was taking proper fill. We pumped same amount of slug 40 bbl of 15.0 ppg and came out of hole. The decision was made by town to trip in hole to do the wiper trip.

Tripping was done with caution and break circulation was conducted every 3000 ft to break the gel. Prior to TD 100 ft, we made up top drive and slowly washed down to TD. While we were circulating, the mud weight out was varied from 13.2 ppg to 14.2 ppg. Additionally, we start losing mud while circulating.

Do you know what went wrong?

The two slugs in the well cause us trouble.

How do you know that?

Let’s do the math. We have total volume in the system (downhole volume and surface volume) of 700 bbl and we have total of 80 bbl of 15.00 pgg of slug.

The following formula is used to determine the final mud weight in the system.

(V1 x D1) + (V2 x D2) = VF x DF

Where;

V1 = volume of fluid 1 (bbl, gal, etc.)

D1 = density of fluid 1 (ppg,lb/ft3, etc.)

V2 = volume of fluid 2 (bbl, gal, etc.)

D2 = density of fluid 2 (ppg,lb/ft3, etc.)

VF = volume of final fluid mix

DF = density of final fluid mix

EXAMPLE   :

V1 = 80 bbl (Total volume of slug in the well)

D1 = 15.0 ppg (density of slug)

V2 = 700 bbl (Total system volume)

D2 = 13.0 ppg (density of drilling mud)

VF = V1 + V2 = 80 + 700 = 780

(80 x 15) + (700 x 13) = 780 x DF

DF = 13.2 ppg

From the equation above, it tells us that we have too much slug in the well which can increase 0.2 ppg of total mud weight. The increment of mud density will directly affect the equivalent circulating density. This is the reason why we have losses while circulating.

Total Bit Revolution in When Running A Drilling Mud Motor and Rotating At Surface

There are two components contributing in the total revolution of downhole drilling bit using mud motor.

1. The rotor RPM of drilling mud motor– each drilling mud motor has its specific performance data which will tell you how many revolutions per flow rate flowing through it.

For example (see the image below), it shows Rev/Gal which means the rotor will turn 0.147 turn/min per one gallon/min of drilling mud passing thought.

2. The rotary speed from top drive or rotary table.

The total drilling bit revolution downhole:

The total drilling bit revolution is equal to summation of the rotor RPM at specific flow rate and the rotary speed on surface.

Please follow the example below to determine the bit revolution down hole.

The performance data of mud motor -

Rev/Gal = 0.147

Flow rate = 320 gpm

Rotary speed = 220 rpm

The rotor rpm based on 320 gpm = 0.147 x 320 = 47 rpm

Total bit revolution down hole = 47 + 220 = 267 revolution per minute (rpm)

How To Determine Hole Size By Fluid Caliper

This article will demonstrate you how to determine hole size by fluid caliper. First of all, you need to know what is the fluid caliper is. The fluid caliper is one simple way to calculate hole diameter. The concept is based on hole volume, annular capacity, and inner capacity. You pump any substance and when you seen it on surface, you back calculate the hole size based on strokes pumped. It is quite tricky to explain so I would like you to see the example below which it will make you clear about what I say.

Assumption for the calculation: No losses

 

7” casing shoe is set at 5000’MD/4500’TVD

7” casing 23 ppf, ID 6.33 inch

4” drill pipe is used to drill the well and its ID is 3.34 inch.

4-3/4” drill collar is used as BHA and its ID is 2.5 inch. The length of drill collar is 500 ft.

This hole section is used water based mud and the bit size is 6-1/8”

Drill to section TD at 10,000’MD/9,000’TVD then drop carbide once TD is reached.

Pump output is 0.1 bbl/stroke.

Surface line from pumps to rig floor is 20 bbl.

Carbide is detected by a gas sensor on surface after 4,000 strokes pump.

With the given information, determine what is the hole size of the open hole section.

The basic drilling formulas that you need to know are inner capacity and annular capacity calculation.

1st step – Determine Inner Capacity and Annular Capacity of All Parts

Inner capacity of 4”DP

Inner capacity of 4”DP = 0.01084 bbl/ft

Inner capacity of 4-3/4”DC

Inner capacity of 4-3/4”DC = 0.00607 bbl/ft

Annular capacity between 4” DP and 7” Casing

Annular capacity between 4” DP and 7” Casing = 0.02383 bbl/ft

I assign “d” is the open hole diameter.

Annular capacity between 4” DP and open hole

Annular capacity between 4 3/4” DP and open hole

2nd step – Determine Volume of All Parts

Volume in 4”DP = Inner capacity of 4”DP x length of 4” DP

Volume in 4”DP = 0.01084 x 9,500 = 103 bbl

Volume in 4-3/4”DC = Inner capacity of 4-3/4”DC x length of 4-3/4”DC

Volume in 4-3/4”DC = 0.00607 x 500 = 3 bbl

Volume between 4”DP and 7” Casing = Annular capacity between 4” DP and 7” Casing x length of 4” DP inside 7” casing

Volume between 4”DP and 7” Casing = 0.02383 x 5,000 = 119 bbl

Volume between 4”DP and open hole = Annular capacity between 4” DP and open hole x length of 4” DP inside open hole

Volume between 4-3/4”DC and open hole = Annular capacity between 4-3/4”DC and open hole x length of 4-3/4”DC inside open hole

3rd step – determine hole size. The total volume pump is equal to total volume in the system.

With the pump output of 0.1 bbl/stroke, 4000 strokes equate to 400 bbl.

In the following equation, I account for the surface volume from the pump to the rig floor.

Solve the equation to get d, d is equal to 6.97”.

Answer: You will get the hole size based on the fluid caliper of 6.97 inch.

I wish this article will give you idea on how to calculate hole size based on the fluid caliper.

MARGIN OF OVERPULL IN DRILLSTRING

Margin of overpull is additional tension to be applied when pulling the stuck drill string without breaking the tensile limit of the drill string. This is the difference between maximum allowable tensile load of drill string and hook load.

The formula for margin of overpull is described below;

Margin of Overpull = Ta – Th

Where;

Ta is the maximum allowable tensile strength, lb.

Th is the hook load (excluding top drive weight), lb.

The ratio between Ta and Th is safety factor (SF).

SF =Ta ÷ Th

Example: The drill string consists of the following equipment:

5” DP S-135, 4-1/2” IF connection, adjusted weight of 23.5 ppf = 8,000 ft

5” HWDP S-135, 4-1/2” IF connection, adjusted weight of 58 ppf = 900 ft

Mud motor and MWD, weight 20 Klb, = 100 ft

Expected hook load at TD = 270 Klb

Tensile strength of 5” DP S-135 (premium class) = 436 Klb

Tensile strength of 5” HWDP S-135 (premium class) = 1,100 Klb

90% of tensile strength is allowed to pull without permission from town.

Determine the margin of overpull from the information above.

Maximum tension will happen at the surface so 5”DP will get the most tension when pulling and since only 90% of tensile strength is allowed. The allowable tensile (Ta) is as follows;

Ta = 0.9 x 436 = 392 Klb

Th = 270 Klb at TD

Margin of over pull = 392 – 270 = 122 Klb

Safety Factor = 392 ÷ 270 = 1.45

Margin of overpull is additional tension to be applied when pulling the stuck drill string without breaking the tensile limit of the drill string. This is the difference between maximum allowable tensile load of drill string and hook load.

The formula for margin of overpull is described below;

Margin of Overpull = Ta – Th

Where;

Ta is the maximum allowable tensile strength, lb.

Th is the hook load (excluding top drive weight), lb.

The ratio between Ta and Th is safety factor (SF).

SF =Ta ÷ Th

Example: The drill string consists of the following equipment:

5” DP S-135, 4-1/2” IF connection, adjusted weight of 23.5 ppf = 8,000 ft

5” HWDP S-135, 4-1/2” IF connection, adjusted weight of 58 ppf = 900 ft

Mud motor and MWD, weight 20 Klb, = 100 ft

Expected hook load at TD = 270 Klb

Tensile strength of 5” DP S-135 (premium class) = 436 Klb

Tensile strength of 5” HWDP S-135 (premium class) = 1,100 Klb

90% of tensile strength is allowed to pull without permission from town.

Determine the margin of overpull from the information above.

Maximum tension will happen at the surface so 5”DP will get the most tension when pulling and since only 90% of tensile strength is allowed. The allowable tensile (Ta) is as follows;

Ta = 0.9 x 436 = 392 Klb

Th = 270 Klb at TD

Margin of over pull = 392 – 270 = 122 Klb

Safety Factor = 392 ÷ 270 = 1.45

Drill Collar Weight Calculation To Prevent Drill Pipe Buckling

Drill collar provides weight to the bit for drilling and keep the drill string from buckling. Additionally, drill pipe should not run in compression because it can get seriously damaged therefore we need to know weight of drill collar that is enough to provide weight to the bit.

 

Drill pipe buckle due to insufficient of drill collar

Drill pipe straight because of sufficient drill collar weight

Drill collar weight in a vertical well

The following formula is used to determine required drill collar weight to obtain a desired weight on bit for a vertical well.

WDC = (WOB x SF) ÷ BF

Where

WDC is drill collar weight in air, lb.

WOB is a required weight on bit, lb.

SF is a safety factor.

BF is mud buoyancy factor.

Drill collar weight in a deviated well

In a deviated well, the drill collar weight will not directly transfer to the bit because of well inclination which has direct affect on weight on bit.

The following formula is used to determine required drill collar weight to obtain a desired weight on bit for a deviated well.

WDC = (WOB x SF) ÷ (BF x COS (θ))

Where

WDC is drill collar weight in air, lb.

WOB is a required weight on bit, lb.

SF is a safety factor.

BF is mud buoyancy factor.

θ  is inclination of the well.

Example: The deviated well has inclination of 30 degree in tangent section and planned mud weight is 12.0 ppg. Safety factor for this case is 25%.

What is the drill collar weight to obtain the desired WOB of 50 Klb?

Buoyancy Factor = (65.5 – 12.0) ÷ 65.5 = 0.817

SF @ 25 % = 1.25

WDC = (50,000 x 1.25) ÷ (0.817 x COS (30))

WDC = 88,333 Klb

In this case, drill collar weight in the air should be 88.3 Klb. In reality, the BHA does not only have the drill collar so you need to adapt this figure. For instant, the BHA consists of mud motor, stabilizer, LWD and HWDP which have a total weight of 30 Klb. Therefore, the actual drill collar weight is just only 58.3 Klb (88.3 – 30).

Volumetric Well Control – When It Will Be Used??

Volumetric well control method is a special well control method which will be used when the normal circulation cannot be done. It is not a kill method but it the method to control bottom hole pressure and allow influx to migrate without causing any damage to the well.

There are several situations where you cannot circulate the well as follows:

• Pumps broken down

• Plugged drill string/bit

• Drill string above the kick

• Drill string is out of the hole completely

With the volumetric method, the volume of gas influx will allow migrating and casing pressure will increase till a certain figure then a specific amount of mud will bleed off to compensate the increase in casing pressure. The volumetric method will allow the kick to surface while the bottom hole pressure is almost constant. Successful use of volumetric method requires personnel understand three basic concepts –

1. Boyle’s Law – Boyle’s law states that at constant temperature, the absolute pressure and the volume of a gas are inversely proportional in case of constant temperature within a closed system. The illustration below demonstrates volume and pressure as per Boyle’s Law.

In term of mathematical relationship, Boyle’s Law can be stated as

P1 x V1 = P2 x V1

Where;

P1 = pressure of gas at the first condition

V1 = volume of gas at the first condition

P2 = pressure of gas at the second condition

V2 = volume of gas at the second condition

2. Hydrostatic pressure – Hydrostatic pressure is pressure created by column of fluid. Two factors affecting hydrostatic pressure are height of fluid and density of fluid.

Pressure at the bottom hole equals to hydrostatic pressure plus surface pressure

Pressure (bottom hole) = Hydrostatic Pressure + Surface Pressure

We will apply this concept to see how the gas bubble will increase the bottom hole pressure.

If the gas bubble is not allowed to expanded, the gas bubble in the well migrates up will act on the mud column below and increase bottom hole pressure. Increasing in the bottom hole pressure equates to hydrostatic pressure below the bubble.

Bottom hole pressure = Gas bubble pressure + Hydrostatic pressure below the bubble

 

If we don’t want increase in bottom hole pressure, mud need to be bled off the well while the gas migrating up and the casing pressure must increase to compensate loss of hydrostatic pressure from bleed off.

In the volumetric control, there are two ways to control bottom hole pressure while allowing the gas migrating up to surface.

1. Wait and let gas migrate. The migration of gas will increase bottom hole pressure and casing pressure.

2. Bleed off mud from the annulus. Mud that is bled off must be equal to the increase in bottom hole pressure.

Both steps above must be carefully performed perform in a sequence. We will go to the detailed procedures in later post.

3. Relationship of height and fluid volume as determined by annular capacity – In order to determine volume of mud that equates to required hydrostatic pressure, we need to understand annulus capacity. It tells us how many bbl per foot in annulus and it can be calculated by this following formulas:

Annular Capacity Factor (ACF) = (OD2-ID2) ÷ 1029.4

Where;

ACF = Annular Capacity Factor in bbl/ft

OD = Outside Diameter of Annular in inch

ID = Inside Diameter of Annular in inch

Once the ACF is know, we can determine Mud Increment (MI) which is the volume of mud bled off from the annulus to reduce the annular hydrostatic pressure by the amount of the pressure required.

Mud Increment (MI) can be calculated by this following equation:

Mud Increment (MI) = (PI x ACF) ÷ (0.052 x MW)

Where;

PI = Pressure Increment in psi

ACF = Annular Capacity Factor in bbl/ft

MW = Mud Weight in the well in ppg

Critical Flow Rate – Drilling Hydraulics

Critical flow rate is the flow rate at the transition point between laminar and turbulent flow. The first step of the critical flow rate determination is to figure out the critical velocity and then substitute it into the annular flow rate.

To get the point at the transition period, the critical Reynold Number for laminar flow must be around 3470 – 1370na. With this relationship, we can determine the critical velocity by rearranging the Reynold Number and Effective Viscosity equation.

The effective viscosity equation for critical velocity is listed below:

The Reynolds number equation for critical velocity is listed below:

The critical annular velocity equation is listed below:

Where;

Vc = critical velocity in ft/sec

Rec = Reynold Number in the annulus

µea = effective viscosity in the annulus, centi-poise

na = power law constant (flow behavior index)

Ka = power law constant (consistency factor)

W = mud weight, ppg

Dh = Diameter of wellbore, inch

Do = Outside Diameter of tubular, inch

After we get the critical velocity, we can figure out the critical flow rate by the following equation.

Where;

Qc = Critical flow rate, gpm

Vc = critical velocity in ft/sec

Dh = Diameter of wellbore, inch

Do = Outside Diameter of tubular, inch

Please follow the step-by-step calculation to learn how to determine the critical flow rate.

Power law constant (flow behavior index), na = 0.51

Power law constant (consistency factor), Ka = 6.63

Mud weight = 10 ppg

Hole diameter = 12-1/4”

Drill pipe size = 5”

Determine the critical annular velocity by substituting factors into this equation.

Vc = 3.82 ft/sec

Then we can figure out the critical flow rate.

Qc = 1172 GPM

Power Law Constants (n and K) Calculation

Many types of drilling mud are classified as non-Newtonian fluid which is the behavior between Newtonian fluid model and Bingham Plastic model. The relationship between shear rate and shear stress is defined by the power law model shown below:

Where

 = shear rate

K = consistency factor

 = shear rate

n = flow behavior index

n and K can be calculated from any two value of shear rate and shear stress. The method of reading shear rate on the rig comes from a V-G meter. Typically, 600 rpm, 300 rpm and 3 rpm are obtained from every mud test and we can use those reading to determine n and K.

The following equations are used to get the Power law constants (n and K).

Where

n = flow behavior index, dimensionless

K = consistency factor, poise

θ300 = reading at 300 rpm

θ3 = reading at 3 rpm

Example for Power Law constants (n and K) calculation

θ300 = 32

θ3 = 3

n= 0.514

K=0.514

Cutting Carrying Index – Simple Tool To Determine Hole Cleaning

To get the good idea on how good of the hole cleaning is, the another method called Cutting Carrying Index (CCI) is utilized. The CCI is an empirical relationship from real data and the equation is below:

Where;

AV is annular velocity in ft/min.

MW is mud weight in ppg.

K is a Power Law Constant.

The Power Law constant (K) can be calculated from the equation below:

Where;

PV is plastic viscosity in centipoises.

YP is yield point in lb/100sqft

n is flow behavior index.

The flow behavior index (n) can be determined by the following equation:

Where;

PV is plastic viscosity in centipoises.

YP is yield point in lb/100sqft

How will the CCI tell you about hole cleaning?

If CCI is equal to 0.5 or less, the hole cleaning is poor and the hole problem may be seen.

If CCI is equal to 1.0 or greater, it indicates that the hole cleaning is good.

Please follow the example to determine CCI.

Mud Weight = 9.2 ppg

Annular Velocity = 140 ft/min

Plastic viscosity = 17 cp

Yield Point = 15 lbs/100 sq ft

Determine “n”

n= 0.61472

Determine “K”

K = 353.71

CCI=(353.71(140)(9.2))/400,000

CCI = 1.14

Cutting Slip Velocity Calculation Method 1

Cutting slip velocity is velocity of cutting that naturally falls down due to its density. In order to effectively clean the hole, effect of mud flow upward direction and mud properties must be greater than cutting slip velocity (settling tendency of cuttings). Otherwise, cutting will fall down and create cutting bed.

You can learn more detail about it via this topic -> Cutting Slip Velocity

This calculation will show annular velocity, cutting slip velocity and net velocity so you can use as a reference for you hole cleaning indication.

There are 2 calculation methods and I will show the first method via this topic.

1. Determine annular velocity with following equation:

Where;

AV is annular velocity in ft/min.

Q is flow rate in gpm (gallon per minute).

Dh is diameter of hole in inch.

Dp is diameter of drill pipe in inch.

 

2. Determine cutting slip velocity with following equation:

Where;

Vs is cutting slip velocity in ft/min.

PV is plastic viscosity in centi-poise.

MW is mud weight in ppg.

Dp is diameter of cutting in inch.

DenP is cutting density in ppg.

 

3. Determine net rise velocity with following equation:

Net rise velocity = AV – Vs

Where;

AV is annular velocity in ft/min.

Vs Cutting Velocity in ft/min.

 

This figure indicates that cuttings are being lifted by mud or are still falling down.

If net rise velocity is positive, it means that you have good flow rate which can carry cutting in the wellbore.

On the other hand, If net rise velocity is negative , your current flow rate is NOT engough to carry cutting s.

Example: Please use the following information to determine annular velocity, cutting slip velocity, net rise velocity, and tell us if the flow rate is good for hole cleaning.

Information

Flow rate = 300 gpm

Hole Diameter = 6.3 inch

Drillpipe OD = 4 inch

PV = 15 cps

MW = 10 ppg

Diameter of cutting = 0.20 inch

Density of cutting = 20.0 ppg

 

1. Determine annular velocity:

AV = 310.3 ft/min

 2. Determine cutting slip velocity:

Vs = 35.4 ft/min

3. Determine net rise velocity with following equation:

Net rise velocity = AV – Vs

Net rise velocity = 310.3 – 35.4 = 274.9 ft/min

 Conclusion: This flow rate is GOOD

Cutting Slip Velocity Calculation Method 2

This is another method to determine cutting slip velocity. The process of calculation is quite different from the first method however it is still straight forward calculation. It still gives you the following answers: annular velocity, cutting slip velocity and net velocity.

 

Let’s get started with this calculation method.

 

1. Determine n

Where;

n is the power law exponent.

Θ600 is a value at 600 viscometer dial reading.

Θ300 is a value at 300 viscometer dial reading.

2. Determine K

Where;

K is the fluid consistency unit

Θ300 is a value at 300 viscometer dial reading.

n is the power law exponent.

 

3. Determine annular velocity with following equation:

Where;

AV is annular velocity in ft/min.

Q is flow rate in gpm (gallon per minute).

Dh is diameter of hole in inch.

Dp is diameter of drill pipe in inch.

 

4. Determine cutting slip velocity with following equation:

Where;

µ is viscosity in centi-poise.

V is annular velocity in ft/min.

Dh is diameter of hole in inch.

Dp is diameter of drill pipe in inch.

n is dimensionless number from the equation 1

K is dimensionless number from the equation 2

 5. Slip Velocity (Vs) in ft/min

Where;

Vs is slip velocity in ft/min.

DensC is cutting density in ppg

DiaC is cutting diameter in inch

µ is viscosity in centi-poise given from equation#4.

 

6. Determine net rise velocity with following equation:

Net rise velocity = AV – Vs

Where;

AV is annular velocity.

Vs Cutting Velocity

 

This figure indicates that cuttings are being lifted by mud or are still falling down.

If net rise velocity is positive, it means that you have good flow rate which can carry cuttings in the wellbore.

On the other hand, If net rise velocity is negative, your current flow rate is NOT enough to carry cuttings.

 

Example: Please use the following information to determine annular velocity, cutting slip velocity, net rise velocity, and tell us if the flow rate is good for hole cleaning.

 

θ300 = 25

θ600 = 40

Flow rate = 750 gpm

Hole size = 12.0 inch

Drill pipe diameter = 5 inch

Cutting diameter = 0.3 inch

Cutting density = 22 ppg

Mud weight = 10 ppg

1. Determine n

n = 0.678

2. Determine K

K = 0.365

3. Determine annular velocity 

AV = 154.4 ft/min

 4. Determine cutting slip velocity

µ = 53.88 centi-poise

5. Determine slip velocity (Vs)

Vs = 35.78 ft/min

6. Determine net rise velocity 

Net rise velocity = 154.4 – 35.78 = 118.6 ft/min

 Conclusion: This flow rate is good for hole cleaning because annular velocity is more than cutting slip velocity.

Dilution of Mud System to Control Low Gravity Solid by Adding Mud

Adding bbl of  drillingfluid can help control low gravity solid (LGS) in mud system. However, this is different from the way to control LGS by adding base fluid as base oil or water because mud that is added into system has some Low Gravity Solid (LGS). Hence, when we calculate it, we need to account for Low Gravity Solid (LGS) of new mud into the calculation as well. This post will demonstrate you how to determine barrels of drilling fluid required to achive the desired Low Gravity Solid (LGS).

Formula, used to calculate dilution of mud system, is listed below;

Vwm = Vm x (Fct – Fcop) ÷ (Fcop – Fca)

Where; Vwm = barrels of dilution water or base fluid

Vm = total barrels of mud in circulating system

Fct = percent low gravity solids in system

Fcop = percent total low gravity solids desired

Fca = percent low gravity solids bentonite and/or chemicals added in mud

Example: Determine how much barrels of oil base mud to diluate total 2000 bbl of mud in system from total LGS = 7 % to desired LGS of 3.5 %. The oil base mud has 2% of bentonite slurry.

Vwm = Vm x (Fct – Fcop) ÷ (Fcop – Fca)

Vwm = 2000 x (7 – 3.5) ÷ (3.5-2)

Vwm = 4667 bbl

In order to dilute total of 2000 bbl of the original mud with 7% LGS down to 3.5% LGS, 4667 bbl of mud that has 2% bentonite is requied to add into the system.