Part 1
Introduction
By
Prof. Dr. Abdel-Alim Hashem
Contents
Definitions
Basic Concepts
Liquid Hydrostatics
Definitions
What is a Kick?
An unscheduled entry of formation fluids into the
wellbore of sufficient quantity to require shutting
in the well
What is a Blowout?
Loss of control of a kick
Why does a kick occur?
Pressure in the wellbore is less than the
pressure in the formation
Permeability of the formation is great enough
to allow flow
A fluid that can flow is present in the
formation
5
How do we prevent kicks?
We must maintain the pressure in the
wellbore greater than formation pressure
But,
We must not allow the pressure in the
wellbore to exceed the fracture pressure
This is done by controlling the HSP of the
drilling fluid, and isolating weak formations
with casing
HSP = HydroStatic Pressure
Hydrostatic Pressure, HSP
HSP
= 0.052
* MW** MW
TVD
HSP
= 0.052
* TVD
HSP = Hydrostatic Pressure, psi
MW = Mud Weight (density), ppg
TVD = Total Vertical Depth, ft
7
HSP
10 ppg mud
TVD
HSP
HSP
HSP
8
Problem # 1
Drive the HSP equation
Calculate the HSP for each of the following:
10,000 ft of 12.0 ppg mud
12,000 ft of 10.5 ppg mud
15,000 ft of 15.0 ppg mud
Solution to Problem # 1
Consider a column of fluid:
Cross-sectional area = 1 sq.ft.
Height = TVD ft
Density = MW
Weight of the fluid = Vol * Density
3
= 1 * 1 * TVD ft * 62.4 lb/ ft * MW ppg/8.33
= 62.4 / 8.33 * MW * TVD
10
Solution, cont.
This weight is equally distributed over an area
of 1 sq.ft. or 144 sq.in.
Therefore,
Pressure = Weight / area
= 62.4 MW * TVD/(8.33*144)
HSP = 0.052 * MW * TVD
F = PA
11
Solution, cont.
HSP = 0.052 * MW * TVD
HSP1 = 0.052 * 12 * 10,000 = 6,240 psi
HSP2 = 0.052 * 10.5 * 12,000 = 6,552 psi
HSP3 = 0.052 * 15.0 * 15,000 = 11,700 psi
12
Terminology
Pressure
Pressure gradient
Formation pressure
(Pore)
Overburden pressure
Fracture pressure
Pump pressure
(system pressure loss)
SPP, KRP, Slow
circulating pressure,
kill rate pressure
Surge & swab
pressure
SIDPP & SICP
BHP
13
U-Tube Concept
400
600
HSP = 5,200 psi
400
600
Mud HSP
=4,800 psi
HSP =
5,200 psi
Mud HSP
=4,800 psi
Influx HSP
=200 psi
Influx HSP
=200 psi
5,600
5,600
5,600
14
More Terminology
Capacity of:
Rig Pumps
casing
Duplex pump
hole
Triplex pump
drillpipe
Annular capacity
KWM, kill weight mud
Fluid Weight up
Displacement of:
Drillpipe
Drill collars
15
Problem # 2
Calculate the mud gradient for 15.0 ppg mud
G15 = 0.052 * MW = 0.052 * 15
= 0.780 psi/ft
Calculate the HSP of 15,000 of 15 ppg mud
HSP = 0.780 * 15,000 = 11,700 psi
16
Problem # 3
The top 6,000 ft in a wellbore is filled with
fresh water, the next 8,000 with 11 ppg
mud, and the bottom 16,000 ft is filled with
16 ppg mud.
1.
2.
3.
4.
5.
What is the BHP?
What is the pressure 1/2 way to bottom?
Plot the mud density vs. depth
Plot the mud gradient vs. depth
Plot the pressure vs. depth
17
Problem # 3 solution
1. BHP = 0.052 * [(8.33 * 6,000) + (11 * 8,000) +
(16 * 16,000)]
= 20,487 psi
2. Pressure 1/2 way down (at 15,000 ft)
= 0.052 * [(8.33 * 6,000)
+ (11 * 8,000) + (16 * 1,000)]
= 8,007 psi
18
Problem # 3 solution
Mud Density, ppg
0 5 10 15 20
3. Plot MW vs. Depth
0
5,000
D 10,000
e 15,000
p
t 20,000
h
25,000
8.33
11.0
16.0
30,000
19
Problem # 3 solution
Mud Gradient, psi/ft
0 0.2 0.4 0.6 0.8 0.9
4. Plot mud
gradient vs. Depth
0
Depth
Gradient
ft
psi/ft
0-6,000
0.433
6,000-14,000 0.572
14,000-TD
0.832
5,000
D 10,000
e 15,000
p
t 20,000
h
25,000
0.433
0.572
0.832
30,000
20
Problem # 3 solution
5. Plot HSP vs.
Depth
ft
psi
@ 6,000
2,599
@14,000 7,175
@ 30,000 20,487
Mud Pressure, kpsi
8 5 10 15 20
5,000
2,599 psi
10,000
7,175 psi
D 15,000
e
p 20,000
t 25,000
h
30,000
20,487 psi
21
Addition of Weight Material
The amount of barite
required to raise the
density of one barrel
of mud from MW1 to
MW2, ppg
1,490 MW2 MW1
WB
35.4 MW2
where
WB Barite Required,lb/bbl
MW1 Old Mud Density,ppg
MW 2 New Mud Density,ppg
1,490 Wt. of 1 bbl of Barite, lbs
35.4 Wt. of 1 gal of Barite, lbs
22
Problem # 4, Derive Barite Eq.
Consider one bbl of mud of density, MW1, add
WB lbs of barite to increase the mud density
to MW2.
Wt, lb
Old Mud
Barite
Mixture
42 * MW1
WB
WB + 42 MW1
Vol, bbl
1
(WB lbs / 1,490 lb/bbl)
1 + (WB / 1,490)
Density of Mixture = total weight / total volume
23
Problem # 4
New Density = Weight / Volume
MW2
= (WB+42 MW1 lbs) / {[1+(WB/1,490)bbl]*42 gal/bbl}
42 MW2 [1+(WB/1,490)] = WB+42 MW1 lbs
WB [(MW2 / 35.4) -1] = 42 MW1 42 MW2
WB(MW2 - 35.4) = (42 * 35.4) * (MW1 - MW2)
1,490 MW2 MW1
WB
35.4 MW2
24
Stopping an Influx
1. Increase Pressure at Surface
2. Increase Annular Friction
3. Increase Mud Weight
25
Depth
Stopping an Influx
Mud Hydrostatic
Pressure
Pressure
26
Depth
Stopping an Influx Soln.1
Mud
Hydrostatic
Pressure
Pressure
27
Depth
Stopping an Influx Soln.2
Mud
Hydrostatic
Pressure
Pressure
28
Depth
Stopping an Influx Soln.3
Mud
Hydrostatic
Pressure
Pressure
29
End
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