Part 1

Introduction

By

Prof. Dr. Abdel-Alim Hashem

Contents

• Definitions

• Basic Concepts

• Liquid Hydrostatics

3

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

4

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

6

HSP = HydroStatic Pressure

Hydrostatic Pressure, HSP

HSP = 0.052 * MW * TVD

– HSP = Hydrostatic Pressure, psi

– MW = Mud Weight (density), ppg

– TVD = Total Vertical Depth, ft

7

HSP = 0.052 * MW * TVD

HSP

8

TVD

10 ppg mud

HSP = HSP = HSP

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

9

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

= 1 * 1 * TVD ft3

* 62.4 lb/ ft3 * MW ppg/8.33

= 62.4 / 8.33 * MW * TVD

10

Solution, con’t.

• 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

11

W

F = PA

Solution, con’t.

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

14

HSP = 5,200 psi

5,600 5,600 5,600

400

400 600 600

HSP =

5,200 psi

Mud HSP

=4,800 psi

Mud HSP

=4,800 psi

Influx HSP

=200 psi

Influx HSP

=200 psi

More Terminology

• Capacity of:

– casing

– hole

– drillpipe

• Annular capacity

• Displacement of:

– Drillpipe

– Drill collars

• Rig Pumps

– Duplex pump

– Triplex pump

• KWM, kill weight mud

• Fluid Weight up

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. What is the BHP?

2. What is the pressure 1/2 way to bottom?

3. Plot the mud density vs. depth

4. Plot the mud gradient vs. depth

5. 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

3. Plot MW vs. Depth

19

D

e

p

t

h

0

5,000

10,000

15,000

20,000

25,000

30,000

Mud Density, ppg

0 5 10 15 20

8.33

11.0

16.0

20

Problem # 3 solution

4. Plot mud

gradient vs. Depth

Depth Gradient

ft psi/ft

0-6,000 0.433

6,000-14,000 0.572

14,000-TD 0.832

D

e

p

t

h

0

5,000

10,000

15,000

20,000

25,000

30,000

Mud Gradient, psi/ft

0 0.2 0.4 0.6 0.8 0.9

0.433

0.572

0.832

21

Problem # 3 solution

5. Plot HSP vs.

Depth

ft psi

@ 6,000 2,599

@14,000 7,175

@ 30,000 20,487

D

e

p

t

h

0

5,000

10,000

15,000

20,000

25,000

30,000

Mud Pressure, kpsi

8 5 10 15 20

2,599 psi

7,175 psi

20,487 psi

Addition of Weight Material

The amount of barite

required to raise the

density of one barrel

of mud from MW1 to

MW2, ppg

22

lbsBarite,ofgal1ofWt.35.4

lbsBarite,ofbbl1ofWt.1,490

ppgDensity,MudNewMW

ppgDensity,MudOldMW

lb/bblRequired,BariteW

where

2

1

B

2

12B

MW35.4

MWMW1,490W

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 Vol, bbl Old Mud 42 * MW1 1 Barite WB (WB lbs / 1,490 lb/bbl) Mixture WB + 42 MW1 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)

2

12B

MW35.4

MWMW1,490W

24

Stopping an Influx

1. Increase Pressure at Surface

2. Increase Annular Friction

3. Increase Mud Weight

25

Stopping an Influx

26 Pressure

Dep

th

Mud Hydrostatic

Pressure

Stopping an Influx – Soln.1

27 Pressure

Dep

th

Mud

Hydrostatic

Pressure

Stopping an Influx – Soln.2

28 Pressure

Dep

th

Mud

Hydrostatic

Pressure

Stopping an Influx – Soln.3

29 Pressure

Dep

th

Mud

Hydrostatic

Pressure

End

30