Resouce for work -> stuff I wrote down.

- Area

Circles

Use the cross sectional area of the cable multiplied by well pressure to determine tool weight needed to run in hole (RIH) with wireline.

  • Note: does not take bouyancy into account.

$ \space\space D^2 * .7854 * \textnormal{Well pressure} $

  • Greaseless line example:

$ \space\space .359in * .359in * .7854 = .1012in^2 $

$ \space\space .1012in^2 * \textnormal{Well pressure} = \textnormal{Weight loss} $

$ \space\space .1012in^2 * 4200psi = 425.04lbs $

  • 9/32” line example:

$ \space\space .292in * .292in * .7854 = .067 in^2$

- Volume

Rectangles - Tanks

  • The next formula, units are inches.

If you multiply length in inches * width in inches, divided by 231, divided by 42, you get bbls per inch.

$\space\space \dfrac{\textnormal{ Length }in * \textnormal{ Width }in * \textnormal{ Height }in}{231 in^3/gal} $

  • In feet:

$ \space\space \dfrac{\textnormal{ Length }ft* 12in * \textnormal{ Width }ft *12in * \textnormal{ Height }ft * 12in} {231 in^3/gal} $

$ \space\space \dfrac{gallons}{42gal/bbl} = bbls \textnormal{ Total volume in } Bbls$

  • Easier method using feet for the dimension:

1728 is 12in * 12in * 12in

$ \space\space \dfrac{1728in^3 * \textnormal{ Length }ft * \textnormal{ Width }ft * \textnormal{ Height }ft}{231 in^3/gal} $

$ \space\space \dfrac{gallons}{42gal/bbl} = bbls \textnormal{ volume} $

Cylinders - Casing

  • Quick way:

$ \space\space \dfrac{D^2}{1029.4} = \textnormal{bbls/ft} $

  • Long way:

$ \space\space \dfrac{\Big(\dfrac{D^2in * .7854 * 12in}{231in^3}\Big)}{ 42 bbls/gal} $

Why do we memorize 1029.4?

What is 1029.4?

$ \space\space \dfrac{1}{.00097143} $

What is .00097143 then? Answer:   It is a constant that represents a diameter of 1 squared times 1/4 pi (area of a 1” circle) times 12 (one foot) divided by 231 (cubic inches per gallon) divided by 42 (gallons per barrel).   It is easier to remember to divide by 1029.4 than it is to remember to multiply by .00097143. Search the Internet for “working memory” and “digit span test” for additional info.

$ \space\space \dfrac{ \dfrac{(1 * 1) * .7854 * 12in} {231in^3} } { 42gal/bbl } = .00097143 $

You can then use this constant based on a diameter of 1, and substitute the actual diameter that you are working with to get barrels per ft of volume.

  • 5.50” 20.0# with a 4.778” ID, casing example:

  • Quick way:

$ \space\space \dfrac{4.778^2}{1029.4} = 0.02217 bbls/ft$

  • Long way:

$ \space\space \dfrac{\Big(\dfrac{4.778^2in * .7854 * 12in}{231in^3}\Big)}{ 42 bbls/gal} = 0.02217 bbls/ft$

Frac

Volume Factor

1 + (prop con x absolute volume factor)

Absolute Volume Factor

(1/22.1)

1/(SG x Grams to Lbs x CM3/gal)

1/(SG x .0022046 x 3785.412)

Slurry Density = Base Fluid Density + (PPG/Volume Factor)

Clean Fluid Ratio = 1/(PPA x AVF +1)

Take casing capacity / Volume Factor to get total sand in the wellbore

PPA Volume Factor Density Hydrostatic Added @ 6000’
.25 1.0113 8.5872 77
.50 1.0226 8.8289 153
.75 1.0339 9.0654 226
1.0 1.0453 9.2967 298
1.25 1.0566 9.5231 369
1.50 1.0679 9.7447 438
1.75 1.0792 9.9616 506
2.0 1.0905 10.1740 572
2.5 1.1131 10.5859 701
3.0 1.1358 10.9814 824
3.5 1.1584 11.3615 943
4.0 1.1810 11.7270 1057
4.5 1.2036 12.0787 1166
5.0 1.2263 12.4175 1272

Internal Yield

$ \space\space P = 0.875 * (2 Y_{p} * T)/D $

P = Internal Yield Psi Yp = Yield Strength (p110 = 110,000) T = Nominal Wall Thickness D = OD in inches

Balloning Effect

$ \Delta F_B = .6(\Delta P_{ia} A_{i} - \Delta P_{oa} A_{o} ) $