part ix engineering calculations

8/10/2019 Part IX Engineering Calculations

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Advanced Mud School

Part IX Engineering Calculations

Presented By:

Jeff Imrie

August 2006



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Engineering Calculations

• Mud engineers must be capable of makingvarious calculations including:

– capacities and volumes of pits, tanks,

pipes and wellbores

– circulation times

– annular and pipe mud velocities – other important calculations.



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Engineering Calculations - Volumes

• Rectangular tank – Volume(bbl) = length × width × height

5.6146



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• Vertical cylindrical tank – Volume(bbl) = (diameter)2 × height

7.1486



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• Horizontal cylindrical tank (half full orless)

Volume (bbl) =

(0.3168 d x h + 1.403 h2 - 0.933 x(h3/d)) × length

5.6146

• h is the height of the fluid level, ft

• d is the diameter of the tank, ft

• All diameters are expressed in inches; section

lengths are expressed in feet.



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• Horizontal cylindrical tank (more thanhalf full)

Volume (bbl) = (diameter)2 × length -

7.1486

(0.3168 d x h + 1.403 h2 - 0.933 x(h3/d)) × length

5.6146

• h is the height of the fluid level, ft

• d is the diameter of the tank, ft

• All diameters are expressed in inches; section

lengths are expressed in feet.



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• Drillpipe or drill collar capacity anddisplacement

– You can use calculations or look the data up in atable

Capacity (bbl/ft) = (inside diameter)2

1029.4

Displacement (bbl/ft) =

(outside diameter)2 - (inside diameter)2

1029.4



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• Capacity of a long cylinder

bbl/100 ft = 0.0972 D2

bbl/inch = 0.000081 D2

bbl/1,000 ft = 0.972 D2

ft/bbl = 1029 ÷ D2

– Where D is the diameter of the cylinder, in



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• Inside diameter of a steel cylinder

ID = OD2

- 0.3745W

• OD is the outside diameter, in

• W is the weight, lb/ft



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• Pump Output – Normally found in tables

– Duplex Pump (bbl/stroke)

Output =

(2 x liner 2 - rod diameter 2) × stroke x Eff

6176.4

– Triplex Pump (bbl/stroke)

Output =

(liner inside diameter)

2

× 0.000243× stroke length



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Engineering Calculations –

Annular Velocity

• Annular Velocity (commonly referred toas AV) is the average rate at which fluidis flowing in an annulus.

• A minimum annular mud velocity isneeded for proper hole cleaning.

• This minimum annular velocity dependson a number of factors, including rate of

penetration, cuttings size, hole angle,

mud density and rheology.



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Annular Velocity

• Annular velocity Annular velocity (AV), ft/min:

AV = 1029.4 × POBPM

ID2HOLE - OD2DP

– POBPM is the pump output in barrels per minute

– IDHOLE is the diameter of hole or inside diameterof casing in inches

– ODDP is drillpipe outside diameter in inches



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Circulation Time

• Total circulation time is the time (ornumber of strokes) required for mud tocirculate from the pump suction down the

drillstring, out the bit, back up theannulus to the surface, through the pitsand arrive at the pump suction onceagain.



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Circulation Time

• Total circulation time

Total circulation time (min) =VSystem

VPump Output

– VSystem= Total system volume (active) (bbl)

– VPump Output = Pump output (bbl/min)



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Circulation Time

• Bottoms-up time is the time (or numberof strokes) required for mud to circulate

from the bit at the bottom of the hole

back up the annulus to the surface.



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Circulation Time

• Bottoms-up timeBottoms-up time (min) =

V Annulus

VPump Output

– V Annulus= Annular volume (bbl) – VPump Output= Pump output (bbl/min)



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Hydrostatic Pressure

• Hydrostatic pressure (P HYD ) is the pressureexerted by the weight of a liquid on its“container” and is a function of the density of

the fluid and the True Vertical Depth (TVD) asshown by the equation below.

• In a well, this is the pressure exerted on the

casing and open hole sections of the wellboreand is the force that controls formation fluidsand prevents wellbore collapse.



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Hydrostatic Pressure

• Formula for Hydrostatic pressure

PHYD (lb/in.2) = Mud weight (lb/gal) x TVD (ft) x 0.052

Conversion factor 0.052 =

12 in./ft

231 in.3/gal

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Mass Volume Balance

• The ability to perform a material balanceis essential in drilling fluids engineering.

• Solids analysis, dilutions, increasingdensity and blending equations are allbased on material balances.

• To solve a mass balance, first determinethe known and unknown volumes anddensities and identify as component or

product.

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Mass Volume Balance

• In general, the following steps lead to solving for the unknown:

– Step 1. Draw a diagram.

– Step 2. Determine components and products,mark volumes, and densities as known or

unknown.

– Step 3. Develop mass and volume balance.

– Step 4. Substitute one unknown into mass

balance and solve equation.

– Step 5. Determine second unknown and calculate

material consumption.

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Mass Volume Balance

• Volume balanceVTotal = V1 + V2 + V3 + V4 + …

• Mass balanceVTotal r Total = V1r 1 + V2r 2 + V3r 3 + V4r 4 + …

V = Volume

r = Density

• These simple formula's as the basis of volume

and mass balance

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Mass Volume Balance

• Example build a weighted mud – Determine the quantities of materials to

build 1,000 bbl (159 m3) of 16.0 lb/gal

(1.92 kg/l) mud with 20 lb/bbl (57 kg/m3)Bentonite use Barite as weighting

agent.

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Mass Volume Balance

• Step 1. Draw a diagram.• Step 2. Determine densities and volumes with

known and unknown.

Components r (lb/gal) V (bbl)

Water 8.345 ?

Bentonite 21.7 22 (see below)Barite 35.0 ?

Mud 16.0 1,000

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Mass Volume Balance

VGel = 20 lb/bbl x 1,000 bbl21.7 lb/gal x 42 gal/bbl

= 22 bbl• Step 3. Develop mass and volume

balance.

VMud r Mud = VWater r Water + VGel r Gel + VBar r Bar

VMud = VWater + VGel + VBar

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Mass Volume Balance

• At this point the mass balance has twounknowns (VBar and VWater) that can bedetermined by using both equations. Solve the

volume balance for one unknown and thensubstitute it into the mass balance.

1,000 bbl = VWater + 22 bbl + VBar

VBar (bbl) = (1,000 – 22) – VWater

VBar (bbl) = 978 – VWater

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Mass Volume Balance

• Step 4. Substitute one unknown into massbalance and solve equation.

V Mud r Mud = V Water r Water + V Gel r Gel + V Bar r Bar

1,000 x 16 = VWater x 8.345 + 22 x 21.7 + (978 – VWater) x 3516,000 = VWater x 8.345 + 477.4 + 34,230 – VWater x 35

VWater (35 – 8.345) = 477.4 + 34,230 – 16,000 = 18,707.4

VWater= 18,707.4

26.655

VWater= 702 bbl

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Mass Volume Balance

• Step 5. Determine second unknown andcalculate material consumption. Thevolume of barite is derived from the

volume balance.VBar = (978 – VWater) = 978 – 702 = 276 bbl

lbBar = 276 bbl x (35 lb/gal x 42 gal/bbl)

= 276 bbl x 1,470 lb/bbl

= 405,720 lb

Or 4057 sacks (100lb).

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Engineering Calculations – Solids Analysis

• The final use of material balance to bediscussed is determining solids analysis.

• Two cases are discussed, an unweighted freshwater system without oil and aweighted system containing salt and oil.

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• The material balance and volumeequation are as follows:

VMudr Mud = VWater r Water + VLGSr LGS

VMud = VWater + VLGS

VMud = Volume of mud

VWater = Volume of water VLGS = Volume of Low-Gravity Solids

r Mud= Density of mud or mud weight

r Water = Density of water

r LGS= Density of Low-Gravity Solids

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• The density of water, low-gravity solidsand mud are all known. If the volume ofmud is 100% and the mud weight is

known, the volume of the LGS can bedetermined.

• First, the volume of water must be solved for in the volume equation.

%VWater = 100% – %VLGS

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• Then this equation must be substitutedinto the material balance.

100% rMud = (100% – %VLGS) rWater + %VLGS rLGS

• Solving for the percent volume of low- gravity solids the following equation isobtained:%VLGS = 100% x (rMud – rWater)

(rLGS – rWater)

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• Example with un-weighted mud – An unweighted freshwater mud has a

density of 9.2 lb/gal. Determine the

percent of low-gravity solids in the system.%VLGS = 100% x (rMud – rWater)

(rLGS – rWater)

%VLGS = 100% x (9.2 – 8.345)

(21.7 – 8.345)

%VLGS = 6.4%

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• Example with weighted saltwater mud – The second case is a weighted system

containing sodium chloride and oil. This

material balance is one of the morecomplicated material balance evaluations

encountered in drilling fluids engineering.

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• For this example, the following is given:

– Mud weight 16.0 lb/gal

– Chlorides 50,000 mg/l

– Oil (%) 5 (7.0 lb/gal)

– Retort water (%) 63 – Weight material Barite (35.0 lb/gal)

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• Step 1. Draw a component diagram.• Step 2. Determine the known and unknown

variables and label the components. Use theappropriate density for the HGS, LGS and oil.

Components r (lb/gal) V (%)

HGS 35.0 ?

LGS 21.7 ?

Oil 7.0 5%Salt ? ?

Water 8.345 63%

Mud 16.0 100%

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• Step 3. Write the material balance andvolume equations.

VMud rMud = VHGS rHGS + VLGS rLGS + VSW rSW

+ VOil rOil

VMud = VHGS + VLGS + VSW + VOil = 100%

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• The volume of saltwater cannot be determineddirectly. The retort measures the quantity ofdistilled water in the mud sample (VWater). The

volume of salt (VSalt) can be calculated aftermeasuring the chloride concentration of thefiltrate (saltwater).

• The volume of saltwater is equal to the retortwater volume plus the calculated salt volume:

VSW = VWater + VSalt

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• The equations are changed to use thesevariables.

VMud rMud = VHGS rHGS + VLGS rLGS + (Vwater +

VSalt) rSW + VOil rOil

VMud = VHGS + VLGS + (Vwater + VSalt) + VOil =

100%

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• Step 4. Develop the correspondingequations to solve for the unknowns.

– The density of the saltwater (r SW) can be

calculated from the chloride concentration.The following equation is a curve fit of

density-to-chloride concentration for

sodium chloride.SGSW = 1+1.166 x 10–6 x (mg/l Cl– ) – 8.375 x 10–13 x (mg/l Cl– ) 2

+1.338 x 10–18 x (mg/l Cl– )3

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SGSW = 1+1.166 x 10–6

x (50000) – 8.375 x 10–13

x (50000)2

+1.338x 10–18 x (50000)3= 1.0564

rSW (lb/gal) = 1.0564 x 8.345 = 8.82 lb/gal

• The weight percent sodium chloride of thesaltwater is calculated by the followingexpression:

% NaCl (wt) = mg/l Cl– x 1.65

SGSW x 10,000

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• Substitute% NaCl (wt) = 50000 x 1.65

1.0564 x 10,000

= 7.81%

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• The volume percent salt of the mud (VSalt)can be calculated from the specific gravityand weight percent sodium chloride of

the saltwater by the following equation:VSalt = VWater 100 -1

SGSW (100 – % NaCl (wt))

VSalt =63% 100 -11.0564 (100 – 7.81))

VSalt = 1.69%

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• Frequently this salt concentration isreported in pounds per barrel using thefollowing conversion:

NaCl (lb/bbl)= (VWater + VSalt) x mg/l Cl– x 1.65 x 3.5

10,000 100

NaCl (lb/bbl) = (63 + 1.69) x 50000 x 1.65 x 3.510000 100

= 18.68 lb/bbl

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• Step 5. Use the material balance andvolume equations to solve for theremaining unknowns.

• VHGS and VLGS are the only remainingunknowns. First the volume equation issolved for VLGS in terms of VHGS and

substituted into the material balanceequation to obtain:VMud rMud = VHGS rHGS + VLGS rLGS + (VWater + VSalt) rSW

+ VOil rOil

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VHGS rHGS = VMud rMud – (100 – VWater – VSalt – VOil – VHGS) rLGS – (VWater +VSalt)rSW – VOil rOil

VHGS = 100 rMud – (100 – VWater – VSalt – VOil) rLGS – (VWater + VSalt)rSW – VOil rOil

rHGS – rLGS

VHGS =16 x 100 – (100 – 63 – 1.69 – 5) x 21.7 – (1.69 + 63) x 8.8 – 7 x 5

(35 – 21.7)

VHGS

= 25.41%

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• This concentration is converted to lb/bbl unitsas follows:

HGS (lb/bbl) = VHGS x rHGS

100

HGS (lb/bbl) = 25.41% x (35 lb/gal x 42 gal/bbl)100

HGS (lb/bbl) = 373.5 lb/bbl

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Next, VLGS can be determined using thevolume equation:

VLGS = 100% – VWater – VSalt – VOil – VHGS

VLGS = 100% – 63% – 1.69% – 5% – 25.41%= 4.9%

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• This concentration is converted to lb/bbl unitsas follows:

LGS (lb/bbl) = VLGS x rLGS

100

LGS (lb/bbl) = 4.9% x (21.7 lb/gal x 42 (gal/bbl))

100LGS (lb/bbl) = 44.7 lb/bbl

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Volume (%)V H20 63

V OIL

5

V SALT 1.69

V HGS 25.41

V LGS 4.9Total 100.0

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Engineering Calculations – Solids Analysis Weight (lb/bbl)

H2O [0.63 x 350] 220.5Oil [0.05 x 7 x 42] 14.7

NaCl 18.7

HGS 373.5

LGS 44.7 .

Total 672.1

rMud (lb/gal) = 672.1 = 16.0 lb/gal

42

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• End

part ix engineering calculations

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