Retort analysis is the method to determine solid and liquid components in the drilling fluid. In this article, we will adapt mass balance and retort analysis data to determine solid density in the mud.
Mass balance for mud is listed below;
We can rearrange the Euqation#1 in order to determine the solid density
Coring is a special process to recover wellbore rock in the well.
Figure 1 – Core from the well. Credit – Wikipedia
This article will demonstrate how to calculate coring cost per footage recovered.
Coring cost per footage recovered is expressed below;
While drilling, cuttings are generated every footage drilled and this topic will demonstrate how to determine volume of cutting entering into the wellbore.
Figure 1 – Cutting Generated While Drilling
The following formula is used to calculate cutting volume generated while drilling;
Where;
Vc is volume of cutting in bbl/hr.
Ø is formation porosity (%).
D is wellbore diameter in inch.
ROP is rate of penetration in feet per hour.
Vc can be presented in several unit as follows;
Buoyancy Factor is the factor that is used to compensate loss of weight due to immersion in drilling fluid and you can find more information from this article > buoyancy factor calculation . In that article, it demonstrates the buoyancy formula only for one fluid in the wellbore. However, this time, we will have the details about buoyancy factor when inside and outside fluid are different.
Buoyancy factor with different fluid inside and outside of tubular is listed below;
Where;
Ao is an external area of the component.
Ai is an internal area of the component.
ρo is fluid density in the annulus at the component depth in the wellbore.
ρi is fluid density in the component depth in the wellbore.
ρs is steel weight density. Steel density is 65.4 ppg.
If you can the same mud weight inside and outside, the equation 1 will be like this
This is the same relationship as this article buoyancy factor calculation.
Let’s take a look at the following example to get more understanding.
Example
13-3/8” casing shoe was at 2,500’MD/2,000’TVD
9-5/8” casing was run to 6,800’MD/6,000 TVD.
9-5/8” casing weight is 40 ppf and casing ID is 8.835 inch.
Current mud weight is 9.5 ppg oil based mud.
The well bore diagram is show below (Figure 1).
Figure 1 – Wellbore Diagram
The well is planned to cement from shoe to surface and the planned cement weight is 14.0 ppg. The displacement fluid is drilling mud currently used.
Please determine the following items.
Air weight of casing string
Air weight of casing string, lb = length of casing, ft × casing weight, lb/ft
Air weight of casing string, lb = 6,800 × 40 = 272,000 lb
Buoyed weight of casing in drilling mud
Figure 2 – Buoyed Weight When Submersed In Drilling Mud
Buoyed weight = Buoyancy Factor (BF) × Air Weight of Casing
Buoyancy Factor (BF) = 0.855
Buoyed weight = 0.855 × 272,000 = 232,489 lb
Buoyed weight of casing when cement is inside casing and drilling mud is outside casing
Figure 3 – Buoyed weight of casing when cement is inside casing and drilling mud is outside casing
We will apply the Equation-1 for this case.
Ao is an external area of the component.
Ao = π × (Outside Diameter of casing)2 ÷ 4
Ao = π × (9.625)2 ÷ 4 = 72.76 square inch
Ai is an internal area of the component.
Ai = π × (Inside Diameter of casing)2 ÷ 4
Ai = π × (8.835)2 ÷ 4 = 61.31 square inch
ρo = 9.5 ppf (mud in the annulus)
ρi = 14.0 ppg (cement inside casing)
ρs = 65.4 ppg.
Buoyancy Factor (BF) = 1.22
Buoyed weight = 1.22 × 272,000 = 331,840 lb
Buoyed weight of casing when cement is outside casing and drilling mud is inside casing
Figure 4 – Buoyed weight of casing when cement is outside casing and drilling mud is inside casing
We will apply Equation-1 for this case as well.
All the calculation parameters are the same.
Ao = π × (9.625)2 ÷ 4 = 72.76 square inch
Ai = π × (8.835)2 ÷ 4 = 61.31 square inch
ρo = 9.5 ppg (mud in the annulus)
ρi = 14.0 ppg (cement inside casing)
ρs = 65.4 ppg.
Buoyancy Factor (BF) = 0.42
Buoyed weight = 0.42 × 272,000 = 114,240 lb
Conclusion: At different stage of the well, you may have different buoyed weight depending on density of fluid inside and outside of the component and it is not always that buoyed weight is less than air weight.
Reference book => 
Formulas and Calculations for Drilling Operations
In the hydro-static pressure formula, if you use an oil field unit, you must see the hydro-static formula like this
Pressure, psi = 0.052 x Mud Weight (ppg) x TVD (ft)
I got the question asking about how the constant 0.052 is figured out.
What is 0.052?
0.052 is a constant used for converting ppg-ft into pound per square inch (lb/in2)
Do you know how it comes from? Let’s get started
First of all, the unit conversions that you need to know are as follows:
1 gallon = 231 cubic inch (inch3)
1 ft = 12 inch
Convert density from ppg (pound per gallon) to lb per cubic inch
Basic hydrostatic pressure
The constant can be determined by substituting the unit conversion.
The constant is 0.051948 and if we round it up, it will be 0.052.
