# Wellsite math

Series Development Geology Reference Manual Methods in Exploration Wellsite methods Wellsite math Greg Dunn Web page AAPG Store

A few simple mathematical formulas are used to make key calculations during the drilling of a well. The results from these calculations increase knowledge of well behavior and facilitate communication with drilling personnel. Definitions of the mathematical variables and abbreviations used in this chapter are given in Table 1.

Table 1 Definitions of terms and abbreviations used in this chapter
Term Definition Unit
P Hydrostatic pressure at depth d psi
MWmuddy water Mud weight lb/gal
D Vertical depth ft
Vbpf Volume bblbarrels/ft
D1 Larger diameter in.
D s Smaller diameter in.
Vh Hole volume; the volume of the open or cased hole bblbarrels/ft
Va Annular volume; the volume of the annulus, the area between the outside of the drillpipe or collar and the open or cased hole bblbarrels/ft
Vd Displacement volume; the volume displaced by the steel volume of drillpipe or collar. Displacement is the volume between the outside diameter and inside diameter of a drillpipe or collar. bblbarrels/ft
Vc Capacity volume; the volume contained inside a drill pipe or collar bblbarrels/ft
W Weight of collars or casing lb/ft
od Outside diameter of pipe in.
id Inside diameter of pipe in.
Dt Triplex mud pump output bblbarrels/stroke
Dd Duplex mud pump output bblbarrels/stroke
Ls Length of pump stroke in.
D1 Diameter of pump liner in.
Dr Diameter of pump rod (duplex only) in.
AV Annular velocity ft/min
GPM Mud pump output gal/mi n
Dh Hole diameter in.
An Jet nozzle area in.2
J1 … Jn Size of jet nozzles, “32nd” omitted 32nd in.
JNV Jet nozzle velocity ft/sec
THhp Total hydraulic horsepower hp
Pp Mud pump pressure psi
JNPL Jet nozzle pressure loss psi
% Hhpb Percentage of total horsepower expended at bit  %
BHhp Hydraulic horsepower at the bit hp
Hhp/in.2 Hydraulic horsepower per square inch area of bit hp/in.2
Ob Bit diameter in.
JIF Jet impact force lb
FFPfinal flowing pressure Formation fracture pressure psi
T% Hhpb Total percentage hydraulic horsepower at the bit  %

## Hydrostatic pressure in liquid columns

Mud weight is the key pressure gauge used in drilling since it responds directly to formation pressure. The hydrostatic pressure at any point in a wellbore filled with liquid drilling fluid (mud) under static conditions (that is, not circulating and/or not moving the drill string up or down) is a function of only two variables:

• The height of the liquid column (depth)
• The density of the liquid (mud weight)

The total volume of liquid or shape of the hole have no influence on hydrostatic pressure, but the height of the liquid column (depth) must be measured in the same direction as the force due to gravity, that is, true vertical. This consideration is important in highly deviated and horizontal wellbores. The hydrostatic pressure (P) at any depth in a wellbore is calculated as follows:

${\mbox{P}}=0.052\times {\mbox{MW}}\times {\mbox{D}}$

Example: If a 10,000-ft-deep hole contains drilling fluid with a weight of 11.5 ppg (pounds per gallon), the hydrostatic pressure at the bottom of the hole is

${\mbox{P}}=0.052\times 11.5\times 10{,}000=5980{\mbox{ psi}}$

### Mud weight

If the equation for hydrostatic pressure is solved for mud weight (MWmuddy water), then the equivalent mud weight can be calculated, which has several important wellsite applications. Solving for MWmuddy water, the equation becomes

${\mbox{MW}}={\frac {{\mbox{P}}}{(0.052\times {\mbox{D}})}}$

Example: Assume a formation at 10,000 ft3,048 m has a known hydrostatic pressure of 6292 psi. The mud weight needed to drill this formation “balanced” (hydrostatic pressure equal to formation pressure) is calculated as follows:

${\mbox{MW}}={\frac {6292}{(0.052\times 10{,}000)}}=12.1{\mbox{ ppg}}$

### Kill weight of mud

In a well control situation after a kick has been taken, it is necessary to increase the mud weight to balance the formation pressure. The kill weight of the mud can be calculated by observing the stabilized shut-in drill pipe pressure. The shut-in drill pipe pressure records the excess formation pressure over the hydrostatic pressure of the drilling fluid in the hole. Shut-in drill pipe pressure is converted to an equivalent mud weight. This weight is added to the mud weight currently in the hole to obtain the kill weight needed. The kill weight of the mud can also be converted back to pounds per square inch to obtain an estimation of the formation pressure.

Example: Assume a well has taken a kick at 10,000 ft3,048 m while drilling with 11.5-ppg mud. The well is shut-in, and the drill pipe pressure reads 312 psi. The excess in pounds per gallon mud weight equivalent is calculated as follows:

${\mbox{MW}}={\frac {312}{(0.052\times 10{,}000)}}=0.6{\mbox{ ppg}}$

Adding the calculated excess of 0.6 ppg to the current mud weight of 11.5 ppg results in 12.1 ppg necessary to control the well. The mud weight is typically increased beyond the calculated kill weight to allow for the negative pressure (swab) exerted on the hole when tripping the drill string.

### Formation fracture pressure

The formation fracture pressure (FTPflowing tubing pressure) can be determined by the use of a leak-off or formation integrity test. In this test, the well is shut-in and drilling fluid is pumped slowly into the well. Wellbore pressure increases linearly up to the point where the formation starts to take drilling fluid. The pressure at this point is the leak-off pressure and is used to estimate formation fracture pressure. Like shut-in drill pipe pressure, leak-off pressure is added to the pressure of the mud in the hole.

Example: Assume a leak-off test conducted in a 10,000-ft wellbore containing 11.5-ppg mud with a leak-off pressure of 1040 psi. The formation fracture pressure is estimated as

${\mbox{MW}}={\frac {1040}{(0.052\times 10{,}000)}}=2.0{\mbox{ ppg}}$

The formation fracture pressure is equivalent to an 11.5 + 2.0 = 13.5 ppg mud weight, or

${\mbox{FFP}}=0.052\times 13.5\times 10{,}000=7020{\mbox{ psi}}$

Therefore, loss of circulation may be a problem in this wellbore if mud weights exceed about 13.5 ppg. This procedure is sufficiently accurate for quick field calculations, but it slightly overestimates the formation fracture pressures because the effective compressibility of the drilling fluid and frictional pressure losses are not accounted for.

## Wellbore volumes

All important wellbore volumes are calculated with a single formula:

${\mbox{V}}_{{{\rm {bpf}}}}={\frac {({\mbox{D}}_{{{\rm {L}}}}{}^{{2}}-{\mbox{D}}_{{{\rm {S}}}}{}^{{2}})}{1029.4}}$

Volumes are reported in units of barrels and can be calculated for an in-gauge hole by using this equation to determine volume in barrels per foot, then multiplying that value by the length of the section of hole in feet. (Note: washouts and thick mud cake can substantially alter hole volume.)

Example: If a 10,000-ft-deep hole with an 8.0-in.-diameter contained pipe with an outside diameter of 5.0 in.0.127 m
0.417 ft
and an inside diameter of 4.0 in.0.102 m
0.333 ft
, the volume calculations are as follows:

${\mbox{V}}_{{{\rm {h}}}}=[(8.0^{{2}})/1029.4]\times 10{,}000=621.7{\mbox{ bbl}}$
${\mbox{V}}_{{{\rm {a}}}}=[(8.0^{{2}}-5.0^{{2}})/1029.4]\times 10{,}000=378.9{\mbox{ bbl}}$
${\mbox{V}}_{{{\rm {d}}}}=[(5.0^{{2}}-4.0^{{2}})/1029.4]\times 10{,}000=87.4{\mbox{ bbl}}$
${\mbox{V}}_{{{\rm {c}}}}=[(4.0^{{2}})/1029.4]\times 10{,}000=155.4{\mbox{ bbl}}$

Although this solution is correct mathematically (that is, the sum of annular volume, displacement, and capacity equals the hole volume), the basic formula results in an underestimate for displacement of drill pipe. Since drill pipe (not collars) has tool joints that are welded onto the pipe, calculation of drill pipe displacement involves an additional step to account for the additional volume displaced by the tool joints. For most calculations with commonly used sizes and weights of drill pipe (4.5 and 5 in.), the addition of the quantity 0.001 to the barrel per foot volume calculation for displacement accounts for the additional tool joint volume with sufficient accuracy. Using the data from the previous example, we obtain the following the displacement calculation:

${\mbox{V}}_{{{\rm {d}}}}=([5.0^{{2}}-4.0^{{2}})/1029.4]+0.001)\times 10{,}000$
$=97.4{\mbox{ bbl}}$

Notice that when drill pipe tool joints are considered, the displacement is 10.0 bblbarrels greater. Again, this additional step is not necessary when calculating the displacement of collars since collars do not have welded-on tool joints.

When volume calculations are made at the wellsite, it is important to identify each hole geometry in the wellbore. This is best accomplished by drawing a schematic diagram of the wellbore and identifying all casing, open hole, drill pipe, and collar dimensions. Several handbooks have extensive tables listing pipe and collar dimensions[1]. Calculations must then be made for each annular displacement capacity and hole volume and then summed for a total.

## Weight of collars or casing

The weight (W) of collars or casing, in pounds per foot, is estimated by

${\mbox{W}}=2.67\times ({\mbox{od}}^{{2}}-{\mbox{id}}^{{2}})$

Often the outside diameter (od) and the weight of the collars are the only dimensions known. In this case, the previous equation can be solved for the inside diameter (id) of the collars:

${\mbox{id}}=[{\mbox{od}}^{{2}}-({\mbox{W}}/2.67)]^{{1/2}}$

Example: To estimate the inside diameter of drill collars having an outside diameter of 6.5 in.0.165 m
0.542 ft
and weighing 96 lb/ft, the following calculation is made:

${\mbox{id}}=[6.5^{{2}}-(96/2.67)]^{{1/2}}=2.5{\mbox{ in}}.$

## Mud pump output

The theoretical pump outputs for triplex and duplex type pumps are calculated by the following formulas:

${\mbox{D}}_{{{\rm {t}}}}=({\mbox{L}}_{{{\rm {s}}}}\times {\mbox{D}}_{{l}}{}^{{2}})/4118$
${\mbox{D}}_{{{\rm {d}}}}={\mbox{L}}_{{{\rm {s}}}}\times [(2\times {\mbox{D}}_{{{\rm {l}}}}{}^{{2}})-{\mbox{D}}_{{{\rm {r}}}}{}^{{2}}]/6176$

Example: The output of a triplex mud pump with an 11-in. stroke length and 6-in. liners installed is

${\mbox{D}}_{{{\rm {t}}}}=(11\times 6^{{2}})/4118=0.0962{\mbox{ bbl/stroke}}$

This is a theoretical calculation based on the pump working at 100% efficiency. Mud pumps are rarely 100% efficient, and this calculation is multiplied by some volumetric efficiency factor, typically 0.90 to 0.99 for triplex pumps and 0.80 to 0.90 for duplex pumps. The actual efficiency of a pump is determined by measuring the suction pit, isolating the suction pit so that mud is pumped out of that pit only for a certain number of pump strokes, then calculating the volume pumped out of the pit and dividing by the total number of pump strokes.

In addition to barrels per stroke, mud pump output can also be expressed in gallons per stroke, barrels per minute, or gallons per minute. Once barrel per stroke is determined, the remaining parameters are easily calculated as follows:

${\mbox{Bbl/min}}={\mbox{Bbls/stroke}}\times {\mbox{Strokes/min}}$
${\mbox{Gal/stroke}}={\mbox{Bbls/stroke}}\times 42$
${\mbox{Gal/min}}={\mbox{Gal/stroke}}\times {\mbox{Strokes/min}}$

Example: If the triplex pump previously discussed was operating at 95% volumetric efficiency at 98 strokes/min, the following calculations can be made:

${\mbox{Bbl/stroke}}=0.0962\times 0.95=0.0914{\mbox{ bbl/stroke}}$
${\mbox{Bbl/min}}=0.9814\times 98=8.96{\mbox{ bbl/min}}$
${\mbox{Gal/stroke}}=0.0914\times 42=3.84{\mbox{ gal/stroke}}$
${\mbox{Gal/min}}=3.84\times 98=376{\mbox{ gal/min}}$

## Circulation or lag time

Bottoms-up circulation, or lag time, is the time for samples or gas created at the bit to arrive at the surface (via the drilling fluid) for examination. This calculation is dependent on the volume and rate.

After calculating wellbore volumes and mud pump output, several parameters for the circulating time of the drilling fluid can be estimated. The most routinely used of these parameters are

• surface-to-bit pump strokes,
• surface-to-bit time,
• bit-to-surface pump strokes (lag strokes), and
• bit-to-surface time (lag time).

Because drilling fluid is pumped down the inside of the drill string to the bit, surface-to-bit (s-to-b) calculations are made as follows:

${\mbox{S-to-b strokes}}={\frac {{\mbox{Total capacity in bbl}}}{{\mbox{pump output in bbl/stroke}}}}$

Drilling fluid is then pumped back up the hole to the surface via the annulus, therefore bit-to-surface (b-to-s) or lag calculations are made as follows:

${\mbox{B-to-s strokes}}={\frac {{\mbox{Total annular volume in bbl}}}{{\mbox{pump output in bbl/stroke}}}}$
${\mbox{B-to-s min}}={\frac {{\mbox{B-to-s strokes}}}{{\mbox{pump rate in strokes/min}}}}$

Example: Using the previously calculated data, we can calculate circulating time as follows:

${\mbox{S-to-b strokes}}={\frac {155.4{\mbox{ bbl}}}{{\mbox{0.914-in. bbl/stroke}}}}=1700{\mbox{ strokes}}$
${\mbox{S-to-b min}}={\frac {1700{\mbox{ strokes}}}{98{\mbox{ strokes/min}}}}=17.35{\mbox{ min}}$
${\mbox{B-to-s strokes}}={\frac {378.9{\mbox{ bbl}}}{{\mbox{0.0914-in. bbl/stroke}}}}=4146{\mbox{ strokes}}$
${\mbox{B-to-s min}}={\frac {4146{\mbox{ strokes}}}{98{\mbox{ strokes/min}}}}=42.30{\mbox{ min}}$

Lag time can also be measured at the wellsite by placing calcium carbide in the pipe when a connection is made. The carbide reacts with the water in the mud and forms acetylene gas that is readily detected by the chromatograph. The stroke counter is reset when the carbide is “dropped” down the pipe, and the number of strokes needed for the acetylene return is noted. The actual and calculated lags are then compared, and an interpretation is made.

Comparison of measured and calculated lag time and a knowledge of volume and mud pump calculations have several other problem-solving applications at the wellsite. These include the location of a washout (crack or hole) in the drill string, where to spot oil when the drill string is stuck, and the density of the intruding kick fluid.

## Bit hydraulics

Hydraulics is a branch of science that deals with practical applications of fluids in motion. Bit hydraulics is important because hole cleaning and drilling efficency are directly affected by horsepower expended at the bit.

### Annular velocity

Annular velocity is the average speed at which the drilling fluid is moving back up the annular space as the well is circulated. Although the mud pump output remains constant, annular velocities vary at different points in the wellbore due to changes in pipe, collar, and hole sizes. Annular velocity (AV) can be calculated as

${\mbox{AV}}={\frac {24.5\times {\mbox{GPM}}}{{\mbox{Dh}}^{{2}}-{\mbox{od}}^{2}}}$

Example: If mud is circulated at 400 gal/min in an 8.5-in. hole containing a 4.5-in. drill pipe and 6.5-in. collars, the annular velocities are

${\mbox{Drill pipe: AV}}={\frac {24.5\times 400}{8.5^{2}-4.5^{2}}}=188.5{\mbox{ ft/min}}$
${\mbox{Collars: AV}}={\frac {24.5\times 400}{8.5^{2}-6.5^{2}}}=326.7{\mbox{ ft/min}}$

### Jet nozzle area

A conventional rotary drilling bit typically has two to four jet nozzles installed to impart a jetting action on the mud to clean the bottom of the hole. Nozzle size is variable and is measured in 32nds of an inch. Note that when jet nozzle sizes are given, the “32nd” is usually omitted. Thus, a bit with “three 13′s” installed means the bit has three 13/32-in. nozzles installed. The total area (An) of the jet nozzles is calculated as

${\mbox{An}}=0.000767+({\mbox{J}}_{{1}}{}^{{2}}+{\mbox{J}}_{{2}}{}^{{2}}+{\mbox{J}}_{{n}}{}^{{2}})$

Example: The area of jet nozzles for a bit with three 13/32-in. nozzles installed is

${\mbox{An}}=0.000767\times (13^{{2}}+13^{{2}}+13^{{2}})=0.3889{\mbox{ in.}}^{{2}}$

### Jet nozzle velocity

Jet nozzle velocity (JNV) is the velocity of the mud exiting the jet nozzles of the bit and is estimated as

${\mbox{JNV}}={\frac {0.32086\times {\mbox{GPM}}}{{\mbox{An}}}}$

Example: The jet nozzle velocity for a bit with three 13's jets installed and a circulation rate of 400 GPM is

${\mbox{JNV}}={\frac {0.32086\times 400}{0.3889}}=330{\mbox{ ft/sec}}$

#### Total hydraulic horsepower

The total hydraulic horsepower (THhp) available for drilling hydraulics is defined by the circulation rate and pressure of the mud pump. Total hydraulic horsepower is calculated as

${\mbox{THhp}}={\frac {{\mbox{Pp}}\times {\mbox{GPM}}}{1714}}$

Example: The total hydraulic horsepower available if circulating at 400 gal/min with a pump pressure of 2000 psi is

${\mbox{THhp}}={\frac {2000\times 400}{1714}}=467{\mbox{ hp}}$

### Jet nozzle pressure loss

Pump pressure is the total pressure expended throughout the circulating system's surface equipment (such as the standpipe, kelly hose, and kelly), drill string bore, jet nozzles, and annulus. Only the pressure expended through the jet nozzles accomplishes useful work for drilling. The remaining pressure losses are referred to as parasitic pressure losses. Jet nozzle pressure loss is estimated as follows:

${\mbox{JNPL}}={\frac {{\mbox{MW}}\times {\mbox{GPM}}^{2}}{10{,}858\times {\mbox{An}}^{2}}}$

Example: The pressure lost through three 13′s jet nozzles while circulating a 12.0-ppg mud at 400 gal/min is

${\mbox{JNPL}}={\frac {12\times 400^{2}}{10{,}858\times 0.3889^{2}}}=1169{\mbox{ psi}}$

### Hydraulic horsepower at the bit

The hydraulic horsepower at the bit (BHhp) is calculated as for total hydraulic horsepower (THhp), but the mud pump pressure (Pp) is replaced by the jet nozzle pressure loss (JNPL):

${\mbox{BHhp}}={\frac {1169\times 400}{1714}}=273{\mbox{ hp}}$

The percentage of the total hydraulic horsepower expended at the bit is an important parameter to determine and is calculated in two ways:

$\%{\mbox{ Hhpb}}={\frac {{\mbox{BHhp}}}{{\text{THhp}}}}\times 100$

or

${\mbox{T}}\%{\mbox{ Hhpb}}=({\mbox{JNPL/Pp}})\times 100$

Example: Using the data from the previous examples, we can calculate the percentage hydraulic horsepower at the bit as follows:

$\%{\mbox{ Hhpb}}=(273/467)\times 100=58\%$

or

$\%{\mbox{ Hhpb}}=(1169/2000)\times 100=58\%$

Another key hydraulic parameter that indicates the magnitude of power expended at the bit is hydraulic horsepower per square inch of bit area. It is calculated by

${\mbox{Hhp/in.}}^{{2}}={\mbox{BHhp}}/(0.7854\times {\mbox{Ob}}^{{2}})$

Example: Using the previous data with an 8.5-in. bit, we can calculate the hydraulic horsepower per square inch of bit area as

${\mbox{Hhp/in.}}^{{2}}=273/(0.7854\times 8.5^{{2}})=4.8{\mbox{ hp/in.}}^{{2}}$

### Jet impact force

One of the theories of “optimum” drilling hydraulics holds that the cleaning action of the bit on the hole bottom is maximized by maximizing the jet impact force. Jet impact force (JIF) is estimated by

${\mbox{JIF}}=0.000516\times {\mbox{MW}}\times {\mbox{GPM}}\times {\mbox{JNV}}$

Example: Using the previous data, we can calculate the jet impact force as follows:

${\mbox{JIF}}=0.000516\times 12\times 400\times 330=817{\mbox{ lb}}$