Part IX Engineering Calculations

Advanced Mud School Part IX Engineering Calculations Presented By: Jeff Imrie

August 2006

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Engin ng ine eering ri ng Calcul lc ula ation ti ons s

• Mud engineers must be capable of making various calculations including: – capacities and volumes of pits, pi ts, tanks, pipes and wellbores – circulation times – annular and pipe mud velocities vel ocities – other important calculations.

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Engine ngin eering Calcula lcul ations tion s - Volume olum es

• 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 or less) Volume (bbl) = (0.3168 d x h + 1.403 h 2 - 0.93 .933 x(h3/d)) /d)) × leng length th 5.6146

• h is the height of the fluid level, ft • d is the diameter of the tank, ft • All diameters are expressed expressed in inches; section lengths are expressed in feet.

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Engineering Calculations - Volumes

• Horizontal cylindrical tank (more than half full) Volume (bbl) = (diameter) 2 × length 7.1486 (0.3168 d x h + 1.403 h 2 - 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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Engineering Calculations - Volumes • Drillpipe or drill collar capacity and displacement – You can use calculations or look the data up in a table 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 to as AV) is the average rate at which fluid is flowing in an annulus. • A minimum annular mud velocity is needed for proper hole cleaning. • This minimum annular velocity depends on a number of factors, including rate of penetration, cuttings size, hole angle, mud density and rheology.

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Engineering Calculations – 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 diameter of casing in inches – ODDP is drillpipe outside diameter in inches

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Engineering Calculations – Circulation Time • Total circulation time is the time (or number of strokes) required for mud to circulate from the pump suction down the drillstring, out the bit, back up the annulus to the surface, through the pits and arrive at the pump suction once again.

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Engineering Calculations – 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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Engineering Calculations – Circulation Time • Bottoms-up time is the time (or number of 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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Engineering Calculations – Circulation Time • Bottoms-up time Bottoms-up time (min) = V Annulus VPump Output

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

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Engineering Calculations – Hydrostatic Pressure • Hydrostatic pressure (P HYD ) is the pressure exerted 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) as shown by the equation below. • In a well, this is the pressure exerted on the casing and open hole sections of the wellbore and is the force that controls formation fluids and prevents wellbore collapse.

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Engineering Calculations – 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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Engineering Calculations – Mass Volume Balance • The ability to perform a material balance is essential in drilling fluids engineering. • Solids analysis, dilutions, increasing density and blending equations are all based on material balances. • To solve a mass balance, first determine the known and unknown volumes and densities and identify as component or product.

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Engineering Calculations – 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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Engineering Calculations – Mass Volume Balance • Volume balance VTotal = V1 + V2 + V3 + V4 + …

• Mass balance VTotal 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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Engineering Calculations – 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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Engineering Calculations – 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

Barite

35.0

?

Mud

16.0

1,000

? 22 (see below)

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Engineering Calculations – Mass Volume Balance VGel = 20 lb/bbl x 1,000 bbl 21.7 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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Engineering Calculations – Mass Volume Balance • At this point the mass balance has two unknowns (VBar and VWater) that can be determined by using both equations. Solve the volume balance for one unknown and then substitute 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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Engineering Calculations – Mass Volume Balance • Step 4. Substitute one unknown into mass balance 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 35 16,000 = 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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Engineering Calculations – Mass Volume Balance • Step 5. Determine second unknown and calculate material consumption. The volume 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 be discussed is determining solids analysis. • Two cases are discussed, an unweighted freshwater system without oil and a weighted system containing salt and oil.

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• The material balance and volume equation 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 solids and mud are all known. If the volume of mud is 100% and the mud weight is known, the volume of the LGS can be determined. • 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 substituted into the material balance. 100% rMud = (100% – %VLGS) rWater + %VLGS rLGS

• Solving for the percent volume of low gravity solids the following equation is obtained: %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 more complicated 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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Engineering Calculations – Solids Analysis • Step 1. Draw a component diagram. • Step 2. Determine the known and unknown variables and label the components. Use the appropriate density for the HGS, LGS and oil. Components HGS LGS Oil Salt Water Mud

r (lb/gal) 35.0 21.7 7.0 ? 8.345 16.0

V (%) ? ? 5% ? 63% 100%

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• Step 3. Write the material balance and volume equations. VMud rMud = VHGS rHGS + VLGS rLGS + VSW rSW + VOil rOil VMud = VHGS + VLGS + VSW + VOil = 100%

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Engineering Calculations – Solids Analysis • The volume of saltwater cannot be determined directly. The retort measures the quantity of distilled water in the mud sample (VWater). The volume of salt (VSalt) can be calculated after measuring the chloride concentration of the filtrate (saltwater). • The volume of saltwater is equal to the retort water volume plus the calculated salt volume: VSW = VWater + VSalt

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• The equations are changed to use these variables. 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 corresponding equations 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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Engineering Calculations – Solids Analysis SGSW = 1+1.166 x 10–6 x (50000) – 8.375 x 10–13 x (50000)2 +1.338 x 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 the saltwater is calculated by the following expression: % 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 gravity and weight percent sodium chloride of the saltwater by the following equation: VSalt = VWater

VSalt =63% VSalt = 1.69%

100 SGSW (100 – % NaCl (wt)) 100 1.0564 (100 – 7.81))

-1

-1

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• Frequently this salt concentration is reported in pounds per barrel using the following 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 3.5 10000 100 = 18.68 lb/bbl

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• Step 5. Use the material balance and volume equations to solve for the remaining unknowns. • VHGS and VLGS are the only remaining unknowns. First the volume equation is solved for VLGS in terms of VHGS and substituted into the material balance equation to obtain: VMud rMud = VHGS rHGS + VLGS rLGS + (VWater + VSalt) rSW + VOil rOil

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Engineering Calculations – Solids Analysis 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 units as 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 the volume 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 units as follows: LGS (lb/bbl) = VLGS x rLGS 100 LGS (lb/bbl) = 4.9% x (21.7 lb/gal x 42 (gal/bbl)) 100 LGS (lb/bbl) = 44.7 lb/bbl

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Volume V H20 V OIL V SALT V HGS V LGS Total

(%) 63 5 1.69 25.41 4.9 100.0

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Engineering Calculations – Solids Analysis Weight H2O [0.63 x 350] Oil [0.05 x 7 x 42] NaCl HGS LGS Total

rMud (lb/gal) = 672.1 42

(lb/bbl) 220.5 14.7 18.7 373.5 44.7 . 672.1

= 16.0 lb/gal

Part IX Engineering Calculations

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