Calculations of drilling fluid
P=pressure of mud column Ib/in2
ρ=Mud density Ib/gal
H=well depth ft.
Mud density of water-base = between 8.33 and 118.32 Ib/gal when add the
barite.
Mud density of oil-base = below 8.33Ib/gal.
Must the densities of the Muds which selected are above 0.25 -0.41Ib/gal
of the value of density of the fluids which exist in the drilled layers that
cause the hydrostatic pressure.
Mud Balance: is used to test the drilling fluid density, Ib/gal, Ib/ft3, gm/cc,
….etc. at 21C0.
Most calculation of drilling fluid is (volume and density) due to the
additions of solid and liquid materials to the system.
- The first step of calculation is the system volume, which equals the
summation of liquid in well, surface pit, and tanks.
This stage is difficult and its need the experience of the drilling engineer.
The assumptions are:
1. All volumes of substance are able to summation
2. All weights of substance are able to summation
Vs+Vm1=Vm2
ρsVs+ρm1Vm1= ρm2Vm2
Vs=volume of solid substance
1
Vm1=initial volume of drilling fluid
Vm2=final volume of drilling fluid
ρs= density of solid substance
ρm1=initial density of drilling fluid
ρm2=initial density of drilling fluid
……………………………………. 1
Since the measuring of the volume solid powder is difficult we using the
equation below:
……………………………………….. 2
If Vm2 is exist using equ.1
If Vm1 is exist using equ.2
To calculate the weight of solid in the liquid using:
……………………………………….3
When using these equ. We can use the field units.
For example 1 sack of barite = 100Ib = 100/(62.4X4.3)=0.373ft3/sack.
Since 1bbl= 5.61ft3
sack = ‫كيس‬
bbl = ‫برميل‬
lb = ‫باون‬
Water density =62.4 lb/ft3
Specific gravity of barite=4.3
1sack=0.373/5.61=0.0665bbl/sack and 1bbl=15sack.
Let SB = No. of sack of barite to increase the density of 100bbl of Mud from
pm1 to ρm2 , equ.2 become with barite density = 35.8Ib/gal.
2
In same way we can calculate No. of Bentonite sack (specific weight=2.5)
1 sack of bentonite = 100Ib = 100/(62.4X2.5)=0.641ft3/sack.
1sack=0.641/5.61=0.114bbl/sack and 1bbl=8.75sack.
No. of sack of bentonite to increase the density of 100bbl of Mud from
pm1 to ρm2 , equ.2 become with bentonite density = 20.8Ib/gal.
But when reduce the density of mud by added water or oil, equ.1 becomes
when using water, density =8.33Ib/gal.
Note: for lab. Calculations can convert field units to lab. units
Example1: The volume and density of drilling fluid (water+ bentonite) is
500bbl and 9 Ib/gal respectively. Amount of barite was added to this mud,
which changes the mud density to 12 Ib/gal. Find the (volume and weight)
percent’s of the solid substances.
3
Sol.:
To find volume of solid substances:
The volume percent of solid substance:
The volume percent of bentonite :
The volume percent of barite :
The total volume percent of solid substances are:
11%+5.3%= 16.3%
To calculate the weight of solid in the liquid using:
The weight percent of solid substances:
Bentonite percent:
4
Barite percent:
The total weight percent of solid substances are:
12.5%+33%=45.5%
Example2: the volume of drilling fluid is 800 bbl was prepared from water
and Bentonite with 5% volume percent, amount of barite was added that
change the density to 12.5 Ib/gal. find the barite added in tons.
Sol.:
Ib/gal
bbl
To find the barite weight in tons:
Example3:
A- Find the amount of water must be add to 1000 bbl of drilling fluid 12
Ib/g to reduce the density to 10 Ib/gal.
B- What is the new volume?
5
Sol.:
The new volume
1000+1197.6=2197.6 bbl
Example4: a well is drilling with 15.5inch (volume=0.15bbl/ft) and drilling
speed 23ft/min. the pump discharge of drilling fluid= 720gal/min. the mud
density at the entrance of well 9.6Ib/gal. the density of cuttings rock
=20.8Ib/gal. Calculate the density of mud in the annular.
Volume of cuttings rock is
23 x 0.15 = 3.45 bbl/min.
Weight of cuttings rock is
3.45 x 42 x 20.8 = 3013.92Ib/min.
Final volume in min=pump disch./min + vol. of cuttings rock/min
Vm2=720+(3013.92/20.8) = 864.9 gal/min
6
Example5: a drilling operation of well reach 7300ft depth with 10 inch
diameter. The volume of surface connection=23ft3, dimensions of mud pool
=20x14x12 ft.
The composition drilling fluid which used during the operation is (water+
Bentonite+ substance of filtration reducing has specific weight 1.8). the
percent of Bentonite weight and filtration reducer are 7% and 0.75%
respectively, also is necessary convert the drilling fluid to emulsion by
added diesel oil (s. w. 0.80), which make the mud density=8.4Ib/gal.
Find amounts of diesel oil and water in bbl, also Bentonite and treatment
substance in tons in the emulsion, where the emulsion volume is equal the
volume of system.
Sol.:
System volume=12x14x40+23+(52xπ/144)x7300=7362ft3=1312bbl
gm/cc ‫دائما وحدات هذا العالقة‬
Assume diesel vol. Vo and diesel density ρo
Where:
7
gm/cc ‫لغرض التحويل من وحدات‬
lb/gal ‫الى‬
8.33= (350/42) ‫نضرب‬
the volume of (water+ Bentonite+ substance of filtration)=1312201.06=1110.94bbl.
And its weight =1110.94x8.726x42=407103lb.
Water in emulsion =
Bentonite in emulsion =
Filtration substance in emulsion =
8