Activity: Well Control

ANSWERS

1. What is the MAASP prior to the kick?

$\begin{array}{l} MAASP = CSG.TVD*(FG - MG)\\ MAASP = 9000*(0.8 - 0.6)\\ MAASP = 1800psi \end{array}$

with

CSG.TVD : the casing set total vertical depth

FG : formation pressure gradient

MG : Mud pressure gradient

2. What is the mud weight required for killing the well?

$\begin{array}{l} FP = HP + SIDPP\\ FP = 0.6*12000 + 500\\ FP = 7700psi\\ MG = \frac{{FP}}{{TVD}}\\ MG = \frac{{7700}}{{12000}}\\ MG = 0.642psi/ft \end{array}$

with

FP : Formation Pressure

HP : Hydrostatic Pressure of Drilling fluids

SIDPP: Shut in Drill pipe Pressure

MG: Mud weight gradient required to control the well

3. What is the kill rate pressure if pressure stabilize at Pst = 1000 psi at the start of the well kill operation.

$\begin{array}{l} ICP = KRP + SIDPP\\ KRP = ICP - SIDPP\\ KRP = 1000 - 500\\ KRP = 500psi \end{array}$

with

ICP : Initial circulating pressure

KRP : Kill rate pressure

4. What is the correct value of pressure of choke Pch at the start of the well kill operation

$\begin{array}{lllllllllllllll} {{P_{ch}} = SICP}\\ {{P_{ch}} = 900psi} \end{array}$

5. What is final circulating pressure?

$\begin{array}{l} FCP = ICP*\frac{{KMW}}{{OMW}}\\ FCP = 500*\frac{{0.642}}{{0.600}}\\ FCP = 535psi \end{array}$
with

KWM: Kill mud weight

OMW: Original mud weight

6. What is the BHP prior to and after the kick?

$\begin{array}{l} BH{P_{before}} = HP\\ BH{P_{before}} = 0.6*12000\\ BH{P_{before}} = 7200psi \end{array}$

$\begin{array}{l} BH{P_{after}} = HP + SIDPP\\ BH{P_{after}} = 0.6*12000 + 500\\ BH{P_{after}} = 7700psi \end{array}$

7. How many strokes will take the kill mud to reach the bit? How many strokes bit to shoe? Why do we want to know this value?

Mud Pump Out Put

${P_{output}} = 0.000243*Stroke.length*Liner.Diamete{r^2}*efficiency$

$\begin{array}{l} {P_{output}} = 0.000243*12*{6^2}*0.97\\ {P_{output}} = 0.102bbl/strk \end{array}$

String Volume

$\begin{array}{l} String.volume = Drillpipe.volume + Collar.volume\\ String.volume = Drillpipe.Length*Drillpipe.capacity + Drillcollar.Length*Drillcollar.capacity \end{array}$

$\begin{array}{l} String.volume = 11000*0.01757 + 1000*0.0077\\ String.volume = 201bbls \end{array}$

Strokes to the bit

$\begin{array}{l} Strokes.to.bit = \frac{{String.volume}}{{{P_{output}}}}\\ Strokes.to.bit = \frac{{201}}{{0.102}}\\ Strokes.to.bit = 1971strokes \end{array}$

Bit to Shoe volume

a ). Volume between Drill collars and open hole

$\begin{array}{l} OH/DC = Length.OH.DC*Annular.Capacity\\ OH/DC = 1000*0.0323\\ OH/DC = 32.3bbls \end{array}$

b). Volume between Drill pipe and open hole

$\begin{array}{l} OH/DP = Length.OH.DP*Annular.Capacity\\ OH/DP = 2000*0.0447\\ OH/DP = 89.4bbls \end{array}$

$\begin{array}{l} Strokes.to.Shoe = \frac{{121.7}}{{0.102}}\\ Stokes.to.shoe = 1194strks \end{array}$

Once the TOP influx is inside the shoe the MAASP figure is not more valid!

8. Calculate the influx volume.

$\begin{array}{l} Height.Influx = \frac{{SICP - SIDPP}}{{MG - GG}}\\ Height.Influx = \frac{{900 - 500}}{{0.6 - 0.1}}\\ Height.Influx = 800ft \end{array}$

With

MG : Mud gradient

GG : Gas gradient

Annular drill collar capacity = 0.0323 bbl/ft

$\begin{array}{l} Influx.Volume = 800*0.0323\\ Influx.Volume = 25.84bbls \end{array}$

9. Calculate the maximum kill circulation rate based on a mixing capacity of 450lbs/min.

We calculate the amount of barite required for each barrel of mud to weight it from the original mud weight to the kill mud weight.

$Barite.weight.Increase = 1470*\frac{{KMW - OMW}}{{35.5 - KMW}}$

$\begin{array}{l} KMW = 0.642/0.052 = 12.35ppg\\ OMW = 0.600/0.052 = 11.54ppg \end{array}$

$Barite.weight.Increase = 1470*\frac{{(12.35 - 11.54)}}{{(35.5 - 12.35)}}$

$\begin{array}{l} Maximum.Weighting.rate = \frac{{450}}{{52.48}}\\ Maximum.Weighting.rate = 8.57bbls/\min \end{array}$

$\begin{array}{l} Maximum.kill.rate = \frac{{8.57}}{{0.102}}\\ Maximum.kill.rate = 84strks/\min \end{array}$

10. How much barite is required?

total.mud.volume = surface.volume + downhole.volume

downhole.volume = string.volume + annular.volume

$\begin{array}{l} cased.hole.volume = CSG.DP.Capacity*CSG.Length\\ cased.hole.volume = 0.0478*9000\\ cased.hole.volume = 430.2bbls \end{array}$

$\begin{array}{l} Downhole.Volume = 430.2 + 121.7 + 201\\ Downhole.Volume = 752.9bbls \end{array}$

$\begin{array}{l} Total.Volume = 300 + 752.9bbls\\ Total.Volume = 1052.9bbls \end{array}$

$\begin{array}{l} {{\rm Re}\nolimits} quired.Barite = 1052.9*52.48\\ {{\rm Re}\nolimits} quired.Barite = 55256.2Lbs \end{array}$

11. How much increase in volume would this addition of barite give.

$Volume.Increase.by.100bbl = 100*\frac{{KMW - OMW}}{{35.5 - KMW}}$

$Volume.Increase.by.100bbl = 100*\frac{{12.35 - 11.54}}{{35.5 - 12.35}}$

$\begin{array}{l} Total.Volume.Increase = 3.57*\frac{{1052.9}}{{100}}\\ Total.Volume.Increase = 37.59bbls \end{array}$

12. Calculate the MAASP after killing the well.

$\begin{array}{l} MAASP = (FG - MG)*CSG.TVD\\ MAASP = (0.800 - 0.642)*9000\\ MAASP = 1422psi \end{array}$

With

FG : Fracture pressure gradient

MG : the new mud weight gradient

We will be very glad when receiving your feedback. Any comments are welcome

Activity: Well Control

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