Mud Calculations

 

 Writing expressi  Writing expressions ons for these assumptions :  Volume  V olume basis basis : V s + V i = V f . . . . . . . . . . . . . . . . . (1) f  f . . . . . . .. . . . (2) (2)  Weight  W eight basis : ρs V   V s + ρ V   iV i = ρ V   Where

 Vs = Volume of solid  Vs  Vi = Volume of initial mud mud (or any liquid)  Vf = Final volum volumee of mixture ρs = Density of solid ρi = Density Density of initial initial mud mud ρf = Density of final mud Solving for Vs :  Vss = Vf (ρf  – ρi) . . . . . . . . . . . . . . . (3)  V (ρs – ρi)

 

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The units have to be in consistent set If the density is in lb/gal, the V will be in gal, and so does gm/cc → cc. Since the net volume of a powdered solid is not readily measurable (usually measured by weight), by multiplying the ρs to equation equation (3), the Vs can be calculated :

 ρsV s = ρsV  f  (  ρ f – ρ ) i) . . . . . . . . . . . . (   ρs – ρ ) i)

(4)

 

Example 1  A 9.5 lb/gal mud contains contains cclay lay (SG (SG = 2.5) and fresh water water. Compute (a) the volume % and (b) the weight % clay in this mud. Using Equations (3) & (4) Solution : (a)  Altering eq. (3)  Volume  V olume % solids = Vs × 100%  Vf  = ρf  – ρi × 100%

ρs - ρi

= 9.4 %

 

(b)) Di (b Divi vide de eq. eq. (4) (4) ρfVf ρfVf :  Weight  W eight % solids solids = ρs V s × 100

ρ V  f  f  f  Substitute Subst itute Vs & Vf in terms terms od densities densities : = 20.6%

 

Example 2 

For laboratory purposes, it is i s desired to mix one liter l iter of  bentonitebenton ite- fresh water mud having having a visc viscosity osity of 30 cP:

(a)  What will be the resulting resulting mud density ? (b) How much of each material should be used ?

Hints for Solution :  Assume Wy Wyoming oming bent bentonite, onite, Solid content = 3 % by volume

 

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For certain types of problems it i t is convenient convenient to express eq. (3) in a different form. Suppose thatdecrease) the quantity of solidsof(Vs) necessar necessary y tois increase (or the density an initial mud desired. Then :  Vss = (Vi + Vs) (ρf  – ρi) . . . . . . . . . . (3a)  V

s – ρi  Where ; Vi + V Vss = Vf . . . . . . . . . .ρ . (from eq. 1) Solving for Vs gives :  Vs  Vs = Vi (ρf  – ρi) . . . . . . . . . . . . . . . (5) (ρs – ρf )

 

Example 3 (a) How much weighting material BaSO4, barite, SG =

4.3) should be added to the mud of Example 2 to increase increa se its dens density ity to 10 ppg ? (b)  What will be the resulting resulting volume ? Hint for Solution : Use eq. (5)

 

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Barite is so universally used as a weighting material material,, it is useful to express eq. eq. (5) in field f ield units. Barite Bari te is sold in 10 100 0 lb bags or sack sackss : 1 sack = 100 lb = 0.373 ccu u.ft 4.3 (62.4 (62.4 lb/cu. lb/cu.ft) ft)

= 5.06.13573cuc.uft./ftbbl = 0.0664 bb bal roiftenet Therefore, 1 bbl (net) of barite = 1 bbl = 15 sacks 0.0664 bbl/sack Note : Assuming SG of Barite = 4.3

 

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Let Sb = Sacks of barite necessary to increase the density of   Vi bbl of mud from ρi to ρf .

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Substituting these special conditions into eq. (5): Sb = Vi (ρ (ρf – ρi) 15

4.3 (8. (8.33 33)) – ρf 

Therefore; Sb = 15 Vi (ρf  – ρi) . . . . . . . . . . . . (5a) 35.8 - ρf 

 

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Similarlyy, if clay (SG = 2.5) is sold in 100 Similarl 100 lb bags or sacks : 1 sack =

100 lb 2.55 (62. 2. (62.4 4 lb lb// cu.f cu.ft) t) = 0.641 cu.ft 5.61 5. 6155 cu cu.f .ft/ t/ bb bbll

= 0.641 cu.ft = 0.1142 bbl of net clay  

1 bb b bl ((n net) of clay = 1 bbl = 8.75 sacks 0.1142 bbl/sack Let Sc = Sacks of clay necessary necessary to increase the density of   Vi bbl of mud from ρi to ρf.

 

Substituting these th ese special conditions into eq. eq. (5): Sc = Vi (ρf – ρi) 8.755 22.5( 8.7 .5(8.3 8.33) 3) – ρf  Therefore,

Sc = 8.75 Vi (ρf – ρi) . . . . . . . . . . . . (5b) 20.8 - ρf 

 

Example 4 (a) How many sacks of barite are necessary to increase

the de densi nsity ty of 1000 1000 bbl of mud fr from om 10 to 14 lb/ gal ? (b)  What will be the final mud volume volume ?

Hint for solution :

Use eq. (5a)

 

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To dilu dilute te or reduce mud density :  Vw + Vi = Vf ........V Volume olume basis f  f  f  ρ w Vw + ρ V   iV i = ρ V  . . . . . . . . Weight basis ρ w Vw + ρ Vi  iVi = ρf  (V (Vw w+V Vii) ρ w Vw – ρ Vw f  = ρ Vi f  – ρ Vi  iVi  V  w = Vi (ρf – ρi) ρw–ρf   Vw = Vi (ρi – ρf) ρf ) . . . . . . . . . . . . . . . . . (6) (ρf – ρw)

 

 where,  Vw = bbls of water water neces necessary sary tto o reduc reducee the the density of Vi bbls bb ls in init itia iall mud mud from from ρi to ρf. ρf.

 

Example 5 (a) How How much fresh water must be added tto o 1000 1000 bbl of  12 lb/gal mud to red reduc ucee its density density to 10 10 lb/ga lb/gall ? (b) What will be the resulting volume ? Hints for solution : (a) Use eq. (6)

 

Example 6 (a) How many sacks of barite are required to raise the

mud weig eight of 75 755 bb bbll fro from 77 77 pcf pcf to 92 pc pcff ? W Wha hatt is the resulting volume ? Calculate late the ne new w mud we weigh ightt when when 12 126 6 bbl of oil (b) Calcu (SG= 0.8)is 0. 8)is added to the new syst system. em. What is the resulting volume ? (c) Determine the quantity of barite required to

maintain a mud weight weight of 92 pcf. What is the final  volumee ?  volum

 

Hintss for solu Hint solution tion : Use eq. (5a)