Introduction to Well Cementing

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Assignment 1

Cement Slurry and Post-flush Fluid Volumes

Using the data provided, calculate the quantity of cement slurry and post-flush fluid required to cement the intermediate string of casing in the Assignment Well. Include a 25% cement contingency.

Contents

Assignment 1 Cement Slurry and Post-flush Fluid Volumes Assignment 2 Fluid Density Assignment 3 Fluid Density Assignment 4 Solution to A:Solution to B:Solution to C:Assessment

Casing String	Hole Diameter D_hole (in)	Hole Depth* (feet)	Casing Shoe Depth* L (feet)	Depth to Float Collar* (feet)	Casing OD (in)	Casing ID (in)
Conductor [c]	30	40	L₄ = 80	N/A	20	19
Surface [s]	17.5	1510	L₃=1505	1465	13.375	12.615
Intermediate [i]	12.25	4015	4008	3968	9.625	8.835

* All depths measured from Rotary Kelly Busing (RKB)
Summary Table

Cement Slurry and Post-flush Fluid Volumes

Solution

Answers:

Volume of cement, ft³: 1546 cf

Post-flush fluid volume, ft³: 1689 cf

Solution:

1. Volume of casing/opening hole annulus (V_Annulus−1):

$V_{Annulus-1} = 0.005454 \left [(D_{HOLE}^2 - OD_i^2)\times L_1 \times (Excess)\right ]$

$V_{Annulus-1} = 0.005454 \left [(12.25^2 - 9.625^2) \times (4008 - 1505) \times (1.25)\right ] = 980\ cf\ or\ 175\ Bbl$

2. Volume of casing/opening hole annulus (V_Annulus−2):

$V_{Annulus-2} = 0.005454 \left [(ID_{s}^2 - OD_i^2)\times L_3 \right ]$

$V_{Annulus-2} = 0.005454 \left [(12.615^2 - 9.625^2) \times 1505 \right ]= 546\ cf\ or\ 97\ Bbl$

3. Volume of cement left inside the casing (V_csg):

$V_{csg} = 0.005454 \left [(ID_{i}^2)\times L_2 \right ]$

$V_{csg} = 0.005454 \left [(8.835^2) \times 40 \right ]= 17\ cf\ or\ 3\ Bbl$

4. Volume of cement in open hole below casing (V_OH):

$V_{OH} = 0.005454 \left [(D_{HOLE}^2) \times (TD - (L_1 + L_3) ) \times (Excess) \right ]$

$V_{csg} = 0.005454 \left [(12.25^2) \times 7 \times 1.25 \right ]= 7\ cf\ or\ 1\ Bbl$

Total slurry volume: 1550 cf or 276 Bbl

Displacement volume:

$V_{displ} = 0.005454 \left [(ID_i^2)\times (L_1+L_3-L_2 ) \right ]$

$V_{displ} = 0.005454 \left [(8.835^2) \times (4008 - 40) \right ]= 1689\ cf\ or\ 301\ Bbl$

Assignment 2

Fluid Density

Using the information in the table below and the relationships covered in the subject, calculate the missing values.

Fluid	$\tfrac{lb_m}{ft^3}$	ppg
Water	62.4
Salt Water		8.95
Cement		15.6

Solution

Using this relationship: $7.48 \tfrac{US\ gal}{ft^3}$

Fluid	$\tfrac{lb_m}{ft^3}$	ppg
Water	62.4	8.34 (62.4/7.48)
Salt Water	66.95 (8.95 x 7.48)	8.95
Cement	116.69 (15.6 x 7.48)	15.6

Assignment 3

Fluid Density

Calculate the pressure gradient ( $\tfrac{psi}{ft}$ ) and bottomhole pressure (psi) in the intermediate string of casing of the Assignment Well when filled with the fluids shown in the table.

Depth = 2015 ft

Calculate the missing values in the following table:

Fluid	Pressure Gradient ( $\tfrac{psi}{ft}$ )	Bottomhole Pressure (psi)
Water (62.4 $\tfrac{lb}{ft^3}$ )
Salt Water (8.95 ppg)
Cement (15.6 ppg)

Solution

Data:

Depth(h): 2015 ft

P_atm: 14.7 psi

Density:

Water: 62.4 $\tfrac{lb}{ft^3}$
Salt water: 8.95 ppg
Class A cement slurry: 15.6 ppg

Formula:

$Pressure\ Gradient (\tfrac{psi}{ft}) = fluid\ density / 144 \tfrac{in^2}{ft^2}$

$BHP\ (psia) = P_{atm}(psi) + Gradient(\tfrac{psi}{ft}) \times h(ft)$

$Unit\ converstion: 7.48 \tfrac{gal}{ft^3}$

Fluid	Pressure Gradient ( $\tfrac{psi}{ft}$ )	Bottomhole Pressure (psi)
Water (62.4 $\tfrac{lb}{ft^3}$ )	0.433 = (62.4/144)	887.2 = (14.7+0.433×2015)
Salt Water (8.95 ppg)	0.465 = ((8.95/144)×7.48)	951.7 = (14.7+0.465×2015)
Cement (15.6 ppg)	0.810 = ((15.6/144)×7.48)	1646.9 = (14.7+0.810×2015)

Assignment 4

Use the data shown here to answer the questions.

A. What is the absolute pressure in the brine column at point A just above the float shoe?

B. What is the absolute pressure at point B, the bottom of the hole, within the cement column?

C. What is the pressure at point C just below the shoe, at 1990 ft?

Solution

Solution to A:

Since the fluid down to point A is brine, the weight of the fluid above it is 8.95 ppg

$P_A = P_{atm} + (0.052 \times \textrm{fluid density} \times h )$

$P_A = 14.7 + (0.052 \times 8.95 \times 1990)$

Answer: $P_A (psia) = 940.8$

Solution to B:

Since the cement fills the annulus down to point B is cement, the weight of the fluid above it is 15.6 ppg

$P_B = P_{atm} + (0.052 \times \textrm{fluid density} \times h )$

$P_B = 14.7 + (0.052 \times 15.6 \times 2015)$

Answer: $P_B (psia) = 1649.3$

Solution to C:

Since the fluid at point C is also in the cement column, the weight of the fluid above it is 15.6 ppg

$P_C = P_{atm} + (0.052 \times \textrm{fluid density} \times h )$

$P_C = 14.7 + (0.052 \times 15.6 \times 1990)$

Answer: $P_C (psia) = 1629.0$

Assessment

1. Identify customary oilfeld units of volume that are used in cementing operations. (Select all that apply.)

A .Barrel ✔
B .Cubic foot ✔
C .Cubic inch
D .Cubic yard
E .Gallon

2. The most basic process for accomplishing a primary cementing job employs what method?

A .Two-wiper plug method ✔
B .Three-wiper plug method
C .One-wiper plug method

3. Which of these statements regarding cement volume calculations are TRUE? (Select all that apply.)

A .Displacement fluid volumes should be calculated to ensure that the cement slurry is displaced to the top of the float shoe.
B .It is uncommon to cement surface casing all the way to surface.
C .It is important to leave enough cement in and around the bottom of the casing to ensure the integrity of the cement bond. ✔
D .A contingency factor is often included in cement volume estimates to account for borehole diameters that are larger than the nominal bit diameter. ✔

4. As with most regulatory agencies, the Pennsylvania Department of Environmental Protection requires that job logs be available for onsite inspection. What cementing-related items are typically required on a job log? (Select all that apply.)

A .Compressive strength of cement
B .Core analysis
C .Pumping rates, pressures, and duration ✔
D .Mix water pH and temperature ✔
E .Cement density ✔

5. What are the main reasons for primary cementing? (Select all that apply.)

A .To seal off lost circulation zones during drilling
B .To repair damaged casing
C .To abandon a depleted producing zone
D .To protect the casing connections
E .To restrict fluid movement between subsurface formations ✔
F .To anchor the casing to the formation ✔

6. In a typical primary cementing job, cement slurry is pumped into the ________ and displaced into the ___________, from the bottom to at least the top of the producing zones, and sometimes to the surface.

A .formation; annulus
B .casing; annulus ✔
C .annulus; casing

7. ____________ is the expected pressure in the wellbore formations created by the continuous presence of water in interconnected pores from the formation to the surface.

A .Overburden pressure
B .Wellbore pressure
C .Fracture pressure
D .Hydrostatic pressure ✔

8. The presence of salt in water __ its density.

A .has no effect on
B .increases
C .decreases

9. Because the density of sedimentary rock is about 2.5 x that of water, the pressure gradient imposed by the weight of the overburden is estimated to be about _ psi/ft of depth.

A .4
B .3
C .2
D .1 ✔

10. Pressure is the force exerted on a unit area. In what units can pressure be expressed? (Select all that apply.)

A .psf ✔
B .psi ✔
C .bar ✔
D .kPa ✔
E .ppg