Up Hole Air Velocity Calculator

Annular Velocity

The annular velocity (Vann) is defined as the actual flow rate divided by the annulus area (Wines, 2004):(5-44)Vann=QaAannwhere Aann is the cantankerous-sectional annular area defined as the cross-exclusive surface area of the housing without coalescers minus the area of the coalescer end caps:(5-45)Aann=πRh2−NπRc2where Rh is the radius of the housing;

From: Handbook of Natural Gas Transmission and Processing , 2012

Basic Equations

William Lyons , ... Norton J. Lapeyrousse , in Formulas and Calculations for Drilling, Production, and Workover (3rd Edition), 2012

1.11 Annular Velocity V_an (ft/min)

Formula ane

AV = pump output, bbl/min ÷ annular chapters, bbl/ft

Case:

pump output = 12.six   bbl/min

annular capacity = 0.1261   bbl/ft

AV = 12.6   bbl/min ÷ 0.1261   bbl/ft

AV = 99.92   ft/min

Formula 2

$\text{AV},\text{ft}/\text{min}=\frac{\mathrm{24.v}(\text{Q})}{\text{D}{\text{h}}^{\mathrm{ii}}-\text{D}{\text{p}}^{\mathrm{ii}}}$

where:

Q = circulation rate, gpm

Dh = inside diameter of casing or hole size, in.

Dp = exterior diameter of piping, tubing, or collars, in.

Example:

Q = 530 gpm

Dh = 12¼ in.

Dp = four½ in.

$\begin{array}{l}\text{AV}=\frac{\mathrm{24.v}(530)}{{12.25}^2-{\mathrm{four.5}}^2}\\ \text{AV}=\frac{12,985}{129.8125}\\ \text{AV}=100\text{ft}/\text{min}\end{array}$

Formula 3

$\text{AV},\text{ft}/\text{min}.=\frac{\text{P}\text{O},\text{bbl}/min(\mathrm{1029.four})}{\text{D}{\text{h}}^2-\text{D}{\text{p}}^2}$

Instance:

pump output = 12.vi   bbl/min

pigsty size = 12¼ in. pipe

OD = 4½ in.

$\begin{array}{l}\text{AV}=\frac{12.6\text{bbl}/\text{min}(\mathrm{1029.iv})}{{12.25}^{\mathrm{two}}-{\mathrm{4.five}}^2}\\ \text{AV}=\frac{12970.44}{129.8125}\\ \text{AV}=99.92\text{ft}/\text{min}\end{array}$

Annular velocity (AV), ft/sec

$\text{AV},\text{ft}/\text{sec=}\frac{17.16\times \text{PO},\text{bbl}/\text{min}}{\text{D}{\text{h}}^{\mathrm{ii}}-\text{D}{\text{p}}^{\mathrm{two}}}$

Case:

pump output = 12.6   bbl/min

hole size = 12¼ in.

pipe OD = 4½ in.

$\begin{array}{l}\text{AV}=\frac{\mathrm{17.sixteen}\times 12.6\text{bbl}/\text{min}}{{12.25}^2-{4.5}^2}\\ \text{AV}=\frac{216.216}{129.8125}\\ \text{AV}=\mathrm{i.6656}\text{ft}/\text{sec}\end{array}$

Metric Calculations

Annular velocity, m/min = pump output, liter/min ÷ annular volume, fifty/chiliad

Annular velocity, m/sec = pump output, liter/min + 60 ÷ annular volume, l/m

SI Unit Calculations

Annular velocity, m/min = pump output, 1000 ³/min ÷ annular volume, one thousand³/m

Pump output, gpm, required for a desired annular velocity, ft/min

$\text{Pump output},\text{gpm}=\frac{\text{AV},\text{ft}/\text{min}(\text{D}{\text{h}}^2-\text{D}{\text{p}}^2)}{\mathrm{24.five}}$

where:

AV = desired annular velocity, ft/min

Dh = within diameter of casing or hole size, in.

Dp = outside diameter of pipe, tubing, or collars, in.

Example:

desired annular velocity = 120   ft/min

hole size = 12¼ in.

pipe OD = iv½ in.

$\begin{array}{l}\text{PO}=\frac{120({12.25}^2-{\mathrm{four.5}}^2)}{24.5}\\ \text{PO}=\frac{120\times 129.8125}{\mathrm{24.v}}\\ \text{PO}=\frac{15577.5}{24.5}\\ \text{PO}=\mathrm{635.eight}\text{gpm}\end{array}$

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Phase Separation

Saeid Mokhatab , ... John Y. Mak , in Handbook of Natural Gas Manual and Processing (Fourth Edition), 2019

5.vii.three.three Annular Velocity

The annular velocity (V _ann) is defined as the bodily flow rate divided past the annulus area (Wines, 2004):

(5.thirteen) ${\text{V}}_{\text{ann}}=\frac{{\text{Q}}_{\text{a}}}{{\text{A}}_{\text{ann}}}$

where, A_ann is cross-sectional annular area divers equally the cross-sectional area of the housing without coalescers minus the area of the coalescer end caps:

(5.fourteen) ${\text{A}}_{\text{ann}}=\mathrm{\pi }{\text{R}}_{\text{h}}^2-\text{N}\mathrm{\pi }{\text{R}}_{\text{c}}^2$

where R_h is radius of the housing, R_c is radius of coalescer end cap, and N is number of coalescers.

The enlarged droplets leaving the coalescer media pack can exist assumed to be equally big every bit possible for the given menses conditions when complete coalescence has occurred. Therefore, the coalesced droplet diameter will be the same for whatever specific design of the coalescer cartridge as long as complete coalescence has been achieved. If complete coalescence is not achieved, the calculation of the coalesced droplets must have into business relationship the degree of coalescence.

In near industrial applications, the coalesced aerosol will range in size from 0.5 to 2.2   mm and will be mostly influenced by the interfacial tension, which is significantly affected past the liquid density, system temperature, and system pressure. As the force per unit area is increased, the gas density will increase while the liquid density is just slightly affected. The solubility of the gas in the liquid is enhanced with increasing pressure. This leads to a substantial decrease in interfacial tension with increasing pressure level and consequently to significantly smaller coalesced droplets at the higher pressures.

Once the coalesced droplet size has been estimated, the side by side step is to make up one's mind the maximum annular velocity that can be sustained without reentrainment. In full general, the coalesced aerosol volition produce Reynolds numbers (Re) outside of the creeping flow regime (<0.i) and Stokes law. Instead, a force remainder is used between the liquid droplets settling past gravity and the drag strength of the gas flowing upward in the opposite direction.

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Solutions to chapter bug

Samuel Bridges , Leon Robinson , in A Practical Handbook for Drilling Fluids Processing, 2020

Part 1

The lowest annular velocity for this well is 60  ft/min. For the drilling fluid properties listed below, make up one's mind which wells will have a problem with hole cleaning.

Fluid #	MW, ppg	PV, cp	YP, lb/100   sqft	K, eff cp	CCI
1	9.0	10	10	340	0.46
2	ix.0	5	5	140	0.19
3	9.0	five	x	580	0.78
4	15.0	30	five	60	0.14
v	15.0	30	20	370	0.83
half dozen	10.0	10	fifteen	740	i.xi

$\text{CCI}=\mathrm{Grand}\left(\text{MW}\right)\left(\text{AV}\right)/\mathrm{400,000}$

With the annular velocity then low (as it might exist in a large diameter pigsty or in a riser), none of these fluids can send solids from the hole equally they should be, except for the last fluid in the list.

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Cuttings transport

Samuel Bridges , Leon Robinson , in A Practical Handbook for Drilling Fluids Processing, 2020

Problem 6.1

Part 1

The lowest annular velocity for this well is 60  ft/min. For the drilling fluid properties listed below, determine which wells will have a problem with hole cleaning.

Fluid #	MW, ppg	PV, cp	YP, lb/100 sqft	Yard, eff cp	CCI
i	nine.0	x	10
two	9.0	5	5
3	9.0	5	10
four	15.0	30	5
five	15.0	30	20
6	10.0	x	15

Part two

The yield signal of a drilling fluid can be inverse without a meaning change in the PV. What should the YP be to make certain that all drilling fluids are cleaning the borehole?

Fluid #	Grand needed, eff cp	YP, lb/100 sqft
1
2
3
four
five
6

Sample calculation:

$K=\frac{\mathrm{400,000}}{(\text{MW},\text{ppg})(\text{AV},\text{ft/min})}$

From the K-value graph (Fig. 6C.ane), read the YP needed.

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Period Drilling: Underbalance Drilling with Liquid Single-Phase Systems

Pecker Rehm , ... Arash Haghshenas , in Underbalanced Drilling: Limits and Extremes, 2012

2.43 Cutting Ship

The primary function of the drilling fluids is to clean the hole effectively. An efficient hole cleaning reduces operational bug and reduces drag and torque in the wellbore. Slip velocity of the drilled cuttings tin can be estimated using diverse equations. The skid velocity of the drilled cuttings is a function of the density difference betwixt the drilled cuttings and the density of the drilling fluid, viscosity of the drilling fluid, and the size and shape of the drilled cuttings. For efficient wellbore cleaning, the velocity of the drilling fluid in the annulus should exist at to the lowest degree twice the slip velocity of the cuttings.

An easier and field proven method is to apply the carrying capacity alphabetize (CCI) which is valid for wells with an inclination angle of 35 degrees or less. CCI is the ratio of a constant number over fluid properties and annular velocity of the drilling fluid. If CCI equals 1 or greater the hole cleaning is constructive.

(two.56) $\text{CCI=}\frac{\text{MW×Grand×AV}}{400,000}$

where

MW = Density of the drilling fluid, ppg

One thousand = Power constabulary parameter of the fluid rheology, cp

AV= Boilerplate velocity of the drilling fluid, ft/min

The Chiliad value in this equation is non the aforementioned as the K values calculated for the ability-law and the Herschel-Bulkley models. The unit of K is in cp and calculated using Eq. (ii.57).

(two.57) $K={511}^{\left(1-\mathrm{due\; north}\right)}{R}_{300}$

Equally indicated in Eq. (2.56), three parameters of fluid density, the One thousand value (rheological parameter), and the average velocity of the drilling fluid touch cutting transportation to the surface. The density of the drilling depends on the pore pressure regime in the wellbore and should be kept equally low as possible to enhance the drilling rate. The average velocity of the drilling fluid is selected based on the size of nozzles for optimizing hydraulic bear upon force or hydraulic horse power at the bottomhole. Other limitations of average fluid velocity are wellbore washout, bottom-hole assembly, and pump capabilities. Therefore, but rheological properties are available for adjustment of cuttings send efficiency.

The Grand factor represents the rheological properties of the drilling fluid. Figure 2-19 illustrates Yard factor every bit function of the Bingham plastic model parameters, plastic viscosity, and yield point. Increasing values of yield point for a constant plastic viscosity results in higher values of Thou factor. However, for a constant yield point, increasing the plastic viscosity reduces the value of K. Usually only small quantity of additive is required to increase the yield point and improve cutting transport efficiency. Values of yield point and plastic viscosity can be back-calculated from Figure two-19 for field application (come across Example 2). More discussion on cutting transport efficiency is available in Drilling Fluids Processing Handbook, 2004 .

Figure 2-19. Thousand factor is shown every bit a function of viscosity and yield point of the drilling fluid (Drilling Fluids Processing Handbook, 2004)

ii.43.one Instance 2

Decide a minimum annular velocity required to make clean a well if the drilling fluid is a ten.0 ppg fluid with a plastic viscosity of xx cp and yield point of 15 lbf/100 ft ^two. If the average velocity in the wellbore is express to 90 ft/min, what should be the minimum yield bespeak of the drilling fluid to ensure efficient pigsty cleaning?

(ii.58) $R_{600}=2{\text{μ}}_p+{\text{τ}}_y$

(2.59) $R_{600}=2\left(\mathrm{xx}\right)+15=55$

(2.lx) $R_{300}={\text{μ}}_p+{\text{τ}}_y$

(2.61) $R_{300}=20+\mathrm{xv}=35$

From the Eq. (2.13) the n parameter of the fluid is

(2.62) $n=3.32\times \log \left(\frac{55}{35}\right)=0.652$

The K factor in the Eq. (two.55) is in cp. The K factor in cp is

(ii.63) $\mathrm{Yard}={511}^{\left(\mathrm{i}-\mathrm{north}\right)}\times 35=307\text{cp}$

The minimum velocity for constructive pigsty cleaning occurs when CCI is 1. The Eq. (two.56) is rearranged to discover the minimum velocity.

(2.64) $\text{A}V\text{=}\frac{400,000}{\mathrm{x}\times 307\times \mathrm{ane}}=130\text{ft}/\min$

If the average velocity is express to 80 ft/min, and then the minimum required Thou value of the drilling fluid is

(2.65) $K=\frac{400,000}{\mathrm{ten}\times 80\times \mathrm{ane}}=500\text{cp}$

From Figure ii-19 for plastic viscosity for K of 500 cp and plastic viscosity of 20 cp, the yield point of the drilling fluid should be 20 lb_f/100 ft^two to ensure efficient cut ship to the surface.

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More Steady Catamenia Applications

Wilson C. Mentum Ph.D. , in Managed Pressure Drilling, 2012

Dynamic constraints

The concrete properties of the annular velocity field are introduced through a suitable constitutive stress-strain relationship. We assume the classical Herschel-Bulkley model with n, Grand, and τ _yield values every bit shown in Equation 5.11a. For the outer annulus, where dU/dr<0, nosotros write

(5.11a) $\mathrm{\tau }={\mathrm{\tau }}_{\mathrm{yield}}+{(-\mathrm{K}\mathrm{dU}\mathrm{/}\mathrm{dr})}^{\mathrm{north}}$

where "−dU/dr" and τ are both positive, and so that all of the ( ) brackets are positive. Nosotros substitute this into Equation 5.8b—that is, τ(r)=−i/2 dp/dz r+C/r—to obtain

(5.11b) ${\mathrm{\tau }}_{\mathrm{yield}}+{(-\mathrm{Chiliad}\mathrm{dU}\mathrm{/}\mathrm{dr})}^{\mathrm{due\; north}}=-1/2\mathrm{dp}\mathrm{/}\mathrm{dz\; r}+\mathrm{C}\mathrm{/}\mathrm{r}$

from which

(five.12a) $\mathrm{dU}(\mathrm{r})\mathrm{/}\mathrm{dr}=-(1\mathrm{/}\mathrm{K}){(-1/2\mathrm{dp}\mathrm{/}\mathrm{dz\; r}+\mathrm{C}\mathrm{/}\mathrm{r}-{\mathrm{\tau }}_{\mathrm{yield}})}^{1\mathrm{/}\mathrm{northward}}$

In social club for solutions to be, the quantity within the brackets must be positive, so that

(5.12b) $-\mathrm{i}/\mathrm{two}\mathrm{dp}\mathrm{/}\text{dz r}+\mathrm{C}\mathrm{/}\mathrm{r}-{\mathrm{\tau }}_{\mathrm{yield}}\ge 0$

When this constraint is satisfied, Equation five.12a can be integrated over (r, R_o) to give

(five.12c) $\mathrm{U}(\mathrm{r})=+(\mathrm{i}\mathrm{/}\mathrm{M})\underset{\mathrm{r}}{\overset{{\mathrm{R}}_{\mathrm{o}}}{\int }}{(-\mathrm{ane}/2\mathrm{dp}\mathrm{/}\mathrm{dz\; r}+\mathrm{C}\mathrm{/}\mathrm{r}-{\mathrm{\tau }}_{\mathrm{yield}})}^{1\mathrm{/}\mathrm{n}}\mathrm{dr}$

for r₊<r<R_o, where we take used the outer no-skid axial velocity status U(R_o)=0.

For the inner annulus, we crave Equation five.11a in a class suitable for dU/dr>0 and τ<0. This is achieved past taking "−τ=τ_yield+(K dU/dr)ⁿ" so that all of the ( ) brackets are positive. If we substitute "τ(r)=−1/ii dp/dz r+C/r," we obtain

(5.13a) $\mathrm{dU}\mathrm{(}\mathrm{r}\mathrm{)}\mathrm{/}\mathrm{dr}=+(\mathrm{ane}\mathrm{/}\mathrm{K}){(\mathrm{one}/\mathrm{ii}\mathrm{dp}\mathrm{/}\mathrm{dz}\mathrm{r}-\mathrm{C}\mathrm{/}\mathrm{r}-{\mathrm{\tau }}_{\mathrm{yield}})}^{\mathrm{ane}/\mathrm{n}}$

for which we crave

(five.13b) $\mathrm{ane}/2\mathrm{dp}\mathrm{/}\mathrm{dz}\mathrm{r}-\mathrm{C}\mathrm{/}\mathrm{r}-{\mathrm{\tau }}_{\mathrm{yield}}\ge 0$

If this is satisfied, we integrate over (R_i, r) and apply the no-slip axial velocity status U(R_i)=0, to obtain

(5.13c) $\mathrm{U}(\mathrm{r})=+(\mathrm{one}\mathrm{/}\mathrm{K})\underset{{\mathrm{R}}_{\mathrm{i}}}{\overset{\mathrm{r}}{\int }}{(1/\mathrm{two}\mathrm{dp}\mathrm{/}\mathrm{dz\; r}-\mathrm{C}\mathrm{/}\mathrm{r}-{\mathrm{\tau }}_{\mathrm{yield}})}^{\mathrm{ane}/\mathrm{n}}\mathrm{dr}$

for R_i<r<r₋. Now the plug moves with a constant speed U_plug in r₋<r<r₊. Its value from Equation 5.12c at r=r₊ must equal that using Equation v.13c at the location r=r₋ if at that place is no slippage kinematically. In other words,

(5.14) $\begin{array}{l}\underset{{\mathrm{R}}_{\mathrm{i}}}{\overset{{\mathrm{r}}_{-}}{\int }}{(1/\mathrm{two}\mathrm{dp}\mathrm{/}\mathrm{dz}\mathrm{r}-\mathrm{C}\mathrm{/}\mathrm{r}-{\mathrm{\tau }}_{\mathrm{yield}})}^{\mathrm{one}\mathrm{/}\mathrm{due\; north}}\mathrm{dr}\\ -------------------------=\mathrm{ane}\\ \underset{{\mathrm{r}}_{+}}{\overset{{\mathrm{R}}_{\mathrm{o}}}{\int }}{(-1/\mathrm{ii}\mathrm{dp}\mathrm{/}\mathrm{dz}\mathrm{r}+\mathrm{C}\mathrm{/}\mathrm{r}-{\mathrm{\tau }}_{\mathrm{yield}})}^{1\mathrm{/}\mathrm{n}}\mathrm{dr}\end{array}$

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Hole Cleaning

Mark Due south. Ramsey P. Due east. , in Practical Wellbore Hydraulics and Hole Cleaning, 2019

3.9.2 Exercise with Cuttings Carrying Index

Using the Yard–PV–YP chart nearby, find the K value for the PV and YP values in the table. And then use the equation for CCI shown below to notice the value of CCI for the following mud properties:

Part I: Calculating the CCI directly

$\mathrm{CCI}=\frac{\mathrm{MW}\times \mathrm{AV}\times K}{400,000}$

No.	PV	YP	1000	AV	MW	CCI
ane	10	20	1160	100	12
two	20	20		100	12
3	20	10		150	12
4	thirty	25		180	18
5	40	xx		180	15

PV, plastic viscosity; YP, yield betoken; AV , annular velocity; MW, mud weight; CCI, Cuttings Carrying Index.

Part Two: Summate the needed K viscosity then wait upwardly the required YP on the graph. Hint: To observe the K viscosity needed to make the CCI=1.0, 2.5, or something in-betwixt, the CCI equation is rearranged as shown beneath:

$K=\frac{\mathrm{CCI}\times 400,000}{\mathrm{MW}\times \mathrm{AV}}$

No.	PV	YP	K	AV	MW	CCI
6	20			lxxx	9	1.0
vii	15			100	ten	2.5

PV, plastic viscosity; YP, yield point; AV, annular velocity; MW, mud weight; CCI, Cuttings Carrying Index.

Substitute one.0 in no. 6 or 2.v in no. vii for CCI and the MW and AV in guild to calculate One thousand viscosity. Then observe that K viscosity on the left centrality of the nautical chart and go beyond until you cross the curved PV line. (PV cannot be chop-chop adjusted since it is due almost entirely to solids content.) Then drop down vertically to read the YP needed on the horizontal axis.

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Well Hydraulics

William C. Lyons , ... Norton J. Lapeyrouse , in Formulas and Calculations for Drilling, Product, and Workover (Fourth Edition), 2016

6.6 Skid Velocity of Cuttings in the Annulus

These calculations provide the slip velocity of a cut of a specific size and weight in a given fluid. The annular velocity and the cutting net ascent velocity are also calculated.

Method i:

Annular velocity, ft./min:

(6.60) $\mathrm{A}\mathrm{V}=\frac{\mathrm{24.v}\times Q}{{D_{\mathrm{h}}}^2-{D_{\mathrm{p}}}^2}$

Cutting slip velocity, ft./min:

(6.61) $\mathrm{V}\mathrm{s}=0.45\left(\frac{\mathrm{P}\mathrm{V}}{\left(\mathrm{M}\mathrm{W}\right)\left(D_{\mathrm{p}}\right)}\right)\left[\sqrt{\frac{36,800}{{\left(\frac{\mathrm{P}\mathrm{V}}{\left(\mathrm{Thousand}\mathrm{W}\right)\left(\mathrm{D}\mathrm{p}\right)}\right)}^2}\times \left(D_{\mathrm{p}}\right)\left(\frac{\mathrm{DenP}}{\mathrm{M}\mathrm{Due\; west}}-1\right)+1}-1\right]$

Where:

Fives   =   Skid velocity, ft./min

PV   =   Plastic viscosity, cps

MW   =   Mud weight, ppg

D _p  =   Diameter of particle, in.

DenP   =   Density of particle, ppg

Case: Using the following data, determine the annular velocity, ft./min; the cuttings slip velocity, ft./min, and the cutting internet rise velocity, ft./min:

Information: Mud weight (MW)   =   xi.0   ppg

  Plastic viscosity (PV)   =   thirteen   cps

  Diameter of particle   =   0.25   in.

  Density of particle   =   22.0   ppg (8.33   ppg   ×   specific gravity, 2.64)

  Flow rate   =   520   gpm

  Diameter of pigsty   =   12¼   in.

  Drill pipe OD   =   v.0   in.

Annular velocity, ft./min:

$\begin{array}{c}\hfill \mathrm{A}\mathrm{V}=\frac{\mathrm{24.v}\times 520}{{12.25}^2-{\mathrm{five.0}}^2}\hfill \\ \hfill \mathrm{A}\mathrm{5}=102\mathrm{ft}./min\hfill \end{array}$

Cut slip velocity, ft./min:

$\begin{array}{c}\hfill \mathrm{5}\mathrm{s}=0.45\left(\frac{13}{\left(11.0\right)\left(0.25\right)}\right)\left[\sqrt{\frac{36,800}{{\left(\frac{13}{\left(11.0\right)\left(0.25\right)}\right)}^2}\times \left(0.25\right)\left(\frac{22.0}{\mathrm{eleven.0}}-\mathrm{ane}\right)+1}-1\right]\hfill \\ \mathrm{V}\mathrm{s}=0.45\left(\mathrm{iv.727}\right)\left[\sqrt{\frac{36,800}{{\left(4.727\right)}^2}\times \left(0.25\right)\left(\mathrm{ane}\right)+\mathrm{ane}}-1\right]\hfill \\ \mathrm{V}\mathrm{south}=2.12715\left(\sqrt{412.6839}-1\right)\hfill \\ \mathrm{V}\mathrm{south}=2.12715\times 19.3146\hfill \\ \mathrm{V}\mathrm{s}=41.085\mathrm{ft}./min\hfill \end{array}$

Cutting net rise velocity:

$\begin{array}{l}\mathrm{Annular}\mathrm{velocity}=102\mathrm{ft}./min\hfill \\ \mathrm{Cut}\mathrm{slip}\mathrm{velocity}=\underset{¯}{-41\mathrm{ft}./min}\hfill \\ \mathrm{Cutting}\mathrm{northward}\mathrm{e}\mathrm{t}\mathrm{rise}\mathrm{velocity}=61\mathrm{ft}./min\hfill \end{array}$

Method 2:

ane.

Determine n:

(6.62) $n=\mathrm{three.32}log\frac{\theta 600}{\theta 300}$

2.

Determine G:

(6.63) $G=\frac{\theta 300}{{511}^n}$

3.

Determine annular velocity, ft./min:

(six.64) $\mathrm{A}\mathrm{V}=\frac{24.5\times Q}{{D_{\mathrm{h}}}^2-{D_{\mathrm{p}}}^2}$

four.

Determine viscosity (μ):

(six.65) $\mu ={\left(\frac{2.4v}{D_{\mathrm{h}}-D_{\mathrm{p}}}\times \frac{2n+\mathrm{one}}{3\mathrm{northward}}\right)}^n\times \left(\frac{200\mathrm{Chiliad}\left(D_{\mathrm{h}}-D_{\mathrm{p}}\right)}{v}\right)$

five.

Slip velocity (Vs), ft./min:

(6.66) $\mathrm{V}\mathrm{s}=\frac{{\left(\mathrm{DenP}-\mathrm{M}\mathrm{W}\right)}^{0.667}\times 175\times \mathrm{DiaP}}{{\mathrm{M}\mathrm{W}}^{0.333}\times {\mu }^{0.333}}$

Classification:

due north  =   Dimensionless

K  =   Dimensionless

ten  =   Dimensionless

θ600   =   600 viscometer dial reading

θ300   =   300 viscometer dial reading

Q  =   Circulation rate, gpm

D _h  =   Pigsty diameter, in.

D _p  =   Pipe or collar OD, in.

v  =   Annular velocity, ft./min

μ  =   mud viscosity, cps

DensP   =   Cutting density, ppg

DiaP   =   Cutting diameter, in.

Example: Using the data listed below, determine the annular velocity, cuttings slip velocity, and the cut internet rise velocity:

Data: Mud weight (MW)   =   11.0   ppg

  Plastic viscosity (PV)   =   xiii   cps

  Yield point (YP)   =   x   lb/100   sq.ft.

  Bore of particle   =   0.25   in.

  Density of particle   =   22.0   ppg

  Hole diameter   =   12¼   in.

  Drill pipage OD   =   5.0   in.

  Circulation rate   =   520   gpm

ane.

Make up one's mind n:

$\begin{array}{c}\hfill \mathrm{north}=3.32log\left(\frac{36}{23}\right)\hfill \\ \hfill n=0.64599\hfill \end{array}$

2.

Determine K:

$\begin{array}{c}\hfill K=\frac{23}{{511}^{0.64599}}\hfill \\ \hfill \mathrm{Thousand}=0.4094\hfill \end{array}$

3.

Determine annular velocity, ft./min:

$\begin{array}{c}\hfill \mathrm{five}=\frac{\mathrm{24.v}\times 520}{{12.25}^{\mathrm{two}}-{\mathrm{v.0}}^2}\hfill \\ \hfill v=\frac{12,740}{125.06}\hfill \\ \hfill v=102\mathrm{ft}./min\hfill \end{array}$

4.

Decide mud viscosity, cps:

$\begin{array}{c}\hfill \mu ={\left(\frac{\mathrm{2.four}\times 102}{12.25-5.0}\times \frac{\mathrm{ii}\left(0.64599\right)+1}{3\times 0.64599}\right)}^{0.64599}\times \left(\frac{200\times 0.4094\times \left(12.25-\mathrm{v.0}\right)}{102}\right)\hfill \\ \hfill \mu ={\left(\frac{\mathrm{244.viii}}{7.25}\times \frac{2.92}{1.938}\right)}^{0.64599}\times \left(\frac{593.63}{102}\right)\hfill \\ \hfill \mu ={\left(33.6\times \mathrm{one.1827}\right)}^{0.64599}\times \mathrm{five.82}\hfill \\ \hfill \mu =10.82\times 5.82\hfill \\ \hfill \mu =63\mathrm{c}\mathrm{p}\mathrm{southward}\hfill \end{array}$

5.

Determine cut slip velocity, ft./min:

$\begin{array}{c}\hfill \mathrm{V}\mathrm{southward}=\frac{{\left(22.0-11.0\right)}^{0.667}\times 175\times 0.25}{{11.0}^{0.333}\times {63}^{0.333}}\hfill \\ \hfill \mathrm{5}\mathrm{s}=\frac{\mathrm{iv.95}\times 175\times 0.25}{2.222\times 3.97}\hfill \\ \hfill \mathrm{V}\mathrm{s}=\frac{216.56}{\mathrm{viii.82}}\hfill \\ \hfill \mathrm{V}\mathrm{s}=24.55\mathrm{ft}./min\hfill \end{array}$

six.

Determine cutting net rise velocity, ft./min:

$\begin{array}{l}\mathrm{Annular}\mathrm{velocity}=102.00\mathrm{ft}./min\hfill \\ \hfill \mathrm{Cutting}\mathrm{slip}\mathrm{velocity}=-24.55\mathrm{ft}./min\hfill \\ \hfill \mathrm{Cutting}\mathrm{n}\mathrm{east}\mathrm{t}\mathrm{rise}\mathrm{velocity}=\overline{77.45\mathrm{ft}./min}\hfill \end{array}$

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Petroleum Related Rock Mechanics

Erling Fjær , ... Rasmus Risnes , in Developments in Petroleum Science, 2021

AE	Acoustic Emission
API	American Petroleum Institute
ASR	Anelastic Strain Recovery
AV	Annular Velocity
AVO	Amplitude Versus Offset
BHA	Bottom Hole Assembly
BHTV	Borehole TeleViewer
CEC	Cation Exchange Chapters
CHOPS	Cold Heavy Oil Product with Sand
CMP	Common Midpoint Gather
CMS	Constant Mean Stress exam
CSL	Critical State Line
CT	Reckoner Tomography
CWT	Continuous Wave Technique
CU	Consolidated Undrained
DEM	Discrete Chemical element Method
DIF	Drilling Induced Fracture
DWVA	Differential Wave Velocity Analysis
ECD	Equivalent Circulating Density
EGS	Enhanced Geothermal Systems
EOR	Enhanced Oil Recovery
ESD	Equivalent Static Density
FBP	Fracture Breakup Pressure
FCP	Fracture Closure Pressure level
FEM	Finite Element Method
FIP	Fracture Initiation Pressure
FIT	Formation Integrity Test
FPP	Fracture Propagation Pressure
GOR	Gas Oil Ratio
GPS	Global Positioning System
HPHT	High Pressure High Temperature
ISIP	Instantaneous Shut-In Pressure
ISRM	International Society for Rock Mechanics
KGD	Khristianovitch–Geertsma–de Klerk

LCM	Lost Apportionment Fabric
LEDO	Low Energy Drilling Operations
LEFM	Linear Elastic Fracture Mechanics
LOP	Leak Off Pressure
LOT	Leak Off Test
LVDT	Linear Variable Differential Transformer
MHF	Massive Hydraulic Fracturing
MPD	Managed Pressure Drilling
MSL	Mean Sea Level
MWD	Measurement While Drilling
NMO	Normal MoveOut
PFC	Particle Flow Lawmaking
PKN	Perkins–Kern–Nordgren
PWD	Pressure While Drilling
REV	Representative Simple Volume
ROP	Rate Of Penetration
RMS	Root Mean Foursquare
RT	Rotary Tabular array (drillfloor level)
SEM	Scanning Electron Microscope
SG	Specific Gravity; density relative to water
SRV	Stimulated Reservoir Volume
sst	Sandstone
TI	Transverse Isotropy
TIF	Thermally Induced Fracturing
TWC	Thick-Walled Cylinder
TWT	Ii-Mode Traveltime
UCS	Uniaxial (Unconfined) Compressive Strength
UST	Uniaxial Strain Test
VSP	Vertical Seismic Profiling
WIFF	Wave-Induced Fluid Flow
XLOT	Extended Leak-Off Exam
XRD	X-Ray Diffraction
YP	Yield Point (mud holding)

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https://www.sciencedirect.com/science/article/pii/B9780128221952000267

Air and Gas Drilling (Drilling Dry and with Mist)

Bill Rehm , ... Arash Haghshenas , in Underbalanced Drilling: Limits and Extremes, 2012

5.5.four Floating Bed

Where the drill collar string and the drillpipe meet, the annular volume increases and the annular velocity is reduced. This is an area (along with upper hole washouts), where floating beds form. The larger cuttings float on the air column because the velocity below is great enough to lift them, simply the velocity in the larger annular area is not great plenty to acquit the cuttings upward. The larger cuttings circulate around in the floating bed until they are broken upwards by the action of the drillpipe and continue up pigsty, or until the air is close off for a connection and the cuttings fall back down around the drill collars. The larger cuttings show up as fill up after the connection or, in the worst case, they stick to the pipage (see Figure 5-7).

Effigy five-7. Floating bed and mud band may occur at the tiptop of drill collars where gas velocity drops as it enters a larger geometry; height of drill collars are usually considered a critical bespeak of hole cleaning requirement to avoid these issues

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URL:

https://www.sciencedirect.com/science/commodity/pii/B9781933762050500127

Up Hole Air Velocity Calculator,

Source: https://www.sciencedirect.com/topics/engineering/annular-velocity

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