Tag Archives: force

Piston Force on Closed-Ended Tubular (Plugged Tubular)

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According to the previous post (piston force on open-ended tubular), applied surface pressure will reduce tensile force on surface. In this article, this is an analysis on the piston force on a plugged tubular string and the details are shown below;

Tubing Detail

  • 5” Tubing
  • ID of tubing = 3.696”
  • Packer seal bore OD = 5.25”
  • Weight per length = 17.7 lb/ft
  • Total Length = 10,000 ft
  • Plugged tubing depth = 10,000 ft
  • Fluid density = 10 ppg
  • Tubing is free to move in the packer
  • Applied surface pressure = 5,000 psi

Figure 1 shows the wellbore schematic. Applied pressure (5,000 psi) will cause a piston effect to push the tubing. Therefore, at the bottom of tubing buoyancy and piston force will act in an upwards direction (compression).

Figure 1 - Wellbore Schematic

Figure 1 – Wellbore Schematic

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Piston Force on Open-Ended Tubular

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Piston force is a load caused directly by changes in pressure acting on the exposed cross sectional area of pipe. This results in changing in length of tubular and force acting against tubular. In this article, it generally demonstrates force distribution based on a simple tubular diagram.

Piston-Force-on-Open-Ended-Tubular

Tubing Detail

  • 5” Tubing
  • ID of tubing = 3.696”
  • Weight per length = 17.7 lb/ft
  • Total Length = 10,000 ft
  • Fluid density = 10 ppg
  • Tubing is free to move in the packer
  • Applied surface pressure = 5,000 psi

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Pressure and Force Relationship and Its Application

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Relationship of pressure, force and cross sectional area is one of the most commonly used concepts in oilfield and we will review this concept and its application. Pressure is force divided by cross section area (see an image below).

Pressure = Force ÷ Area

We normally use pressure in many units such as psi (pound per square inch), Pascal, kg/m3, etc.

In drilling operation, we mostly use circular area so area can be calculated by this formula;

Area = π x (radius)2 or π x (diameter)2÷ 4

Where  π= 22/7 = 3.143, so we can write a formula above in easy way

Area = 3.143 x (radius)2 or 0.7857 x (diameter)2

Pressure calculation based on the relationship above is shown below;

Pressure = force ÷ (3.143 x (radius)2) or force ÷ (0.7857 x (diameter)2)

Example : For this example, we will use the oilfield unit so force is in lb, diameter is in square inch (in2), and diameter is in inch.

Let’s try to apply pressure and force relationship in drilling operation. We plan to bullhead well and we still have drill string in the hole.

Drill string weight in the air = 45,000 lb
Mud weight in hole = 12.0 ppg
Bit size = 8.5”
Drill pipe size = 5″

What is the maximum pressure at surface you can apply before drilling string will be hydraulically pushed out due to bull heading pressure?

Solution

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

Buoyed weight of drill string = 45,000 x 0.817 = 36,765 lb

Area = 0.7857 x (diameter)2= 0.7857 x (8.5)2= 56.77 square inch

Pressure = 36,765 lb ÷ 56.77 square inch= 647 psi.

In order to perform safe bullheading operation with drill string in hole, you need to apply bullheading pressure less than 647 psi on surface.

Bull Heading

Bull Heading

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.