Pressure and Friction Losses
The hydrostatic pressure is the force exerted by the column of the drilling fluids at a determined depth at a static status. It is related to the vertical depth and the density of the fluid. It is not related to the volume or the shape of the well (vertical, deviated or horizontal). At the dynamic status (while circulating) there are other friction losses which contribute in the level of the pressure in the wellbore depending on the direction of the flow. These friction losses are due to the internal friction of the fluids and the contacts between the drilling fluids wellbore walls.
Problem 1:
Case 1:
Phyd= 0.052 x 10 x 5000
Phyd= 2600 psi
In this case the bottom pressure is the same as the hydrostatic pressure
Pbottom=2600 psi
In this case the bottom pressure is the same as the hydrostatic pressure
Pbottom=2600 psi
Case 2:
Related to the case 1, the Phyd=2600 psi
In this case of circulation (Q=2000 lpm), the bottom pressure equals to the sum of the hydrostatic pressure and the annular friction losses:
Pbottom=2600 + 200
Pbottom=2800 psi
Problem 2:
The friction losses are related to the flow rate Q, density MW, any change of one of these parameters or both of them can affect the level of the friction losses.
In the case when the mud weight is changed, the new friction losses becomes:
In the case when the flow rate is changed, the new friction losses becomes:
In the case when both of them are changed, the new friction losses becomes:
The followed examples illustrate which parameter can affect more the losses.
Related to the case 1, the Phyd=2600 psi
In this case of circulation (Q=2000 lpm), the bottom pressure equals to the sum of the hydrostatic pressure and the annular friction losses:
Pbottom=2600 + 200
Pbottom=2800 psi
Problem 2:
The friction losses are related to the flow rate Q, density MW, any change of one of these parameters or both of them can affect the level of the friction losses.
In the case when the mud weight is changed, the new friction losses becomes:
The followed examples illustrate which parameter can affect more the losses.
Case 1: Change of flow rate
Q1=2000 lpm, Q2=1500 lpm
Phyd=0.052 x 10 x 5000
Phyd=2600 psi
Pbottom=2600+200
Pbottom=2800 psi
There is a change of the pressure due to the change of the flow rate:
Pannu2=200 x (1500 / 2000)²
Pannu2=112 psi
Q1=2000 lpm, Q2=1500 lpm
Phyd=0.052 x 10 x 5000
Phyd=2600 psi
Pbottom=2600+200
Pbottom=2800 psi
There is a change of the pressure due to the change of the flow rate:
Pannu2=200 x (1500 / 2000)²
Pannu2=112 psi
The bottom pressure becomes in the new situation:
Pbottom=2600+112
Pbottom=2712 psi
And the new stand pipe pressure becomes:
SPP2=2000 x (1500 / 2000)²
SPP2=1125 psi
Case 2: Change of mud weight
Pbottom=2600+112
Pbottom=2712 psi
And the new stand pipe pressure becomes:
SPP2=2000 x (1500 / 2000)²
SPP2=1125 psi
Case 2: Change of mud weight
MW1=10 ppg, MW2=12 ppg
Phyd2=0.052 x 12 x 5000
Phyd2=3120 psi
The new friction losses is:
Pannu2=200 x (12 / 10)
Pannu2=240 psi
So the bottom pressure is:
Pbottom=3120 + 240
Pbottom= 3360 psi
And the stand pipe pressure becomes:
SPP2=2000 x (12 / 10)
SPP2= 2400 psi
Problem 3: Are you going to get a kick?
In this example, the possibility of getting a kick is investigated.
Que question is: in both situations, in static or with circulation, can the formation fluids come into the wellbore?
Case1:
Phyd=0.052 x 12.50 x 9400
Phyd=6110 psi
In static Pbottom=Phyd
With this situation: the well will not flow in static, because the hydrostatic pressure Phyd is greater than the pore pressure (Pbottom > Ppores)
Case2:
Pbottom=Phyd+Pannu
Pbottom=6110+220
Pbottom=6330 psi
Obviously, the well will not flow while circulation because the bottom pressure is more than the pore pressure.
Pbottom=Phyd+Pannu
Pbottom=6110+220
Pbottom=6330 psi
Obviously, the well will not flow while circulation because the bottom pressure is more than the pore pressure.
In this next example, the pore pressure is Ppores=6200psi.
According to calculation performed previously, you can notice that in static the bottom pressure is less than the pore pressure (Phyd=6110 psi < Ppores=6200 psi). So, the well will flow, but in the dynamic mode (with circulation) the bottom pressure becomes greater than the pore pressures (Pbottom=6330 psi > Ppores=6200 psi) so the well will not flow.
Exercise 4: What is your pressures after closing on a kick?
In this example, the shut in drill pipe pressure and the casing pressure are calculated.
1. SDPP:
Phyd= 0.052 x 12.50 x 11600
Phyd=7540 psi
Pbottom=Phyd (Static mode)
SDPP=Ppores-Pbottom
SDPP=8100-7540
SDPP=560 psi
2. SICP
The annulus contains two column of two different fluids: the drilling fluids and the formation fluids, so the pressure on the bottom is the combination of these two columns.
Z=Zinflux+Zdrilling_fluids
Zinflux is the height of the kick and it is equals to Hinflux=650ft
Zdrilling_fluids is the height of the drilling fluids in the annulus
Zdrilling_fluids=Z-Hinflux
Zdrilling_fluids=11600-650 ,
Zdrilling_fluids=10950 ft
The bottom pressure in the annulus is sum of the exerted pressure by the columns of drilling fluids and the influx.
Pbottom=(0.052 x 12.50 x 10950) + (0.052 x 2.50 x 650)
Pbottom=7202 psi
ICP=Ppores-Pbottom
ICP=8100-7202
ICP=898 psi
The bottom pressure in the annulus is sum of the exerted pressure by the columns of drilling fluids and the influx.
Pbottom=(0.052 x 12.50 x 10950) + (0.052 x 2.50 x 650)
Pbottom=7202 psi
ICP=Ppores-Pbottom
ICP=8100-7202
ICP=898 psi
nice
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