شرح حساب سعه مواسير الحفر-Drill Pipe Capacity Calculations

n this composition, we shall bandy Drill Pipe Capacity computations( Fill- Up, relegation Volumes in both open- end & bull plugged). At the end of the composition, you can also download a sample exceed distance for Drill pipe capacity computations

How To Perform Drill Pipe Capacity computations

Capacity may be a term constantly used interchangeably with volume. As employed in the oilfield, it’s the volume that a particular length of pipe will hold. Knowing the form of the pipe is round; the volume is frequently calculated by hand.


In this part, we will bandy filler- up and relegation computations. Capacity computations are one among the colorful feathers of fine problems which will be greatly simplified by using the Halliburton Cementing Tables( Red Book). Section 210( Capacity) lists capacity factors for colorful sizes of drill pipe, tubing, covering, and open hole. presently, these are listed in terms of gallons per bottom, barrels per bottom, and boxy bases per bottom(Fig. 1).

To use the Capacity tables, detect the correct table for the type of pipe you ’re dealing with drill pipe, tubing, covering, or open hole. Next, detect the confines and weight of pipe within the two left columns.
Note For tubing, there are four columns.) also find the volume units you want across the top. Read the capacity factor where the columns cross.

Sample Problem

What's the capacity, in gallons, of 1000 ft of2-7/8 clout,10.4 lb/ ft internal worried drill pipe? Use the sample of Section 210 shown in Figure5.1 to help your computations.

result

Find the applicable capacity factor( in girl/ ft) in Figure 1 & multiply it by the drill pipe length.
Capacity Factor = 0.1888 girl/ ft
Capacity = 0.1888 gal/ ft × 1000 ft = 188.8 gal

A) Fill- Up Drill Pipe Capacity computations

Fill- up of pipe is defined because the length of pipe a specified volume will fill. Fill- up factors also are listed in Section 210 of the Red Book.

Sample Problem

What's the length in bases of3.5 clout,15.50 lb/ ft internal worried drill pipe will 25 barrels of oil painting fill? Use Figure 1 to help in your computations.

result

Fill- up Factor = 152.05 ft/ bbl
Fill- up = 152.05 ft/ bbl × 25 bbl = 3801.25 ft

B) Pipe Displacement computations

In the oilfield, the terms relegation and displace may make confusion. The term displace frequently refers to pumping the fluid inside any pipe out of the pipe, as in displacing Hi- Vis Pill with water( check also slush relegation procedures). To do this, the volume of fluid pumped is generally equal to the capacity of the pipe.

Bull- Plugged Unrestricted End Drill Pipe Displacement

Figure 2 illustrates the volume of fluid displaced when the bull- plugged pipe or unrestricted- end drill pipe is run in the hole. This volume is equal to the outside periphery’s flat face area multiplied by the length of the pipe
displacement = OD Area × Length
or
displacement = 0.7854 × OD × OD × Length

Sample Problem

Calculate the relegation, in barrels, for the bull- plugged pipe shown in Figure 2?

result

Convert 3 elevation to bases|3/12 elevation/ ft = 0.25 ft
relegation = 0.7854 ×0.25 ft ×0.25 ft × 1000 ft = 49.09 ft3
Conversion factor = ( located in( Red Book) section 240, runner 85) = 0.1781 bbl/ ft
relegation = 49.09 ft 3 3 ×0.1781 bbl/ ft = 8.74 bbl

Open- Ended Drill Pipe Displacement

When the drill pipe is open- concluded( that is, some opening permits the pipe to fill up on the inside as it's run into the hole), it ’ll displace less fluid than the bull- plugged pipe. In Figure 3, you can see open- concluded drill pipe will displace a volume equal only to the volume of sword placed in fluid. This relegation is frequently estimated by multiplying thecross-sectional area by the length
Displacement = Cross-sectional Area × Length

Sample Problem

Calculate the relegation, in barrels, for the open- concluded pipe shown in Figure 3?

result

= 3 in. = 0.25 ft, ID = 1.5 in. = 0.125 ft

OD Area = 0.7854 × 0.25 ft × 0.25 ft = 0.0491 ft^2

ID Area = 0.7854 × 0.125 ft × 0.125 ft = 0.0123 ft^2

Cross-sectional Area = 0.0491 ft^2 – 0.0123 ft^2 = 0.0368 ft ^2

Displacement = 0.0368 ft^2 × 1000 ft = 36.8 ftيمكنك التحميل من اويل دوت كوت اقوي الكورسات والكتب والبرامج والاخبار الحصريه وفرص العمل الخاصه بمجال البترول oil.com