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Writing expressions for these assumptions :

Volume basis :Vs + Vi =Vf . . . . . . . . . . . . . . . . . (1)

Weight basis : sVs + iVi = fVf . . . . . . .. . . . (2)Where

Vs = Volume of solidVi = Volume of initial mud (or any liquid)

Vf = Final volume of mixture

s = Density of solid

i = Density of initial mud

f = Density of final mudSolving for Vs :

Vs = Vf (f i) . . . . . . . . . . . . . . . (3)(s i)

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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 sodoes gm/cc cc.

Since the net volume of a powdered solid is not readilymeasurable (usually measured by weight), bymultiplying the s to equation (3), the Vs can becalculated :

sVs =sVf (fi) . . . . . . . . . . . . (4)(s i)

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Example 1A 9.5 lb/gal mud contains clay (SG = 2.5) and fresh water.Compute (a) the volume % and (b) the weight % clay in thismud.

Using Equations (3) & (4)

Solution :

(a) Altering eq. (3)

Volume % solids = Vs 100%

Vf

= f i 100% = 9.4 %s - i

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(b) Divide eq. (4) fVf :

Weight % solids = sVs 100

fVfSubstitute Vs & Vf in terms od densities := 20.6%

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Example 2 For laboratory purposes, it is desired to mix one liter of

bentonite- fresh water mud having a viscosity of 30 cP:

(a) What will be the resulting mud density ?

(b) How much of each material should be used ?

Hints for Solution :

Assume Wyoming bentonite,

Solid content = 3 % by volume

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For certain types of problems it is convenient to express eq.(3) in a different form.

Suppose that the quantity of solids (Vs) necessary toincrease (or decrease) the density of an initial mud isdesired. Then :

Vs = (Vi + Vs) (f i) . . . . . . . . . . (3a)

s iWhere ; Vi + Vs = Vf . . . . . . . . . . . (from eq. 1)

Solving for Vs gives :Vs = Vi (f i) . . . . . . . . . . . . . . . (5)

(s f)

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

4.3) should be added to the mud of Example 2 toincrease its density to 10 ppg ?

(b) What will be the resulting volume ?Hint for Solution :

Use eq. (5)

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Barite is so universally used as a weighting material, it isuseful to express eq. (5) in field units.

Barite is sold in 100 lb bags or sacks :

1 sack = 100 lb = 0.373 cu.ft

4.3 (62.4 lb/cu.ft)

= 0.373 cu.ft = 0.0664 bbl of net5.615 cu.ft/bbl barite

Therefore, 1 bbl (net) of barite = 1 bbl = 15 sacks0.0664 bbl/sack

Note : Assuming SG of Barite = 4.3

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Let Sb = Sacks of barite necessary to increase the density ofVi bbl of mud from i to f.

Substituting these special conditions into eq. (5):

Sb = Vi (f i)15 4.3 (8.33) f

Therefore;

Sb = 15 Vi (f i) . . . . . . . . . . . . (5a)35.8 - f

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Similarly, if clay (SG = 2.5) is sold in 100 lb bags orsacks :1 sack = 100 lb = 0.641 cu.ft

2.5 (62.4 lb/ cu.ft)= 0.641 cu.ft = 0.1142 bbl of net clay

5.615 cu.ft/ bbl

1 bbl (net) of clay = 1 bbl = 8.75 sacks

0.1142 bbl/sack

Let Sc = Sacks of clay necessary to increase the density ofVi bbl of mud from i to f.

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Substituting these special conditions into eq. (5):

Sc = Vi (f i)

8.75 2.5(8.33) f

Therefore, Sc = 8.75 Vi (f i) . . . . . . . . . . . . (5b)

20.8 - f

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

the density of 1000 bbl of mud from 10 to 14 lb/ gal ?

(b) What will be the final mud volume ?

Hint for solution :

Use eq. (5a)

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To dilute or reduce mud density :

Vw + Vi = Vf . . . . . . . . Volume basis

wVw + iVi = fVf . . . . . . . . Weight basis

wVw + iVi = f(Vw + Vi)

wVw fVw = fVi iVi

Vw = Vi (f i)

wf

Vw = Vi (i f) . . . . . . . . . . . . . . . . . (6)

(f w)

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where,

Vw = bbls of water necessary to reduce the density of Vibbls initial mud from i to f.

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Example 5(a) How much fresh water must be added to 1000 bbl of12 lb/gal mud to reduce its density to 10 lb/gal ?

(b) What will be the resulting volume ?

Hints for solution :

(a) Use eq. (6)

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

mud weight of 755 bbl from 77 pcf to 92 pcf ? Whatis the resulting volume ?

(b) Calculate the new mud weight when 126 bbl of oil(SG= 0.8)is added to the new system. What is theresulting volume ?

(c) Determine the quantity of barite required tomaintain a mud weight of 92 pcf. What is the finalvolume ?

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Hints for solution :

Use eq. (5a)