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Understand Hydrostatic Pressure

In a well, any pressure created by a static column of fluid is called ‘Hydrostatic Pressure’ (HP); at any given True Vertical Depth (TVD). With ‘hydro’ representing water, which exerts pressure, ‘static’ means it has no movement. Any pressure developed by a column of fluid that isn’t moving, therefore, can be considered hydrostatic pressure; fluid in this sense can be either liquid or gas.

The relationship of hydrostatic pressure is shown in the equation below.

HP (Hydrostatic Pressure) = density x g (gravity acceleration) x h (True Vertical Depth, TVD)

In oilfield term, the formula above is modified so that people can use it easily. The formulas are as follows:

HP (Hydrostatic Pressure) = Constant x MW (Mud Weight or Mud Density)  x TVD (True Vertical Depth)

HP (psi)  = 0.052 x MW (ppg) x TVD (ft) ** Most frequent used in the oilfield **

HP (psi) = 0.007 x MW (pcf) x TVD (ft)

HP (kPa) = 0.00981 x MW (kg/m3) x TVD (m)

Depending on which unit is used for calculation, there are several conversion factors such as 0.052, 0.007, 0.00981 for instant as you can see from the equations above.

According to the equations above, Hydrostatic Pressure is not a function of hole geometry. Only mud weight and True Vertical Depth (TVD) affect on Hydrostatic Pressure. For example (a picture below); well A and well B have the same vertical depth. With the same mud density in hole, the bottom hole pressure due to hydrostatic pressure is the same. The only different between Well A and Well B is mud volume.

This concept is basic and very important for many aspects such as well control, balance cementing, u-tube, etc.

You can learn more about hydrostatic pressure calculation from the following article – Hydrostatic Pressure Calculation

Pressure in a well

In a static condition

  • Pressure at any depth = Hydrostatic Pressure (HP) + Surface Pressure (SP)
  • Pressure between 2 points is HP between these points

The diagram below demonstrates the relationship of pressure in a well.

At point 1, Pressure@1 = Surface Pressure (SP) + Hydrostatic Pressure @ 1 (HP1)

At point 1, Pressure@2 = Surface Pressure (SP) + Hydrostatic Pressure@1 (HP1) + Hydrostatic Pressure@2 (HP2)

Ref books: 

Lapeyrouse, N.J., 2002. Formulas and calculations for drilling, production and workover, Boston: Gulf Professional publishing.

Bourgoyne, A.J.T., Chenevert , M.E. & Millheim, K.K., 1986. SPE Textbook Series, Volume 2: Applied Drilling Engineering, Society of Petroleum Engineers.

Mitchell, R.F., Miska, S. & Aadny, B.S., 2011. Fundamentals of drilling engineering, Richardson, TX: Society of Petroleum Engineers.

Corrected D exponent

The original “d” exponent is good for constant mud weight but in reality several drilling operations drill with various mud weights in hole due to weight up. In order to account for mud weight variation, so modification of d exponent, called “corrected d exponent”, has been made to correct for mud weight changes.

The corrected d-exponent is listed below.

dc = log (R ÷ 60N) ÷ log (12W ÷ 1000D) x (MW1 ÷ MW2)

Where;

dc = corrected “d” exponent

R = penetration rate in feet per hour

d = exponent in drilling equation, dimensionless

N = rotary speed in rpm

W = weight on bit in kilo pound

D = bit size in inch

MW1 = initial mud weight in ppg

MW2 = actual mud weight in ppg

 

Example: Determine the corrected d-exponent from following information.

Rate of penetration (R) = 90 ft/hr

Rotary drilling speed (N) = 110 rpm

Weight on bit (W) = 20 klb

Bit Diameter (D) = 8.5 in

MW1 = 9.0 ppg

MW2 = 12.0 ppg

Solution: dc = log [90÷ (60 x 110)] ÷ log [(12 x 20) ÷ (1000 x 8.5)] x (9.0 ÷ 12.0)

dc = 1.20 x 0.75

dc = 0.9

** Please remember that single d exponent or corrected d exponent valve does not help identify abnormal pressure. The trend of d exponent will help drilling personnel detect high formation pressure zones while drilling.

Please find the excel sheet for calculating the corrected D Exponent

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

D Exponent Calculation

D exponent is an extrapolation of drilling parameters to get a trend while drilling into over-pressured zones. Usually, mud logger will correct all data, calculate d-exponent and plot the d exponent valve on the curve. The d-exponent can be utilized to detect transition from normal pressure regime to abnormal formation pressure. While drilling, if the change of trend is observed, rig supervisors must be cautious about this situation because this is one of the possible well control indications.

The “d” exponent described from the equation below:

d = log (R ÷ 60N) ÷ log (12W ÷ 1000D)

Where; R = penetration rate in feet per hour

d = exponent in drilling equation, dimensionless

N = rotary speed in rpm

W = weight on bit in kilo pound

D = bit size in inch

** Note: this equation is is valid for constant drilling fluid weight.

Example: Determine the d-exponent from following information.

Rate of penetration (R) = 90 ft/hr

Rotary drilling speed (N) = 110 rpm

Weight on bit (W) = 20 klb

Bit Diameter (D) = 8.5 in.

Solution: d = log [90÷ (60 x 110)] ÷ log [(12 x 20) ÷ (1000 x 8.5)] d = 1.20

Please find the Excel Sheet for calculating d-exponent.

References

(2010). Well Control for Completions and Interventions. 1st ed. Texas: Gulf Publishing.

Grace, R. (2003). Blowout and well control handbook [recurso electrónico]. 1st ed. Paises Bajos: Gulf Professional Pub.

Grace, R. and Cudd, B. (1994). Advanced blowout & well control. 1st ed. Houston: Gulf Publishing Company.

Watson, D., Brittenham, T. and Moore, P. (2003). Advanced well control. 1st ed. Richardson, Tex.: Society of Petroleum Engineers.

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.

Please find the excel sheet used to calculate how much barrel of drilling fluid to control Low Gravity Solid (LGS) in mud system.

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

Directional Drilling Books

Dilution of Mud System to Control Low Gravity Solid

By adding bbl of base fluid required, dilution of mud can help control Low Gravity Solid (LGS) in mud system. This post will demonstrate you how to determine barrels of dilution fluid such as water or base fluid required to achieve the desired low gravity solid.

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

Vwm = Vm x (Fct – Fcop) ÷ (Fcop)

Where; Vwm = barrels of dilution water or base fluid needed

Vm = total barrels of mud in circulating system

Fct = percent low gravity solids in system

Fcop = percent total low gravity solids desired

Example: Determine how much barrels of base oil to dilute total 2000 bbl of mud in system from total LGS = 7 % to desired LGS of 3.5 %.

Vm = 2000 bbl

Fct = 7%

Fcop = 3.5%

Vwm = 2000 x (7 – 3.5) ÷ 3.5

Vwm = 2000 bbl

In order to dilute total of 2000 bbl of the original mud with 7% LGS down to 3.5% LGS, 2000 bbl of base oil is required to add into the system.

Please find the excel sheet used to calculate how much barrel of base fluid to control Low Gravity Solid (LGS) in mud system.

 

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition