SERVICIOS REGULADOS DE ALMACENAMIENTO Y ENTREGA DE AGUA

To find the hook load of different strings of casing, you cannot simply multiply the “weight per foot” by the length of the string. Casing weighs less in fluid than it does in air, where its “weight” rating is calculated. The difference is the buoyant force which is equal to the weigh of the volume of fluid displaced by the metal in the casing.

Some engineering tables give the displacement of the metal in the string along with the displaced fluid per 100 couplings. While this rather exacting method is technically correct, a very close approximation can be made for straight holes by using the hydrostatic head of the fluid at the bottom of the string, acting against the cross sectional end area of the casing. In highly deviated and horizontal completions there are many other varieties affecting string weight, such as friction in areas of high build angle.

Calculations are done by first obtaining the casing inside and outside diameters, from the Engineering tables, and calculating the end area of the string. The hydrostatic head of the Column of fluid is then calculated and the resultant force applied as to lift the string from the well.

40 W50760 – Rev05 – © 2005 – Weatherford. All rights reserved. 1.9.1 Guide Shoe

With a guide shoe, the only area affected is the cross sectional area of the casing itself; and the casing is being “pushed” upward, or lightened, by the hydrostatic pressure at the bottom of the shoe acting on this end area. Everything inside the shoe will cancel out. To calculate casing weight, multiply the # ft by the depth of that weight casing, then subtract your estimate of the buoyant force. The buoyant force is calculated by multiplying the end area of the casing by the hydrostatic pressure at the shoe. If casings of different weight (and ID’s) are used, you can add the number of feet casing times the weight per foot for each section.

However, changes in casing ID affect the buoyant force and its calculation. The required calculation method will be covered later in this section.

1.9.2 Float Shoe

When a float shoe is run on the end of the casing string, none of the wellbore fluid is allowed to enter the casing and the string acts as if it were plugged at the bottom. Casing weight is estimated by calculating the end area of the casing using casing OD then multiplying by the hydrostatic pressure at the shoe. Any fluid inside the casing (sometimes the rig crew will “fill” the casing using a mud hose) should be taken into account by first estimating the amount of fluid inside the casing in (ft), then estimating the hydrostatic pressure at the shoe (remember the fluid inside the casing may be of a different density than that outside), and multiplying by the inside area of the casing. Remember to add the casing weight in the air to obtain correct hook load. When different weights of casing are used, the same rules apply as those used in the Guide Shoe example. Also, remember that the buoyant force estimated cannot exceed the weight of casing in air, that is, there can be no negative hook load.

1.9.3 Differential Fill Equipment

The general rule for running differential fill float equipment states that when one piece of differential equipment is used, the casing will be 90 0/0 full,

with two pieces (collar and shoe) 810/0 full. Practice is to calculate the

casing weight when casing has stopped filling and the valve has closed. At this point, your calculations should be made exactly as you would a float shoe with the casing partially filled. The only difference is that you have a somewhat more accurate estimate of the height of the fluid column inside the casing; you should know that the fluid is the same as the fluid outside the casing.

Guide Shoe Calculations:

Using the well schematic on the right: • 8700 feet of 5 ½ in 23 # casing • Mud at 12.6 # gal

Example calculation of hook load:

Calculate fluid gradient = 12.6 x 0.052 = 0.655 psi/ft Calculate the hydrostatic pressure = 8700 x 0.655

= 5698.5 psi

Calculate the end area of casing = 0.7854 x (5.52 – 4.6702) = 6.630 in2

Estimate the buoyant force = 5698.5 x 6.630 = 37,781 # ↑ Cal. the weight of casing in air = 23 x 8700

= 200, 100 # ↓ Hook load of casing = 200,100 - 37,781

= 162,319 # ↓ TRY IT

Using well schematic on the right:

42 W50760 – Rev05 – © 2005 – Weatherford. All rights reserved. Float Shoe Calculations

Using the well schematic on the right: • 4800 feet of 7 in 20# casing • Mud at 9.2#/gal

Example calculation of hook load: Calculate the fluid gradient:

= 9.2 x 0.052 = 0.478 psi/ft Calculate the hydrostatic pressure:

= 4800 x 0.478 = 2294 psi Calculate the end area of the casing:

= 0.7854 x 7.0002

= 38.485 in2 Estimate the buoyant force:

= 2294 x 38.485 = 88,285#↑ Calculate the weight of casing in air:

= 20 x 4800 = 96,000#↓ Hook load of casing:

= 96,000 - 88,285 = 7,715#↓ TRY IT

Using well schematic on right

Differential Fill Calculations Using the well schematic on the right

• 4800 feet of 7 in. 20# casing • Mud at 10.8 #/gal

• Differential shoe on bottom Example calculation of hook load Calculate the fluid gradient:

= 10.8 x 0.052 = 0.562 x psi/ft Calculate the depth of fluid inside the casing:

Using a single piece of differential equipment means that 90% of the casing will be filled.

Fluid Depth in casing:

= 0.90 x 4800 = 4320 feet Calculate the hydrostatic pressures:

Inside casing = 4320 x 0.562

= 2428 psi

Outside casing = 4800 x 0.0562

= 2698 psi Calculate the areas of the casing:

Inside casing = 0.7854 x 6.4562

= 32.735 in2 Outside casing = 0.7854 x 7.0002

= 38.485 in2

Calculate the weight of the casing in air: = 4800 x 20 = 96,000 # ↓ Balance forces:

44 W50760 – Rev05 – © 2005 – Weatherford. All rights reserved. Calculation Using More Than One casing Weight in the

String Example:

Using the “Guide Shoe” example from previous page, modify The casing schedule to read:

• 1500 ft of 5½“ 23# • 2500 ft of 5½“ 20# • 4700 ft of 5½“ 15.50# • Mud at 12.6# /gal

Remember: Casing schedules read from the bottom up:

Calculate hydrostatic pressure:

At 4700 ft: = 4700 x 0.655 = 3078 psi At 7200 ft: = 7200 x 0.655 = 4716 psi At 8700 ft: = 8700 x 0.655 = 5699 psi Calculate areas at casing I.D. changes:

At 4700 ft: = 0.854 x (4.9502 – 4.7782) = 1.314 in2

At 7200 ft: = 0.7854 x (4.7782 – 4. 6702) = 0.801 in2

At 8700 ft. – end area of casing:

= 0.7854 x (5.52 – 4.6702) = 6.630 in2

Calculate forces acting at casing I.D. changes: At 4700 ft: = 1.314 x 3078 = 4044 #↓ At 7200 ft: = 0.801 x 4716 = 3776 #↓ At 8700 ft: = 6.630 x 5699 = 37,784 #↑ Calculate weight of casing in air:

= (1500 x 23) + (2500 x 20) + (4700 x 15.50) = 34,500 + 50,000 + 72,850

= 157,350 #↓ Hook load of casing:

= 157,350 + 4044 + 3776 – 37,784 = 127,386 #↓

TRY IT

Using same schematic as on this page. • 1650 ft of 7” 35# casing • 2550 ft of 7” 26# casing • 4850 ft of 7” 20# casing • Mud weight 10.2 ppg