Drilling Calculations
Course
Common Units and symbols
parameter
Unit
Symbol
length
Feet,inches or Mtr
Ft ,inor M
Area
Square feet, Square
inches or square
meters
Volume
Cubic feett, cubic
centimeters cubic
meters or
Cuft. Cu In or Meters
cube
Capacity
US barrels,US
gallons, Cubic ft
bbl, gal(US), cuft
Mass
Pounds,Short tons
lbs, Sh tn
Force (weight)
Pounds-force
lbf
Presure
Ppounds –force per
square inch
psi
Density (mudweight)
Pounds per gallon,
pounds per cubic ft
Ppg ,pcf
torque
Foot pounds
ftlbs
AREAS
Areas
Length
• Area of a square
breadth
Area of a Square = length X breadth
Length
This same expression is applicable to a
rectangle and parallelogram
breadth
base
Area of a triangle =
Height
Height
Area of a Triangle
base
Base x Height
2
Area of a Circle
Diameter
Area of a circle = Π x radius x radius
Where r= radius of the circle
Π = 3.142
Since diameter= 2 x radius
Area of circle= Π x Diameter x Diamater
4
Radius
VOLUMES
Annular area
d
D
Annular or cross sectional area = Π {(D x D)-(d x d)}
4
This can be used to calculate annular areas of the wellbore or
Cross sectional areas of tubulars
Volumes
hieght
Volume of a cube
length
Volume of a cube = length x breadth x height
Height
Cylindrical Capacities
Area
Volume of a cylinder = Area x Height
= Π x Diameter x Diameter x Height
4
This is the basis for all volume the capaciy calculations in well control
Capacity Calculations
Rule of thumb:
Capacity of a cylinder in bbls/ft= Diamater x Diameter
1029.4
Multiply this by the Height of the cylinder ( or length of the hole)
gives the volumetric capacity in barrels
HVolume = Diameter x Diameter x height(or length)
1029.4
This expression is generally used on the rig for volume and capacity
calculations
Annular volume calculations
Top view
d
D
Annular Volume
in bbls
{(D x D)-(d x d)}xlenght
1029.4
Hole volume calculations
ANNULAR VOLUME AROUND DP =
[( D3² - D1² ) / 1029.4] x L1
ANNULAR VOLUME AROUND DC =
[( D3² - D2² ) / 1029.4] x L2
Hole volume calculations
example
Using the following data, calculate the total
annular volume.
L1 = 4000 ft
L2 = 800 ft
D1 = 5 in
D2 = 6.5 in
D3 = 8.5 in
Hole volume calculations
ANNULAR VOLUME AROUND DP =
[( 8.5² - 5² ) / 1029.4] x 4000 = 183.6 bbls
ANNULAR VOLUME AROUND DC =
[( 8.5² - 6.5² ) / 1029.4] x 800 = 23.31 bbls
Total Annular Volume =183.6 + 23.31 =
206.91 bbl
D
Pump displacement volumes
L
The same expression for calculation od hole volumes can
also be used for the calculation of mud pump displacements
Triplex Pump displacement in bbls/ft =
D² x L x e x3
1029.4 x 12 x100
Where D= liner size in inches
L= Stroke length in inches
e= Volumetric efficiency in %
D
Pump displacement volumes
example
L
Where D= 7 inches
L= 11 inches
e = 97%
Pump displacement= 7 x 7 x 97 x 11 x 3
1029.4 x 12 x 100
= 0.127 bbls/ stroke
Weight on bit
• Drill collars and/or heavy wall drill pipes are uses
to put weight on the bit
• Drill collars are also used to keep the string in
tension and prevent it from buckling
• There is a point in the drill collars where
compression stops and tension starts this point
is called the neutral point
• Weight on bit is usually about 75-90%of the
buoyed drill collar weight
compression
Neutral point
Tension
Bouyancy factor
The apparent loss of weight when an object
Is immersed in a fluid
Is calculated using the expression:
65.44-mw or
1 - mw
65.44
65.44
This factor must be multiplied by the weight in air
to calculate the buoyed weight
Buoyed weight= weight in air x buoyancy factor
Apparent or buoyed weight
WB =
WA x fluid density
Steel density
Where:
WB = weight of tubulars in fluid
WA = Weight of tubulars in air
Fluid density = Density of fluid in ppg
Steel Density =Density of Steel in PPG
Example
• A drill string weighs 180000 lbs in air how much
will it weigh in 11.5ppg mud?
Buoyed weight = weight in air X buoyancy factor
Bouyancy factor= 65.44-mw
65.44
= 65.44-11.5
= 0.8242
65.44
Buoyed weight = 180,000 X 0.8242
= 148 356 lbs
A practical application of this is expression is
when calculating the number of drill collars to
give a desired weight on bit
Example
Find the maximum weight on bit if a drill
string has 20 drill collars each 8 inch by
2.5 inch and 30ft in length and weight 154
lbs/ft
The density of the drilling fluid is 13.4ppg
and 90% of the buoyed weight will be
utilized
Answer
Weight of drill collars in air=20x30x154= 92400 lbs
Bouyancy factor= 65.44-13.4 = 0.795
65.44
Effective drill collar wt in fluid
= 92400 x =0.795 =73458lbs
90% of available drill collar weight
=0.9x73458=66112 lbs
Example
Find the maximum weight on bit if a drill
string has 30 drill collars each 6.5 inch by
3 inch and 30ft in length and has a
nominal wt of 88.88 pounds per foot
The density of the drilling fluid is 12.4 ppg
and 80% of the buoyed weight will be
utilized
Example
Find the maximum weight on bit available if
a drill string has 15 drill collars each 9.5
inch by 2.75 inch and 30ft in length
The density of the drilling fluid is 13 ppg and
80% of the buoyed weight will be utilized
WET AND DRY TRIPPING
Tripping Dry
When a length of pipe is
pulled from the hole, the
mud level will fall.
Tripping Dry
The volume of fall is equal
to the volume of steel pulled
from the hole.
The trip tank is then used to
fill up the hole.
If 1 barrel of steel is
removed from the hole, then
using the trip tank, we have
to add 1 barrel of mud.
Tripping Dry
1- Calculate the volume of
steel pulled:
Length x Metal Displacement
Example:
DP Metal Disp = 0.00764 bbls/ft
Length Pulled 93 feet
Volume Of Steel Pulled:
93 x 0.00764 = 0.711 bbls
Tripping Dry
2- Fill up the hole:
You must pump 0.711 barrel of
mud from the trip tank.
You must investigate ( flow
check) if more mud or less mud
is needed.
Tripping Dry
3- NO FILL UP:
If you fail to fill up the hole, the
mud level will drop by the
volume of steel pulled.
It will drop inside the pipe and
in the annulus.
Tripping Dry
3- NO FILL UP:
Example:
Volume Of Steel Pulled:
93 x 0.00764 = 0.711 bbls
DP Capacity: 0.01776 bbl/ft
Annular Capacity: 0.0504 bbl/ft
The mud will drop inside the pipe
and the annular:
0.01776 + 0.0504 = 0.06816 bbl/ft
Tripping Dry
3- NO FILL UP:
Example Cont’d:
The volume of drop is 0.711 bbls and will
drop in a volume of
0.06816 bbl / ft,
then the length of drop will be:
0.711 / 0.06816 = 10.43 feet.
If 93 feet are pulled with no fill up, the
mud level will drop by 10.43 feet.
Tripping Wet
When a length of pipe is
pulled from the hole, the
mud level will fall.
Tripping Wet
The volume of fall is equal
to the volume of steel pulled
from the hole plus the
volume of mud inside this
pipe.
The trip tank is then used to
fill up the hole.
If 3 barrels of steel and mud
are removed from the hole,
then using the trip tank, we
have to add 3 barrels of
mud.
Tripping Wet
1- Calculate the volume of steel
pulled:
Length x Metal Displacement
Example:
DP Metal Disp = 0.00764 bbls/ft
Length Pulled 93 feet
Volume Of Steel Pulled:
93 x 0.00764 = 0.711 bbls
Tripping Wet
2- Calculate the volume of mud
pulled:
Length x DP Capacity
Example:
DP Capacity = 0.01776 bbls/ft
Length Pulled 93 feet
Volume Of Mud Pulled:
93 x 0.01776 = 1.65 bbls
Tripping Wet
3- Calculate the total volume of
steel and mud pulled:
1.65 + 0.711 = 2.36 barrels
Tripping Wet
4- Fill up the hole:
You must pump 2.36 barrels of
mud from the trip tank.
You must investigate ( flow
check) if more mud or less mud
is needed.
Tripping Wet
5- NO FILL UP:
If you fail to fill up the hole, the
mud level will drop by the
volume of steel and mud pulled.
It will drop inside the annulus.
Tripping Wet
5- NO FILL UP:
Example:
Volume Of Steel and Mud Pulled:
93 x (0.00764+0.01776) = 2.36 bbls
Annular Capacity: 0.0504 bbl/ft
The mud will drop inside the
annular by:
2.36 / 0.0504 = 46.84 feet
Mud Weight
The Mud weight is the density ( mass per unit volume
) of a drilling fluid.
Mud Weigth can be expressed in:
Pound per Gallon ( ppg )
Pound per Cubic Feet ( pcf )
Specific Gravity ( sg )
Pressure
The pressure is a force or weight per unit area
exerted by a gas or liquid against the walls of a
vessel, hole, or object inserted into said gas or
liquid.
Pressure can be expressed in:
Pounds per square inches ( psi )
OR
Kilograms per square centimeters ( kg / cm²)
Derivation of 0.052
A cubic foot contain 7.48 US gallons.
If the fluid used to fill the cube weights 1
ppg, then the cube would weight 7.48
pounds.
The base of the cube has: 12x12=144
square inches.
7.48/144=0.05194
A 1 foot column of 1ppg fluid
exerts a pressure of 0.052 psi
Pressure or Mud Gradient
Since the pressure is measured in psi and depth is
measured in feet, it is convenient to convert mud
weights from ppg to a pressure gradient in psi/ft.
The conversion factor is 0.052
Pressure gradient (psi/ft) = Fluid Density (ppg) x 0.052
Example:
A fluid having a mud weight of 10 ppg will have:
10 x 0.052 = 0.52 psi ft as a pressure gradient.
A 1 foot column of 10 ppg mud will have a pressure of
0.52 psi
Hydrostatic Pressure
Hydrostatic Pressure is the pressure exerted by a column
of fluid ( at rest )and is calculated by multiplying the
gradient of the fluid by the True Vertical Depth at which
the pressure is being measured:
Hyd Pressure(psi) = Fluid gradient (psi/ft) x TVD
Example:
What is the hydrostatic pressure at 10,000 ft for a mud
gradient of 0.52 psi /ft ?
Hyd Pressure(psi) = 0.52 x 10,000
Hyd Pressure = 5,200 psi
Hydrostatic Pressure
It is the vertical height/depth of the fluid column that matters,
its shape is unimportant
Hydrostatic Pressure
Using the formula for the pressure gradient and for the hydrostatic
pressure:
Pressure gradient (psi/ft) = Fluid Density (ppg) x 0.052
Hyd Pressure(psi) = Fluid gradient (psi/ft) x TVD
We can write:
Hyd Pressure(psi) = Fluid Density (ppg) x 0.052 x TVD
Hydrostatic Pressure
It is usefull to visualize the well as a U-tube.
One column is for the pipe in the well and the other column is
for the annulus.
If the same fluid is used is both columns, hydrostatic would be
equal and the fluid would be static on both sides of the tube.
Mud Hydrostatic
in the string
Mud Hydrostatic
in the annulus
Well bore hydraulics
Well bore pressure losses
2600 psi
2500 psi
0 psi
100 psi
400 psi
300 psi
80 SPM
2100 psi
500 psi
1600 psi
1300 psi
300 psi
Hydraulic horsepower =
PxQ
1714=
Where P= Pressure loss
V= Quantity of flow in Gallons per minute
Annular velocity= Pump output (bbls/min)
Annular capacity (bbls/ft)
Example
• What is the hydraulic horsepower delivered by
a pump at 400 gallons per minute at 3000psi
3000 x 400
1714
=700HP
Annular Velocity
Annular velocity is the speed with which the
fluid travels up the annulus.
It is an indication of the amount of HHP
available in the annulus
Annular Velocity
Annular velocity = pump output (bbls/min)
Annular capacity (bbls/str)
Pump pressure vs pump rate
Pump pressure increased as pump speed
increases
P2 =P1x SPM2 2
SPM1
Where:
P1=old pump pressure
P2=new pump pressure
SPM1=Old pump rate
SPM2= New pump rate
Pump pressure vs fluid density
• Pump pressure also increases with fluid density
P2= P1 x MW2
MW1
Where P1=old pump pressure
P2=new pump pressure
MW1= Old mud wdensity
MW2 new mud density
Example
At 100 spm the pumping pressure is 2850
psi with 13 ppg
What is the new pressure if the pumps are
decreased to 65 spm?
What is the new pressure if the mud weight
is then decreased to 10.5 ppg?
Solution:
2850 x 65 x 65 x 10.5
100 x 100
13
= 973 psi
The new pump pressure will be 973 psi
approximately
Work Done by the Drilling line
Each time the block are raised and lowered
during the drilling process a certain
amount of work is done by the drilling line
Work done = Force X Distance
= Ton X Miles
Tin mile calculations are used to determine
the actual slip and cut off programs for the
rig
Varieties of Ton-Miles Calculations
•
•
•
•
Round Trip Ton Miles
Drilling Ton miles
Casing Ton miles
Coring Ton miles
Drilling Ton Miles
Dtm=3X(RTtm2-RTtm1)
Where:
Dtm =Drilling Ton Miles between Trips
RTtm2 =Ton miles at depth when bit is run
RTtm1 =Ton miles at depth when bit is pulled
Round trip Ton Mile calculations
RTtm = Wp X (Lp+D) +2XDX(2Wb+Wc)
5280 X 2000
Where:
RTtm =Round trip Ton Miles
Wp= Buoyant weight of drill pipe in mud
D=Depth of the hole
Lp=average stand length
Wb=Weight of travelling block assembly
Wc=Buoyant weight of drill collars minus the buoyant
weight of the same length of drill pipe
Example
Depth 7000 ft. Drilling fluid density 10 ppg
Drill pipe 5 inch diameter x 19.5 lb/ft.
Collars 500 ft of 8 inch OD x 3 inch ID x 147
lb/ft.
Travelling block assembly 15000 lbs.
Calculate the Round Trip ton miles at 7000 ft.
Answer:
From Drilling Data Handbook table the buoyancy factor = 0.848
The WP Buoyant weight of drill pipe = 19.5 x 0.848 = 16.53 lb/ft.
Wc,the Buoyant weight of drill collars =
(500 x 147 x 0,848) - (500 x 16,53)= 62328 - 8265 = 54063 lbs
Take average length of 1 stand = 90 ft.
RTtm = WP x D x ( LP + D ) + 2 x D x ( 2 Wb + Wc )
5280 x 2000
and with the numbers entered into the formula RTt= 16,53 x 7000 x ( 90 + 7000 ) + 2 x 7000 x ( 2 x 15000 x 54063 )
5280 x 2000
= 189 TM
Casing Ton Miles
• Ctm= D xWcx(Lc +D)+4 x D xWb
5280 x2000x2
Where:Ctm= casing ton miles
Wc=Submerged wt of casing in mud (lbs/ft)
Lc=Average lenght of 1 joint f casing
Wb=weight of travelling block assembly
Ton miles for coring
Coring =2x(RT -RT )
tm
tm2
tm1
Where:RTtm1=Ton miles for trip where core barrel was run
:RTtm2=Ton miles for trip where core barrel was pulled
Stuck pipe calculations
Free point determination
L=M xexA
12 xP
Where:
L= lenght of free pipe
M=Modulus of elasticity of steel
e =Strech in drill pipe
A=cross sectional area of drill åpipe in sq inches
P= difference betweem maximum and minimum pull on pipe
Example
A 10000 ft string of 4½" and 16,6 lb/ft drill
pipe is stuck.
Maximum pull of 250000 lbs and a
minimum pull of 180000 lbs gave a stretch
figure of 45 inch. The cross sectional area
of 4½" drill pipe is 4.4 sq.in.
Calculate the position of the free point.
M = 30000000
e = 45 inch
A = 4.4 sq.in
P = 250000 - 180000 = 70000 lbs
Length of free pipe =
L = M x e x A = 30000000 x 45 x 4.4
12 x P
12 x 70000
= 7071 ft below the rotary table
Accummulator Volumetric
requirements
Definitions
• Stored Hydraulic Fluid.
The fluid volume recoverable from the accumulator system between the
maximum designed accumulator operating pressure and the precharge
pressure.
• Usable Hydraulic Fluid. The hydraulic fluid recoverable from the
accumulator system between the maximum accumulator operating
pressure and minimum calculated operating pressure or 200 psi above
precharge pressure whichever is greatest.
• Minimum Calculated Operating Pressure. The minimum calculated
pressure to effectively close and seal a ram-type BOP against a wellbore
pressure equal to the maximum rated working pressure of the BOP
divided by the closing ratio specified for that BOP.
• Component Minimum Operating Pressure Recommended by the
Manufacturer. The minimum operating pressure to effectively close and
seal ram-type or
annular-type preventers under normal operating
conditions, as prescribed by the manufacturer.
Boyles Law
Pressure X Volume is constant
Or P1V1=P2V2
At constant temparature If the volume of an enclosed gas bubble is
doubled them the presure is reduced by half and if the pressure of a
gas bubble is doubled the volume is halved
Example
Accumulator bottle size 10 gallons.
Precharge pressure 1.000 psi
Initial condition with only gas :
Pressure x Volume = Constant
1.000 x 10 = 10.000
The pump system is started and hydraulic fluid is pumped into the accumulator
bottle until maximum operating pressure is reached at 3.000 psi :
P1 x V1 = P2 x V2

1.000 x 10 = 3.000 x V2
V2= 1000 X10
3000
= 3.33 gallon gas
Stored hydraulic fluid = 10 - 3.33 = 6.67 gal
1.000 psi
3.000 psi
1.200 psi
1.500 psi
2.110 psi
10 gal
3.33 gal
8.33 gal
6.66 gal
4.74 gal
6.66 gal
1.66 gal
3.33 gal
5.26 gal
Fig 58 b
Gas
Fig 58 c
Fig 58 d
Fluid
Fig 58 e
Fig 58 a
The pump system is isolated and the BOP’s functioned until accumulator pressure reach
precharge pressure + 200 psi:
P1 x V1 = P2 x V2

1.000 x 10 = 1.200 x V2
V2= 10 X10000 =8.33 gallons gas
1200
Usable hydraulic fluid = 8.33 - 3.33 = 5 gal
If the minimum operating pressure recommended by the manufacture is 1.500 psi as for
Shaffer Annular Preventer with pipe size smaller than 7” the usable hydraulic fluid
would be :
P1 x V1 = P2 x V2

1.000 x 10 = 1.500 x V2
V2= 10 X10000 =6.66 gallons gas
1500
Usable hydraulic fluid = 6.66 - 3.33 = 3.33 gal
Well Control
Mud Weight
The Mud weight is the density ( mass per unit volume ) of
a drilling fluid.
Mud Weight can be expressed in:
Pound per Gallon ( ppg )
Pound per Cubic Feet ( pcf )
Specific Gravity ( sg )
Pressure
The pressure is a force or weight per unit area
exerted by a gas or liquid against the walls of a
vessel, hole, or object inserted into said gas or
liquid.
Pressure can be expressed in:
Pounds per square inches ( psi )
Kilograms per square centimeters ( kg / cm²)
Hydrostatic Pressure
Using the formula for the pressure gradient and for the hydrostatic
pressure:
Pressure gradient (psi/ft) = Fluid Density (ppg) x 0.052
Hyd Pressure(psi) = Fluid gradient (psi/ft) x TVD
We can write:
Hyd Pressure(psi) = Fluid Density (ppg) x 0.052 x TVD
Hydrostatic Pressure
It is usefull to visualize the well as a U-tube.
One column is for the pipe in the well and the other column is
for the annulus.
If the same fluid is used is both columns, hydrostatic would be
equal and the fluid would be static on both sides of the tube.
Mud Hydrostatic
in the string
Mud Hydrostatic
in the annulus
Hydrostatic Pressure
If the pipe or the annulus are not kept full, then the hydrostatic
pressure in reduced.
Pipe is not full
Annulus is not full
HP is reduced
HP is reduced
Balanced
Balanced when Hydrostatic Pressure = Formation Pressure
Under Balanced
Under Balanced when HP < FP
Over Balanced
Over Balanced when HP > FP
Kill Sheet
The kill sheet is a
tool used to kill the
well. It gathers all
the informations
needed and help you
to perform the
calculations.
The first section to
fill is the well data.
Kill Sheet
The pump displacement
must be entered and the
Slow Circulating Rate
must be recorded.
Kill Sheet
This section is used to calculate all the hole volumes, to
convert them in strokes and in time.
Drillpipe Internal volume
Heavy weight dril pipe Internal Volume
Dril Collars Internal Volume
Drill Collar open hole annular volume
Drill pipe- Open hole annular volume
Drill pipe-Casing annular volume
Kill Sheet
The formulas are
used to calculate
the kill mud,
Initial Circulating
Pressure, Final
Circulating
Pressure and to
prepare the
graph.
Kill Sheet
The graph will be followed to maintain a constant Bottom
Hole Pressure.
Pumping a Slug
It is useful to pump a slug before tripping.
The slug weight being heavier than the mud, a length of pipe
will be empty.
The HP is not reduced because the heavier mud
will compensate for the empty pipe.
Pumping a Slug
The total HP is the same on both sides of the pipe.
HP kmw
HP mud
HP mud
Pumping a Slug
Example:
If 20 bbls of 12 ppg slug are pumped in a 10,000 ft hole
containing 10 ppg mud, what will be the height of empty pipe?
DP capacity = 0.01776 bbl/ft
1- Calculate the height of the slug:
20 / 0.01776 = 1126 ft
Pumping a Slug
2- Calculate the HP of the slug:
1126 x 12 x 0.052 = 702.6 psi
702.6 psi
Pumping a Slug
2- Calculate the HP of the mud in the annulus:
10,000 x 10 x 0.052 = 5,200 psi
702.6 psi
5,200 psi
Pumping a Slug
3- The total hydrostatic beeing the same on both sides,
calculate the HP of the mud below the slug:
5,200 - 702.6 = 4497.4 psi
702.6 psi
4497.4 psi
5,200 psi
Pumping a Slug
4- Calculate the height of mud needed to give 4497.4
psi as a HP:
TVD = 4497.4 / ( 10 x 0.052 ) = 8648.8 feet
1,126 ft
8648.8 ft
10,000 ft
Pumping a Slug
4- Calculate the height of empty pipe
10,000 - 8648.8 - 1,126 = 225.2 ft
225.2 ft
1,126 ft
8648.8 ft
10,000 ft
Practical Exercise
3- If 28 bbls of 14 ppg slug are pumped in a 12,000 ft
hole containing 11 ppg mud, what will be the height of
empty pipe?
DP capacity = 0.01776 bbl/ft
MUD WEIGHT CHANGE
• A well is being
drilled using 10
ppg mud. At 80
spm the total
circulating
system pressure
losses are 2600
psi.
• It is decided to
increase the mud
weight to 11 ppg.
2600 psi
80 spm
Mud wt 10 ppg
MUD WEIGHT CHANGE
• It is a good drilling practice to
calculate the new circulating
pressure before changing the
mud weight.
• The way we calculate this
change in pressure is to use the
following formula;
• New Mud ppg
• Old Mud ppg x Old psi.
• 11 ppg
10 ppg x 2600 = 2860 psi
• The new pump pressure would
be approximately 2860 psi.
2860 psi
80 spm
Mud wt 11 ppg
MUD WEIGHT CHANGE
Final Circulating Pressure
• The formula that was just used to
calculate the pressure change due
to a change in mud weight, is also
the formula used to calculate the
Final Circulating Pressure.
Kill Mud ppg
Old Mud ppg x Slow Pump psi.
PUMP STROKE CHANGE
• A well is being
drilled using 10 ppg
mud. At 80 spm the
total circulating
system pressure
losses are 2600 psi.
• It is decided to
increase the pump
speed from 80 spm
to 100 spm.
2600 psi
80 spm
Mud wt 10 ppg
PUMP STROKE CHANGE
• It is a good drilling practice
to calculate the new
circulating pressure before
changing the pump speed.
• The way we calculate this
change in pressure is to use
the following formula;
New SPM 2
Old psi x Old SPM
• 2600 x 100 spm 2
80 spm = 4063 psi
• The new pump pressure
would be approximately
4063 psi.
4063 psi
100 spm
Mud wt 10 ppg
Height of Influx
Pit gain(bbls)
= (ft.)
hi =
annuluscapacity(bbls/ft.)
Gi = G m -
P sia - P sidp
psi / ft
hi
Influx gradient
P sia - P sidp
psi / ft
Gi = G m hi
Where:
Gi=Influx Gradient
Gm= Mud Gradient
hi= Hieght of influx
DEPTH
0
1
Gas grad. 0.07 psi/ft
2
Oil grad. 0.30 psi/ft
3
Fresh W. grad 0.433 psi/ft
4
Salt W. grad 0.465 psi/ft
5
10 ppg grad. 0.52 psi/ft
2500
6
15 ppg grad. 0.7785 psi/ft
7
21 ppg grad. 1.091 psi/ft
5000
1000
2000
3000
4000
5000
PRESSURE